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Contents
1 Fluctuations close to equilibrium 41.1 General discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Prerequisites: Probability theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3 Multivariate Gaussian distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.4 Fluctuations of thermodynamical properties . . . . . . . . . . . . . . . . . . . . . . 11
2 Onsager’s Theory of Linear Relaxation Processes 132.1 Relaxation to equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.2 Relaxation of fluctuations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.3 Microscopic reversibility and symmetry of kinetic coefficients . . . . . . . . . . . . . 192.4 Simple examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.4.1 Example 1: Heat transport . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.4.2 Example 2: Mass transport . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.4.3 Example 3: Thermoelectric phenomena . . . . . . . . . . . . . . . . . . . . . 23
2.5 Continuous systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.5.1 Heat equation and diffusion equation . . . . . . . . . . . . . . . . . . . . . . 242.5.2 The Fokker-Planck equation and the Einstein’s relation . . . . . . . . . . . . 282.5.3 General hydrodynamic description . . . . . . . . . . . . . . . . . . . . . . . 31
3 Fluctuations and dissipation 413.1 Linear response to an external force . . . . . . . . . . . . . . . . . . . . . . . . . . . 413.2 Response function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433.3 Spectrum of fluctuations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
3.3.1 Fluctuation-Dissipation Theorem in Fourier representation . . . . . . . . . . 453.4 More on random processes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
3.4.1 Some definitions concerning random processes: Ergodic hierarchy. . . . . . . 463.4.2 Conditions for ergodicity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 473.4.3 The Wiener-Khintchine Theorem . . . . . . . . . . . . . . . . . . . . . . . . 48
3.5 Fluctuations and noise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
4 Brownian motion 524.1 Historical overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524.2 The Langevin’s original approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524.3 Taylor (Green-Kubo) formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544.4 Overdamped limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 564.5 From Langevin equation to Fokker-Planck equation I . . . . . . . . . . . . . . . . . 57
4.5.1 Generalization to higher dimensions . . . . . . . . . . . . . . . . . . . . . . . 59
5 Another approach to the Fokker-Planck equation 605.1 Einstein’s theory of Brownian motion . . . . . . . . . . . . . . . . . . . . . . . . . . 605.2 Equations for the probability density . . . . . . . . . . . . . . . . . . . . . . . . . . 615.3 Stochastic integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
5.3.1 The problem of interpretation . . . . . . . . . . . . . . . . . . . . . . . . . . 665.4 Transition rates and master equations. . . . . . . . . . . . . . . . . . . . . . . . . . 715.5 Energy diffusion and detailed balance . . . . . . . . . . . . . . . . . . . . . . . . . . 73
2
6 Escape and first passage problems 756.1 General considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 766.2 Mean life time in a potential well . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
6.2.1 The flow-over-population approach . . . . . . . . . . . . . . . . . . . . . . . 776.3 Rates of transitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
6.3.1 The Arrhenius law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 806.3.2 Diffusion in a double-well . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
6.4 Kolmogorov backward equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 826.4.1 Pontryagin-Witt equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
6.5 The renewal approach. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 866.5.1 Example: Free diffusion in presence of boundaries . . . . . . . . . . . . . . . 88
6.6 An underdamped situation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
7 Simple bath model: Kac-Zwanzig bath 92
8 The Boltzmann equation 988.1 The BBGKY-Hierarchy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 988.2 Single-particle distribution function . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
8.2.1 Kinetic models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1028.3 The Boltzmann equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
8.3.1 Maxwell’s transport equations . . . . . . . . . . . . . . . . . . . . . . . . . . 1058.3.2 Collision invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1078.3.3 The H -Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
8.4 Relaxation time approximation for hydrodynamics: The BGK equation . . . . . . . 110
9 Kinetics of quantum systems 1159.1 The Golden Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1159.2 Quantum mechanical Master equations . . . . . . . . . . . . . . . . . . . . . . . . . 1179.3 Electron transfer kinetics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1229.4 The density matrix approaches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
9.4.1 The density matrix and the quantum Liouville equation. . . . . . . . . . . . 1259.4.2 Two level systems in the bath. . . . . . . . . . . . . . . . . . . . . . . . . . . 128
9.5 Master equation from density matrix approach. . . . . . . . . . . . . . . . . . . . . 138
3
1 Fluctuations close to equilibrium
1.1 General discussion
We start our considerations by a discussion of fluctuations in TD systems close to equilibrium,which will then allow us for introducing many important notions and notation, which will be usefulfor the discussion of non-equilibrium situations.
Although in a system in the state of TD equilibrium the TD-parameters (e.g. T, p, V, ...) do notchange on the average or in a longer run, the molecular motion causes fluctuations in the valuesof the corresponding parameters. These fluctuations appear and dissolve, are typically small inthe macroscopic parts of the system, but can be perceptible and important if a small subsystemis considered. As an example let us consider a small cylinder with a piston filled by a gas andimmersed into the bath filled by the same gas at the same temperature and pressure. Due tomolecular impacts the position X of the piston (which translates from the volume of the gas insidethe cylinder) will fluctuate. Assuming the motion of molecules to be “chaotic” (in the Boltzmann’ssense), we can then describe this X(t) as a random process. If we only concentrate on the result ofa single-time measurement, X can be considered as a random variable, and we can put a question,what is the probability (density) to measure some particular value of X.
We discuss the Einstein’s approach to such fluctuations, based on the inversion of the Boltz-mann’s formula for microcanonical entropy. Let us consider our system (or a part thereof), whichwill be called a “body”, to be in contact with a bath; the whole system, consisting of the body andthe bath (which will be called a “compound”) will be considered as isolated. If the fluctuationsare “small” (i.e. the system stays close to equilibrium), its state even in the case of fluctuation canbe described by a set of relevant thermodynamical variables (e.g. S, T, p, V, ...). The existence offluctuations implies that the entropy of the system is not always maximal (i.e. that the system isnot always in its “most probable” state), but that it might depart from it to a less probable one,and then return. This entropy is still the function of thermodynamical parameter of the body andof the bath.
Let X be some parameter of the body, say, its volume, or electric charge, or magnetization,etc. (in our example above it is the coordinate of the piston which is translated from the volume),taking the value X0 in equilibrium. Let us assume that we can define the total entropy of thecomposite Sc as a function of this X. Since the entropy of an isolated system at equilibriumattains its maximum, we have
Sc(X) = Sc(X0) +1
2
∂2Sc∂X2
(X −X0)2, (1)
if we assume that the maximum is a simple quadratic one (no phase transitions!). The correspond-ing second derivative ∂2S
∂X2 is therefore negative. Let us denote it by −γ = −δ2.Since the total entropy is connected with the number W of microstates and therefore with the
probability p ∝ W of the corresponding macrostate as
S = kB lnW = const · kB ln p,
by inverting this relation we get W = exp(Sc/kB). The probability to find a particular (macro)statecompatible with the corresponding restrictions is then proportional to the corresponding W , sothat the probability to find a fluctuation lowering the total entropy by amount ∆Ss is
p ∝ exp
(∆SckB
).
4
with ∆Sc = Sc(X)− Sc(X0), or
p ∝ exp
(− γ
2kB(X −X0)2
),
i.e. is a Gaussian distribution. Here
γ = − ∂2S
∂X2
∣∣∣∣X=X0
> 0 (2)
where we assume that the entropy at equilibrium reaches its simple quadratic maximum. It isimportant to note that the Gaussian nature has to do with the fact that the total entropy isexpanded close to equilibrium, far from the points of phase transitions, and that the value ofX − X0 is small. Gaussian fluctuations are universally observed close to equilibrium under suchrestrictions, but there are no reasons to look for Gaussian fluctuations in non-equilibrium systemsor close to phase transition points. Large deviations also may not follow the Gaussian law. Upto now we have no idea how to calculate γ. Here we follows a somewhat old-fashioned way ofdiscussion; another approach can be found in1.
We assume, that the state of the whole system does not depend on whether the particularstate of the body appeared as a spontaneous fluctuation, or was prepared on purpose as long asthe total energy of the system does not change (it is fixed to be Ec as it is in the case when weconsidered an isolated system). To prepare the corresponding state on purpose we have to perform(reversible) work R on the body and (reversibly) remove the corresponding amount of energy fromthe bath, in the form of heat: ∆Ec = R + Q = 0. Since the system as a whole is at a constanttemperature T (it mostly consists of the bath) we have Q = T∆Sc, so that T∆Sc+R = 0, and thetotal change in entropy is ∆Sc = −R/T . Recalling the connection of the entropy with the numberof microstates and therefore with the probability p ∝ W we have
p ∝ exp
(∆SckB
)= exp
(− R
kBT
). (3)
We now calculate R provided the internal energy of the body, its volume and entropy, as wellas external pressure and temperature are known. According to the First Law we have: ∆E =Abody +Qbody = R − p∆V + T∆S. (All quantities without subscripts are those of the body. Notethat the total work performed with the body is not R, but R− p∆V since the bath also performswork, at least in the case when the volume of the body changes). Thus,
R = ∆E + p∆V − T∆S.
Here p and T are equilibrium values of the pressure and of the temperature of the body, equalto those of the bath. In what follows we denote them by p0 and T0 in order to distinguishthem from probably fluctuating local values2. Note that for more complex bodies there might befurther terms including, say, magnetization, or the number of particles, etc. Our thermodynamicalapproach generates therefore a quite general formalism.
We thus get:
p ∝ exp
(−∆E + p0∆V − T0∆S
kBT0
).
1 See e.g. W. Gopel, H.-D. Wiemhofer, Statistische Thermodynamik, Spectrum Verlag, 20002A question about fluctuations of temperature and other intensive parameters is somewhat tricky. If you want
to learn (much) more, see M. Falcioni et al., Am. J. Phys., 79, 777-785 (2011).
5
We now note that since dE = −pdV + TdS the first non-vanishing contribution in enumeratorappears in the second order in ∆V and ∆S: we expand ∆E = E(S+∆S, V +∆V, ...)−E(S, V, ...)up to the second order in changes of its natural variables and get
p ∝ exp
[− 1
kBT0
(1
2
∂2E
∂V 2(∆V )2 +
∂2E
∂V ∂S∆V∆S +
1
2
∂2E
∂S2(∆S)2
)]. (4)
In general, if (V, S, ...) is a vector of relevant extensive variables describing the body, and x =(∆V,∆S, ...) is a vector containing their deviations from the equilibrium values, the total structureof the probability density of fluctuations is
p ∝ exp(−12xMx) (5)
where the matrix M has elements
mij =1
kBT0
∂2E
∂xi∂xj,
i.e. is real and symmetric. Since the equilibrium situation corresponds to the maximal probability(i.e. maximal entropy) and is stable, the matrix is positively defined. Eq.(5) with the correspondingpositively defined matrix corresponds to a multivariate Gaussian distribution.
In what follows, the matrix of the second derivatives if the total entropy with entries
γij = − ∂2Sc∂Xi∂Xj
∣∣∣∣Xi,j=(Xi,j)eq
=1
T0
∂2E
∂Xi∂Xj
= kBmij, (6)
will be denoted by γ.
1.2 Prerequisites: Probability theory
What you had to know coming to the class is:
• Random variables. Discrete and continuous distributions. Probability P (x) and probabilitydensity p(x). Frequentest interpretation (the “physical” interpretation related to a repeatedexperiment). P (x1 < x < x2) =
∫ x2
x1p(x)dx. A discrete distribution as a “fence” of δ-
functions.
• The probability density function (PDF) p(x) of a proper distribution is normalized:∫p(x)dx =
1.
• Mean values (a mathematician will say “expectation“, again all fine stuff is omitted): A meanof an observable f = f(x) is 〈f〉 =
∫f(x)p(x)dx, provided it exists (the integral converges).
• Moments of distributions: special means Mn = 〈xn〉 =∫xnp(x)dx. Note that due to nor-
malization M0 = 1.
These definitions can be generalized on many variables x1, ..., xn (or vector-valued variable x =(x1, ..., xn)).
The mode of the distribution is a value of xm at which maximum of p(x) is achieved. If thedistribution is monomodal, one can introduce its height as p(xm) and its width on the half-height.
The quantiles of the distribution Qq are the points where∫ Qq−∞ p(x)dx = q. Shown are the lower
and the upper quartiles (Q1/4 and Q3/4) and the median, Q1/2.
6
Figure 1: A sketch of a typical PDF with its mode and tails, and with its robust characteristics.
• The random variables are called (statistically) independent if p(x) can be represented inthe form p(x1, x2, ..., xn) = p1(x1)p2(x2)...pn(xn), i.e. factorizes into a product of functionsdepending on these variables separately. It is necessary to note that statistical independencedoes not mean anything more than this: the variables still may possess a kind of causalconnection.
• The mean of a function of such variables is f(x) =∫f(x)p(x)dx.
• If all p1(x) = p2(x) = ... = pn(x) the variables x1, ..., xn are called independent and identicallydistributed (iid).
The most important mean is given by the tautology p(y) =∫δ(y − x)p(x)dx, i.e.
p(y) = 〈δ(y − x)〉x.
Notation: If there are more than one variable, and we don’t average over all of them, we list thevariables which we average over (i.e. the dumb variables of the corresponding integrations) in thesubscript of the angular bracket. The most important consequence of this statement is the formulafor the change of variables in the PDF. Let pX(x) be known, and let y = f(x) be the property ofinterest. If the function f(x) is invertible, and its inverse is f−1, then
py(y) =
∫δ(y − f(x))px(x)dx = px[f
−1(y)]
∣∣∣∣df−1(y)
dy
∣∣∣∣ .If f is not invertible, but each y admits at most a countable number of roots (i.e. such xi thaty = fi(xi)) then
py(y) =∑i
px(f−1i (y))
∣∣∣∣df−1i (y)
dy
∣∣∣∣7
where xi = f−1i (y).
Another important mean is the characteristic function
f(k) = 〈eikx〉 =
∫eikxp(x)dx.
It does exist (i.e. the integral converges absolutely) for any proper PDF. The characteristic functiondefines the PDF uniquely (in the same sense as a Fourier-transform identifies the correspondingfunction).
Note: If the random variable x can only assume non-negative values, the use of the Laplacetransform, i.e. calculation of 〈e−ux〉 gives an alternative tool.
Note that the characteristic function is the generating function of the moments of the distri-bution: Expand the exponential in the Taylor series and obtain
f(k) =
∫ ∑n
(ikx)n
n!p(x)dx =
∑n
(ik)n
n!
∫xnp(x)dx =
∑n
(ik)n
n!Mn.
The moment Mn is thus (up to a prefactor in) equal to the n-th Taylor coefficient in expansion off(k) around k = 0. If all moments exist f(k) is analytical function.
Often it is convenient to write
f(k) = 〈eikx〉x = eψ(k),
i.e. define ψ(k) = ln f(k). The Taylor expansion of ψ(k) gives us the cumulants κn of thedistribution:
ψ(k) =∞∑n=1
κn(ik)n
n!.
The cumulants may be expressed in terms of the moments
κ1 = M1
κ2 = M2 −M21
κ3 = M3 − 3M1M3 + 2M31
κ4 = M4 − 4M1M3 − 3M22 + 12M2
1M2 − 6M41
...
Note that the distribution does not have to possess moments or cumulants (i.e. f(k) and ψ(k) arenon-analytical) but the corresponding functions f(k) and ψ(k) are still uniquely defined for anydistribution.
The same discussion may be made for multivariate distributions (i.e. for the distributions ofthe vector-valued random variables x = (x1, ..., xn)). The moments are defined as
〈xp1xq2...x
rn〉 =
∫...
∫xp1x
q2...x
rnp(x1, x2, ..., xn)dx1...dxn,
and are generated by the characteristic function of the distribution
f(k) = f(k1, ..., kn) = 〈ei(k1x1+k2x2+...+knxn)〉
=
∫...
∫ei(k1x1+k2x2+...+knxn)p(x1, x2, ..., xn)dx1...dxn :
8
〈xp1xq2...x
rn〉 = (−i)p+q+...+r ∂
p+q+...+rf(k)
∂kp1∂kq2...∂k
rn
∣∣∣∣k=0
.
The logarithm of the characteristic function is the generating function of the cumulants of thedistribution.
The change of variables in multivariate distributions is performed analog to the single-variableones, e.g. if u = u(x, y) and v = v(x, y), and the inverse transformation exists (i.e. one can expressx = x(u, v), y = y(u, v)) then
pu,v(u, v) = px,y(x(u, v), y(u, v))
∣∣∣∣∂(x, y)
∂(u, v)
∣∣∣∣ ,and analog for more variables.
If the multivariate (joint) probability distribution is known, the probability density of theresults of measurement of each particular variable or of several (but not all) variables may beobtained by calculation the marginal PDF
p(x) =
∫p(x, y)dy
(formally x and y can also be vectors of variables). In many cases we are interested in thedistribution of the values of x in experiments giving particular value of y0 as an outcome (i.e.under the condition that y = y0), e.g. in the conditional probability density p(x|y0) (describing,say, the amount of rain during the days with temperature of 20C). By virtue of the Bayes theorem
p(x, y) = p(x|y)p(y)
such conditional density is defined when the joint density of x and y is known (note that p(y) isthe marginal distribution of y).
1.3 Multivariate Gaussian distributions
Let x = (x1, ..., xn) be the vector of random variables. The multivariate Gaussian probabilitydensity function has the form
p(x) = C · exp
(−1
2xMx
)= C · exp
(−1
2
n∑i,j=1
mi,jxixj
),
where C is a normalization constant
C =
(∫...
∫exp
(−1
2xMx
)dnx
)−1
(which we will explicitly calculate now) and M = mij is a real, symmetric matrix.A real symmetric matrix can be diagonalized by an orthogonal coordinate transformation x′ =
Ox: M′ = OMO−1 = diag. The matrix O consists of eigenvectors of M taken as rows, and thematrix O−1 = OT consists of the same eigenvectors taken as columns.
Let the (diagonal) elements of M′ be m′1, ...m′n. Since the Jacobian of an orthogonal transfor-
mation is unity, the integral∫...∫
exp(−12xMx)dnx doesn’t change when passing from x to x′, so
9
that
C−1 =
∫...
∫exp(−1
2xMx)dnx =
∫...
∫exp(−1
2x′M′x′)dnx′
=
∫e−
12m′1x
21dx1...
∫e−
12m′nx
2ndxn
=
√2π
m1
...
√2π
mn
=(2π)n/2√
detM′=
(2π)n/2√detM
.
Therefore
p(x) =
√detM
(2π)n/2· exp(−1
2xMx).
Let us now pay more attention to the meaning of matrix M. Let
cij = 〈xixj〉 = C
∫...
∫xixj exp(−1
2xMx)dnx
be the mean of a product of xi and xj. These means build a matrix C = cij (a matrix ofcovariation coefficients) which is a symmetric and real one. We now note that
cij = C
∫...
∫xixj exp
(−1
2
n∑i,j=1
mijxixj
)
= C · 2 · ∂
∂mij
∫...
∫exp
(−1
2
n∑i,j=1
mijxixj
)
=2√
detM
(2πn/2)
∂
∂mij
(2π)n/2√detM
=1
detM
∂
∂mij
detM.
Let us expand the determinant w.r.t. the column j:
detM =n∑i=1
(−1)i+jmijMij,
where Mij are the corresponding minors. We now see that
∂
∂mij
detM = (−1)i+jMij,
and therefore
cij =(−1)i+jMij
detM=M−1
ij
(7)
i.e. is proportional to the corresponding element of the inverse matrix.We know well the corresponding relation for a univariate distribution,
p(x) =1√2πσ
exp(−x2/2σ2),
when matrix M is essentially a scalar m = 1/σ2 and therefore c = 〈xx〉 = σ2 = m−1.
10
The matrix C also defines the characteristic function of the multivariate Gaussian distribution
f(k) = f(k1, ..., kn) = 〈exp(ikx)〉 =
∫...
∫p(x1, ..., xn)eik1x1+...+iknxn .
We again perform the variable transformation x′ = Ox. Note that the vector k in the scalar prod-uct is essentially a row-vector kT , which is transformed by the matrix OT ≡ O−1. In the followingfew lines I will distinguish between the row and the column vectors and write the scalar productexplicitly as kTx. We note that the inverse matrix is diagonalized by the same transformation asthe initial one, so that (M′)−1 = OM−1O−1, and, respectively M−1 = O−1(M′)−1O.
We have:
f(k) =
√detM
πn/2
∫...
∫exp(−1
2xTMx) exp(ikTx)dx
=
√detM′
(2π)n/2
∫...
∫exp(−1
2x′TM′x′) exp(ik′Tx′)dx′
=
√∏ni=1m
′i
(2π)n/2
∫...
∫exp
(−1
2
n∑i=1
m′i(x′i)
2 + in∑i=1
k′ix′i
)dx′1...dx
′n
=n∏i=1
√m′i2π
∫exp
(−1
2m′i(x
′i)
2 + ik′ix′i
)dx′i
=n∏i=1
exp
(− 1
2m′i(k′i)
2
)= exp
(−1
2
n∑i=1
(m′i)−1(k′i)
2
)
= exp
(−1
2k′T
(M′)−1k′)
= exp
(−1
2kTO−1(M′)−1Ok
)= exp
(−1
2kTM−1k
).
This can be rewritten as
f(k) = exp
(−1
2kTCk
),
and
p(x) =1(√
2π)n√
detCexp
(−1
2xTC−1x
).
These relations will be repeatedly used in what follows.
1.4 Fluctuations of thermodynamical properties
Calculating the second derivatives of the internal energy of the body with respect to all relevantnatural variables and using Eq.(5) we obtain a multivariate Gaussian distribution with the matrixM whose elements have no special reason to vanish (end essentially do not vanish anywhereexcept at the points of phase transitions); the same is true for the elements of the matrix C. As wewill see later, these correlations between the changes in different extensive variables indicate therelation between different transport phenomena, for example the non-vanishing correlator 〈∆S∆N〉
11
indicates the connection between heat transport and diffusion. In general, even to obtain the meanvalue of a fluctuation of a single extensive variable, we have to derive the corresponding matrix Mincluding all relevant extensive variables, invert it to obtain C, and then take the correspondingdiagonal element, representing the mean square of fluctuation. There is a way to circumvent thistedious operation.
Let us return to Eq.(4) and note that (∂E/∂V ) = −p and (∂E/∂S) = T , so that
∂2E
∂V 2∆V +
∂2E
∂V ∂S∆S = − ∂p
∂V∆V − ∂p
∂S∆S ≡ −∆p
∂2E
∂S2∆S +
∂2E
∂S∂V∆V = +
∂T
∂V∆V +
∂T
∂S∆S ≡ ∆T (8)
where ∆p and ∆T describe the fluctuations of pressure and temperature of the body. Eq.(9)therefore takes the form
p ∝ exp
(−−∆p∆V + ∆T∆S
2kBT0
)(9)
which is very simple (in the enumerator we see the products of fluctuations of conjugated thermo-dynamic variables), and it is easy to memorize and to generalize, e.g.
p ∝ exp
(−−∆p∆V + ∆T∆S + ∆µ∆N + ∆σ∆A+ ...
2kBT0
)(10)
where µ is the chemical potential, and N the particle’s number, σ is the surface tension, and A is thesurface area in the system, etc. Moreover, using this equation we can obtain also the fluctuationsof intensive variables, or use a mixture of intensive and extensive variables as independent ones.
Imagine, we are interested in the fluctuation of V in a simple fluid body as described by Eq.(9).Let us consider V and T (and not V and S) as independent observables. Expressing ∆S via ∆Tand ∆V we get:
∆S =
(∂S
∂T
)V
∆T +
(∂S
∂V
)T
∆V.
The first partial derivative is equal to CV /T , and the second one can be expressed via the Maxwellrelation (∂S/∂V )T = ∂2F/∂V ∂T = −(∂p/∂T )V . On the other hand,
∆p =
(∂p
∂T
)V
∆T +
(∂p
∂V
)T
∆V.
Inserting this in Eq.(9) we get
p ∝ exp
[− Cv
2kBT 2(∆T )2 +
1
2kBT
(∂p
∂V
)T
(∆V )2
]:
the distribution of an extensive variable factorizes from the one of intensive variables not conjugatedto the considered extensive one. Thus:
〈(∆V )2〉 = −kBT(∂V
∂p
)T
= kBTV κT , 〈(∆V∆T )〉 = 0, 〈(∆T )2〉 =kBT
2
CV,
where κT = −V −1(∂V/∂p) is the isothermic compressibility.Looking at this example we see that the squared fluctuation of an extensive variable is pro-
portional to the volume (mass, number of particles...), and the corresponding r.m.s. fluctuation
12
to√V (or
√N). The r.m.s. fluctuations of the densities of extensive variables, and the ones of
intensive variables are proportional to 1/√N or 1/
√V (see however3).
To stress the finding, let us consider p and S as independent variables.
∆V =
(∂V
∂p
)S
∆p+
(∂V
∂S
)p
∆S.
and
∆T =
(∂T
∂p
)S
∆p+
(∂T
∂S
)p
∆S =T
Cp∆S +
(∂T
∂p
)S
∆p.
Using the Maxwell relation (∂V/∂S)p = ∂2H/∂p∂S = (∂T/∂p)S, we get
p ∝ exp
[1
2kBT
(∂V
∂p
)S
(∆p)2 − 1
2kBCp(∆S)2
],
so that
〈(∆S)2〉 = kBCp, 〈(∆S∆p)〉 = 0, 〈(∆p)2〉 = −kBT(∂p
∂V
)S
.
The examples above can be generalized and teach us that: If we are interested in the behaviorof as single extensive thermodynamical variable, we take all other independent variables to beintensive ones, and the extensive one of interest decouples. If we are interested in the behaviorof an intensive thermodynamical variable, we (parallel to what is done previously) take all otherindependent variables to be extensive ones, and the intensive one of interest decouples. Integratingover these irrelevant variables leaves us with the Gaussian distribution, say, of the form
p(∆V ) ∝ exp
(− 1
2kBTV κT(V − Veq)2
).
Comparing this with Eq.(1) we find the corresponding λ = 1/TV κT .
2 Onsager’s Theory of Linear Relaxation Processes
Our discussion will be based on the assumption that the further behavior of the system doesnot depend on whether the corresponding change in parameters appeared spontaneously as afluctuation or was introduced on purpose by some preparation procedure. This assumption iscalled the Onsager’s regression hypothesis (1931). It states that the regression (≈ relaxation) ofmicroscopic thermal fluctuations at equilibrium follows the macroscopic law of relaxation of smallnon-equilibrium perturbations.
Since statistical physics (whether in equilibrium or not) typically gives us ensemble means of thecorresponding properties, we concentrate on such means. Thus we either wait until the measuredvalue of, say, the volume V of a body takes the prescribed value V0 in course of fluctuations and
3 Here we return to our footnote 2, and discuss the restrictions on the possibility to measure these fluctuationsof intensive variables. Therefore the deviations of these variables can better be considered as internal parametersof the theory which are linearly connected with the fluctuations of the extensive ones via Eq.(8), and the intensiveparameters themselves, defined as means (e.g. in a classical system the temperature as proportional to the meankinetic energy of the molecules, the pressure as proportional to the mean force acting on the piston, etc.) as non-fluctuating. In any case, when discussing the experimental results for fluctuations of intensive variables, one has toclearly define the measurement operation.
13
follows then the further evolution of V (t), or we prepare the body in a state with volume V0 (andother parameters as in the previous experiment) and follow V (t). The experiment is repeated manytimes and we look at the behavior of 〈V (t)|V0〉, i.e. of the conditional mean of the volume providedthe experiment started at a state when this volume was V0. In course of the time 〈V (t)|V0〉 willtend to the equilibrium value of the volume V . We thus start from a discussion of relaxation tothe equilibrium of the nonequilibrium states introduced on purpose.
2.1 Relaxation to equilibrium
Let us concentrate at the behavior of some macroscopic quantity X characterizing a body ofinterest, and (at the beginning of our discussion) neglect fluctuations. Let the value of X inequilibrium be X0, and let x = X − X0 (note that X, a natural variable of the entropy, is anextensive variable. Therefore x is also extensive!). Then x = 0 corresponds to equilibrium andx(t) 6= 0 to a nonequilibrium state. The equilibrium state corresponds to a maximum of theentropy of the total system (compound), see Eq.(1). Let us consider the entropy of the system asa function of x. In equilibrium Sc(x = 0) = max which in a typical case means:(
∂Sc∂x
)x=0
= 0;
(∂2Sc∂x2
)x=0
≤ 0.
If a state in which x 6= 0 (which is the state with an entropy which is smaller than the entropy of theequilibrium state) is prepared on purpose and then the system is let to establish the equilibrium,the value of x will then change approaching zero in order to maximize the entropy. Thus
d
dtSc(x) =
∂Sc∂x· dxdt≥ 0.
We now follow Onsager and give the physical interpretation of the two factors appearing in thelast expression.
The derivative
Fx =∂Sc∂x
is considered as the driving force of the relaxation to equilibrium. This term is called a thermo-dynamic force (Note that it doesn’t have the dimension of the force, so that the name is not avery proper one! The dimension of the thermodynamic force is [kB][X]−1, i.e. the inverse of thedimension of X multiplied by Joule per Kelvin), the analogy with mechanics is in the fact that the(negative) entropy plays the role of a “potential energy” which tends to be minimal in equilibrium.The second term
Jx =dx
dt
is called thermodynamic flux or thermodynamic flow of variable x4. Further, Onsager consideredsituations in which a linear relation between the thermodynamic force and the flux holds:
Jx = LFx,
4The signs in definitions of F and J are somewhat arbitrary, and some authors (including myself in some works)define both force and flux with “−”: F = −∂S/∂x and J = −dx/dt; none of the further relations changes from suchan exchange of signs. The difference is whether the entropy or its negative is considered as a corresponding potential.Taking “+” makes the −S to be more like some mechanical potential energy which is minimal at equilibrium. Inthe current version of the script I decided to use the “+” the signs.
14
so that the whole ensuing theory is called linear irreversible thermodynamics. The idea behindis, that the thermodynamic force is the cause of the thermodynamic flow and both should dis-appear at the same time, i.e. when equilibrium is reached. The coefficient L is called Onsager’sphenomenological coefficient, or Onsager’s kinetic coefficient. Since, according the Second Law,the entropy grows when the system approaches equilibrium, the Onsager’s coefficient is strictlypositive: Out of equilibrium
0 <d
dtS(x) = Jx · Fx = L · F 2
x .
The time derivative of the entropy P = dS/dt is called the entropy production, in the Onsager’slinear theory it is a bilinear form
P = J · F(where we now omit the subscript x indicating the variable). Close to equilibrium
F =∂S
∂X
∣∣∣∣X=X0+x
=∂2S
∂x2
∣∣∣∣x=0
x = −γx.
Using the previous equations we finally get the following linear relaxation dynamics
x = −Lγx.
Introducingλ = Lγ
being the so-called relaxation coefficient of the quantity x we get finally
x = −λx. (11)
This linear kinetic equation describes the relaxation of a thermodynamic system prepared in anon-equilibrium state to its equilibrium one. Starting with the initial state x(0) the dynamics ofthe variable x(t) is then
x(t) = x(0) exp(−λt) = x(0) exp
(− tτ
). (12)
where τ = λ−1 is a typical decay time of the initial deviation from equilibrium (the relaxationtime).
A similar procedure can be performed when dealing with several thermodynamic variables.Close to equilibrium the entropy as a function of xi, (i = 1...N) is given by
S(x1, ..., xn) = Smax −1
2
∑i,j
γijxixj
where γij are given by Eq.(6)with γ introduced in Eq.(2) 5. Following the Onsager’s ideas we definethe thermodynamic forces
Fi = −∑j
γijxj
and fluxesJi = xi.
5The matrix γ is connected with the matrix M describing fluctuations via γij = kBmij .
15
The generalized linear Onsager-Ansatz now reads
Ji =∑j
LijFj
(note that the flux of a variable i is caused also by forces corresponding to other variables, rememberthe Le-Chatelier principle for indirect changes). The Second Law again requires the positivity ofthe entropy production, and thus
Lijxixj ≥ 0,
where the equality takes place only for xi = 0, (i = 1, ..., n). This corresponds to the requirementof positive definiteness of the matrix Lij. Parallel to above,
xi =∑j
LijFj
and introducing the matrix of relaxation coefficients of the linear processes near equilibrium stateswe end up with
xi = −∑j
λijxj (13)
withλij =
∑k
Likγkj
(i.e. in the matrix form λ = Lγ). Since the matrix γij determines the dispersion of the stationaryfluctuations close to equilibrium, we see that there exists an intimate relation between fluctuationsand relaxation, i.e. we have a fluctuation-dissipation relation for a set of fluctuating and relaxingvariables6.
In the next chapter we will discuss the phenomenological equations of linear nonequilibriumthermodynamics (Fourier equation for heat transport, Ficks equation for diffusion, other, morecomplex cases), but first return to our fluctuations, and discuss what do we learn about theproperties of the kinetic coefficients from what we know about them.
Although our discussion was conducted for thermodynamic variables (i.e. for a small subsystembeing a fluid body), it can be generalized to a broader class of relaxation phenomena. Let usconsider an example of a (macroscopic) harmonic oscillator in the contact with the bath. Eq.(3)p ∝ exp(−R/kBT ) gives the connection of the probability of fluctuation with the reversible worknecessary to create the state of the subsystem observed. For an oscillator it is
R = ∆E =kx2
2+
p2
2m=kx2
2+mv2
2
(in this equation x is the coordinate of the center of mass, and p is the momentum, not theprobability). Therefore the joint probability density to find the position x and the velocity v is
p(v, x) ∝ exp
(− kx2
2kBT− mv2
2kBT
),
6 The matrix L of kinetic coefficients is, of course, real and positively defined, but does not have to be symmetric,as we will see later. Therefore the matrix λ does not have to be symmetric as well. Its eigenvalues may be complex(although the physical fact that the system always relaxes to equilibrium suggests that the real parts of theseeigenvalues are positive). Therefore in the case when the system has to be described by more than one extensivevariable, the relaxation processes may be oscillatory.
16
i.e. is given by the Gibbs-Boltzmann distribution. The entropy reads
S(x, v) = kB lnW = const− kx2
2T− mv2
2T.
Therefore the elements of the matrix γ are:
γ11 =k
T, γ22 =
m
T, γ12 = γ21 = 0.
The vector of thermodynamical forces has then the components
Fx =∂S
∂x= −kx
T=f
T
where f is the mechanical force, and
Fp =∂S
∂p= − v
T= − p
mT.
The fact that according to the upper equation the thermodynamical force is proportional to themechanical force was the reason why I use the non-standard sign convention. Using p instead of vis not important in this equation, but it is p that in a large system that is an extensive variable!
Let us now put down the Onsager’s equations for the thermodynamical flux J = (x, p).
x = L11
(−kxT
)+ L12
(− vT
)p = L21
(−kxT
)+ L22
(− vT
)In the case of vanishing friction we compare this with the Hamiltonian equations of motion:
x =p
m= v
p = −kx
and getL11 = L22 = 0, L12 = −T, L21 = T.
Note the relation L12 = −L21, which is a special case of the Onsager-Casimir relation which willbe discussed later. In the case with friction, when the equation of motion of the damped oscillatorreads mx = −γx− kx we get
L11 = 0, L22 = γT, L12 = −T, L21 = T.
We see therefore that these mechanical cases fall into the general Onsager’s scheme of linearrelaxation processes.
17
2.2 Relaxation of fluctuations
It is interesting to know, what kind of a random process can describe the fluctuations of someproperties of interest x(t) as a function of time7, i.e. what is their temporal dynamics. Since oursystem is at equilibrium (none of its relevant properties changes with time), the properties of thefluctuations do not depend on when the observation started, and the corresponding random processis stationary. This means for example that all properties of the process x(t) at two different times(e.g. the correlation functions 〈x(t1)x(t2)〉) depend only on the difference of the times, i.e. timelag τ = t2 − t1, and not on t1 and t2 separately.
If fluctuations can be neglected, it is reasonable to assume that
xi = −∑j
λijxj,
and (as we will discuss) experiments in many cases support this assumption. The fluctuations canbe neglected if (a) the body is so large that the fluctuations are small and/or (b) the deviation fromthe equilibrium x(t) is much larger that the fluctuation background 〈(∆x)2〉1/2. It is reasonableto assume that when the value of x(t) gets smaller than the correlation background, the timeevolution of the ensemble mean of x(t) is still well-described by Eq.(11) or Eq.(13), i.e. in the lastcase
〈xi(t)〉 = −∑j
λij〈xj(t)〉.
Here again the experiments show that the assumption is quite reasonable in a vast amount of cases.In the case of the system prepared on purpose the means are the ones under given initial condi-tions, and in the case of fluctuations they are the conditional means 〈xi(t)|x(0)〉 and 〈xj(t)|x(0)〉.Experiments show that similar equations hold not only for thermodynamical variables, but for alarger class of variables (e.g. for macroscopic velocities or momenta of bodies in the contact withbath); special cases will be considered later.
Let us consider the correlation functions
Cij(t1, t2) = 〈xi(t1)xj(t2)〉.
Since the process of fluctuations close to equilibrium is stationary, Cij(t1, t2) can only depend onthe time lag τ = t2 − t1: Cij(t1, t2) = Cij(t2 − t1) ≡ Cij(τ). On the same reason t1 can alwaysbe set to zero. Note that the matrix C discussed in §1.2 is a matrix of single-time correlationscij = Cij(0). For a stationary process we have
Cij(τ) = 〈xi(0)xj(τ)〉 = 〈xi(t)xj(t+ τ)〉= 〈xj(t+ τ)xi(t)〉 = 〈xj(t′)xi(t′ − τ)〉 = Cji(−τ).
(in the fourth expression I interchanged the sequence of multipliers; t′ = t+ τ).Since 〈xj(t)|x(0)〉 decays according to a deterministic system of linear differential equations
d
dt〈xj(t)|x(0)〉 = −
∑i
λji〈xi(t)|x(0)〉,
7i.e. what are the properties of the collection of the values of the corresponding observable at different times t
18
we can multiply both sides this equation by xj(0), and then average over the equilibrium distribu-tion of all x(0) to get the corresponding unconditional mean:
〈xi(0)d
dt〈xj(t)|x(0)〉〉x(0) = −
∑i
λjk〈xi(0)〈xk(t)|x(0)〉〉x(0)
ord
dt〈xi(0)xj(t)〉 = −
∑k
λjk〈xi(0)xk(t)〉
(Bayes theorem!), i.e.d
dtCij(t) = −
∑k
λjkCik(t).
There exists a very important relation between the fluctuations of thermodynamical variablesxi and the corresponding forces Fi (which we will often denote by Xi in what follows: a capitalletter denotes a force, the small letter denotes the variable). The distribution of x is a Gaussian
p(x) ∝ exp(−12xTMx),
with M = k−1B γ, so that 〈xixj〉 = Cij(0) = (M−1)ij. On the other hand, the vector of ther-
modynamical forces F (or X) is X = −γx = −kBMx. We note that the square, symmetricmatrix M is invertible (all eigenvalues are positively defined, no infinite fluctuations), we canwrite x = −(1/kB)M−1X. The Jacobian of this transformation is constant (∝ detM) and thus
p(X) ∝ exp(−12k−2B XTM−1MM−1X) = exp(−1
2k−2B XTM−1X).
Therefore 〈XiXj〉 = k2Bmij.
Let us now use Xi = −kB∑
kmikxk and calculate
〈Xixj〉 = −〈kB∑k
mikxkxj〉 = −kB∑k
mik〈xkxj〉 = −kB∑k
mikckj
= −kB∑k
(M)ik(M−1)kj = −kB(MM−1)ij = −kBδij
(where in the last line the expression Eq.(7) is used).We note that in the case when xi corresponds to the deviation from equilibrium of an extensive
thermodynamical variable this result reproduces the decoupling of this extensive variable fromdeviations of non-conjugated intensive ones in the Einstein’s theory of fluctuations.
2.3 Microscopic reversibility and symmetry of kinetic coefficients
Changing τ to −τ means the reversal of the “direction of time”, which in the microscopic languagecorresponds to the reversal of all momenta, electric currents and therefore magnetic fields, etc.Let us now consider what happens with macroscopic variables x. The variables which are thefunctions of coordinates only, and for those which are the quadratic functions of momenta (likethe kinetic or total energy, and therefore for the internal energy as well) stay the same under thistransformation:
xi(t) = xi(−t).
19
Such variables are called even variables. On the other hand, the variables depending linearly onthe momenta or currents (total momentum, magnetization, etc) change their sign,
xi(t) = −xi(−t),
and are called odd variables. This behavior is a consequence of the principle of microscopicreversibility of motion. The property of being even or odd will be called the parity ε of thecorresponding variable: it is equal to εi = 1 for the even variable and εi = −1 for the odd one.Therefore
Cij(τ) = 〈xi(t)xj(t+ τ)〉 = ±〈xi(−t)xj(−t− τ)〉 = ±Cij(−τ)
depending on whether the parity of the of the both variables is the same or different: if it is thesame, the sign is “+”, and when it is different, the sign is “−”, i.e.
Cij(τ) = εiεjCij(−τ).
We now use this notation and put down
d
dτxi(τ) =
∑k
LikXk(τ)
d
dτxj(τ) =
∑k
LjkXk(τ).
We multiply the first equation with xj(0), the second equation with xi(0), and average over therealizations:
d
dτ〈xi(τ)xj(0)〉 =
∑k
Lik〈Xk(τ)xj(0)〉
d
dτ〈xj(τ)xi(0)〉 =
∑k
Ljk〈Xk(τ)xi(0)〉.
The l.h.s. of the corresponding equations are the time derivatives of Cij(−τ) = Cji(τ) and ofCji(−τ) = εjεiCji(τ), so that
d
dτCji(τ) =
∑k
Lik〈Xk(τ)xj(0)〉
εiεjd
dτCji(τ) =
∑k
Ljk〈Xk(τ)xi(0)〉.
Now we go to the limit τ → 0+ and use that 〈Xixj〉 = −kBδij:
Cji(τ)∣∣∣τ→0+
= kB∑k
Likδkj = −kBLij
εiεj Cji(τ)∣∣∣τ→0+
= kB∑k
Ljkδki = −kBLji.
From this we get thatLij = εiεjLji. (14)
Equation Eq.(14) gives us the famous Onsager-Casimir relation defining the symmetry of kineticcoefficients.
20
2.4 Simple examples
While discussing the size of fluctuations of thermodynamical variables we relied on the isothermalsituation of a body or other smaller system in a contact with a heat bath, the existence of the bathwas never assumed when considering the kinetic coefficients and symmetry thereof. We only usedthe fact that at equilibrium the entropy of the isolated composite system reaches its maximum.Therefore we can consider the situation with several (two or more) subsystems of comparable sizebuilding together an isolated system. We suppose that each of them can be characterized by a setof thermodynamical parameters (the assumption of local equilibrium).
2.4.1 Example 1: Heat transport
Let us consider two subsystems forming a composit isolated one. At the beginning of the experi-ment they are separated by an adiabatic wall and have different temperatures T1 and T2. At timet = 0 the adiabatic wall is replaced by a weakly heat conducting one. At each instant of timeeach of the systems will be considered as being at local equilibrium at some temperature T1(t) andT2(t).
T1 T2 → T1(t) T2(t)
adiabatic wall dE1 = −dE2
(no heat transport) heat conduction
During the equilibration
0 < dS =∂S1(E1, ...)
∂E1
dE1 +∂S2(E2, ...)
∂E2
dE2
=1
T1
dE1 +1
T2
dE2
=
(1
T1
− 1
T2
)dE1 :
The heat flows from the warmer to the colder subsystem, which leads to the equilibration oftemperature. The thermodynamic force corresponding to the variable E1 is F = T−1
2 − T−11 . The
phenomenological linear Ansatz than implies
Q ≡ E1 = J =dE1
dt= L(
1
T1
− 1
T2
) ' L
T 2(T2 − T1) = L(T2 − T1)
(J – the heat flux, L – the empirical kinetic coefficient, T =√T1T2 is a (geometric) mean temper-
ature of the reservoirs). Taking dQ = C1dT1 = −C2dT2, (C1, C2 – the specific heat values of thesubsystems) we get
dT1
dt=
L
C1
(T2 − T1)
dT2
dt= − L
C2
(T2 − T1).
21
We note that this system of equations has an integral of motion, the total energy C1T1 + C2T2 =const, so that it can be easily solved by substitution. We get
T1 = [T1(0)− T ] exp
(− LCt
)+ T
T2 = [T2(0)− T ] exp
(− LCt
)+ T
with
C =C1C2
C1 + C2
and
T =C1T1(0) + C2T2(0)
C1 + C2
being the final temperatures of the subsystems. The temperatures relax exponentially to theirfinal values.
2.4.2 Example 2: Mass transport
N1 N2 → N1(t) N2(t)
impenetrable wall dN1 = −dN2
(no mass transport) porous wall
We consider the transport of particles between the two containers of the volumes V1 and V2 (theconcentrations are proportional to the corresponding numbers of particles, and the coefficient ofthis proportionality is the volume). The particles considered are either in an ideal solution (or ina gas phase) so that the chemical potentials are equal to µ1,2 = const+RT lnn1,2, and
S = S0 −R(N1 lnn1 +N2 lnn2).
The total number of particles stays constant: V1n1 + V2n2 = N . Therefore
dS
dt= −RT [V1n1 lnn1 + V1n1 + V2n2 lnn2 + V2n2] = RTV1n1 [lnn2 − lnn1]
(since V2n2 = −V1n1). The particle flux close to equilibrium is
J = n1 = LRTV1(lnn2 − lnn1) ' L(n2 − n1).
Note that the situation considered here is much more the equilibration of concentrations due tothe (osmotic) pressure difference than the true diffusion. True diffusion takes place under equaltotal pressure in both containers and assumes that another component (solvent) is diffusing in theopposite direction.
22
Figure 2: A thermocouple.
2.4.3 Example 3: Thermoelectric phenomena
Let us consider the system of two contacts between two different materials A and B; each contactis placed in its own heat reservoir (“small bath”) at temperatures T1 and T2 (a thermocouple);there is additionally a capacitor C which allows us to define the potential difference between thecontacts. Such a capacitor can be considered to be placed outside of the reservoirs. The wholesystem is isolated.
If the charge of the capacitor q the electrostatic energy is UC = q2/2C = qu(q), with C beingthe capacitance and u = q/C the potential difference. For each bath we have dEi = TidSi; thetotal entropy of the system is S = S1 + S2. The total internal energy reads E = E1 + E2 + UC ,and is conserved:
dE = T1dS1︸ ︷︷ ︸dE1
+T2dS2︸ ︷︷ ︸dE2
+udq = 0.
The change of the entropy under heat and charge transport is given by
dS =dE1
T1
+dE2
T2
=
(1
T1
− 1
T2
)dE1 −
u
T2
dq.
Thermodynamic forces are:
F1 =∂S
∂E1
= T−11 − T−1
2 ≈ T2 − T1
T 2
(T – mean temperature of the two reservoirs) and
F2 =∂S
∂q≈ − u
T.
The following relations hold:
U1 = L11F1 + L12F2 = L11∆T
T 2− L12
u
T
q = L21F1 + L22F2 = L21∆T
T 2− L22
u
T.
Each of the summands has its physical interpretation:
23
• (L11/T2)∆T describes heat conduction. For u = 0 we have Q = U1 = (L11/T
2)∆T ≡ λ∆T ,with λ being the heat conduction coefficient.
• (L22/T )u describes electric conduction. For ∆T = 0 q = −(L22/T )u = −(1/R)u, with Rbeing the electric resistivity.
• The two cross-terms describe thermoelectric effects: the Seebeck-effect (1821) and thePeltier-effect (1834).
a) The Peltier effect. For constant u (and therefore for constant electric current) in course ofthe time a state with different temperatures T1 and T2 is established. The internal energies of thereservoirs stay constant in spite of continuous heat exchange: E1 = 0. The temperature differenceis ∆T = (L12T/L11)u = κu. The coefficient κ is called the Peltier-coefficient.
b) The Seebeck effect. In a stationary state with different temperatures there exists a potentialdifference u corresponding to zero current I = q = 0. This means that L21 = L22T (u/∆T )|I=0
i.e. u = α∆T with α = L21/(L22T ) called the coefficient of thermoelectric potential (or simply athermoelectric coefficient).
The Onsager-Casimir principle implies that L12 = L21. This means that the experimentallyaccessible coefficients λ, R, α and κ are connected by κ = αT/(Rλ). This relation is experimentallyverified.
The descriptions above are archetypical: the corresponding dynamics follows from the linearrelations between the fluxes and forces and from the conservation law.
2.5 Continuous systems
In this section we discuss the most important equations for continuous systems: The Fourierequation for heat transport, the Fick’s equation for diffusion, the “embryonic” Fokker-Planckequation for diffusion in a force field. Then we turn to the discussion of a general framework withinwhich such transport equations can be derived within the linear non-equilibrium thermodynamictheory. Such transport equations will later be discussed in the microscopic framework, startingfrom the Boltzmann equation.
2.5.1 Heat equation and diffusion equation
Let us return to our Example 1 and discuss a continuous system at local equilibrium conditions,i.e. in the case when each part of it can be characterized by some local temperature T (x) (in what
24
follows we consider a one-dimensional setting; generalizations to higher dimensions are straightfor-ward). Let us first discuss the connection between the macroscopic kinetic coefficients with specificproperties of the material our system is made of. This material will be considered as homogeneousthroughout the whole system. We start from a slab of a material of cross-section A. Let us considera thin slice of our system of thickness ∆x, (say placed between x and x+ ∆x) in contact with twoheat baths at some given temperatures, and assume the temperatures on its borders to be T (x)and T (x+ ∆x) = T (x) +∇T∆x, where ∇T is the local temperature gradient. Assuming the heatflux in our system (i.e. the total heat flux from the hotter to the colder bath) to be proportionalto the difference of the temperatures we get the values of the corresponding flux
J = L(∆x)(T (x+ ∆x)− T (x)) = L(∆x)∇T∆x.
Let us now consider what happens in the midpoint of our element (say at x+ ∆x/2). The energyflux through the cut through the midpoint is the same as the total energy (i.e. heat) flux, and,due to the symmetry, the temperature of the midpoint can be assumed to be exactly the meantemperature of the ends (all these plausible assumption hold experimentally for linear media),which means that [L(∆x)∆x]∇T = [L(∆x/2)∆x/2]∇T , i.e. that L is inverse proportional to thelayer’s thickness. Similarly, we can consider several layers placed in parallel, and see that the totalenergy flux for given temperature difference sums up, so that L is proportional to the layer’s areaA. Introducing λ being the value of the corresponding coefficient for a slice of a unit area andunit thickness we get that in a homogeneous system J = λA∇T . After this we can introduce theheat flux density j = J/A and write j = λ∇T . Generalizing the discussion to higher dimensionsand noting that the heat always flows against the temperature gradient we can write down thevectorial equation
j = −λ∇T. (15)
The equations like this are the continuous analogues of the Onsager’s equations of linear nonequi-librium thermodynamics. Here the gradient of temperature plays the role of the thermodynamicalforce, and the heat flux appears as a response to the force. Such phenomenological equations forheat conduction, diffusion, electric conduction etc. were known much earlier than the Onsager’sgeneralizing discussion, and are the true historical basis of the theory.
Let us now consider the energy balance in our system. Let us first consider the medium asimmobile, and look at the slice of the system between x−∆x/2 and x+ ∆x/2. The total change
in the energy of a piece of system under discussion is
dU
dt= Jtot = J(x+ ∆x/2)− J(x−∆x/2)
= −S[j(x+ ∆x/2)− j(x−∆x/2)]
= −S∆xdivj.
25
Considering the internal energy density u = U/V (energy per unit volume) of the correspondingslice we get
du
dt= −divj,
a local conservation law. In the immobile system the total derivative du/dt is essentially the partialderivative quantifying the change of u as an explicit function of the time. Noting now that thelocal temperature is connected with the local internal energy of the system via ∆e = cv∆T , we get
∂T
∂t= − 1
cvdivj,
which together with Eq.(15) gives us the heat equation, a closed equation for the temperature
∂T
∂t= κ∆T,
where ∆ is now not a difference but a Laplace operator and κ = λ/cv is the heat conduction coeffi-cient (“thermal diffusivity”). If heat conduction in a mobile fluid is considered, the correspondingequation for the energy conservation gives us the energy balance for a fluid element (i.e. the changeof internal energy in Lagrange coordinates moving together with the element). We may then wantto change to the immobile (Eulerian) system, i.e. to the laboratory coordinate system, in whichthe velocity of our fluid element is v. We get
d
dtu =
[∂
∂t+ vgrad
]u
and obtain the heat conduction-advection equation
∂T
∂t+ vgradT = κ∆T.
The scheme we used in getting the equation is as follows: One considers the local situation closeto equilibrium and describes this via linear non-equilibrium theory. One changes to the localdescription and introduces flux densities. The equation for the flux density will be called the locallinear response law in what follows. In our case this is the Fourier’s law. One formulates thecorresponding conservation law (in the differential form). The combination of the linear responselaw and the conservation law gives us the transport equation.
According to the same lines we can obtain the diffusion equation (second Fick’s law)
∂n
∂t= D∆n
for the particles’ concentration n, as a consequence of the linear phenomenological equation
j = −Dgradn
(the first Fick’s law), where D is the diffusion coefficient, and local particle conservation8
∂n
∂t= −divj. (16)
8Note that in the case when the diffusion coefficient D(x) depends on the coordinates, this dependence, within thelinear theory, affects only the linear response equation, j(x) = −D(x)gradn(x), and doesn’t change the conservationlaw. The diffusion equation in this case reads ∂
∂tn(x) = ∇D(x)∇n(x), a so-called kinetic form. We have to notethat also other forms of diffusion equations with position- dependent diffusion coefficient are possible and will bediscussed in detail when dealing with random processes and microscopic derivation of the diffusion equation.
26
A note. While in the discussion above we have simply postulated that the particles’ flux is pro-portional to the gradient of the concentration, it still reasonable to return to the initial discussionand note that the corresponding driving force of the particles’ flux is the gradient of chemicalpotential.
Let us return to the diffusion phenomena and note that the particle flux, as we already discussedin the previous section, is caused by the gradient of the chemical potential, i.e. of the Gibbs’ freeenergy per mol or per particle µ. This is indeed true, since the reversible work R necessary tointroduce new particles into the system under isothermic and isobarixc conditions is equal to thechange of the Gibbs’ potential (free enthalpy) of the system. Here we use the molecular values,since n is a molecular (not molar) concentration. The value of µ depends on the partial pressure(and therefore on the concentration) of the corresponding particles, on the temperature and mayalso depend on external potential u(x) acting on the particle. The whole system is isothermal.While in the discussion above we could simply postulate that the particles’ current is proportionalto the gradient of the concentration here it is better to return to the initial discussion and note thatthe reversible work nwecessary to change the concentration, the change in the Gibbs’ potential.The diffusion phenomena (particle transport under isothermal and isobaric conditions) imply thepresence of at least two particle sorts which we call solvent and solute (tracer). This is theredistribution of tracers which we follow. Let us again consider a slice or a small part of thesystem of volume V . In the ideal case (at least when the mixture of the ideal gases is considered)the chemical potential of the component i (the index 1 numbers the tracer, the index 2 the solvent)is given by
µi(T, p,Ni) = gi(T, p) + kBT lnNi
Ntot
with Ntot =∑
iNi. Here gi(T, p) is the Gibbs’ potential of the corresponding particles in thecorrwesponding pure substance for given temperature and pressure, and the second term representsthe entropy of mixing. Under isobaric conditions the total number of particles N in the volumeis constant. In the last term Ni and Ntot can be changed for the corresponding concentrations niand ntot, and under isobaric conditions ntot is constant. The thermodynamic force per unit valumecorresponding to the particle number (i.e. to concentration) of species i = 1 in the absence of theexternal potential is F = −∂(kBT
−1)G/∂n1 = −[const+ lnn(x)] (where the tracer concentration,former n1 is changed for n).
The further discussion is absolutely identical to the one of the heat equation, and we obtainthe equation for the particle flux as a response to the changes in n
j(x) = −l∇ lnn(x) = − l
n(x)∇n(x),
where we associate the prefactor of ∇n(x) with the diffusion coefficient (and assume the changesin n(x) to be small to guarantee the applicability of the linear approximation). However theapproximation we did is quite weak, especially taking into account the unclear behavior in thecase of small concentrations. Experiments show that the diffusion coefficient is well-defined in alldilute solutions for small concentrations of solved particles, and tends to a constant (dependingon temperature) when the solute concentration tends to zero. Therefore it is indeed D = l
n(x)and
not l which is a reasonable macroscopic characteristic of a medium. We will return to this pointlater when we discuss stochastic derivation of the corresponding equations. All in all, the diffusionequation is “much better” than the linear theory.
We also have to note that if a single-component system (say ideal gas) is considered, macro-scopic diffusion is not a mechanism of concentration equilibration which might be observed in such
27
a system. Indeed, if the concentration of the gas is inhomogeneous, then there exists a gradient ofthe pressure, and therefore the main mechanism of transport will be the hydrodynamical motion,and not diffusion. We need at least a two-component system, where differences in the local con-centrations of the both components compensate the possible pressure differences in the case (sayin a gas) when the total concentration stays constant. The way to understand the situation is toconsider a full, two-component system (solute + solvent, or a mixture of two ideal gases) and theflows of both components under the assumption of the conservation of the total amount of parti-cles. Here we immediately see that the nature of of diffusion phenomena is quite involved9. Thetwo-component discussion in the absence of convective motions (the pressure everywhere in thegas is the same, and therefore the total concentration of the atoms is constant, n1(x) +n2(x) = n)however leads to very similar expressions for the minority component.
Analog we can consider the cross-effects between the diffusion and heat conduction (the Du-four and the Soret effects) by assuming for the corresponding flux densities
jq = −λ∇T + l12∇njn = l21∇T −D∇n
with the corresponding conservation laws. The Onsager-Casimir symmetry relations allow us thento find the relations between the cross-coefficients.
The last continuous equations we discuss is the one for the mass transport in the absence ofdiffusion, (e.g. in a one-component system), the Navier-Stokes equation. The force here isthe “normal” mechanical force, the conjugated variable is the “normal” coordinate, so that theflux would be the velocity. The corresponding conservation law is the momentum conservation,maybe in presence of the additional external forces like volume ones (e.g. gravitation) and pressuregradients. The complication now is that the flux density is a one of a vector (i.e. it is a second-rank tensor). The linear response equation assumes that the corresponding force is proportionalto the velocity; this is fluid friction. Assuming incompressibility (∇·v = 0) additionally simplifiesthe situation. Together with the mass conservation law this gives (for incompressible fluid in thestandard notation)
ρd
dtv = µ∇2v −∇p+ f ,
or
ρ
(∂v
∂t+ v∇v
)= µ∇2v −∇p+ f .
One can also mix all possible equations of this type together, with all possible cross-effects, andobtain the general hydrodynamic descriptions of all phenomena of interest amenable to such atheory, see the next chapter where I closely follow the book G. Kluge and G. Neugebauer, Grund-lagen der Thermodynamik, Spektrum Verl. 1994. The most important thing we learn from thisdiscussion is the fact that the linear theory does not necessary lead to linear equations.
2.5.2 The Fokker-Planck equation and the Einstein’s relation
Let us return to the diffusion phenomena and note that the particle flux, as we already discussedin the previous section, is caused by the gradient of the chemical potential, i.e. of the Gibbs’
9See e.g. pedagogical discussion in E. Bringuer “Anatomy of particle diffusion”, Eur. J. Phys. 30 1447-1470(2009) and E. Bringuer “From mechanical motion to Brownian motion, thermodynamics and particle transporttheory ”, Eur. J. Phys. 29 1243-1262 (2008). A lot of interesting information is contained in a book “Simple Brow-nian Diffusion: An Introduction to the Standard Theoretical Models” by Daniel Gillespie and Effrosyni Seitaridou,Oxford Univ. Press, 2013.
28
free energy per mol or per particle µ. This is indeed true, since the reversible work R necessaryto introduce new particles into the system under isothermic and isobarixc conditions is equal tothe change of the Gibbs’ potential (free enthalpy) of the system. Here we use the molecularvalues, since n is a molecular (not molar) concentration. The value of µ depends on the partialpressure (and therefore on the concentration) of the corresponding particles, on the temperatureand may also depend on external potential u(x) acting on the particle. The whole system isisothermal. While in the discussion above we could simply postulate that the particles’ current isproportional to the gradient of the concentration here it is better to return to the initial discussionand note that the reversible work nwecessary to change the concentration, the change in the Gibbs’potential. The diffusion phenomena (particle transport under isothermal and isobaric conditions)imply the presence of at least two particle sorts which we call solvent and solute (tracer). This isthe redistribution of tracers which we follow. Let us again consider a slice or a small part of thesystem of volume V . In the ideal case (at least when the mixture of the ideal gases is considered)the chemical potential of the component i (the index 1 numbers the tracer, the index 2 the solvent)is given by
µi(T, p,Ni) = gi(T, p) + kBT lnNi
Ntot
with Ntot =∑
iNi. Here gi(T, p) is the Gibbs’ potential of the corresponding particles in thecorrwesponding pure substance for given temperature and pressure, and the second term representsthe entropy of mixing. Under isobaric conditions the total number of particles N in the volumeis constant. In the last term Ni and Ntot can be changed for the corresponding concentrations niand ntot, and under isobaric conditions ntot is constant. The thermodynamic force per unit valumecorresponding to the particle number (i.e. to concentration) of species i = 1 in the absence of theexternal potential is F = −∂(kBT
−1)G/∂n1 = −[const+ lnn(x)] (where the tracer concentration,former n1 is changed for n). If there is an external potential u(x) acting on the tracers only, this getsto be a part of g1(T, p), and the thermodynamical force reads F = −[const+u(x)/kBT + lnn(x)].The further discussion is absolutely identical to the one of the heat equation, and we obtain theequation for the particle flux as a response to the changes of n in space
j(x) = l∇F = −l∇[lnn(x) + u(x)/kBT ]
= −l[∇n(x)
n(x)+∇u(x)
kBT
]= − l
n(x)[∇n(x) +
1
kBTn(x)∇u(x)].
If we associate the prefactor of ∇n(x) with the diffusion coefficient, we get
j(x) = −D∇n(x)− n(x)D
kBT∇u(x). (17)
In an electric field −∇u = eE, and the electric current is connected with the particles’ flux (weassume only one type of charge carriers) via je = ej where e is the charge of the charge carrier.Therefore in the absence of the concentration gradient we have je = ej = n(x) D
kBTe2E, i.e. the
Ohm’s law je = σE with σ = n(x)e2D/kBT . Our Eq.(17) is therefore a combination of the firstFick’s law and the Ohm’s law.
The microscopic interpretation of the second term in Eq.(17) is as follows: In the absence of theconcentration gradient the external potential gradient (i.e. the external force) induces the currentdensity proportional to this gradient: j(x) = −n(x) D
kBT∇u(x). Considering our system as being
29
consisting of independent, non-interacting particles (e.g. immersed in a fluid or in the puffer gas),we can assume that this current density is proportional to a mean velocity each particle acquires inthe external field, v = µf = −µ∇u, where µ now is the particle’s mobility. The particles’ currentdue to the potential gradient is then j = −nµ∇u. Comparing this with our previous expressionfor the current we get
µ =D
kBT,
i.e. that the mobility and the diffusion coefficient at the same temperature are connected by asimple relation (Einstein’s relation).
Combining Eq.(17) with the continuity equation Eq.(16) we get
∂n
∂t= ∇D∇n+∇(nµ∇u), (18)
i.e.∂n
∂t= ∇D∇n−∇(µfn).
Historically, D and µ or σ where introduced as independent phenomenological coefficients ofthe corresponding responses and the Einstein’s relation was proposed before the general Onsager’sapproach was formulated. This one can therefore be obtained also on a different way.
Note that the Einsteins relation connects not the off-diagonal terms in the Onsager’s matrixof response coefficients but involves the diagonal term, which in a diffusion equation would be astand-alone one. This is a relation of a very different nature than the Onsager’s symmetry one,and belongs to the class of fluctuation-dissipation relations, which we will later discuss in detailand hold in general for all responses to mechanical forces.
In our picture of independent particles the density n(x, t) is proportional to the probabilitydensity p(x, t) to find a particle at position x at time t, so that, due to the linearity of Eq.(18) theequation for p(x, t) reads
∂p
∂t= ∇D∇p−∇(µfp).
This equation for the probability density to find a particle at some given position is called theFokker-Planck equation.
Let us return to our Eq.(18) and assume that our system does possess an equilibrium state, towhich it will tend in course of the time (this exercise will later be given as a Homework). Thisstate is characterized by vanishing of the particle flux, i.e. by the probability density function(PDF) fulfilling the equation
D∇p+ µp∇u = 0.
The solution for p(x) in equilibrium then reads
p(x) = A exp(− µDu(x)
).
Let us compare this property with what we know about the equilibrium distribution of the particles’coordinates. We know that the corresponding Boltzmann distribution for positions and velocitiesof classical particles is given by the Boltzmann-factor
w(p,q) ∝ exp
(−H(p,q)
kBT
)
30
where H is the Hamilton function of the whole system. The kinetic part depending on the mo-menta always decouples, and (for non-interacting particles) the potential energy factorizes intocontributions of single particles. Integrating over all coordinates and momenta except for one, weobtain
p(x) ∝ exp
(−u(x)
kBT
),
and comparing it with the equation for p above we get µ/D = 1/kBT , i.e. again obtain theEinstein’s equation as a consequence of the local equilibrium.
2.5.3 General hydrodynamic description
As already stated, a transport equation for some quantity (mass, energy, momentum, etc.) isobtained from a corresponding conservation law (which may include production terms e.g. forentropy) and the corresponding linear response equation, which follows from identification of forcesand fluxes in the corresponding entropy production rate, and assuming J = LX (note that J andX may be scalars, vectors, tensors etc., and the mathematical nature of L corresponds to theproperties of these J and X, as it will be discussed below)10.
Let A be some extensive quantity, and a(r, t) the corresponding density, so that the amountof A in the volume V is A =
∫Va(r, t)dr. Let moreover j(r, t) be the flux density of A and s(r, t)
the corresponding production rate of A (a source term in the corresponding equation). Then thegeneral continuity equation reads
∂a
∂t+∇j = s.
Note that a might be scalar, vector or tensor. The energy, the momentum or the angular mo-mentum are exactly conserved. In the non-relativistic case the same is true for the total mass. Inthese cases s = 0. The entropy can be produced but not destroyed. The particle numbers mayboth grow or decay if (chemical) reactions in the system are allowed for. The same is true for themasses of different components of the reacting system.
Balance of mass. In a reaction the corresponding conservation law is given by the reactionscheme
m∑i=1
(−νi)Ci k∑
i=m+1
νiCi
where νi are the stoichiometric coefficients of the reaction. For example in the reaction scheme
2H2 + O2 2H2O
we have m = 2 (two educts of reaction), k = 3 (total number of components: two educts anda product), ν1 = −2, ν2 = −1 and ν3 = 2. As another example we discuss the simple infectionpropagation scheme
I + S→ 2I
with I - infected, S - susceptible. This is a simplest autocatalytic reaction (let me remind youthat the reaction is called catalytic if some of its educts and products coincide). The simplifieddescription of this reaction is S → I catalyzed by I.
10 Here I closely follow the book G. Kluge and G. Neugebauer, Grundlagen der Thermodynamik, Spektrum Verlag,1994.
31
The balance equation for the (mass) density of the component k then reads
∂ρk∂t
+∇jk = Γk
with the production term Γk = ωνimi, where mi is the mass (or the mole mass) of the correspondingparticles, and ω is the reaction rate, the total number of elementary reaction acts per unit time.
If several reactions (R on the total) are running in the system, there are on total R differentproduction terms: Γk =
∑Rr=1 ωrνk,rmk. Here ωr denotes the rate of the corresponding reactions,
the number of the elementary acts of the reaction r per unit volume per unit time. A reversiblereaction like the first one above corresponds to two reaction running in the opposite directions.Summing up all the production terms we get, due to the mass conservation in the non-relativisticcase
K∑k=1
Γk =K∑k=1
r∑r=1
ωrνk,rmk = 0.
Introducing the total density
ρ =K∑k=1
ρk (19)
and the center-of-mass velocity
ρv =K∑k=1
ρkvk (20)
we see that this fulfills the usual continuity equation
∂ρ
∂t+∇(ρv) = 0
without production terms. Using the expression for the material (or substantial) derivative
d
dt=
∂
∂t+ v∇
and the introducing the specific volume (the volume per unit mass) v = 1/ρ we may write
ρdv
dt−∇v = 0.
Now we can introduce the specific values of extensive variables: the properties per unit mass:a = a/ρ. For these we get
∂a
∂t=∂ρa
∂t= ρ
da
dt−∇(av).
From this and from the general balance equation it follows that
ρda
dt+∇(j− av) = s.
The vector J = j − av is the conductive mass flux density; its second term av is called theconvective flux density. If we have k components in our system, the continuity equation for eachof them reads
ρdckdt
+∇Jk =R∑r=1
ωrνk,rmk, (21)
where ck is the concentration of the component k. From Eqs.(19) and (20) it follows that∑K
k=1 Jk =0.
32
Balance of momentum. The momentum is a vectorial property, and its flux density is a secondorder tensor with Cartesian elements −τ (the tensor of momentum flux density is essentiallyconnected with mechanical stress tensor, c.f. electrodynamics). The component-wise balance isgiven by the equation
ρdvαdt−∑β
∂τα,β∂xβ
= fα.
The production term is the mechanical force. In a vector notation we have
ρdv
dt−∇τ = f . (22)
Here f is the overall force (per unit volume) acting on the system. In case of several componentswith different properties (e.g. some of them may be charged and others not) the last term can wewritten as
f =K∑k=1
fk =K∑k=1
ρfk
with fk = fk/ρ.
Balance of energy. As above
ρde
dt+∇je = q. (23)
The production term is the work of external forces per unit volume
q =K∑k=1
vkfk.
The energy of the unit volume of the fluid consists of the corresponding kinetic energy of thisvolume and of its internal energy, so that for the last we have
u = e− 1
2ρv2.
The balance of kinetic energy follows from the momentum balance: multiplying both sides ofEq.(22) by v we get
ρvdv
dt− v∇τ = v
k∑k=1
fk.
The second term in this equation written component-wise reads
v∇τ =∑α
vα∑β
∂τα,β∂xβ
.
The second term in the right hand side can be rewritten using the fact the the stress tensoris symmetric τα,β = τβ,α. This symmetry follows from the angular momentum conservation in(mechanical) equilibrium. Let us introduce another symmetric tensor: the one of deformationvelocities
Vαβ =1
2
(∂vα∂xβ
+∂vβ∂xα
).
33
We first note that ∑α
∑β
vα∂τα,β∂xβ
=∑β
∑α
∂(vατα,β)
∂xβ−∑α
∑β
τα,β∂vα∂xβ
,
and then note that due to the symmetry of the stress tensor∑α
∑β
τα,β∂vα∂xβ
=∑α
∑β
τα,β1
2
(∂vα∂xβ
+∂vβ∂xα
)= τ : V
where A : B denotes the trace of the Hadamard product of the corresponding matrices in the formA : B =
∑α,β aαβbαβ = tr(ABT ) (this is an analogue of scalar product for tensorial quantities).
Now the kinetic energy balance equation reads
ρd
dt
(v2
2
)−∇(τv) = −τ : V + v
k∑k=1
fk.
The balance of the internal energy can be obtained by subtracting this last equation from Eq.(23):
ρdu
dt+∇(q) = τ : V +
K∑k=1
Jk fk
with q = je + τv being the conductive heat flux density.
Balance of entropy. Here we again have the standard balance equation for a scalar, of thetype
ρds
dt+∇Js = σ
where σ is the entropy production density, which is discussed below.
Density of entropy production. For a fluid system, starting from the general form of theentropy
dS =dU + pdV −
∑Ki=1 µidni
T
we get the entropy production
ds
dt=
1
T
du
dt+p
T
dv
dt− 1
T
∑k
µkdckdt
(here µk is chemical potential per unit mass). This equation has to be brought into the form ofthe balance equation for the entropy. To do so we multiply its both parts by ρ and use the balanceequations for masses, energy and momentum to bring it into the form
ρds
dt= − 1
T∇q +
1
T
∑k
µk∇Jk +
p
T∇v +
1
T
(τ : V +
∑k
fk∇Jk
)− 1
T
K∑k=1
R∑r=1
ωrνk,rµkmk.
34
The equation then is rewritten by rearranging the terms and using differential identities to bebrought into the form
ρds
dt= −∇
(q
T−∑k
µkT
Jk
)+ q∇
(1
T
)
−∑k
Jk
(∇ µkT− fkT
)− 1
T(τ + pI) : V −
R∑r=1
ωrArT
with Ar =∑
k µkνkr being the activity of the corresponding component in the given reaction.Assotiating
Js =1
T
(q−
∑k
µkJk
)with the conductive entropy flux density, we get the expression for the entropy production density.Using the fact that
∑Kk=1 Jk = 0 we can exclude the last flux from this expression and get
σ = q∇ 1
T−
K−1∑i=1
Ji
(∇ µi − µK
T− fi − fK
T
)
+(τ + pI) :V
T−
R∑r=1
ωrArT.
The entropy production density has the form σ =∑E
j=1 JjXj where j numbers the physicaleffects. Our equation describes four effects (Greek letters denote the Cartesian components):
• Heat conduction: flux and force are the vectors: J = q, X = ∇(1/T ).
• Diffusion (including transport by external forces): flux and force are vectors J = Jk, X =
−∇ µk−µKT
+ fk−fKT
, there are K − 1 such fluxes on the total.
• Viscosity: flux and force are symmetric tensors: J = τ + pI, X = 12V .
• Reactions: flux and force are scalars: J = wr, X = −Ar/T , there are R such pairs on thetotal.
Liner response. We consider further a fluid system. There exists a general symmetry principlewhich predicts, which flux densities and which forces can at all be connected by phenomenologicalcoefficients (or, in other words, which coefficients are inevitably vanishing); this one is of empiricalnature. It states that the scalar fluxes may only be connected with scalar forces, the vector fluxeswith vector forces, and tensor fluxes with tensor forces of the same symmetry. Its “philosophical”formulation is that the symmetry of the effect is the same as the symmetry of the cause. Thisstatement is called the Curie-principle. If we have a specific microscopic model for the effect,we see that it always hold, but there is no general proof. This means that the phenomenologicalequations for properties of different tensor structure are decoupled.
When considering the scalar forces we must note that the trace of the deformation velocitytensor is also a “linear” scalar force trV = ∇v. The corresponding scalar flux is the trace ofτ + pI, i.e. trτ + 3p.
35
Therefore the equations for the scalar properties read:
ωr =R∑s=1
Lrs
(−AsT
)+ Lωr
(1
3T∇v
)
trτ + 3p =R∑r=1
Lτr
(−ArT
)+ Lτ
(1
3T∇v
)(24)
Lrr ≥ 0, Lτ ≥ 0 and the symmetry relations are
Lrs = Lsr
Lωr = −Lτr
since the activity is an even variable, and the velocity divergence is an odd one.The linear phenomenological relations for the vectorial fluxes are:
q = Lq∇(
1
T
)+
K−1∑k=1
Lqk
[−∇ µk − µK
T+
fk − fKT
]
Jk = LJk
(1
T
)+
K−1∑r=1
Lrk
[−∇ µr − µK
T+
fr − fKT
]with Lq ≥ 0, Lkk ≥ 0 and symmetry relations Lsk = Lks and Lqk = LJk . These equation is written foran isotropic medium, where the “embryonic diffusion coefficients” Lrk are scalars. In an anisotropicmedium they are symmetric tensors of rank 2, and there are more indices: force components andderivatives of the chemical potential with respect to different directions get different couplings.We concentrate here on isotropic media, and disregard all these complications.
The relation between the trace-free parts of stress tensor and the tensor of deformation velocitiesis given by
τα,β −1
3
∑γ
τγγ =L
2T
[∂vα∂xβ
+∂vβ∂xα− 2
3
(∑γ
∂vγ∂xγ
)δαβ
](25)
with L ≥ 0. From Eq.(25) and Eq.(24) it follows that
ταβ = −pδαβ +L
2T
(∂vα∂xβ
+∂vβ∂xα
)+Lτ − 3L
9T
(∑γ
∂vγ∂xγ
)δαβ −
R∑r=1
Lτr3
(ArT
)δαβ.
The tensor ταβ + pδαβ is the friction tensor.
Hydrodynamic equations. As in simple examples before, the closed equations for the timeevolution of the variables of interest are obtained by substitution of the linear response equationsinto the balance ones. For the simplest situation of a single-component system we have a systemof three equations as following from the balance of the momentum, mass and internal energy:
ρdv
dt= −∇p+ η∆v +
(η3
+ ζ)∇∇v + f
(Navier-Stokes-equation),∂ρ
∂t+∇(ρv) = 0
36
(continuity equation), and
ρdu
dt− κ∆T = −p∇v + 2ηV : V −
(2
3η − ζ
)[∇v]2 ,
(the heat equation). The systems of equation is closed by thermodynamics caloric and thermalequations of state
p = p(T, ρ)
u = u(T, ρ).
Here we use the notation for the kinetic coefficients which is usual in hydrodynamics, κ = Lq/T 2
for the heat conduction coefficient, η = L/2T for shear viscosity, and ζ = Lτ/9T for the volumeviscosity which are typically assumed constant in the relevant domains of parameters.
“Classical kinetics”: Linear and non-linear approaches to chemical reactions. Letus consider a multicomponent system, and concentrate only on the part of entropy productionconnected with reactions (i.e. for the time being assume that our system is well mixed, locally atrest, etc.):
σc = −R∑r=1
ωrArT
(the index c stands for “chemical”, inducing all nuclear reactions, biological changes, infections,etc.), cf. Eq. (24). Let us moreover consider a single reaction (not forgetting that a reversiblereaction corresponds to a couple of independent reactions flowing in opposite directions). Let usseparate the activities of the educts and the products and write
σ =ω
T(A+ − A−)
with
A+ = −m∑i=1
νiµi, A− =K∑
i=m+1
νiµi
so that Ar = A− − A+.The linear Ansatz for the reactions is ω = −L (A/T ). The problem here is however that this
Ansatz definitely working close to equilibrium for a particular reaction, may easily break downeven if the system as a whole is close to equilibrium but a particular reaction is not (for examplewe just have mixed the reactants to start the reaction; we note that in many cases the relaxationtimes of all other relevant thermodynamical variables might be much shorter than that of thereactions). In this case one often uses the nonlinear Ansatz
ω = Λ
(eA+kBT − e
A−kBT
)(which will be motivated a bit later). The Ansatz is not bad since it guarantees the non-negativityof the entropy production since for each x and y one has (ex − ey)(x − y) ≥ 0. The expressionabove can be rewritten as
ω = ΛeA−kBT
(eArkBT − 1
).
37
For Ar kBT the expression in the brackets can be expanded into a Taylor series, and we get thelinear Ansatz
ω = −(
Λ
kBeA−kBT
)ArT
with L = (Λ/kB)eA−kBT .
The motivation for the non-linear Ansatz follows from the mass action law. For a reaction inan ideal solution (or in a mixture of ideal gases) we have
µi(T, p,Ni) = gi(T, p) + kBT lnNi
N,
where the first term on the r.h.s. denotes the Gibbs’ potentials (free enthalpies per molecule) ofpure substances at given temperature and pressure, gi = hi − Tsi, and the second term is theentropy of mixing. The Gibbs potential can be expressed via the enthalpies per molecule fromwhich it follows that
µi(T, p,Ni) = hi(T, p0)− T∫ T
T0
cp(T )
TdT + kBT ln
p
p0
+ kBT lnNi
N
= f(T ) + kBT lnp
p0
+ kBT lnNi
N
where T0 and p0 are some reference values for temperature and pressure giving us the integrationconstants, and f(T ) is a function of the temperature only. The pressure p can be expressed viathe thermal equation of state p = NkBT/V . Introducing the concentrations Ci = Ni/V we canwrite
µi = f(T ) + kBT ln kBT + kBT lnN
V+ kBT ln
Ni
N= µVi (T ) + kBTCi,
with µVi being (for gases) the function of the temperature only (in a general case of ideal solutionit may still depend on pressure).
Now we get for the partial affinities
A+ = kBT lnm∏i=1
C−νii −m∑i=1
νiµVi (T )
and
A− = kBT lnK∏
i=m+1
Cνii +
K∑i=m+1
νiµVi (T ),
so that our nonlinear Ansatz takes the form
ω = k+
m∏i=1
C−νii − k−K∏
i=m+1
Cνii
with the reaction rate constants of the forward and the backward reaction
k+ = Λ exp
(−∑m
i=1 νiµVi (T )
kBT
), k− = Λ exp
(∑Ki=m+1 νiµ
Vi (T )
kBT
).
38
From this we obtain the equations for the concentrations of the educts of the reaction. Notethat ω defines the change of the numbers of corresponding of molecules per unit time, so thatdNi/dt = νiω, or for concentrations: dCi/dt = νiω/V = νiω, where ω is the change of numbers ofthe molecules per unit volume per unit time.
The reason to chose the Ansatz has to do with what we know about reactions in the gas phase,namely with the mass action law (MAL) of thermodynamics (Guldberg and Waage 1864, 1867).If we take for the partial pressures pi = CikBT , then in equilibrium (ω = 0) we get
K∏i=1
Cνii =
k+
k−=
K∏i=1
(pikBT
)νii.e.
K∏i=1
pνii = (kBT )∑i νi
(k+
k−
)= Kp(T )
where Kp(T ) is the constant of the MAL, which is the standard thermodynamical formulation ofthe law. Its “backward engineering” gives us our nonlinear Ansatz. Therefore the whole kineticscheme based on the Ansatz is often called a MAL in the general sense.
Here, as a reminder, the derivation of the thermodynamic MAL in equilibrium. Let Ni be theparticle numbers of components. For any change δNi of these numbers in a reaction, the reactionequation implies that
δN1
ν1
=δN2
ν2
= ... =δN1
ν1
= δξ,
where the variable ξ is called the extent of reaction (measured in particle numbers; chemiststypically measure it in moles. Our ω is essentially dξ/dt.). Now we put down the total freeenthalpy of the system
G(T, p,Ni) =∑i
Nigi(T, p) + kBT∑i
Ni lnNi
N
where the first term is the sum of free enthalpies of pure components, and the second one is themixing entropy. The variation of this quantity due to small variations of the particles’ numbersmust vanish in equilibrium:
0 = δG =∑i
δNi
[gi(T, p) + kBT ln
Ni
N
]+ kBT
∑i
Ni
(δNi
Ni
− δN
N
).
The second sum vanishes (since N =∑
j Nj):∑i
Ni
(δNi
Ni
− δN
N
)=∑i
δNi −∑i
Ni
∑j δNj∑j Nj
= 0.
The particles numbers’ changes in the first equation can then be expressed via δξ: δNi = νiδξ, sothat
0 = δG = δξ∑i
νi
[gi(T, p) + kBT ln
Ni
N
],
i.e.
kBT∑i
νi lnNi
N=∑i
νigi(T, p),
39
The further step for ideal gases is to note that Ni/N = pi/p, so that∏i
pνii = exp
[−∑νigi(T, p)
kBT
]p∑j νj ,
and to show that the right hand side is independent from pressure, i.e. the function of thetemperature only.
The nonlinear Ansatz for reactions gives an extremely transparent scheme (“classical kinetics”)leading to systems of ordinary differential equations. For example, for A+B C (recombination)we get
dA
dt= −k+AB + k−C
dB
dt= −k+AB + k−C
and for the reverse reaction (product and educts interchanged)
dC
dt= −k−C + k+AB.
These equations of classical kinetics are extremely simple to understand: for a reaction between Aand B to take place, the A and B particles have to meet, and the number of such close encountersin a well-mixed system is proportional to the product of the concentrations of A and B. This showsthat our initial assumption was correct. The reaction rate k+ depends on how often such encounterstake place (transport rate) and on the probability that the elementary act of the reaction takesplace on the encounter (elementary reaction probability). Later on we will be concerned aboutcalculating these contributions for some particular cases.
Two more complicated examples are famous Brusselator and Oregonator schemes
A→ X
B + X→ Y + C
2X + Y → 3X
X→ D
and
A + Y → X
X + Y → P
B + X→ 2X + Z
2X→ Q
Z→ nY.
describing (within the corresponding minimal models) chemical oscillations, as first observed inBelousov-Zhabotinsky reaction (on the total 11 reactions with 18 elementary acts, 12 species and21 intermediates).
40
Adding the diffusion term leads to systems of reaction-diffusion equations, being (typically)nonlinear partial differential equations of the type11
∂Ci(x, t)
∂t= Di∆Ci(x, t) +Ri(Cj),
where Ri denotes the nonlinear reaction term as following from classical kinetics. The reaction-diffusion equations for the schemes above describe complex structure formation.
A much simpler example, for an irreversible I+S→ 2I reaction with total conservation of parti-cle number leads to a famous FKPP (Fischer-Kolmogorov-Petrovsky-Piskunov) equation. Assum-ing the total concentration I + S = C to be homogeneous in space, and the diffusion coefficientsof particles equal, we get a single closed equation for I:
∂I
∂t= D∆I + kI(C − I)
which will be partily investigated as a homework12.
3 Fluctuations and dissipation
Let us now return to our simple (“discrete”, in the sense “not continuous”) systems and discussin more detail the relations between response coefficients and the properties of the correspondingfluctuations. Here we consider the simplest realization of the corresponding theory. Our aim isto find a relation between the response of the system to the external force, i.e. the equilibriumvalue of x corresponding to given, fixed F , and the properties of fluctuations. Let us assume thatin the absence of external force the equilibrium value of 〈x〉 is 〈x〉 = 0 and consider our systemas prepared in some state characterized by some mean value of 〈x(t = 0)〉 = 〈x0〉 6= 0, whichcan be achieved by applying a constant force F ; the system at a fixed force is in an equilibriumstate. At t = 0 this force is switched off, so that the state prepared seizes to be equilibrium; wenow follow the equilibration of our system to a new equilibrium state corresponding to F = 0 and〈x(t→∞)〉 = 0.
3.1 Linear response to an external force
We consider the system under the influence of our external force (e.g. of a normal mechanical one)which is then switched off at t = 0, F (t) = F0 for t < t0 and F (t) = 0 for t > 0. Keeping the forceconstant we were able to create the initial state of the system characterized by
〈x(t = 0)〉 =
∫p(x|F0)xdx,
11 The corresponding equations follow from the balance equations for the mass, Eq.(21), translated into the oneof the concentrations of each component
∂Ci/∂t+∇ji = ωνi
and from the fact that, according to the Curie principle, there are no cross-effects between the scalar chemical forceand the vectorial force (concentration gradient translated into the gradient of the chemical potential).
12In the discussion of the Homework we see, that the assumptions of homogeneity (“well-mixedness”) leading tothe mixing entropy forms used in MAL can be easily violated in course of the reaction. Such kinetics, not describedby the MAL, are often termed “fluctuation-dominated”.
41
where the corresponding probability density p(x|F0) is proportional to the Boltzmann factor,
p(x|F0) ∝∫
exp
(−H0(...x...) + U(x)
kBT
)dx1, ..., dxndp
where H0(x, p, xi, pi) is the unperturbed Hamilton function of our system, the one in the absenceof the external force, and U(x) = −xF0. The integration goes over all momenta, but skips thecoordinate x of the particle which the force is applied to. We can also consider the vectorial variant,or force acting on many particles, etc., by skipping several coordinates. The overall structure ofthe theory stays the same. We can rewrite our equation as
p(x|F0) ∝ exp
(S(x)
kB
)exp
(xF0
kBT
). (26)
where S(x) is the conditional entropy, i.e. the entropy calculated when keeping the variable xfixed. The behavior of p(x) for t→∞ after switching off the force is given by the same expression,where F0 is set to zero. In this case we assume 〈x(t→∞)〉 = 0. One can moreover consider x tobe some thermodynamical variable and F the corresponding thermodynamical force. Here, we canimmediately start from our expression Eq.(26) which is essentially simply an other way to writeEq.(3). In this case F0 is the value of the thermodynamical force conjugated to x, which is neededto prepare the system in a state with given 〈x(t = 0)〉 which is different from the value 〈x〉 = 0the variable would assume in a system without external action (think of a paramagnetic systemwith x = M being the magnetization and F = H, the external magnetic field necessary to keepthis magnetization constant).
Let us now consider the behavior of the mean value of x at time t for a given initial coordinatex0, i.e. a conditional mean 〈x(t)|x0〉, and assume that the behavior of this mean as a functionof time is known. Since we know the distribution of x0 for given F0 we can also obtain the timeevolution of 〈x(t)〉 after preparation of the initial state by a given force. We can namely write〈x(t)|F0〉 as a double average (the second average is taken over the values of x0 for given F0)
〈x(t)|F0〉 = 〈〈x(t)|x0〉〉 =
∫〈x(t)|x0〉p(x0|F0)dx0. (27)
Let us assume F0 to be small enough and expand the corresponding exponential around theequilibrium value of x = 0,
p(x|F0) =exp
(S(x)kB
)(1 + xF0
kBT+ ...
)∫
exp(S(x)kB
)(1 + xF0
kBT+ ...
)dx.
We now assume that close to the equilibrium the entropy S(x) is a quadratic function of x (with
a maximum at x = 0), so that in the denominator we can take∫
exp(S(x)kB
)(1 + xF0
kBT+ ...
)dx =∫
exp(S(x)kB
)dx + O(F 2
0 ) and neglect the second-order term in force. Therefore for F0 small (and
for (F0/kBT )√〈x2〉eq 1, which guarantees the smallness of the further terms) we get
p(x|F0) 'exp
(S(x)kB
)∫
exp(S(x)kB
)dx
(1 +
xF0
kBT
)= peq(x)
(1 +
xF0
kBT
)
42
where peq(x) is the probability density to find the value of x in equilibrium, without the externalforce. Substituting this in our Eq.(27) we get
〈x(t)|F0〉 =
∫〈x(t)|x0〉
(1 +
x0F0
kBT
)peq(x0)dx0
=F0
kBT
∫〈x(t)|x0〉x0peq(x0)dx0
=F0
kBT〈〈x(t)|x0〉x0〉x0
=F0
kBTCxx(t). (28)
In transition from the first to the second line the integral∫〈x(t)|x0〉peq(x0)dx0 vanishes since it is
equal to the mean value of x(t) measured at equilibrium without the external force, 〈x(t)〉 = 0, inthe line above the last one we recognize the construct we have seen in Sec. 2.2, i.e. the correlationfunction.
Note: Taking t = 0 we see that the l.h.s. of the equation gets to be 〈x|F0〉, i.e. corresponds tothe mean value of the variable x which it takes at equilibrium under fixed force F0 (assumed smallto guarantee the linearity of response) 〈x|F0〉 = 〈x(F0)〉 = KF0, where K is the correspondingresponse coefficient (susceptibility). On the other hand Cxx(0) = 〈x2〉 is the mean squared valueof x in equilibrium in the absence of the force. We see: K = 〈x2〉/kBT .
3.2 Response function
Up to now we only considered a response to a constant force which was then switched off. Letus now assume some general time dependence F (t′). We assume that the system was preparedlong ago, and the the properties of preparation procedure (initial conditions) were forgotten anddo not play any role at the times at which the system is observed. Example of such a procedureis adiabatic switching of the force in a system which was at equilibrium at t → −∞, and thenfollowing the evolution. The dependence of x on the force will still be considered linear. The mostgeneral form of such dependence is given by the integral expression
〈x(t)〉 =
∫ t
−∞ψ(t− t′)F (t′)dt′. (29)
The response function ψ(t) describes the behavior of 〈x(t)〉 to a short (δ-like) pulse of externalforcing F (t) = δ(t− t0): in this case
〈x(t)〉 = ψ(t− t0).
Since the system at t→ −∞ was in equilibrium (and in the absence of force), one has to assumethat for t < t0 we had 〈x(t)〉 = 0, so that the function ψ(t) has to fulfill the causality conditionψ(τ) = 0 for τ < 0.
The same assumptions can be done for continual systems. An example of such relation for acontinual system is e.g. the dielectric response
D(t) =
∫ t
−∞ε(t− t′)E(t′)dt′,
43
where ε(t) is the dielectric constant, which, in media with dispersion, may be time-dependent. Thesame relations hold if the response to the force is not a “coordinate” but a current, e.g.
j(t) =
∫ t
−∞σ(t− t′)E(t′)dt′
is the corresponding generalization of the Ohm’s law13.Let us now return to our Eq.(28) with the force which was switched off at time t = 0, i.e. take
F (t) = F0Θ(−t) where Θ(t) is the Heaviside theta-function. The l.h.s. of Eq.(28) for t > 0 nowreads
〈x(t)|F0〉 = F0
∫ t
−∞ψ(t− t′)Θ(−t′)dt′ = F0
∫ 0
−∞ψ(t− t′)dt′ = F0
∫ ∞t
ψ(t′′)dt′′
with t′′ = t− t′, and the whole Eq.(28) takes the form F0
∫∞tψ(t′′)dt′′ = (F0/kBT )Cxx(t), so that∫ ∞
t
ψ(t′′)dt′′ =1
kBTCxx(t). (30)
Differentiating w.r.t. to t we get
d
dtCxx(t) = −kBTψxF (t). (31)
Since in what follows we will have to do with different response functions, it is reasonable to denoteby subscripts what the variables are, i.e. what variable x and as a function of what force F isconsidered: we already used this notation above, writing ψxF (t).
3.3 Spectrum of fluctuations
From electrodynamics course we know that the relations like the one between D(t) and E(t) aremore conveniently described in the frequency (Fourier-) domain. Therefore, in order to proceedfurther we have to learn how to describe correlations in the Fourier domain.
Passing to the Fourier domain we get for the correlation function of some variables i and j
Cij(ω) =
∫ ∞−∞
dτCij(τ)eiωτ ,
the time- and frequency domains are distinguished by explicitly writing the corresponding variableas an argument. The value of Sij(ω) = 2πCij(ω) is called a spectral density corresponding to thecorrelation function of the variables i and j.
13Note that in spite of similarities of the linear response the cases when the response is a thermodynamicalcoordinate (i.e. if the force acting on the system is constant, the system will reach a state of thermodynamicalequilibrium in the course of time) and when it is a flux (the system reaches a non-equilibrium steady state), thefluctuation-dissipation relations for these cases (said to be of the first or of the second kind) are different. Below,we will concentrate on the first kind of such relations. The Einstein’s relation discussed in the previous section,belongs to the second kind.
44
3.3.1 Fluctuation-Dissipation Theorem in Fourier representation
Let us return to our Eq.(31), multiply its both parts by eiωt and integrate over all t > 0:∫ ∞0
(d
dtCxx(t)
)eiωtdt = −kBT
∫ ∞0
ψxF (t)eiωtdt.
We now perform partial integration in the l.h.s. and get (assuming that Cxx(t) → 0 for t → ∞,i.e. that the fluctuations decouple, as it was the case within the Onsager’s theory) −Cxx(0) −iω∫∞
0Cxx(t)e
iωtdt = −Cxx(0)− iωC+xx(ω), where C+
xx(ω) is a one-sided Fourier transform of Cxx(t).Thus,
kBTψ(ω) = Cxx(0) + iωC+xx(ω)
(note that ψ(t) vanishes identically for t < 0 and therefore its one-sided Fourier transform is simplyits Fourier transform). Now we repeat the procedure multiplying both parts of Eq.(31) times e−iωt,and get kBTψ(−ω) = Cxx(0) + iωC+
xx(−ω), or
kTψ∗(ω) = Cxx(0)− iω[C+xx(ω)]∗.
We now subtract this equation from the previous one:
kBT [ψ(ω)− ψ∗(ω)] = iωC+xx(ω) + [C+
xx(ω)]∗.The r.h.s. of the equation is the sum of two integrals:
C+xx(ω) + [C+
xx(ω)]∗ =
∫ ∞0
Cxx(t)eiωtdt+
∫ ∞0
Cxx(t)e−iωtdt
Noting that Cxx(t) = Cxx(−t) we change the variable of integration in the second integral tot′ = −t and write∫ ∞
0
Cxx(t)e−iωtdt =
∫ 0
−∞Cxx(−t′)eiωt
′dt′ =
∫ 0
−∞Cxx(t
′)eiωt′dt′.
Now both integrals sum up to∫∞−∞Cxx(t)e
iωtdt = Cxx(ω), and the whole equation reads
kBT [ψ(ω)− ψ∗(ω)] = iωCxx(ω).
Since ψ(ω)− ψ∗(ω) = 2Imψ(ω) we may write
Cxx(ω) =2kBT
ωImψ(ω). (32)
We can rewrite this equation in a slightly different form: using the fact that
ψ(ω = 0) =
∫ ∞0
ψ(t)dt
and our Eq.(30) we get:
ψ(ω = 0) =Cxx(t = 0)
kBT≡ 〈x
2〉kBT
we may write kBT = Cxx(t = 0)/ψ(ω = 0) and
Cxx(ω) = Cxx(t = 0)2Imψ(ω)
ωψ(ω = 0).
Note: If the dimension of the physical observable x is [X], then the dimension of Cxx(t) is [X2],and the dimension of Cxx(ω) is [X2 · T], so that the dimension is the previous equations are OK.Once again, Cxx(t) and Cxx(ω) are not only two different functions, but the two functions havingdifferent dimensions!
45
3.4 More on random processes.
Note that here we are speaking about random processes, and not about the systems which generatethem. Therefore the definitions are different from ones in what is called the ergodic theory.
3.4.1 Some definitions concerning random processes: Ergodic hierarchy.
Here are some definitions. A random process x(t) is stationary if 〈x(t)〉 is time-independent:the whole situation is on average invariant with respect to time shift (translation). One canconsider other random processes related to x(t), like x(t)x(t + τ), x(t)x(t + τ1)x(t + τ2), etc.If the process is truly stationary, than all the means, Cxx(t, t + τ) = 〈x(t)x(t + τ)〉 = C(τ),Cxxx = 〈x(t)x(t + τ1)x(t + τ2)〉 = C(τ1, τ2) etc. depend only on the differences of the timearguments (time lags). If this takes place only for some lower correlation functions (provided theyexist) than the process will be said to be second-order stationary, third-order stationary etc.
A constant process x(t) = xi taking different constant values in its different realizations isstationary (assumed all moments of x exist), the process x(t) = sin(t) is non-stationary. Theprocess x(t) = sin(t+ φi) where φi is a random initial phase homogeneously distributed on [0, 2π]is stationary again.
Now let us define a time average
x(t) = limT→∞
1
T
∫ T
0
x(t′)dt′.
For multipoint properties we define the averages as
x(t)x(t+ τ) = limT→∞
1
T
∫ T
0
x(t′)x(t′ + τ)dt′
etc. Some authors (taking into account the experimental data acquisition procedure in a finite runof duration T ) modify the definition to
x(t)x(t+ τ) = limT→∞
1
T − τ
∫ T−τ
0
x(t′)x(t′ + τ)dt′.
The existence (i.e. convergence in probability) of the corresponding limit is called a weak ergodicityproperty in some branches of the theory of random processes. This property means that x(t) orthe corresponding multipoint values tend to a deterministic number.
For a constant process the limit limT→∞1T
∫ T0x(t′)dt′ = xi is still fluctuating. This process is
not even weakly ergodic. For the process x = sin(t) we have limT→∞1T
∫ T0x(t′)dt′ = 0, and this is
a weakly ergodic one. The same is true for x(t) = sin(t+ φi). The process is (first-order) ergodicif it is stationary and weakly ergodic, and if
〈x(t)〉 = x(t)
(assumed that both exist). Note that the stationarity is a prerequisite for the ergodicity becausethe right hand side, if it converges to a limit, is time-independent, while the left hand side istime-dependent for non-stationary processes.
Typically one saves on defining the weak ergodicity property (and starts exactly with thisdefinition), but I introduce it explicitly because it will be later important in my course on disorderedsystems. The same definitions can be given for multipoint quantities. The corresponding quantitieswill define the ergodicity in second-, third-, etc. order. If the process is ergodic in all orders wewill call it truly ergodic.
46
3.4.2 Conditions for ergodicity.
Let us consider the ergodicity of a single point observable x(t) (i.e. the first-order ergodicity). Let
us consider under what conditions the expression limT→∞1T
∫ T0x(t′)dt′ tends to a deterministic
limit. Let us denote x0 = 〈x(t)〉 and consider
∆ = limT→∞
1
T
∫ T
0
x(t′)dt′ − x0 = limT→∞
1
T
∫ T
0
[x(t′)− x0]dt′.
For an ergodic process this has to tend to a deterministic limit of zero. For finite time T in eachspecific realization of x(t) the value of ∆T = 1
T
∫ T0
[x(t′) − x0]dt′ is a random variable, with theensemble mean
〈∆T 〉 =
⟨1
T
∫ T
0
[x(t′)− x0]dt′⟩
=1
T
∫ T
0
〈[x(t′)− x0]〉dt′ = 0.
The values of ∆ are deterministic if the fluctuations around this limit tend to zero as T →∞: Theonly distribution for which 〈∆2〉 = 0 is a δ-distribution p(∆) = δ(∆), i.e. the one corresponding
exactly to ∆ = 0. The fluctuations of ∆T = 1T
∫ T0
[x(t′)− x0]dt′ are given by
∆2T = 〈∆2
T 〉 =
⟨(1
T
∫ T
0
[x(t′)− x0]dt′)2⟩.
Let us introduce the centered variable y(t) = x(t)− x0. Now we use the standard trick and write(∫ T
0
y(t)dt′)2
=
∫ ∫y(t′)y(t′′)dt′dt′′
so that
∆2T =
1
T 2
∫ T
0
∫ T
0
Cyy(t′, t′′)dt′dt′′.
Now we assume that our process is stationary, Cyy(t′, t′′) = Cyy(|t′′ − t′|) = Cyy(t
′′, t′). Using thissymmetry we can rewrite the integral as
∆2T =
2
T 2
∫ T
0
dt′∫ t′
0
Cyy(t′, t′′)dt′′.
and introduce the new variable of integration τ = t′′ − t′:
∆2T =
2
T 2
∫ T
0
dt′∫ t′
0
Cyy(τ)dτ.
Changing the sequence of integration we get the condition for ergodicity in the form
limT→∞
2
T 2
∫ T
0
(T − τ)Cyy(τ)dτ = limT→∞
2
T
∫ T
0
(1− τ
T
)Cyy(τ)dτ = 0. (33)
This is an integral condition which is always satisfied if Cyy(t) → 0 for t → ∞, i.e. whenever thecorrelations decouple (to see this change to the integration variable ξ = τ/T ). The processes forwhich Cyy(t)→ 0 when t→∞ (as in the case discussed in Sec. 3.3.1) are called mixing processes.The mixing condition is stronger than ergodicity.
47
3.4.3 The Wiener-Khintchine Theorem
The Wiener-Khintchine theorem connects the spectral density of the ergodic random process withthe properties of the (reasonably defined) Fourier transform of the process itself. The problem withthe Wiener-Khintchine theorem (1930/1934) is that its formulation in different publications isdifferent, and sometimes wrong. It may seem that the WK-theorem is a theorem of a harmonicanalysis, which it is not, and in some cases authors manipulate with mathematical objects whichare ill-defined. Here we consider a relatively weak version, which, however, will be enough for whatfollows.
We consider here the WK-theorem on the level of ensemble averages (one can also proof thatit is valid for single realizations of the process). The aim of this discussion is not so much the WKtheorem itself but the understanding of the correct notion of the spectral density of the processand its connection with its Fourier transform, which otherwise stays obscure.
Let us consider the empirical (auto)correlation function of a real variable x(t)
RT (t) =1
T
∫ T
0
x(τ)x(t+ τ)dτ
and its limitR(t) = lim
T→∞RT (t).
We assume that this limit exists for any t. The process x(t) is assumed to be centered, i.e. ithas a zero mean. For an ergodic process x(t) the limit R(t) of the empirical correlation functionis equal to C(t) = 〈x(τ)x(t + τ)〉, the autocorrelation function (ACF) of the process. For astationary random process (note that stationarity is a prerequisite for the ergodicity) we have〈x(t1)x(t2)〉 = 〈x(0)x(t2 − t1)〉 = C(t2 − t1). For T large enough RT (t) can be approximated byR(t) and therefore by C(t) with a sufficient accuracy. Note that this C(t) is an even function of t,and so is also R(t) (provided the corresponding limit exists).
Now let us consider the Fourier transform of C(t)
C(ω) =
∫ ∞−∞
C(t)eiωtdt.
We will assume that there exists a characteristic decay time τ of the ACF, i.e. the convergenceof this Fourier integral is fast enough, i.e. the corresponding integral between −T1 and T2 ispractically equal to C(ω) for all T1, T2 τ (the assumption done is essentially that our processis mixing, and it is much more restrictive than necessary, but simplifies the argumentation. Thenecessary condition is ergodicity.).
Note: the empirical correlation function is not equal to
Q(t) =
∫ ∞−∞
x(τ)x(t+ τ)dτ
which is infinite whenever R(t) is finite, and can only be finite when R(t) vanishes. The finedifference between the two definitions will be considered below. Using Q(t) is a reason for appear-ing of awkward and useless δ-functions in many literature sources dealing with Wiener-Khinchintheorem, or a reason for the fact that it simply looks like an easy convolution statement. Not onlythis: working with non-existent objects sometimes leads to errors.
48
Let us fix T large but finite. We define
xω =1
T
∫ T
−Tx(t)eiωtdt
withω =
π
Tn,
n = 0, 1, 2, ... (note that the factor 2 is omitted in the denominator in the prefactor of the integral,as it is common in the theory of Fourier series). Alternatively we could define
X(ω) =
∫ T
−Tx(t)eiωtdt,
i.e. consider the Fourier integral, but in finite limits. Let us consider T large enough and consider
x∗ωxω′ =1
T 2
∫ T
−T
∫ T
−Tx(t1)x(t2)e−iωt1eiω
′t2dt1dt2.
Averaging this over the realizations of the process we get
〈x∗ωxω′〉 =1
T 2
∫ T
−T
∫ T
−T〈x(t1)x(t2)〉e−iωt1eiω′t2dt1dt2.
The expression in the angular brackets is C(t2−t1). Now let us note that e−iωt1eiω′t2 = ei(ω
′−ω)t1eiω′(t2−t1)
and rewrite the integral as follows:
〈x∗ωxω′〉 =1
T 2
∫ T
−T
∫ T
−TC(t2 − t1)eiω
′(t2−t1)ei(ω′−ω)t1dt1dt2.
Changing the variables of integration to t1 and t = t2 − t1 (the Jacobian of the transformation isunity) we get
〈x∗ωxω′〉 '1
T 2
∫ T
−Tdt1e
i(ω′−ω)t1
∫ T−t1
−T−t1C(t)eiω
′tdt.
For moderate t1 the internal integral is practically C(ω′); the differences from this value are onlyperceptible for t1 in the τ -vicinity of ±T . Therefore for T large we can with a considerable accuracyassume that
〈x∗ωxω′〉 '1
TC(ω)
[1
T
∫ T
−Tei(ω
′−ω)t1dt1
].
The expression in the square brackets is proportional to the Kronecker-Delta:
1
T
∫ T
−Tei(ω
′−ω)t1dt1 =
∫ π
−πei(n−n
′)zdz = 2πδn,n′ .
This is different from zero only for n = n′ and therefore for ω = ω′. Therefore for all ω′ 6= ω wehave 〈x∗ωxω′〉 = 0, i.e. the Fourier-components of our random signal x(ω) and x∗(ω′) with differentfrequencies are uncorrelated. The component with the same frequency fulfill
〈x∗ωxω〉 '2π
TC(ω).
49
Note that this expression does not have a true limit for T →∞: the left hand side namely vanishes.If we use the X(ω) instead, we get
〈X∗ωXω〉 ' 2πTC(ω),
and the l.h.s. diverges. We want however something which goes to a well-prescribed limit S(ω)when T →∞. Now we know how to define this:
S(ω) = T 〈x∗ωxω〉 =1
T〈X∗ωXω〉.
This S(ω) is the spectral density of the process, discussed above. Now we can also return to thesingle realizations, introduce the empirical spectral density
ST (ω) =1
T
∣∣∣∣∫ T
−Tx(t)eiωtdt
∣∣∣∣2 ,and show that
limT→∞
ST (ω) = S(ω) = 2πC(ω).
This is the Wiener-Khinchin theorem.
3.5 Fluctuations and noise
Let us consider an example: a harmonic oscillator under influence of fluid friction or its electricanalogue, an RLC circuit. This system is in a contact with the heat bath. Our system is assumedto be linear one. We moreover suppose that its mean response to the external force is given by theknown equations which describe the behavior of the corresponding system when fluctuations aredisregarded.
The equation for the charge of the capacitor reads
Q+R
LQ+
Q
LC= U(t)/L.
The equation for the coordinate of the damped mechanical harmonic oscillator in a gas has thesame structure:
x+ (γ/m)x+ ω20x = F (t)/m.
Here F (t) is the external force and U(t) is the external voltage acting on the system. Since weassumed that the response of the mean coordinate to the deterministic force is described by thesame response function as a response of the coordinate to the deterministic force in our setupwithout fluctuations, which is given by the deterministic equations of motion a couple of linesabove, we can calculate the response function. This response function can be easily evaluated viathe deterministic equations of motion when passing to the Fourier domain. For the oscillator
−ω2x(ω)− iωγx(ω) + ω20x(ω) =
1
mF (ω),
or, respectively,
−ω2Q(ω)− iωRLQ(ω) + ω2
0Q(ω) =1
LU(ω).
50
Thus, in the mechanical case,
x(ω) =1
m
1
(ω20 − ω2)− iω(γ/m)
F
and
ψ(ω) =1
m
1
(ω20 − ω2)− iω(γ/m)
,
and in an electric circuit we simply have to change m → L, γ → R and ω20 → 1/LC. Using our
Eq.(32) we get
Cxx(ω) =2kBT
ωImψ(ω) =
2kBTγ
m2
1
(ω2 − ω20)2 + ω2(γ/m)2
(34)
The form of Cxx(t) follows by an inverse transform.Now let us discuss some physics behind the expression. What is the reason for the fluctuations?
In the case of a mathematical pendulum in the gas it is clear that the source of the fluctuationsare the random forces resulting from the random impacts of the molecules of the gas collidingwith the oscillating mass. Of course, the friction also stems from the interaction from the gas,but let us assume for a while that this is indeed described by the constant friction coefficient γ.We note that the interaction with the bath can only take place through the part of the systemwhich can emit heat, and this takes place via friction. In equilibrium the heat transmitted to thebath on the average is compensated by the work of this random forces. In the electric circuit theinteraction with the bath takes place in the resistor (again, energy exchange with the bath canonly take place via the part of the system which can emit and therefore also absorb heat, and itis only the resistor which can emit heat in our case). The thermal fluctuations in contact withthe bath (which can be considered as moving electrons back and forth through the resistor) canbe considered as random voltage (of a so-called noise source, a small random battery switched ina sequence with the resistor and being an essential source of the fluctuations of the current andcharge). In any case, the fluctuations are entering the system via dissipative element. Thus, it isreasonable to assume that the fluctuations of coordinate / charge are caused by the fluctuatingforce / voltage. Such random forces or voltages will be called noise in what follows.
Assuming now that the same response function that gives us the response of deterministiccoordinate (charge) to deterministic force (voltage) gives the response of fluctuating coordinate tothe fluctuating force, we can write x(ω) = ψ(ω)F (ω). This relation also holds if the process isobserved on the finite time interval 2T . Then we get,
〈x(ω)x(ω′)〉 = ψ(ω)ψ(ω′)〈F (ω)F (ω′)〉,and introducing the proper normalization according to the Wiener-Khintchine theorem we get
Cxx(ω) = |ψ(ω)|2CFF (ω). (35)
Using Eq.(35) and Eq.(34) we now obtain
CFF (ω) =Cxx(ω)
|ψ(ω)|2= 2kBTγ.
In the case of the electric contour we get
CUU(ω) = 2kBTR.
The last expression is called the Nyquist-theorem (1928), and was originally derived in a muchmore elegant way. The corresponding spectral density differs from the Fourier-representation ofthe correlation function by a factor 2π. The hot resistor in an electric circuit is a source of a whitenoise.
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4 Brownian motion
4.1 Historical overview
The methods of modern kinetics stem to a large extent from different variants of the theory ofBrownian motion14. The first approach due to Einstein was based on a random walk (RW) scheme(1905) and contained in a nutshell many approaches which then were elaborated within the Fokker-Planck equations. Einstein was not aiming onto giving the theory of a Brownian motion, but wasinterested in consequences of fluctuations at equilibrium. Only after the publication of the workit was pointed out that the effect Einstein described was Brownian motion known since the worksof Ingenhousz (1785) and Brown (1827) (and thoroughly investigated by many scientists sincethen). Brownian motion is an unceasing motion of small but macroscopic (micron-sized) particlesimmersed in a quiescent fluid, which can be easily observed in a microscope. Its connectionwith the thermal motion of the molecules of fluid was postulated rather early, but there wereno theoretical predictions which could be compared with the experiment. The Einstein’s theorydiscussed the force-free behavior and the behavior under a constant force (gravitation) and providedsuch predictions which could be (and were) proved experimentally. The experiments of Perrin(1908 / 1909) and Seddig (1908) closely following the Einstein’s scheme (and many others, usingdifferent setups) confirmed the predictions of the theories proposed up to the time15.
Parallel to experimental work, theorists worked on different, more transparent and mathemati-cally rigorous variants of the theory. An alternative, combinatorial approach, based on the similarRW scheme was given by Smoluchowski in 1906. Smoluchowski claimed that his approachis more straightforward and therefore simpler than the one of Einstein. Two years later (1908)Langevin formulated an approach which is very similar to the one discussed in the previous sec-tion, and was essentially its predecessor. Langevin claimed his approach to be infinitely simplerthan the Einstein’s one. Indeed, the approach seems to be more transparent (but partly onlybecause the difficulties here are hidden), and opened a new domain of mathematics: the one ofstochastic differential equations.
The main difference between our discussion in this section and the discussion in Sec.3, is thatthe Brownian motion (being of course a fluctuation effect) does not correspond to an equilibriumsituation. Concentrating on three degrees of freedom (center of mass position of the particle)starting at some given position in the fluid we have to do with an initial condition problem.Although the description adopted by Langevin is very close to the one we used in the last partof the previous section, the non-stationarity of x(t) process on timescales of interest makes theproblem more rich and more tricky. After building the theory of Brownian motion we will return tothe systems discussed in Sec. 3.5 and see, how do these results reappear within more transparentand more detailed approaches.
4.2 The Langevin’s original approach
Einstein’s discussion (which we will reproduce later) showed that the probability to find a particleat distance x from its initial position (I use one-dimensional notation here, but the generalization to
14For a short overview of historical development of theoretical concepts one may see W. Ebeling, E. Gudowska-Nowak and I.M. Sokolov, On stochastic dynamics in physics - Remarks on history and terminology, Acta PhysicaPolonica B 39 1003-1018 (2008)
15The whole story is much more involved and dramatic than typically discussed in textbooks, see R. Maiocchi,The case of Brownian Motion, BJHS 23 257-283 (1990) for the review of all early experimental work.
52
higher dimensions is trivial) fulfills the diffusion equation. Moreover Einstein was able to connectthe diffusion coefficient in this equation with other properties of the system. Solving the diffusionequation (for the force-free case and in an unbounded medium) one gets the (Gaussian) PDF ofdisplacements, and the mean squared displacement 〈x2(t)〉 = 2Dt in x-direction (or 〈r2(t)〉 = 2Ddtfor the absolute value of displacement in a d-dimensional space). This is exactly this result for themean squared displacement which Langevin’s theory explains in a simple and elegant way.
It was Langevin, who postulated that the interaction of a macroscopic but light particle withthe bath can be separated into the mean (friction) force and random impacts (noise), as we alreadydiscussed in Sec.3.5. The initial Langevin’s approach started from the following postulates (herein 1d notation):
• The equation of motion of the Brownian particle is
mx = −γx+ f(x) + ξ(t) (36)
where −γx is the friction force, f(x) is an external (say gravitational) force acting on theparticle (will be set to zero in what follows) and ξ(t) is the noise. In our present discussionwe put f(x) = 0, but we will also consider nonzero forces later on.
• The noise force is independent on the particle’s position, and possesses zero mean 〈ξ(t)〉 = 0.
• The (properties of the friction force and the noise are such that the) mean squared velocityof the particle fulfills the equipartition theorem m〈x2〉/2 = kBT/2.
Since at time of Langevin no regular way to solve Eq.(36) was known, he had to invent a trick,leading to the final result without solving the equation. Following his trick we proceeded as follows:
• Multiply both parts of Eq.(36) by x:
mxx = −γxx+ xξ(t).
• Average over the realizations of the process. Note that ξ(t) is independent on x and that〈ξ(t)〉 = 0. We get:
m〈x(t)d2
dt2x(t)〉 = −γ〈x(t)
d
dtx(t)〉 (37)
• Note that 〈x(t) d2
dt2x(t)〉 = d
dt〈x(t) d
dtx(t)〉 − 〈
(ddtx(t)
)2〉. Rewrite the equation above (37) as
md
dt〈x(t)x(t)〉 −m〈(x(t))2〉+ γ〈x(t)x(t)〉 = 0.
• Use the equipartition theorem to rewrite the second term and put
md
dt〈x(t)x(t)〉 − γ〈x(t)x(t)〉 − kBT = 0.
Solve this simple linear inhomogeneous ODE for y(t) = 〈x(t)x(t)〉 under initial conditionx(0) = 0 leading to y(0) = 0. The solution reads:
〈x(t)x(t)〉 =kBT
γ
[1− exp
(− γmt)].
53
• Note that 〈x(t) ddtx(t)〉 = 1
2ddt〈x2(t)〉. Integrate the relation
d
dt〈x2(t)〉 = 2
kBT
γ
[1− exp
(− γmt)]
with initial condition 〈x2(0)〉 = 0 and get:
〈x2(t)〉 = 2kBT
γ
t− m
γ
[1− exp
(− γmt)]
.
For large t this behavior corresponds to the diffusive one, 〈x2(t)〉 ' 2Dt with
D = kBT/γ (38)
(or D = kBTµ where µ = 1/γ is the mobility of the particle). Here we re-derived the Einstein’s re-lation discussed in Sec.2.5.2. The smaller the mass of the particle, the earlier the linear dependenceof 〈x2(t)〉 establishes itself. The characteristic time after which this linear dependence dominatesis the relaxation time of the system τ = m/γ.
Note that for macroscopic spherical particles at small Reynolds numbers one can assume thatγ = 6πηa (Stokes’ friction law), with η being the viscosity of the fluid (known) and a being theradius of the particle (known), therefore measurement of D at given temperature gives us thevalue of kB (and therefore of the Avogadro number NA if the value of the universal gas constantR = NAkB is known).
The whole discussion looks much simpler than it really is. Let us clarify some important pointsof it: what is universal, what is assumed ad hoc, what are physical assumptions and what aremathematical facts.
4.3 Taylor (Green-Kubo) formula
First let us discuss the relation between the mean squared displacement and the properties ofthe velocity of the process. Let us assume the process does possess the velocity, i.e. x(t) is adifferentiable function of time (we can give it up for the sake of simplicity, but giving this up isan approximation, since the correct microscopic equation of motion is Newton’s law, according towhich the process even possess acceleration!). Then, for a particle starting at x(0) = 0
x(t) =
∫ t
0
v(t′)dt′,
and
x2(t) =
(∫ t
0
v(t′)dt′)2
=
∫ t
0
∫ t
0
v(t′)v(t′′)dt′dt′′.
Averaging over the realizations of the process, we get
〈x2(t)〉 =
∫ t
0
∫ t
0
〈v(t′)v(t′′)〉dt′dt′′ =∫ t
0
∫ t
0
Cvv(t′, t′′)dt′dt′′.
Moreover, due to symmetry Cvv(t′, t′′) = Cvv(t
′′, t′) we can integrate in t′′ only from t′ to t anddouble the result. If v(t) is a stationary random process (for the normal Brownian motion it really
54
is), one can assume Cvv(t′, t′′) = Cvv(|t′′ − t′|) (it has to be an even function of the time lag).
Changing the second variable of integration from t′′ to τ we get
〈x2(t)〉 = 2
∫ t
0
dt′∫ t′
0
Cvv(τ)dτ. (39)
If the integral I(t′) =∫ t′
0Cvv(τ)dτ as a function of t′ tends to a limit D = limt′→∞
∫ t′0Cvv(τ)dτ ,
we get for t large〈x2(t)〉 = 2Dt,
and define the diffusion coefficient
D =
∫ ∞0
Cvv(τ)dτ.
Let us now perform some “reverse engineering” and differentiate Eq.(39) twice with respect tot. The first differentiation gives
d
dt〈x2(t)〉 = 2
∫ t
0
Cvv(τ)dτ,
and the second one givesd2
dt2〈x2(t)〉 = 2Cvv(t).
We now return to the Langevin’s result
〈x2(t)〉 = 2kBT
γ
t− m
γ
[1− exp
(− γmt)]
,
and differentiating it twice get
d2
dt2〈x2(t)〉 = 2
kBT
mexp
(− γmt),
so that
Cvv(t) =kBT
mexp
(− γmt)≡ 〈v2〉 exp
(− γmt),
where we again applied the equipartition theorem. Here we obtained Cvv(t) for t > 0, but we knowthat Cvv(t) is an even function of t. We can pass to the Fourier domain and write
Cvv(ω) = 2kBT
m
∫ ∞0
exp(− γmt)
cos(ωt)dt =2kBTγ
m2ω2 + γ2.
Now we return to the Langevin’s (i.e. Newton’s) equation for velocity
mv + γv = f(t) (40)
(where in our case above the force f(t) = ξ(t) was stochastic), pass to the Fourier representation
iωmv(ω) + γv(ω) = f(ω)
and get v(ω) = 1/(iω + γ)f(ω)
ψvf =1
imω + γ.
55
This function gives us the response of velocity to an external perturbation f(t) (it is not theresponse function of the class we discussed before, connecting the force with the coordinate). Nowwe again assume that there is no difference in response between deterministic forcing f(t) and thestochastic one ξ(t) and write for the velocity-velocity correlation function
Cvv(ω) = |ψ(ω)|2Cξξ(ω).
Solving the equation for Cξξ(ω) we get:
Cξξ(ω) = (m2ω2 + γ2)2kBTγ
m2ω2 + γ2= 2kBTγ (41)
Coming back to time domain we get
Cξξ(t) = 2kBTγδ(t).
Our noise force is therefore δ-correlated (and the same as we obtained in the previous section).It is interesting to discuss where this assumption was hidden in the initial Langevin’s approach(lecture!).
4.4 Overdamped limit
Let us consider the limiting case of very small masses when the correlation time m/γ gets verysmall. This essentially means that we can simply neglect the inertial term in our Langevin equationEq.(36), and write
γx = f(x) + ξ(t), (42)
orx = γ−1f(x) + γ−1ξ(t).
The first term is a terminal velocity v(x) of a particle (with mobility µ = 1/γ) moving in the viscousfluid under the influence of the external force f(x) in the absence of the noise. This equation iscalled the overdamped limit of our previous Langevin equation (the word “overdamped” meansthat the damping, i.e. friction prevails over inertia). Assuming that the external force vanishes,we get x = γ−1ξ(t), so that
Cvv(t) = γ−2Cξξ(t) =2kBT
γδ(t)
Using the Taylor-Green-Kubo formula we obtain the diffusion coefficient
D =
∫ ∞0
Cvv(τ)dτ =kBT
γ,
exactly the same as the terminal value of the diffusion coefficient
D = limt→∞
〈x2(t)〉t
given by Eq.(38) (Note that the integration boundary at 0 “cuts” the δ-function just in the middle.This gets clear when one performs the corresponding limiting transition from the case of non-vanishing mass as discussed above. The same effect will appear later, when discussing the transitionfrom the Langevin equation to the Fokker-Planck one). We note also that the equation (40) for thevelocity has the same form as our equation (42). These first order equations are of course simplerthan the initial second-order Langevin equation we started from. Therefore in the next sectionwe start from these equations and discuss the following question: How can one pass from theproperties of single trajectories, described by the Langevin’s equation to the ensemble propertiesdescribed by the probability densities of the corresponding coordinates.
56
4.5 From Langevin equation to Fokker-Planck equation I
This is our first, “quick and dirty” approach to relation between the Langevin and the Fokker-Planck equations16.
Let us consider the random process x(t) governed by the equation of motion
dx
dt= v(x) + ξ(t) (43)
(where I now redefined ξ′(t) = γ−1ξ(t) and omitted the prime, i.e. introduced the rescaled noiseintensity; the noise now has the dimension of velocity). I simply don’t want to carry the prefactorsthrough the whole (lengthy) calculation. The only important thing to note is that the correlationfunction for this new noise is
〈ξ(t)ξ(t′)〉 = 2Dδ(t− t′)
with D discussed above.We can formally integrate Eq.(43) and get for a given realization of the noise
x(t) =
∫ t
0
[v(x(t′)) + ξ(t′)] dt′
and formally put down the probability density P (x|t) to find a particle at position x at time t as
P (x|t) = δ(x− x(t))
where x(t) is a known function of time given by the previous equation. Note that this x(t) is adifferent object than the variable x as an argument of P (x|t). We now note that for a whateverfunction f(x, t) which can be written in a form f(x, t) = f(x− y(t)) we gave
∂
∂tf(x, t) = −dy
dt
∂
∂xf(x, t)
and therefore in our case∂P
∂t= − ∂
∂x
(dx(t)
dtP
),
where we have interchanged the sequence of derivatives in x and in t. This is the local conservationlaw (the continuity equation) for the probability density P . Replacing the time derivative of x onthe right-hand side of this equation by the expression given by Eq.(43) we get:
∂P
∂t= − ∂
∂x[(v(x) + ξ(t))P ] . (44)
This is a (stochastic) differential equation giving P (x, t) in a particular realization of the process;it is the full equivalent of the initial Langevin equation.
Eq.(44) is a linear differential equation, whose formal solution can be obtained as follows. Letus introduce the linear operator L corresponding to the deterministic part of the behavior:
LP =∂
∂x(v(x)P ) .
16In this section I follow more or less the discussion in: R. Zwanzig, Nonequilibrium Statistical Mechanics”, OxfordUni. Press, 2001
57
Eq.(44) then reads:∂P
∂t= −LP − ∂
∂x[ξ(t)P ] . (45)
We formally solve this equation as we would formally solve any inhomogeneous linear differentialequation:
P (x, t) = e−LtP (x, 0)−∫ t
0
dse−L(t−s) ∂
∂x[ξ(s)P (x, s)] . (46)
Differentiate Eq.(46) and you see that it is indeed a solution of Eq.(45)! This solution showsexplicitly that P (x, t) depends on the noise at all times s < t. We now substitute this result intothe last term in the right-hand side of Eq.(45) and get:
∂
∂tP (x, t) = − LP (x, t)− ∂
∂xξ(t)e−LtP (x, 0)
+∂
∂xξ(t)
∫ t
0
dse−L(t−s) ∂
∂x[ξ(s)P (x, s)] . (47)
We now note that if an ensemble of realizations of some random variable x is considered, thePDF of x is the mean value of the δ-function over the realizations:
p(y) = 〈δ(y − x)〉x.
Therefore, in our case the ensemble average of the probability density P (x, t) (i.e the PDF of par-ticle positions for different realizations of the noise) gives us the sought PDF p(x, t) = 〈P (x, t)〉 =〈δ(x− x(t))〉ξ(t). We now apply this ensemble (noise) average to our Eq.(47) term by term.
The term on the left hand side and the first term in the right-hand side, containing the P (x, t)only, are averaged easily. The second one, containing only the initial condition (independent onthe realization of the noise) and ξ(t) vanishes under the averaging since 〈ξ(t)〉 = 0, so that onlythe third one, with the paired noise factors, needs to be averaged explicitly:
∂
∂tp(x, t) = −Lp(x, t) +∇
∫ t
0
dse−L(t−s)∇ [〈ξ(t)ξ(s)P (x, s)〉] .
Now, for s 6= t ξ(t) is uncorrelated with ξ(s) for all s < t (since the noise is δ−correlated) andthus not correlated with the state of the system P (x, s), given by the values of the noise at timesearlier than t. Due to the symmetry of the noise (i.e. to the fact that 〈ξ(t)〉 = 0), the integrandis zero for all s 6= t, so that 〈ξ(t)ξ(s)P (x, s)〉 = 0 for s 6= t. On the other hand, P (x, s) givenby an integral Eq.(46) over the times s′ < s = t, is dominated by the values of the noise at theprevious times, and thus can be considered as independent from ξ(t) exactly at time t, so that〈ξ(t)ξ(s)P (x, s)〉 = 〈ξ(t)ξ(s)〉 〈P (x, s)〉 ≡ 2Dδ(t− s)p(x, t). Thus,
∂
∂tp(x, t) = −Lp(x, t) +
∂
∂x
∫ t
0
dse−L(t−s) ∂
∂x2Dδ(t− s)p(x, s)
= − ∂
∂x(v(x)p) +D
∂2
∂x2p (48)
which is a Fokker-Planck equation we looked for. The factor 2 in front of D disappeared since theupper boundary if integration (at t) “cuts” the δ-function just in the middle.
We note that this is a typical derivation on the physical degree of the accuracy: we didn’t bothertoo much about the properties of e−L(t−s) and about other subtleties, a mathematician should dothis. Thus the overdamped Langevin equation for a single trajectory generates the Fokker-Planckequation for the ensemble dynamics. Rewriting v(x) = µf(x) we get the form of our Eq.(18).
58
4.5.1 Generalization to higher dimensions
The generalizations to the multidimensional case is trivial: we consider the multidimensionalrandom process xi(t) governed by the equations of motion
dxidt
= vi(x1, ..., xn) + ξi(t)
where ξ(t) = (ξ1(t), ..., ξn(t)) is the vector of random variables (noises) with zero means 〈ξi(t)〉 = 0and correlation properties
〈ξ(t)ξ(t′)〉 = 2Bδ(t− t′).
We are looking for the probability distribution function of the values of xi at time t. The localconservation law now reads
∂P
∂t+∑i
∂
∂xi
(dxi(t)
dtP
)= 0,
and the analogue of Eq.(44) reads
∂P
∂t= −
∑i
∂
∂xi[(vi(x1, ..., xn) + ξi(t))P ] .
This is a stochastic differential equation giving p(x, t) in a particular realization of the process.The linear operator L corresponding to the deterministic part of the behavior reads
LP =∑i
∂
∂xi(vi(x1, ..., xn)P ) ,
Performing the same steps as before we get
∂p
∂t= −∇v(x)p+∇B∇p
(note that the corresponding matrix B is standing between the differential operators, so that (forconstant B) the diffusion term reads ∇B∇p =
∑i,j Bij
∂2
∂xi∂xjp). Now we know how to obtain
a Fokker-Planck equation if the Langevin equation is known. We however do not yet have anyfeeling about what is behind our formal derivation. Before going further let us discuss what is theFokker-Planck equation for our initial (not overdamped) Langevin equation with inertial term. Toget it we rewrite it for a vector x = (v, x):
v =f(x)
m− γ
mv +
1
mξ(t)
x = v
and get the “velocity vector”
v =(
f(x)m− γ
mv v
)and the diffusivity matrix
B =
(kBTγ/m
2 00 0
)
59
(since 〈ξ(t)ξ(s)〉 = 2kBTγδ(t − s), see the previous section, and therefore 〈[ξ(t)/m][ξ(s)/m]〉 =(2kBTγ/m
2)δ(t− s); the prefactor 2 disappears in B since, as previously discussed, the δ-functionis “cut in the middle” by the boundary of integration), and immediately put down
∂
∂tp(v, x, t) = − ∂
∂xvp(v, x, t)− ∂
∂v
(f(x)
m− γ
mv
)p(v, x, t)
+kBTγ
m2
∂2
∂v2p(v, x, t). (49)
This particular form of the Fokker-Planck equation is called the Klein-Kramers equation.To check that we got things right we can first integrate both parts of this equation over all
values of x, and get the equation for the marginal distribution of velocity:
∂
∂tp(v, t) =
∂
∂v
(f(x)
m− γ
mv
)p(v, t) +
kBTγ
m2
∂2
∂v2p(v, t)
(the integral of the first term in the r.h.s. of the Klein-Kramers equation vanishes since p(v, x, t)→0 for x→ 0). Now let us consider the force-free case f = 0 and look for the equilibrium solution,i.e. the one of the equation
− γmvp(v, t) +
kBTγ
m2
∂
∂vp(v, t) = 0
which is
p(v) = const · exp
(− mv2
2kBT
),
a Maxwell distribution.
5 Another approach to the Fokker-Planck equation
5.1 Einstein’s theory of Brownian motion
Starting from the physical analysis of the (Brownian) motion of particles suspended in a quiescentfluid Einstein has connected this motion with diffusion and showed that this diffusive behaviorfollows from the three postulates. First, the particles are assumed not to interact with each other:their trajectories are independent. Second, one assumes that the motion of the particles lacks long-time memory: one can choose such a time interval τ , that the displacements of the particle duringtwo subsequent intervals are independent. Third, the distribution of a particle’s displacements sduring the subsequent time intervals φ(s) possesses at least two lower moments. Moreover, for theforce-free situation φ(s) is symmetric. The displacement of the particle can thus be considered asa result of many tiny, independent, equally distributed steps: the whole stochastic process of theparticle’s motion is described as a random walk scheme. The further line of his reasoning is veryclose to the Kramers-Moyal expansion discussed in some detail in the next section.
For the simplicity, following Einstein, we analyze a one-dimensional problem: The concentrationof particles n in vicinity of point x is proportional to the probability density p(x, t) to find oneparticle at this point. Comparing the probabilities at time τ and at time t+ τ we get (due to theindependence of the new displacement of the previous position and to the fact that x(t + τ) =x(t) + s)
p(x, t+ τ) =
∫p(x− s, t)φ(s)ds.
60
Now, since both τ and s are both small compared to the time- and space-scales of interest, onecan expand the function f in Taylor series on both sides of the equation. On the left-hand side itis enough to expand up to the first order in t, on the right-hand side we need the second order ins. We get:
p+
(∂p
∂t
)τ + ... = p+
(∂p
∂x
)∫sφ(s)ds+
1
2
(∂2
∂x2p
)∫s2φ(s)ds+ ... .
The integral∫sφ(s)ds vanishes due to the symmetry. In the lowest order we thus get:
∂p
∂t=σ2
2τ
∂2
∂x2p,
where σ2 =∫s2φ(s)ds. Here we recognize a diffusion equation, and associate σ2/2τ with the
diffusion coefficient D. The derivation of the diffusion equation by A. Einstein was the very firststep of statistical physics into the new domain of non-equilibrium phenomena. Note however, thatit was not the derivation of the diffusion equation, which seemed to Einstein to be the main topicof this work: The discussion of the diffusion of particles in the solution gave the way to determinethe Avogadro number NA through the macroscopic measurements on large particles, the mea-surement performed by Perrin some 3 years later. Such measurements were necessary to providesolid basement for atomistic theory of matter. The corresponding theoretical considerations weresummarized in Einstein’s PhD thesis ”Eine neue Bestimmung der Molekuldimensionen” presentedon April 30, 1906 to the University of Zurich.
5.2 Equations for the probability density
We note that the initial approach of Einstein was based on the discussion of the deterministic equa-tions for the probability densities, while the Langevin’s one emphasized the stochastic nature ofsingle realizations. The generalization of the Einstein’s approach leads to the Fokker-Planck equa-tion, whose first derivation was proposed by M. Smoluchowski in the work ”Uber BrownscheMolekularbewegung unter Einwirkung außerer Krafte und deren Zusammenhang mit der verall-gemeinerten Diffusionsgleichung” (Smoluchowski, 1915). This derivation follows the lines of theEinstein’s discussion and starts from the general Markovian assumption. One considers the processcharacterized by the transition probabilities p(x, t|x′, t′), the probability for a particle to be at xat time t provided it started at x′ at time t′ < t. We note that x here has not necessarily to beconsidered as a scalar, but for the simplicity we use the scalar notation. The Markovian propertymeans that the distribution of the particle’s position at t is fully determined by its position at t′
and does not depend on its prehistory.The Markovian property leads us to the assumption that the probability to be at x, at time t
when we started at x0 at time t0 can be expressed through the integral
p(x, t2|x0, t0) =
∫dx′p(x, t2|x′, t1)p(x′, t1|x0, t0) (50)
with t0 < t1 < t2, which essentially doesn’t mean anything else than the statement of the factthat at time t1 the particle has to be found at some place it really could get to. Depending on thecommunity, this equation is termed as a Smoluchowski, or as a Chapman-Kolmogorov equation.
Now let us use this equation for the case when the difference ∆t = t2 − t1 is small, set t0 = 0,and redefine our former t1 as t and former t2 as t+ ∆t:
p(x, t+ ∆t|x0, 0) =
∫dx′p(x, t+ ∆t|x′, t)p(x′, t|x0, 0).
61
Subtracting p(x, t|x0, 0) from the both parts of the integral equation we get
p(x, t+ ∆t|x0, 0)− p(x, t|x0, 0) =∫dx′p(x, t+ ∆t|x′, t)p(x′, t|x0, 0)− p(x, t|x0, 0). (51)
For ∆t being small the left-hand side can be taken as (∂/∂t)p(x, t|x0, 0)∆t. We assume that typicalchanges in the coordinates during this short time are also to some extent small. Let us denote r =x−x′ and represent the transition probabilities as functions of r: w(y, r; t,∆t) = p(y+r, t+∆t|y, t).Now we rewrite Eq.(51) in the form
∂
∂tp(x, t|x0, t0)∆t =
∫drw(x− r, r; t,∆t)p(x− r, t|x0, 0)− p(x, t|x0, t0) (52)
and expand the integrand in powers of r. We namely use the fact that
w(x− r, r; t,∆t)p(x− r, t|x0, t0) = w(x, r; t,∆t)p(x, t|x0, 0)
−r ∂∂xw(x, r; t,∆t)p(x, t|x0, 0) +
+1
2r2 ∂
2
∂x2w(x, r; t,∆t)p(x, t|x0, 0) + ...
which lets us rewrite the first integral in the form
∂
∂tp(x, t|x0, t0)∆t = p(x, t|x0, t0)
∫drw(x, r; t,∆t)
− ∂
∂xp(x, t|x0, t0)
∫drrw(x, r; t,∆t) +
+∂2
∂x2p(x, t|x0, t0)
∫drr2w(x, r; t,∆t) + ...
−p(x, t|x0, t0).
Noting that due to the normalization∫drw(y, r; t,∆t) = 1 we see that the first and the last terms
of the expression in the right-hand side cancel. Performing the limiting transition ∆t→ 0 one has
∂
∂tp(x, t|x0, t0) =
∞∑n=1
(−1)n
n!
∂n
∂xnKn(x, t)p(x, t|x0, t0) (53)
where the transition moments
Kn(x, t) = lim∆t→0
1
∆t
∫dr rnw(x, r; t,∆t).
The expression Eq.(53) corresponds to the Kramers-Moyal expansion. For the case of theBrownian motion and other diffusive processes one finds that only the first two transition momentsare different from zero: these are
A(x, t) = lim∆t→0
1
∆t
∫dr r w(x, r; t,∆t) (54)
and
B(x, t) = lim∆t→0
1
∆t
∫dr r2w(x, r; t,∆t). (55)
62
All higher moments vanish under the limiting transition ∆t→ 0 since∫dr rnw(x, r; t,∆t) = o(∆t) for n ≥ 3,
see the discussion below. In this case the Kramers-Moyal expansion of the Chapman-Kolmogorovequation leads to the Fokker-Planck equation for p(x, t|x0, t0) 17
∂
∂tp(x, t|x0, t0) = − ∂
∂xA(x, t)p(x, t|x0, t0) +
1
2
∂2
∂x2B(x, t)p(x, t|x0, t0).
Comparing this general equation with our phenomenological Fokker-Planck equation, we see thatit also has a form of a continuity equation18
∂
∂tp =
∂
∂x
[−Ap+
1
2
∂
∂xBp
]. (56)
In the case of the overdamped motion the coefficients A and B can be easily obtained from thediscussion of the overdamped Langevin motion: The transition probability density w(x, r; t,∆t) isGaussian
w(x, r; t,∆t) =1√
4πD∆texp
(−(r − µf∆t)2
4D∆t
)(57)
from which we get that A(t) = µf and B = ∆t−1[(µf∆t)2 + 2D∆t
]→ 2D for ∆t→ 0, and that
the higher transition moments (being the combinations of the two lower ones) are of a higher orderin ∆t (note that for small ∆t we can neglect the fact that the force f may depend on the coordinate,since the typical displacements are tiny)19. Hence, the resulting Fokker-Planck equation reads
∂
∂tp =
∂
∂x
[−µfp+D
∂
∂xp
], (58)
(with D and µ connected by the fluctuation-dissipation relation D = kTµ). Eq.(58) is exactly theone following from our initial phenomenological considerations. The Fokker-Planck equation forthe overdamped situation is often called the Smoluchowski equation.
17 According to the Pawula’s theorem either the first two moments (given by Eqs.(54) and (55)) are enough forthe full description, or the whole infinite series has to be used. Truncating the series after any other term than thesecond one leads to equations which do not guarantee the non-negativity of their solutions, which therefore cannotbe interpreted as probability density functions. In this case the integral equation representations cannot be reducedto anything considerably more simple.
18This is an equation of a somewhat different, more general form than the phenomenological Fokker-Planckequation; it, for example, contains an additional term of the form p(∂2/∂x2)B(x, t) which is absent in the phe-nomenological equation. The phenomenological equation follows from the general one under additional assumptionsintroduced by the detailed balance principle, which will be discussed in Sec. 5.5.
19This Gaussian character can be seen when subdividing the ∆t interval into smaller intervals of duration δt. Sincethe displacements are small, the deterministic force can be considered as constant. The Einstein’s random walkis then a discrete process in which particle makes independent, identically distributed steps si, and its coordinateafter N such steps is given by xn =
∑ni=1 sn. Random walks allow for a genuinely probabilistic description not
relying on the instruments which we introduce in this course, see e.g. J. Klafter and I.M. Sokolov, First Steps inRandom Walks, Oxford, 2011. In the case when the mean squared displacement exists, the Central Limit Theorempredicts that the distribution of xn tends to a Gaussian with the mean 〈x〉 = n〈s〉 and dispersion σ2
n = n〈s2〉. Forthe simple random walk, when one assumes that each step takes time δt to be performed, n can be translated intot: n = ∆t/δt. The Langevin equation is nothing else than the limiting description of this scheme when δt→ 0 butD = 〈s2〉/δt = const. The fact that the corresponding limiting transition leads to Gaussian displacement is thenpostulated in the definition of the Langevin equation by saying that the noise ξ(t) in this scheme is Gaussian whitenoise with zero mean.
63
We note that taking the limits in Eq.(54) and (55) corresponds essentially to the derivatives ofthe corresponding transition moments. The first one can be then interpreted as the mean velocityat point x, 〈v(x)〉, and the second one as the time derivative of the mean squared displacementfrom x, connected to a local diffusion coefficient.
The form of the Fokker-Planck equation stays the same if one considers the variable x as avector in the coordinate space or in the phase space: The only important assumptions are theMarkovian nature of transitions and the existence of transition moments.
5.3 Stochastic integration
For Markov processes, the knowledge of the Fokker-Planck equation (FPE) allows in principle foranalytical evaluation of many important properties which typically come as moments of the PDFswhich are the solutions of the corresponding equation under various boundary conditions. Althoughthere are many analytical methods of such solutions, and many quite reasonable approximations forthe case the solution cannot be expressed in a closed form20, in many cases it is easier to simulatethe trajectories of the process, and then make the statistical analysis of the corresponding results.All the physically interesting properties are then the functionals of the corresponding trajectoriesx(t), are calculated for each trajectory and averaged.
Let us consider the Langevin equation
dx
dt= f(x) +
√2Dξ(t) (59)
with Gaussian white noise ξ(t) (〈ξ〉 = 0, 〈ξ(t)ξ(t′)〉 = δ(t− t′)), or, as it is often written,
dx = f(x)dt+√
2DdW (60)
where dW is the infinitesimal increment of the Wiener process W (t) =∫ t
0ξ(t)dt (a Gaussian
process with stationary increments, and with the variance growing as t). Note that the Wienerprocess can be considered as a limit of a random walk, and is a process which is continuous (inthe probabilistic sense) but nowhere differentiable.
The Wiener process is considered as a limit of a Gaussian random walks with independentsteps. In each δt interval the corresponding change δW of the value of W (t) is a Gaussian randomvariable with zero mean and with dispersion δσ2 = δt. Then, the absolute value of the change|δW | = |W (t + δt) −W (t)| is distributed according to a “half-Gaussian” with the correspondingdispersion, which goes to zero as
√δt. For each ε > 0 one can chose δt so small that |δW | < ε
with probability P which is arbitrarily close to unity. Therefore one says that a Wiener process iscontinuous with probability one (almost surely).
Let us now consider the finite interval of duration ∆t and subdivide it into n shorter δt intervals.Let δWi be the change in W (t) in i-th such interval. The sum
∑n−1i=0 (δWi)
2 is a sum of squares ofindependent random variables, and it converges to the corresponding dispersion:
∑n−1i=0 (δWi)
2 →nδσ2 = n∆t
n= ∆t almost surely (i.e. with probability one). Now let us consider the typical value
of the derivative of the Wiener process, and look at |δW/δt|. Consider
1
δt∆W 2 =
1
δt
n−1∑i=0
(δWi)2 =
n−1∑i=0
|δWi||δWi|δt
.
20 The standard source is: H. Risken, The Fokker-Planck Equation: Methods of Solution and Applications,Springer, 1996 (3rd edition).
64
Now we make a simple estimate for the mean absolute value of dW/dt, |W ′| = 1n
∑n−1i=0
|δWi|δt
. Notingthat 1
n1δt
∆W 2 = 1n
1δt
∆t = 1 we can put down
1 =1
n
n−1∑i=0
|δWi||δWi|δt≤ max
i|δWi|
(1
n
n−1∑i=0
|δWi|δt
)= max
i|δWi||W ′|.
We can estimate the maximal value maxi |δWi| of n i.i.d. Gaussian random variables with zeromean and dispersion σ2 = ∆t/n each (this will be given as a Homework), we will see that its mostprobable value is
√2 lnnσ =
√2∆t lnn/n and tends to zero for n → ∞. Therefore |W ′| has to
tend to infinity in this limit.One can improve the estimates, and show that this takes place not only in the most probable
case, but with probability unity. Here the following almost surely holding statement may be used:
limε→0
sup|W (ε)|√
2ε ln ln(1/ε)= 1.
A side note. As already stated, in many cases analytical treatment of the problems is hard. Inthis case one resorts to numerical simulations. Let us integrate Eq.(60) over a finite time interval:
∆x =
∫ t+∆t
t
f(x(t′))dt′ +√
2D∆W. (61)
This ∆W is a Gaussian process with zero mean and with dispersion given by σ2(∆t) = ∆t. Nowlet us fix ∆t and consider our equation above (61) as the forward Euler scheme for solving Eq.(59)
• generate ∆W as a Gaussian random variable with zero mean and with dispersion given byσ2(∆t) = ∆t (say, by the Box-Muller algorithm).
• Update x: x(t+ ∆t) = x(t) + f(x(t))∆t+√
2D∆W
• Repeat the previous steps until the upper time limit is reached.
This is a “quick and dirty” method which has all deficiencies of forward Euler algorithms, forexample needs relatively small time steps ∆t. Note that while in the absence of the stochasticforcing the Euler algorithm has an accuracy of the order of ∆t the stochastic Euler scheme has theaccuracy (∆t)1/2, since the typical displacement due to noise grows as σ, and therefore is simplybad. This is because at very short times the noise term has a larger amplitude (O((∆t)1/2)) thanthe deterministic one (O(∆t)).
The simplest way to speed up the calculation is to use the Heun algorithm (which can be inter-preted as a predictor-corrector method, or as a member of the family of Runge-Kutta integrators).In what follows I will introduce the notation: ti = t0 + i∆t, and will denote by f(ti) the actualvalue of the drift force at ti: f(ti) ≡ f(x(ti)). After generating the increment of the Wiener processfor the i-step ∆Wi one performs two substeps:
• Euler predictor substep: x(ti+1) = x(ti) + f(ti)∆t+√
2D∆Wi
• Corrector substep: x(ti+1) = x(ti) + 12[f(x(ti+1)) + f(ti)]∆t+
√2D∆Wi.
The method is second order in ∆t for integration of the deterministic part, and on the total of theorder of (∆t)3/2. In most cases this is quite sufficient.
Thus the reasonable way to proceed is: Not to use the Euler algorithm. Use Heun algorithmas a simple all-purpose instrument. If the results are still not good enough, analyze the problemcarefully, read literature, and find a specific algorithm best suited for a particular problem at hand.
65
5.3.1 The problem of interpretation
The really new problems arise when the position-dependent diffusion coefficient come in play. Nowwe want to integrate the equation
dx
dt= µ(x)f(x) + g(x)ξ(t). (62)
Here g(x) is some function of x, and g2(x)/2 (positive) is interpreted as a diffusion coefficient. Thecombination µ(x)f(x), the deterministic drift, will be denoted as v(x) in what follows. Such a SDEis called a SDE with multiplicative noise (since the noise multiplies some function of the dependentvariable). As we proceed to show, this stochastic differential equation is not a differential equationat all, and the problems with this lay in the definition of stochastic integrals.
Let us consider a formal integral of Eq.(62).
x(t)− x(t0) =
∫ t
t0
dx
dtdt =
∫ t
t0
v(x(t′))dt′ +
∫ t
t0
g(x(t′))ξ(t′)dt′.
We now consider the time interval of integration short, and concentrate on the stochastic part,containing the noise (the deterministic part doesn’t make any problems). Thus, let us consider
I =
∫ b
a
g(x(s))ξ(s)ds ≡∫ b
a
g(x(s))dW (s).
Such constructs, containing integration over the increments of a Wiener process are called stochas-tic integrals. In order to estimate the integral let us expand g(x(s)) in a Taylor series in vicinityof x(a): g(x(s)) = g(x(a)) + g′(x(a)) dx
ds
∣∣a
(s− a):
I = g(x(a))
∫ b
a
ξ(s)ds+ g′(x(a))dx
ds
∣∣∣∣a
∫ b
a
ξ(s)(s− a)ds+ ... .
The first term is “simply” the noise, but the second one is much less trivial. The value of dxds
∣∣a
can be obtained from the Langevin equation, dxds
= f(s) + g(x)ξ(s), and is dominated by the noiseterm: in the lowest order
g′(x(a))dx
ds
∣∣∣∣a
∫ b
a
ξ(s)ds ' g′(x(a))g(x(a))ξ(a)
∫ b
a
ξ(s)(s− a)ds.
Taking into account that ξ(s) is unbounded since 〈ξ2(t)〉 = limt→0〈ξ(t)ξ(t′)〉 =∞, we see that theintegral in the last term is preceded by an unbounded prefactor. One can try to circumvent theproblem by the partial integration, but runs into exactly the same difficulty some steps afterward.The problem does not appear only for constant D, when the derivative of g(x) vanishes identically.
Thus, the stochastic integral is not a normal integral, and the SDE is not an ODE. Essentially,an SDE as written is not an equation but a pre-equation21. The expression “pre-equation” meansthat we need something in addition to it to be able to obtain a reasonable result. We namely havefirst to prescribe the exact rule for evaluation of the stochastic integral. The problem does notarise only if dg
dx≡ 0, i.e. for additive noise.
21See the discussion in N. van Kampen, Stochastic processes in Physics and Chemistry, North Holland, e.g. 3rdEdition, 2007.
66
Let us first define the corresponding integrals as (limits of) finite sums and consider the standard
definition for the integral∫ baf(x)dx (here f stands for an arbitrary function of the coordinate).
We subdivide the integration interval into n subintervals of length h = (b−a)/n starting at pointsxi = hi with i = 0, 1, ..., n− 1. Than the (Riemann) integral sum approximating the integral canbe written as
∑n−1i=0 hf(xi), and will tend to the true value of the corresponding integral if n→∞
or h→ 0. Therefore we approximate the integral by∫ b
a
f(x)dx 'n−1∑i=0
hf(xi) ≡n−1∑i=0
f(xi)[xi+1 − xi].
This defines the rectangle rule for the integration. Note that the last expression does not contain theintegration step h explicitly. We can also evaluate f(x) not at the beginning of the correspondingsubinterval but at its end: ∫ b
a
f(x)dx 'n−1∑i=0
f(xi+1)[xi+1 − xi],
which means that we start summation not from the lower but from the upper limit, and evaluate∫ baf(x)dx = −
∫ abf(x)dx using the same method. We can also take∫ b
a
f(x)dx 'n−1∑i=0
f(xi+1) + f(xi)
2[xi+1 − xi],
using the trapezoid integration scheme. For smooth functions the difference between the results ofapplication of each of these prescriptions is of the order of h2, but evaluating the prefactor of this h2
we will need the values of derivatives of f(x) at evaluation points (since f(xi+1) ≈ f(xi) +hf ′(xi).If f(x) is nowhere differentiable, the prefactor diverges!
If we want to calculate∫ baf(x)dg(x) we use the same rules and write e.g.∫ b
a
f(x)dg(x) 'n−1∑i=0
f(xi)[g(xi+1)− g(xi)]
for the first, and the corresponding expressions for the other methods.For stochastic integration we can use the same variants of rectangle or trapezoid rules, and the
three standard prescriptions are:
• Ito prescription:∫ baG(t)dWt ≡
∑n−1i=0 G(ti)[W (ti+1)−W (ti)] (rectangular, initial-point inte-
gration scheme),
• Stratonovich prescription:∫ baG(t)dWt ≡
∑n−1i=0
G(ti)+G(ti+1)2
[W (ti+1) −W (ti)] (trapezoid in-tegration scheme),
• Klimontovich (Hanggi-Klimontovich, or kinetic) prescription. Here∫ baG(t)dWt ≡
∑n−1i=0 f(Gi+1)[W (ti+1)−
W (ti)] (rectangular, end-point integration scheme),
with G(t) = g(x(t)). One often says that the Ito scheme is the only non-anticipating one (we doneed only the values of G(t) at the beginning of the integration sub-interval to get the value atthe end), but it is not clear whether this is a boon (discussion in the lecture).
67
To get some motivation let us return to our Langevin equation and look for∫ b
a
g(x(s))ξ(s)ds '∫ b
a
g(x(a))dt+
∫ b
a
g′(x(a))[x(s)− x(a)]dW (s)
and note that x(s) − x(a) ≈ f(x(a))(s − a) + g(x(a))[W (s) − W (a)] (which is sufficient if theinterval (b− a) is taken very short). The integrals not containing W , or containing only dW makeno difficulties, and the only thing we have to look at is the integral∫ b
a
W (s)dW (s).
Note, that if W were a differentiable function, the schemes would give the same result, but W isnot. Now they lead to different results, and the reason is that the contribution of the noise is ofthe order of h1/2, so that the second order in the noise is of the same order of magnitude as thedeterministic drift, and interferes with this. If W (s) were a “normal” function than the integral
should be∫ baF (s)dF (s) = 1
2[F 2(b)− F 2(a)], but now it is not always the case.
Let us denote Wi = W (ti), dWi = W (ti+1)−W (ti) and look what all three prescriptions giveus for the integral above. To do so we use the analogue of integration by parts formula for thesums: Suppose fk and gk are the two sequences. Then
n∑k=m
fk(gk+1 − gk) = fn+1gn+1 − fmgm −n∑
k=m
gk+1(fk+1 − fk). (63)
It is important to note that the index of g in the sum on the r.h.s. is k + 1.
• Ito prescription.∫ baW (s)dW (s) = limn→∞
∑n−1i=0 WidWi. Using the partial summation for-
mula we get:
n−1∑i=0
WidWi = W 2n −W 2
0 −n−1∑i=0
Wi+1dWi
= W 2n −W 2
0 −n−1∑i=0
WidWi −n−1∑i=0
dWidWi
= W 2n −W 2
0 −n−1∑i=0
WidWi −n−1∑i=0
(dWi)2
where in the second line we used the fact that Wi+1 = Wi + dWi. Since the left hand sideequals the third term in the r.h.s. of the last expression taken with the opposite sign we get:
n−1∑i=0
WidWi =1
2[W 2
n −W 20 ]− 1
2
n−1∑i=0
(dWi)2.
Now we take the limits of both parts for n→∞. The sum∑n−1
i=0 (dWi)2 is the sum of squared
of the i.i.d. Gaussian random variables with zero mean, and it tends to the total variance:limn→∞
∑n−1i=0 (dWi)
2 = (b−a) and is deterministic. Therefore in the Ito prescription we have∫ b
a
W (s)dW (s)(I) =1
2[W 2(b)−W 2(a)]− 1
2(b− a). (64)
68
We note that this prescription defines the new kind of calculus (Ito stochastic calculus) whichdiffers considerably from the standard calculus. The theoretically positive property of thedefinition is that the Ito integral is a martingale, which makes possible many mathematicalproofs.
• Klimontovich prescription. Using exactly the same method (but considering Eq.(63) as anequation for the last term on its r.h.s.) we get∫ b
a
W (s)dW (s)(K) =1
2[W 2(b)−W 2(a)] +
1
2(b− a). (65)
• Stratonovich prescription. Taking the mean value of the last two expressions we get theStratonovich result: ∫ b
a
W (s)dW (s)(S) =1
2[W 2(b)−W 2(a)]. (66)
It is only in the Stratonovich interpretation that the stochastic integral behaves as an integral,and the stochastic differential equation as an ODE. This means that if we do know that our functionξ(t) is a limit of a continuous process (whose changes however are very fast, and whose correlationtime is much smaller than all other characteristic times of the problem) the Stratonovich calculus(i.e. “normal” calculus) should be used. However the situation is not as simple because the processis Markovian only in the limit.
One can put down the general formula: if the prescription is∫ b
a
G(t)dWt ≡n−1∑i=0
[(1− α)G(ti) + αG(ti+1)][W (ti+1)−W (ti)]
with any 0 ≤ α ≤ 1, then∫ b
a
W (s)dW (s)(α) =1
2[W 2(b)−W 2(a)] +
(α− 1
2
)(b− a).
The procedure of fixing the meaning of the stochastic integral is called its interpretation, α isthe interpretation parameter, and α = 0, 1/2 and 1 for the Ito, Stratonovich and Klimontovichinterpretations, respectively.
Let us now turn to the Fokker-Planck equations (FPEs) for the corresponding processes. Todo so we have to use the Langevin equations to obtain the drift and diffusion coefficients. Fromthe Kramers Moyal expansion we know that
∂
∂tp(x, t) =
∂
∂x
[−A(x, t)p(x, t) +
1
2
∂
∂xB(x, t)p(x, t)
],
with
A(x, t) = lim∆t→0
1
∆t
∫drrw(x, r; t,∆t)
and
B(x, t) = lim∆t→0
1
∆t
∫drr2w(x, r; t,∆t)
being the first and the second transition moments. To get them we need only to know the meandisplacement and the mean squared displacement of a particle during a short time interval ∆t.
69
First we assume our process be time-homogeneous, and neglect the possible explicit t-dependenceof f and g. The coefficients A(x) and B(x) would in this case be time-independent. Now we takeour Eq.(62) and write
x(∆t)− x(0) = v(x(0))∆t+
∫ ∆t
0
g(x(s))dW (s) (67)
and use the fact that g(x(s)) = g(x(0)) + g′(x(0))(W (s)−W (0)) (with W (0) = 0):
x(∆t)− x(0) = v(x(0))∆t+ g(x(0))W (∆t) + g′(x(0))g(x(0))
∫ ∆t
0
W (s)dW (s).
Thus,
x(∆t)− x(0) = v(x(0))∆t+ g(x(0))W (∆t)
+g′(x(0))g(x(0))
[W 2(∆t)
2+
(α− 1
2
)∆t
].
The drift and diffusion coefficients are the means
A(x) =1
∆t〈x(∆t)− x(0)〉
B(x) =1
∆t〈[x(∆t)− x(0)]2〉.
Now we note that W 2(∆t) is a random variable with the mean equal to the accumulated dispersionof the noise at time interval ∆t and with higher moments which are o(∆t). Therefore in the order∆t this term is deterministic: W 2(∆t) = ∆t. The term g(x)W (∆t) is of the order of (∆t)1/2,but vanishes under averaging since 〈W (∆t)〉 = 0, and all other terms contain the first and higherorders in ∆t. Therefore for small ∆t we have
A(x) = v(x) + g′(x)g(x)
[1
2+ α− 1
2
]= v(x) + αg′(x)g(x).
On the contrary, B(x) is dominated by the second term 〈[g(x)W (∆t)]2〉 = g2(x)∆t. All otherterms are of the order of ∆t3/2 or smaller. Therefore
B(x) = g2(x).
Our FPE reads:
∂
∂tp(x, t) =
∂
∂x
[−v(x)− αg′(x)g(x) +
1
2
∂
∂xg2(x)
]p(x, t),
The second term, which is present only for non-Ito interpretations is called the spurious drift. Letus rewrite this equation as follows:
∂
∂tp(x, t) =
∂
∂x
[−v(x) + (1− α)g′(x)g(x) +
g2(x)
2
∂
∂x
]p(x, t),
or∂
∂tp(x, t) =
∂
∂x
[−µ(x)f(x) + (1− α)D′(x) +D(x)
∂
∂x
]p(x, t). (68)
70
For time-dependent force and diffusion coefficient we return to Eq.(67) and take the lower limit ofintegration to be t and the upper one t+ ∆t. Repeating the steps of calculation of the moments ofx(t+ ∆t)−x(t) we see that the whole scheme still works, with the only difference that now v(x, t)and g(x, t) get to be explicit functions of time.
Let us return to the structure of Eq.(68). The second term vanishes only for the kineticinterpretation, and only in this interpretation the equilibrium is of the Boltzmann type if the FDT(i.e. the Einstein’s relation D(x) = kBTµ(x)) is locally fulfilled. Each form of these equations canbe considered as an Ito form, and each corresponding process as the Ito process, but only after theredefinition of the deterministic force which, for non-Ito cases, should include the spurious drift22.
Note also that the Euler integrator (provided the small enough time step is used) will give usthe approximation to the Ito result, while the Heun integrator will give the Stratonovich result. Ifthe diffusion coefficient is x-dependent, then the predictor step in the Heun algorithm is x(ti+1) =x(ti)+µf(x(ti))∆t+g(x(ti))∆Wi, with g(x) =
√2D(x) and the corrector step is x(ti+1) = x(ti)+
12[µf(x(ti+1))+µf(x(ti))]∆t+
12[g(x(ti+1))+g(x(ti))]∆Wi ≈ x(ti)+ 1
2[µf(x(ti+1))+µf(x(ti))]∆t+
12[g(x(ti+1)) + g(x(ti))]∆Wi.
5.4 Transition rates and master equations.
The Kramers-Moyal expansion is a formal trick which allows for a rigorous derivation of a Fokker-Planck equation; this trick however is physically not quite transparent. In some situations, bothin classical and in quantum systems, the transition probability between the states of the systemcan be described by the corresponding rates. In these cases a very transparent situation emerges,the one having immediate thermodynamical implications and deep connections with the canonicalformalism of statistical physics.
Let us assume the following structure for the transition probability during the time ∆t, p(x, t+∆t|x′, t):
p(x, t+ ∆t|x′, t) = w(x|x′)∆t+
[1−∆t
∫dx′′w(x′′|x′)
]δ(x− x′) +O(∆t2)
(here the argument t is suppressed in all w-functions). The meaning of this assumption is asfollows: Imagine that at time t the system was in state x′; during the time ∆t it made a transitionto a state x. The probability of such a transition is now expressed through the transition ratew(x|x′); the function w(x|x′) (or wx′→x) is the probability (density) of the transition from x′ to xper unit time. The second term follows from the normalization condition
∫dxp(x, t+ ∆t|x′, t) = 1
which must be fulfilled for any ∆t. This is the probability of not changing the state, it decays as∆t grows. We now insert this expression into our Eq.(50),
p(x, t+ ∆t|x0, 0) =
∫dx′p(x, t+ ∆t|x′, t′)p(x′, t′|x0, 0)
22Having in mind that the Langevin equation describes the limiting case of some random walk, we note that thenecessity for interpretation lays in the fact that in the initial scheme we had two parameters, 〈s2〉 and δt which arenow functions of x, and are amalgamated into a single one, D(x). The resulting dynamics however still depends on〈s2〉 and δt separately and in a different way, and thus it plays a role for this overall dynamics whether 〈s2〉, or δt,or both change. The procedure which defines this separate dependence when the amalgam D(x) is fixed is calledinterpretation. This question will be discussed in more detail when considering Homework 5.
71
and get:
p(x, t′ + ∆t|x0, 0) =
[∫dx′w(x|x′)p(x′, t|x0, t0) +
−∫dx′(∫
dx′′w(x′′|x′))δ(x− x′)p(x′, t|x0, t0)
]∆t
+p(x, t′|x0, 0) +O(∆t2).
Taking the limit ∆t→ 0, we get the following equation for p(x, t|x0, 0):
∂
∂tp(x, t|x0, 0) =
∫dx′w(x|x′)p(x′, t|x0, 0) (69)
−∫dx′w(x′|x)p(x, t|x0, 0),
where in the second line the integration variable is renamed from x′′ to x′ after performing allcalculations. Eq.(69) is called a master equation, and has a transparent physical meaning: Thechange in the probability p(x, t|x0, 0) per unit time is due to the two competing processes: tothe inflow due to the fact that the system may changed its state from some other state x′ to thestate x and to the outflow, due to the fact that it may have already been in state x and mayhave left it for another state. Of course, we can obtain further a Fokker-Planck equation from themaster equation, assuming that the transitions are only very short-ranged, so that only the twofirst moments of r = x − x′ play the role. The Fokker-Planck equation follows from the masterequation in a usual way (by expanding the integrands in r)23 leading to
A(x) =
∫ ∞−∞
dr rw(x+ r|x)
B(x) =
∫ ∞−∞
dr r2w(x+ r|x)
If the second moment diverges, we have to do with a jump-process, whose description lies outsideof the range of applicability of a Fokker-Planck equation, but which still can be treated within themaster equation formalism.
For a discrete system, where the states are numbered by integers n instead of a continuousvariable x the situation gets even more transparent:
∂
∂tp(n, t|n0, 0) =
∑n′
wn′→np(n′, t|n0, 0)−
∑n′
wn→n′p(n, t|n0, 0) (70)
where wn′→n is the transition rate from the state n′ to the state n. The coefficients wn′→n form amatrix W . Such discrete equations are in many cases more reasonable instruments of descriptionthan the continuous ones, especially in the cases when n can be interpreted not as a coordinate,but as a number of particles (birth-death processes, chemical reactions, etc.), or as a number ofa discrete quantum state. In this quantum case the transition rates are explicitly given by theFermi’s Golden rule.
23An alternative discussion will also be given in Homework 8.
72
5.5 Energy diffusion and detailed balance
Master equations are often introduced on a quite formal basis, and are intimately connected withthe thermodynamical formalism. Let us assume a closed system in a contact with the heat bathat temperature T . According to the Zeroth Law, whatever nonequilibrium the initial state ofthe system is, in course of the time the system will tend to an equilibrium one, if all externalperturbations are switched off. Let ν enumerate the states of this system. Let moreover Eν be theenergy of the state ν. Then, the final, equilibrium state of the system will be characterized by theBoltzmann distribution
peq(ν) =1
Zexp
(−EνkT
)with Z being the partition function. We now consider the relaxation of our system from a nonequi-librium state characterized by the some distribution function p(ν) to the equilibrium state. Wenote that during this evolution the states themselves (which depend on the external conditionswhich are assumed not to change) do not change; the relaxation is fully due to the change ofthe probabilities for a system to be in a state ν. Let us assume these probabilities to followthe master equation with constant transition rates. This assumption follows from the one of thetime-independent nature of the states and of the heat bath.
We note that the assumption of the time-independent states fully characterized by their energymay have quite different physical implications. The simplest situation is that the states Eν are thelocalized quantum states (say of an electron at different impurities in a doped semiconductor), andthe incoherent transitions between them are caused by the interaction with the heat bath. Just assimple are the Ising-like systems without any internal dynamics. Here the only possible changesare the flips of single spins or their clusters. Each such discrete event may (or may not) changethe energy of the system. A similar situation takes place at a coarse-grained level of descriptionin a classical system corresponding to a particle in a rugged energy landscape. An underdampedsituation corresponds to a much more complex situation: Here, even at a constant energy, acomplex dynamics corresponding to the ”microcanonical” (i.e. purely Hamiltonian) evolution ofthe system takes place. In order to be able to assume that the state is fully characterized bythe value of its energy, we have to consider a ”state” as a microcanonical ensemble of systems,equally populating the energy surface H(x, p) = Eν . The interaction of the system with the heatbath introduces the transitions between different energy surfaces, the rates of such transitions arewν′→ν . We will return to this situation a bit later.
The equilibrium state is a stationary (time-independent) solution of the master equation:
0 =∑ν′
wν′→νpeq(ν′)−
∑ν′
wν→ν′peq(ν).
Thus, the Zeroth Law requires that∑ν′
[wν′→νpeq(ν′)− wν→ν′peq(ν)] = 0.
This requirement is tightened by the Second Law: one namely has to assume that not only thesum, but each term separately vanishes, so that
wν′→νpeq(ν′)− wν→ν′peq(ν) = 0. (71)
It follows thatwν→ν′
wν′→ν=peq(ν
′)
peq(ν)≡ exp
(−Eν
′ − EνkT
). (72)
73
This assumption is called the principle of detailed balance. If the external conditions are time-dependent, the detailed balance principle has to hold at each time for time-dependent rates, aslong as Markovian dynamics holds. The same assumptions are of course valid in the continuouscase.
The meaning of the principle is as follows: Imagine that the transition rates between the states1 and 2 of the system do not follow Eq.(71), so that the number of the transitions from state 1 to thestate 2 is not balanced by the backwards transitions. For example, let us assume that in equilibriummore transitions take place immediately from 1 to 2 than back from 2 to 1: the back flow from2 to 1 follows through some intermediate state(s) 3, so that a perpetual current 1 → 2 → 3 → 1flows. (The perpetual superconductive currents or the probability currents discussed above arenot the currents between the states, but currents within a state characterized by a given energy;they are not forbidden by the following consideration). If the energy of the state 2 is lower thanthe energy of state 1, then at first step the energy is dissipated to the heat bath, and the secondstep, 2→ 3→ 1, is thermally activated: the energy is taken from the bath. The equilibrium stateis stable, so that small external perturbation wouldn’t change the state considerably. Thus, if wewould be able to use the unbalanced 1→ 2 current for producing work against weak external force,this work will be produced on the cost of cooling the only one heat reservoir, which is explicitlyforbidden by the second law. We refrain here from the discussion of possible constructions of suchperpetuum mobiles. The discussion of the thermodynamical implications of detailed balance wasfirst given by Bridgman, 1928.
The following discussion is different from the one given in the lecture, and, I hope, free of gravemistakes!
Now let us return to our master equation in the case when it can be reduced to a Fokker-Planckone and show that the Einstein’s relation follows in general from the detailed-balance principle.Note that in the case when the master equation can be reduced to a Fokker-Planck one, i.e. in thecase when w(x + r|x) decays sufficiently fast as a function of r. The force f(x) and the value ofw(x+ r|x) will be considered as changing only slowly in the typical scale of decay of w(x+ r|x) asa function of r. The last property allows us for taking
w(x− r|x) = w(x|x+ r)− r ddxw(x|x+ r) + o(r). (73)
The detailed balance requires the connection between the transition moments: In the overdampedcase one can assume E(x) = U(x) = −
∫f(x)dx, so that the energy difference between the state
x+ r and the state x is E(x+ r)− E(x) = −∫ x+r
xf(x′)dx′, so that Eq.(72) implies
w(x|x+ r)
w(x+ r|x)= exp
(E(x+ r)− Ex
kBT
)= exp
(− 1
kBT
∫ x+r
x
f(x′)dx′)' 1− rf(x)
kBT, (74)
i.e.
w(x|x+ r) = w(x+ r|x)
(1− rf(x)
kBT
)+ o(r).
Substituting this in Eq.(73) we thus get
w(x− r|x) = w(x+ r|x)
(1− rf(x)
kBT
)− r d
dxw(x+ t|x) + o(r). (75)
74
The integral for A in the lowest order in r then reads
A(x) =
∫ ∞−∞
dr rw(x+ r|x) =
∫ ∞0
r [w(x+ r|x)− w(x− r|x)] dr
'∫ ∞
0
r
[w(x+ r|x)
(1− 1 + r
f(x)
kBT
)+ r
d
dxw(x|x+ r)
]dr
=f(x)
kT
∫ ∞0
w(x+ r|x)r2dr +d
dx
∫ ∞0
w(x+ r|x)r2dr.
Note that the terms of the zeroth order vanishes. The lowest-order approximation for the coefficientB is zeroth order in r and reads
B(x) =
∫ ∞−∞
dr r2w(x+ r|x) =
∫ ∞0
r2 [w(x+ r|x) + w(x− r|r)] dr
=
∫ ∞0
w(x+ r|x)r2[1 + 1 +O(r)]dr ' 2
∫ ∞0
w(x+ r|x)r2dr.
Thus, one readily infers that
A(x) =f(x)
2kBTB(x) +
1
2
d
dxB(x) =
f(x)
kBTD(x) +
d
dxD(x).
where we remembered that B/2 is exactly the diffusion coefficient D. Now we get the Kolmogorov’sforward equation
∂
∂tp(x, t) = − ∂
∂xA(x)p(x, t) +
∂2
∂x2D(x)p(x, t)
=∂
∂x
[−A(x)p(x, t) +
∂
∂xD(x)p(x, t)
]=
∂
∂x
[− D
kBTf(x)p(x, t)−
(d
dxD(x)
)p(x, t) +
(d
dxD(x)
)p(x, t) +D
∂
∂xp(x, t)
]=
∂
∂x
[− D
kBTf(x)p(x, t) +D(x)
∂
∂xp(x, t)
]=
∂
∂x
[−µ(x)f(x)p(x, t) +D(x)
∂
∂xp(x, t)
],
i.e. the phenomenological Fokker-Planck equation with the particles’ mobility µ(x) connected toD(x) via the Einstein relation D = kBTµ.
In our discussion we could also assume that the transition rates are time-dependent but thecondition Eq.(74) if fulfilled at each instant of time. The result will be the same, with the onlydifference that we should write D = D(x, t), µ = µ(x, t) provided D(x, t) = kBTµ(x, t).
6 Escape and first passage problems
The possibility to calculate the probability density function or its moments, like the mean squaredisplacement, does not exhaust the whole class of problems of stochastic theory. One of the mostimportant classes of other problem settings are the first passage problems for a stochastic process.
75
Mathematically, the task is to calculate the probability density ψ(t) (as a function of time) for aprocess x(t) to reach a point x for the first time (a typical one-dimensional problem position), orto cross a boundary of a spatial domain in spatial dimension more than one. Physical problemsleading to this mathematical formulation are abundant. The genuine first-passage problems areoften pertinent to reaction kinetics, where the two particles interact if they approach each other atdistance a. These situations correspond to crossing the boundary from the outside. An oppositesituation of the crossing a boundary from inside, emerges when describing a decay of a boundedstate in some given potential (a problem emerging in the description of a dissociation of a complexmolecule). This situation is often termed an escape problem. In ecology, the first passage throughsome prescribed value, might mean the extinction of the population, or the start of the epidemicoutbreak. The literature discussing the first passage problems both from the mathematical, andfrom the physical point of view is abundant24. In this Chapter we consider only some simplesituations.
6.1 General considerations
Let us first concentrate on one-dimensional problems and look at realizations of the correspondingrandom process. The physical situation here can be formulated as follows: At time t = 0 a particleis introduced into a system at x = 0. As soon as particle reaches the place x0 the realization isstopped, and the time necessary to reach the boundary (the first passage time) is recorded. Then,we can obtain the distribution of these times, and use it for further calculations. It is importantto note that we are interested in the frequencies of the realizations of a random process, i.e. inthe trajectories (paths) of the particle. In order to get the distribution of the first passage times,we have to know, in how many realizations the path of the particle crossed the point x = x0 attimes t′ < t. In order to count them, it is enough to assume an absorbing boundary conditionat x = x0: Particles having touched the point disappear and their trajectories are disregarded atfuture times. We assume the overdamped regime of motion, where the probability density to finda particle’s at point x is governed by Smoluchowski equation
∂p(x, t)
∂t= − ∂
∂xµf(x, t)p(x, t) +D
∂2p(x, t)
∂x2. (76)
Here the force f = −dU/dx. In cases corresponding to the systems considered in thermodynamics(i.e. ones possessing true thermodynamical equilibrium) the mobility µ and the diffusion coefficientD are connected to each other through the Einstein’s relation D = µkBT . The equation, Eq.(76)has to be solved under the boundary condition p(x0, t) = 0 and for the initial condition p(x, 0) =δ(x), so that the corresponding solutions are essentially the Green’s functions of the Fokker-Planckequation, p(x, t) = G(x, t|0, 0). It is important to stress here that the equivalence of our assumptionthat all particles touching x0 disappear and the actual condition that the trajectories crossing x0
have to be disregarded at larger times assumes the continuity of the trajectories, which is the casefor Brownian motion, but might be violated for some other processes, so that care has to be takenwhen generalizing the results discussed here to the processes other than Fickian diffusion.
After solving the equation for all times, we can calculate the overall probability for the particleto stay within the interval, which is
P (t) =
∫ x0
−∞p(x, t)dx,
24 A simple physical introduction into the topic is given in the book S. Redner A Guide to First-Passage Processes,Cambridge Univ. Press, 2001
76
and note that the change in P (t) between the times t and t+ dt is exactly the probability to leavethe interval during dt. This is,
ψ(t) = −dP (t)
dt. (77)
By noting thatdP (t)
dt=
∫ x0
−∞
∂p(x, t)
∂tdx
and using Eq.(76) we get
ψ(t) =
∫ x0
−∞
[− ∂
∂xµfp(x, t) +D
∂2p(x, t)
∂x2
]dx.
Applying partial integration, and using the natural boundary condition p(x, t) → 0 and for x →−∞ we get:
ψ(t) = −D ∂p(x, t)
∂x
∣∣∣∣x=x0
,
i.e. that the first passage time probability density is equal to the diffusion flux through theabsorbing boundary. If we are interested in the case of two absorbing boundaries at the ends ofan interval the corresponding currents give us the probability per unit time to leave the intervalthrough the corresponding end.
This approach based on the solution of the forward equation assumes the knowledge of itstime-dependent solution, and is the typical ”physicist’s” approach to the problem, which is notnecessarily the simplest or the most elegant one. However, this is the one which works and,moreover, the one which works reliably even when the coefficients in the Fokker-Planck equationare time-dependent.
6.2 Mean life time in a potential well
One of the most important results following from the theory based on the Fokker-Planck equationis the typical life-time in a potential well, i.e. the time necessary to overcome a potential barrier.Since this result will be repeatedly used in what follows, we shall discuss it here in some detail.
The situation considered here is depicted in Fig.3. Imagine, a particle starts at the minimumof the potential, at point x = 0. We say that the particle overcame a barrier if it arrived for thefirst time at a maximum of the potential curve at x = x0. The value of the potential energy ofthe particle there is U(x0) while the minimum of the potential energy at x = 0 is taken for thereference point, U(0) = 0.
6.2.1 The flow-over-population approach
In order to find the mean first passage time, we don’t need to solve the whole time-dependentproblem.
Let us return to the physical formulation of the problem, and again discuss the experimentwith putting the particles into a system. In order to obtain the result, it is not necessary to putthe particles one by one, and wait until the corresponding realization terminates. Let us imaginethat at the point x = 0 a constant current of strength I is flowing into the system, see Fig.4. Atthe beginning, after switching on such current, the concentration of particles in the system willgrow, and the output current, leaving the system at point x0, will be smaller than the input current
77
Figure 3: The typical potential discussed in the escape problem: a potential well with a simplequadratic minimum and quadratic maximum maximum.
I. Eventually, the steady state is reached, the input and the output currents equilibrate, and thesteady-state concentration profile establishes itself. Now, if the mean first passage time, i.e. thetime a particle on the average spends within the system, is τ , the mean number of particles Nwithin the system will be exactly Iτ . Therefore
τ =N
I.
The concentration profile n(x) under the application of the current is given by the stationarysolution of the Fokker-Planck equation (for concentration, not for the probability density, howeverthese two differ only in normalization) with corresponding additional conditions. Thus, we have:
τ = I−1
∫ x0
−∞n(x)dx (78)
where n(x) is the steady-state solution of the Fokker-Planck equation (76)25.Note that the Fokker-Planck equation is essentially the continuity equation for the probability
(particles’) flux, and that
I = µ(x)f(x)n(x)−D(x)d
dxn(x). (79)
Here we remind that f = −dU/dx. This is an ordinary linear differential equation, which can berewritten in the form:
dn(x)
dx+ n(x)
µ(x)
D(x)
d
dxU(x) = − I
D(x)(80)
Eq.(80) is a linear differential equation, whose formal solution reads:
n(x) = exp(−V (x))
[n(0)−
∫ x
0
I
D(x)exp(V (x′))dx′
](81)
25 In the lecture we have discussed how this approach follows from the Green’s function solution of the Fokker-Planck equation when its coefficients are not time-dependent.
78
Figure 4: The flow over population approach: We assume the particles are introduced at x = 0and leave the system through the absorbing boundary at x = x0. This corresponds to the constantcurrent of particles (or probability) through the system.
with V (x) = µ(x)U(x)/D(x) = U(x)/kBT . The boundary condition n(x0) = 0 leads to theexpression for n(0):
n(0) = exp(−V (0))
∫ x0
0
I
D(x)exp(V (x′))dx′. (82)
From here on we assume for simplicity that the mobility and the temperature are constant through-out the whole system, and D = kBTµ. According to our choice of the reference energy in Fig.4 wehave exp(−V (0)) = 1, which simplifies the expressions. Note that the probability current to theleft from point x = 0 vanishes, so that to the left of this point the steady-state solution coincideswith the equilibrium solution n(x) = n(0) exp(−V (x)). We thus have:
n(x) =I
Dexp(−V (x))
∫ x0
0
exp(V (x′))dx′ for x ≤ 0
n(x) =I
Dexp(−V (x))
∫ x0
x
exp(V (x′))dx′ for 0 < x < x0
(83)
We thus have the following expression for the mean first passage time in the system:
τ =1
D
∫ 0
−∞exp(−V (x′))dx′
∫ x0
0
exp(V (x′))dx′ +
+1
D
∫ x0
0
[exp(−V (x′′))
∫ x0
x′′exp(V (x′))dx′
]dx′′, (84)
the expression that can be rewritten in a form
τ =1
D
∫ x0
0
dy′∫ y′
−∞dy′′ exp[V (y′)− V (y′′)]. (85)
Let us now concentrate on the case of a deep well or low temperatures. In this case we can makesimple estimates for our integrals which now depend only on the properties of the potential closeto the points x = 0 and x = x0.
79
Figure 5: A symmetric double-well potential U(x) = x4 − 2x2 + 1
6.3 Rates of transitions
6.3.1 The Arrhenius law
Let us assume that close to the maximum at point x = x0 one has V (x) ' V (x0)−(2kBT )−1U ′′(x0)(x0−x)2. In this case the integral
∫ x0
0exp(V (x′))dx′ in the first line of the last expression and the integral∫ x0
x′′exp(V (x′))dx′ (which, for x′′ not too close to x0, is practically a constant) are approximately
equal to each other and can be evaluated using the Laplace method:∫ x0
0
exp(V (x′))dx′ '√πkBT√
2|U ′′(x0)|exp[V (x0)]. (86)
If we assume that the potential has a simple quadratic minimum close to x = 0, i.e. that V (x) '(2kBT )−1U ′′(0)x2, the same type of an approximation for∫ 0
−∞ exp(−V (x′))dx′ and for∫ x0
0exp(−V (x′))dx′ again shows that the integrals are approximately
equal to each other and can be approximated by∫ x0
0
exp(−V (x′))dx′ '√πkBT√2U ′′(0)
exp[−V (0)] (87)
(in our case we have V (0) = 0). The overall expression for τ the reads
τ ' πkBT
D√U ′′(x0)U ′′(0)
exp
(U(x0)
kBT
): (88)
the mean first passage time depends exponentially on the height of the barrier relative to thethermal energy of particles. This exponential growth is often termed as the Arrhenius law.
6.3.2 Diffusion in a double-well
As the next example let us consider the behavior of a particle in a double-well potential consistingof two wells of depths U1 and U2 measured with respect to the top of potential barrier. Then themean first passage time from one well to another one is determined by the corresponding well’sdepths. Fig.5 shows a symmetric potential well U(x) = x4−2x2 +1 (having two minima at x = ±1separated by a barrier of height 1 with the maximum at x = 0. A trajectory of the diffusive processx(t) for µ = 1 and for T = 0.15 is shown in Fig.6. Note that the behavior of the trajectory shown
80
Figure 6: A realization of diffusion in a double-well potential shown in Fig.5 for µ = 1 and forT = 0.15.
in Fig.6 suggests its possible description as a sequence transitions between the two well-definedstates, in which the particle is localized either in the left well (around x = −1) or in the rightone (around x = 1). Let us now identify the position of the particle to the left from the top ofthe barrier with the L-state of the system, and the position to the right from the top with theR-state. The average life-time in either state is 2τ (since after reaching the top of the potentialbarrier the particle can either change the well or return to the initial one, with equal probability),which defines the transition rate between the wells, k, which corresponds to the mean number oftransitions in either direction per unit time. For the wells of different depth, two different rates,k+ for the transition in positive direction and k− for the transition in negative direction; each ofthem being equal to a half of the inverse of the corresponding mean first passage time to the topof the barrier.
The transitions between the L and the R state of the system can be considered as a simplestreversible isomerization reaction,
L R. (89)
and the probabilities PL and PR to find a system in either state are given by the system of ordinarydifferential equations,
dPLdt
= −k+PL + k−PR
dPRdt
= −k−PR + k+PL. (90)
This coarse-grained approximation corresponds to the formal kinetics approach to the situation.If we consider the ensemble of such two-state systems, than the number of systems being in statesR and L are proportional to the corresponding probabilities, and their sum equals to the overallnumber of systems N .
The equilibrium probabilities following from Eq.(90) are PL = k−/(k++k−) and PL = k+/(k++k−) so that
PLPR
=k−k+
. (91)
81
Expressing the transition rates through the first passage times and using the expression for thefirst passage time derived above we get
PLPR
=
√U ′′(xL)
U ′′(xR)exp
(U(xR)− U(xL)
kBT
). (92)
Note that this is exactly the result which one would obtain if one simply uses the equilibriumBoltzmann distribution p(x) = A exp(−U(x)/kBT ) with A being the normalization constant and
define PL =∫ 0
−∞ p(x)dx and PR =∫∞
0p(x)dx. The corresponding integrals are exactly of the type
of Eq.(87) so that
PL = A√πkBT/U ′′(xL) exp[−U(xL)/kBT ] (93)
andPR = A
√πkBT/U ′′(xR) exp[−U(xR)/kBT ], (94)
and their quotient is exactly given by Eq.(92). This connection between the rates and the equi-librium populations is rather universal and holds whenever rates exist. It is closely related to themass action law of equilibrium thermodynamics.
6.4 Kolmogorov backward equation
Our flow-over-population approach seems to look a bit too “physical”, i.e. lacks the mathematicalrigor. It however has a rigorous mathematical background footing on a Kolmogorov backwardequation, which was not considered so far. Here we discuss some details of the approach.
The transition probability densities p(x, t|y, s) are the functions of four arguments: the finalposition x, the final time t, the initial position y and the initial time s. In the forward equationwe treat x and t as arguments of the function, and y and s as its parameters, defining the initialconditions. In the backward equation the situation is reversed: y and s are the arguments, andthe final position and the final time serve as parameters.
Let us start from the Chapman-Kolmogorov equation
p(x, t|y, s) =
∫dx′p(x, t|x′, t′)p(x′, t′|y, s).
Now, parallel to the derivation of the forward equation we introduce W (y, r, s,∆s) = p(y + r, s+∆s|y, s) and write
p(x, t|y, s) =
∫drp(x, t|y + r, s+ ∆s)W (y, r, s,∆s),
or
p(x, t|y, s) =
∫drp(x, t|y + r, s′)W (y, r, s, s′ − s)
(W (y, r, s, s′ − s) is not yet a transition rate but still the transition probability, normalized in itsr-variable:
∫W (y, r, s, s′− s)dt = 1 for all y, s and s′). Now we expand p(x, t|y+ r, s′) in a Taylor
series
p(x, t|y + r, s′) = p(x, t|y, s) +∂p(x, t|y, s)
∂s(s′ − s) +
∂p(x, t|y, s)∂y
r +1
2
∂2p(x, t|y, s)y2
r2 + ...
82
and substitute this into the Chapman-Kolmogorov equation:
p(x, t|y, s) = p(x, t|y, s)∫W (...)dr +
∂p
∂s(s′ − s)
∫W (...)dr
+∂p
∂y
∫rW (...)dr +
1
2
∂2p
∂y2
∫r2W (...)dr + ... .
Now we denote s′ − s = ∆s and note that∫W (...)dr = 1. We therefore obtain
0 =∂p
∂s+
(1
∆s
∫rW (...)dr
)∂p
∂y+
1
2
(1
∆s
∫r2W (...)dr
)∂2p
∂y2+ ...
Now we pass to the limit ∆s → 0 and recognize in the expressions in brackets the first twoKramers-Moyal coefficients. Neglecting all terms of higher order we write
−∂p∂s
=
(A∂
∂y+B
2
∂2
∂y2
)p,
the Kolmogorov backward equation with the operator L+ in the r.h.s. conjugated to the Fokker-Planck operator
L =∂
∂xA ·+ ∂2
∂x2
B
2·,
the one we have already seen in our Homework 3.26.For time-independent drift and diffusion coefficients A(x, t) = A(x) and B(x, t) = B(x) the
transition probabilities are stationary, i.e. depend only on t − s: p(x, t|y, s) = G(x, y, t − s). Inthis case the backward equation reads
∂
∂tG =
(A∂
∂y+B
2
∂2
∂y2
)G. (95)
6.4.1 Pontryagin-Witt equations
Let us consider the first passage time probability density to the point x0 when starting at y:
ψ(t|x0, y, s) = − d
dt
∫ x0
−∞p(x, t|y, s)dx. (96)
Note that ψ is obtained form p by action of the integrodifferential operator O = − ddt
∫ x0
−∞ dx·,namely ψ = Op, and this operator does not act on y and on s. Applying this integrodifferentialoperator to the both parts of the Kolmogorov backward equation
−∂p∂s
=
(A∂
∂y+B
2
∂2
∂y2
)p
we get the equation for ψ:
− ∂ψ
∂s=
(A∂
∂y+B
2
∂2
∂y2
)ψ = L+ψ. (97)
26The Kolmogorov backward equation (KBE) is not the equation which answers the question on what was thedistribution of the particle’s position at time s given its position at time t: the analogy with quantum mechanicsis misleading. The equation contains the same information on forward propagation as the FPE. Several renownpeople ran into grave mistakes using the KBE for back-propagating the PDF in time.
83
Thus, the first passage density obeys the KBE with respect to its variables corresponding to initialconditions.
In what follows we fix x0, the absorbing point, and consider ψ(t|x0, y, s) as a function of twovariables, the starting point y and the starting time s, and concentrate on the case of time-independent drift and diffusion coefficients. In this case ψ(t|x0, y, s) is only the function of t− s.
Let us now discuss the n-th moment of the first passage time, and consider it as an explicitfunction of y:
〈τn(y)〉 =
∫ ∞s
(t− s)nψ(t|x0, y, s)dt.
We multiply both parts of Eq.(97) times (t− s)n and integrate it over t:
−∫ ∞s
(t− s)n∂ψ∂sdt =
∫ ∞s
(t− s)nL+ψdt
From here on we use the stationarity of the transition probabilities, from which it follows thatψ(t|x0, y, s) = ψ(t− s|x0, y, 0) ≡ ψ(t− s|x0, y):
−∫ ∞s
(t− s)n ∂∂sψ(t− s|x0, y)dt =
∫ ∞s
(t− s)nL+ψ(t− s|x0, y)dt.
Now we change the variable of integration to τ = t−s and note that ∂∂sψ(t−s|x0, y) = − ∂
∂τψ(τ |x0, y).
We get: ∫ ∞0
τn∂
∂τψ(τ |x0, y)dτ =
∫ ∞0
τnL+ψ(τ |x0, y)dτ = L+
∫ ∞0
τnψ(τ |x0, y)dτ.
In the l.h.s. we now perform the partial integration
τnψ(τ |x0, y)|∞0 − n∫ ∞
0
τn−1ψ(τ |x0, y)dτ = L+
∫ ∞0
τnψ(τ |x0, y)dτ
and note that if the n-th moment of the FPT exists, the first term on the l.h.s. must vanish:in order for the n-th moment to exist the function ψ(τ) must decay faster than τn at infinity.From this we get the hierarchy of the equations for the moments of the FPT (the Pontryagin-Wittequations):
L+〈τn(y)〉 = −n〈τn−1(y)〉. (98)
The first equation for the mean FPT τ(y) is(A∂
∂y+B
2
∂2
∂y2
)τ(y) = −1. (99)
Possible boundary conditions are:
• absorbing boundaries at a and at b when starting within the interval: τ(a) = τ(b) = 0 sincethe mean first passge time to the boundary when being at the boundary is zero.
• reflecting boundary at a, absorbing at b: ∂∂yτ(y)
∣∣∣y=a
= 0, τ(b) = 0. The first one follows
from the properties of the transition probability close to the boundary. A simple explanation:Assume that the coefficients A and B close to the boundary do not diverge or vanish. Then,in the close vicinity of the boundary they may be considered constant (i.e. we may assume thespacial homogeneity), and locally there is no difference between the forward and backwardoperator. Then, for y → x0, p(x, t|y, s) ≈ p(x0 − y, t − s). Since ∂
∂xp(x, t|y, s) vanishes for
x→ x0, so does also ∂∂yp(x, t|y, s).
84
In the case of the Kramers problem with the potential U(x) which grows sufficiently fast forx→ −∞ one can assume that this serves as a reflective boundary at infinity.
• free boundary at −∞, absorbing boundary at x0 = b: τ(b) = 0, ∂∂yτ(y)
∣∣∣y=→−∞
= 0. There
are also other kinds of “qualitative boundary conditions”: for example if one knows that thesecond moment exists, one can assume that p and therefore τ decay sufficiently fast at −∞.
The solution of Eq.(99) exactly reproduces the result of the Flow-over-Population approach.Let us consider the case of local detailed balance µ(y) = D(y)/kBT , introduce the force f =−(dU(y)/dy) and the dimensionless potential V (y) = U(y)/kBT , and use a shorthand notationτ(y) = 〈τ(y)〉. Our equation now reads:
D(y)
(−dVdy
d
dyτ +
d2
dy2τ
)= −1.
Let us introduce
z(y) =dτ
dy
and use the boundary conditions
τ(x0) = 0,dτ
dy
∣∣∣∣y→−∞
= 0.
Our equation for z(y) now reads:dz
dy− dV
dyz = − 1
D.
The general solution to this equation reads:
z(x) = e−F (x)
[η −
∫ x
ξ
1
D(x)eF (x′)dx′
]with η and ξ defining the integration constant: z(ξ) = η and with
F (x) = −∫ x
ξ
dV (x′)
dx′dx′ = V (ξ)− V (x).
Now we can take ξ → −∞ and use the boundary condition z(−∞) = 0 to obtain the finalexpression
z(x) = −∫ x
−∞
e[V (x)−V (x′)]
D(x′)dx′.
Now we integrate z(x) to get the final solution τ(y) =∫z(x)dx using the boundary condition
τ(x0) = 0:
τ(y) = −∫ y
x0
dx′′∫ x′′
−∞
e[V (x′′)−V (x′)]
D(x′)dx′
=
∫ x0
y
dx′′∫ x′′
−∞
e[V (x′′)−V (x′)]
D(x′)dx′,
which is our Flow-over-Population result.We note that higher Pontryagin-Witt equations Eq.(98) give a possibility to calculate higher
moments of the waiting time, which play important role in several applications. We also note thatthe corresponding equations can be generalized to the multidimensional situations.
85
6.5 The renewal approach.
The renewal approach to the first passage problem uses explicitly the continuity of sample pathsand the Markovian nature of the problem. It reduces the solution of a problem with an absorbingboundary condition to the one for the free problem, the one with “natural” boundary conditions,which is sometimes much simpler, especially in the one-dimensional case. Like the standard Flow-over-Population or Pontryagin-Witt approaches, it explicitly relies on the stationarity of transitionprobabilities.
Let us start from the simple case of a single boundary situated at x = x0 > 0. We considerthe process starting at t = 0 at x = 0. Imagine that at some time tf > 0 the particle is found tothe right of the boundary, at some point xf > x0. The probability of this event is given by thetransition probability density function of the ”free” process p(xf , tf |0, 0).
Due to the continuity of trajectories, the process has to have passed the point x0 before reachingx at some time t < tf (and might have passed the point x0 one or several times after this). Thusthe realizations of the process leading from x to xf may be uniquely classified with respect to thetime they first crossed x0. Let us denote by ψ(t, x0) the first passage time distribution throughpoint x0, and concentrate on all realizations of the process contributing to ψ(t, x0).
Here the Markovian nature of the process comes in play. The probability that the particlereaches x = xf at t = tf provided it was at x0 at time t depends only on x0 and t but not onthe history of the process at t < t0 and is given by the transition probability density function ofthe free process p(xf , tf |x0, t0). This means that the probability density to cross the point x0 attime t and then to reach xf at time tf is simply a product p(xf , tf |x0, t)ψ(t, x0). Summing overall possible times t < tf one gets
p(xf , tf |0, 0) =
∫ tf
0
p(xf , tf |x0, t)ψ(t, x0)dt. (100)
This is an exact equation determining the first passage time for a Markovian process.A next step can be done if the process x(t) in a “free” motion is stationary (i.e. whenever
the coefficients in our Fokker-Planck equation are time-independent). In this case the transitionprobability density depends only on the difference between its time arguments: p(x2, t2|x1, t1) =G(x2, x1, t2 − t1) and is equal to the Green’s function solution of the Fokker-Planck equationwith time-independent coefficients. The integral in the r.h.s. of Eq.(100) has now a form of aconvolution. Another simplification stems from the fact that for diffusion processes one can takea limit xf → x0 (since the corresponding Green’s function is non-singular in this limit), so that
G(x0, 0, tf ) =
∫ tf
0
G(x0, x0, tf − t)ψ(t, x0)dt. (101)
This is a Volterra integral equation, which can be easily solved numerically. The simplest wayto its analytical solution is to take Laplace transforms of the both sides of the equation, so thatG(x2, x1, u) =
∫∞0G(x1, x2, t)e
−utdτ . Applying this transform we get G(x0, 0, u) = G(x0, x0, u)ψ(u, x0),i.e.
ψ(u, x0) =G(x0, 0, u)
G(x0, x0, u).
Of course, in many situations, the inverse Laplace transform has to be performed numerically.
However, the expression for the mean first passage τ =∫∞
0tψ(t, x0)dt = d
duψ(u, x0)
∣∣∣u=0
easily
follows analytically.
86
The renewal approach can also be applied for calculating splitting probabilities. Let us con-sider a particle moving between the two absorbing boundaries. Here there are two first passageprobability densities, ψ(t, xL) for first crossing the left boundary situated at xL, and ψ(t, xR) offirst crossing the right boundary situated at xR. Parallel to our previous consideration one cansay that if the particle which started at x = 0 at t = 0 is found at time t to the right from theright boundary (at some xf > xR) it might either have first crossed the right boundary withouttouching the left one at some time t′ < t and then made the way from xR to xf or crossed first theleft boundary and then made the way from xL to xf . The probability to touch the left boundaryfor the first time without previously touching the right one between times t and t + dt is givenby the function ψ(t, xL)dt and the probability to first touch the right boundary (without touchingthe left one before) is given by ψ(t, xR)dt. Note that ψ(t, xL) and ψ(t, xR) are not proper proba-bility density functions: both integrals PL =
∫∞0ψ(t, xL)dt and PR =
∫∞0ψ(t, xR)dt (representing
the probabilities to leave the interval through its left or through its right boundary, the so-calledsplitting probabilities) are in general less than unity. The normalization condition for the splittingprobabilities is given by PL + PR = 1. Following the same scheme as before we obtain
G(xf , t|0, 0) =
∫ t
0
G(xf , t|xR, t′)ψ(t′, xR)dt′ +
∫ t
0
G(xf , t|xL, t′)ψ(t′, xL)dt.′ (102)
This is an equation determining both ψ(t, xL) and ψ(t, xR) since it has to be valid for any xf . Toobtain the explicit equations we use two different xf : one taken to lie to the left from the leftboundary and another one to lie to the right of the right boundary, and then make the limitingtransitions, taking the corresponding xf ’s to tend to the boundaries of the interval from the outside.Since for stationary case the integrals in both such equations are of the convolution type we canmake the Laplace-transform of the both and get
GRRψR +GRLψL = GR0 (103)
GLRψR +GLLψL = GL0,
where the following shorthand notation is introduced for the Green’s functions: GRR = G(xR, xR;u),GLL = G(xL, xL;u), GRL = G(xR, xL;u), GLR = G(xL, xR;u), GR0 = G(xR, 0;u), and GL0 =G(xL, 0;u). Moreover, ψL = ψ(u, xL) and ψR = ψ(u, xR). The solution of the system of equations(103) is
ψL =GR0GLL −GL0GRL
GRRGLL −GLRGRL
(104)
and
ψR =GRRGL0 −GLRGR0
GRRGLL −GLRGRL
(105)
so that the probabilities can be found explicitly. The splitting probabilities PL and PR are simplygiven by the limiting values of the corresponding functions at u→ 0.
The importance of this approach lays in the fact that it may still give reasonable results innon-Markovian cases. In some cases these results are exact, in other cases they give reasonableapproximations.
Thus the renewal equation still holds for ”semi-Markovian” situations like continuous-time ran-dom walk (”Disordered Systems” course). In this case the renewal approach is effective and prefer-able because the probability density for free motion might be obtained by the alternative meth-ods, without explicitly solving the corresponding (non-Markovian, fractional, integro-differential)
87
Fokker-Planck equations. Eq.(101) does not in general hold for non-Markovian processes. How-ever, for non-Markovian process with continuous trajectories Eq.(101) might still be a reasonableapproximation: it simply assumes that having arrived to a position of the absorbing boundary, theparticle has practically forgotten initial conditions, so that its further behavior can be describedby a new initial condition problem. In this case the renewal approximation based on Eq.(101)is equivalent to the so-called Wilemski-Fixman approximation, often used for describing chemicalreactions involving polymers. It can be considered as taking into account only the lowest order ofthe corresponding formal perturbation series.
6.5.1 Example: Free diffusion in presence of boundaries
As an example let us consider the first passage time distribution for free diffusion (Brownianmotion), where G(x2, x1, t) = (4πDt)−1/2 exp [−(x2 − x1)2/4Dt], so that
G(x2, x1, u) =1√
4Duexp
(−|x2 − x1|
√u
D
). (106)
This delivers ψ(u, x0) = exp(−|x0|√u/D) from which it follows that
ψ(t, x0) =|x0|√
4πDt3/2exp
(− x2
0
4Dt
). (107)
This distribution of the first passage time to a boundary is called Smirnov, Levy-Smirnov orsometimes ”inverse Gaussian” distribution, and decays for long times as ψ(t, x0) ∝ t−3/2 so thatit does not have the first moment: the mean first passage time diverges. Note that the functionψ(t, x0) is a proper probability density function, so that
∫∞0ψ(t, x0)dt = 1. This means that the
particle in 1d is eventually captured at the boundary. This fact has to do with the recurrence ofthe one-dimensional Wiener process, which visits any point on the line with probability 1 at longertimes.
We note that our consideration here are based on the fact that the trajectories of the processare continuous. The result that at long times ψ(t, x0) ∝ t−3/2 is, however, valid also for a largeclass of jump processes, i.e. the ones with discontinuous trajectories. However in the case ψ(t, x0)has to be considered not as the distribution of first passage times, but as the distribution of timesat which the particle is for the first time found on the other side of x0 than its initial positionx = 0 was27.
Let us now consider splitting probabilities for a particle starting at X = 0 on the intervalwith absorbing boundaries at xL < 0 < xR. Using equations (104) and (105) and our Laplace-transformed Green’s function, Eq.(106) we get after expanding the exponential up to the first orderin their arguments:
PR = 1− PL =1
2+|xL| − |xR|2|xR − xL|
. (108)
The same result should be obtained in Homework 6 by an alternative approach.
27This statement is one of the important consequences of the Sparre-Andersen theorem from the theory ofrandom walks, see W. Feller, An Introduction to Probability Theory and its Applications, Wiley, N.Y. 1991
88
6.6 An underdamped situation
This is only the overdamped motion in 1d (described by the Smoluchowski equation) for whichmany beautiful exact solutions are readily available. For more general cases, say the ones describedby the Klein-Kramers equation, Eq.(49),
∂
∂tp(v, x, t) = − ∂
∂xvp(v, x, t)− ∂
∂v
(F (x)
m− γ
mv
)p(v, x, t)
+kBTγ
m2
∂2
∂v2p(v, x, t) (109)
(with F (x) being the position-dependent force) exact expressions are not known (in the sense thatyou have first to solve equation to get the result). Here a lot of nice numerical work is done, andseveral approximations based on special properties of the process are known. However, the limitingcase of extremely low friction is also amenable for theoretical investigation. The treatment heremight be based on the idea of energy diffusion discussed in the previous section.
For γ → 0 (small friction) the Klein-Kramers equation is reduced to the Liouville equation:We have now to do with the microcanonical ensemble with conserved energy E = mv2/2 + U(x),defining the states of the system. The external noise causes the transitions between the differentstates. In this case we can approximately reduce the two-dimensional Klein-Kramers equation toa one-dimensional one using the idea of energy diffusion.
The instantaneous probability of transitions between the two states characterized by differentenergy E depends on where the phase point is situated during the transition. Therefore the nec-essary step is then averaging over all such transitions to get the corresponding second momentgiving the diffusion coefficient in the corresponding Smoluchowski equation. This procedure, in-troduced by Kramers in 1940, leads to the closed equation for the probability density p(E, t)(which in what follows will be denoted by g(E, t), the letter p will be reserved for the particles’momentum. The equation for g(E, t) we obtain at the end is a one-dimensional equation for whichthe discussions of our previous section hold. 28. Using the flow-over-population approach or thePontryagin-Witt equation we the can get an explicit result for the mean life time in a well.
Here we give some details of the procedure which also elucidates some details on the energydiffusion which were not discussed in §5.5. We start from the Klein-Kramers equation for the PDFin the phase space p(x, v, t), Eq.(49), which we first rewrite as an equation for the probabilitydensity of the phase space variables x and p = mv which will be denoted by f(x, p, t)
∂
∂tf(p, x, t) = − ∂
∂x
p
mf(p, x, t)− ∂
∂p
(F (x)− γ p
m
)f(p, x, t)
+kBTγ∂2
∂p2f(p, x, t). (110)
Then we put it into the following form:
∂f
∂t= −L0p+ γkBT
∂
∂pfeq
∂
∂p
f
feq
(111)
where L0 is the Liouville operator
L0 =p
m
∂
∂x− dU
dx
∂
∂p
28 A simple derivation given here stems from the Robert Zwanzig’s book “Nonequilibrium Statistical Mechanics”,Oxford University Press, 2001, §4.3; this is not the only possibel approach.
89
and
feq = N exp
(−H(x, p)
kBT
)is the Boltzmann-Gibbs distribution with N being the normalization constant, and H the particle’sHamilton function
H(x, p) =p2
2m+ U(x).
The fact that Eq.(111) reproduces Eq.(110) can be checked by substitution.The distribution of the energy, g(E, t) can be found by the standard trick:
g(E, t) =
∫ ∫dxdpδ(H(x, p)− E)f(x, p, t).
Taking Eq.(111), multiplying its both parts by δ(H(x, p) − E) and performing the inegration weget
∂
∂tg(E, t) = γkBT
∫ ∫dxdpδ(H(x, p)− E)
∂
∂pfeq
∂
∂p
f
feq
since the term including the Liouvillian vanishes (due to energy conservation) since L0H = 0.The approximation transforming the Klein-Kramers equation into the equation for the energy
diffusion corresponds to the assumption that f(x, p, t) on the r.h.s. can be replaced by the functionof E or H only, which we will call φ(E, t). This is given by the requirement that∫ ∫
dxdpδ(H − E)φ(H, t) =
∫ ∫dxdpδ(H − E)f(x, p, t)
= g(E, t) = φ(E, t)
∫ ∫dxdpδ(H − E).
The last integral is the microcanonical partition function Ω(E) =∫ ∫
dxdpδ(H − E). Thereforeφ(E, t) and g(E, t) differ only in normalization, and
φ(E, t) =g(E, t)
Ω(E)
(note that g and φ have different dimensions). Now
f(x, p, t)
feq(x, p)=g(E, t)
geq(E)
and therefore∂
∂tg(E, t) = γkBT
∫dx
∫dpδ(H − E)
∂
∂pfeq(H)
∂
∂p
g(H, t)
geq(H).
Noting that g is a function of H = p2/2m + U(x) we first fix x and perform partial integration
over the momenta (denoting F (H) = feq(H) ∂∂p
g(H,t)geq(H)
):∫dpδ(H − E)
∂
∂pF (H) = δ(H(x, p)− E)F (H)|∞−∞ −
∫dpF (H)
∂
∂pδ(H − E)
90
(note that H = H(x, p) and the first term appearing under the partial integration vanishes becausep→ ±∞ corresponds to E →∞). Now, for fixed x, we replace ∂/∂p by (p/m)∂/∂H in the innerintegral, and note that ∂
∂Hδ(H − E) = − ∂
∂Eδ(H − E) to obtain∫
dpδ(H − E)∂
∂pF (H) = −
∫dp
p
mF (H)
∂
∂Hδ(H − E)
=∂
∂E
∫dp
p
mδ(H − E)F (H).
Now one performs a similar transformation ∂∂p→ p
m∂∂H
in the expression for F (H) = feq(H) ∂∂p
g(H,t)geq(H)
:
∂
∂tg(E, t) = γkBT
∂
∂E
∫dx
∫dp( pm
)2
δ(H − E)feq(H)∂
∂H
g(H, t)
geq(H)
and notes that due to the presence of the δ-function F (H) is essentially evaluated at H = E.Therefore our equation for g (still containing feq) gets to be
∂
∂tg(E, t) = γkBT
∂
∂E
∫dx
∫dp( pm
)2
δ(H − E)feq(E)∂
∂E
g(E, t)
geq(E).
Now we remember that
feq(E) =geq(E)∫ ∫
dxdpδ(H − E)
and put the last equation into the form of the Smoluchowski equation
∂
∂tg(E, t) =
∂
∂ED(E)geq(E)
∂
∂E
g(E, t)
geq(E)
(although we didn’t use this form before, this follows immediately from the standard form for theoverdamped case under the detailed balance assumption). The diffusion coefficient is given by
D(E) = γkBT
∫ ∫dxdp(p/m)2δ(H − E)∫ ∫
dxdpδ(H − E).
Integration over momenta is performed using the identity∫dzf(z)δ(z2 − a2) =
f(a) + f(−a)
2|a|,
(where z = p/√
2m) and gives
D(E) =γkBT
m
2∫dx[E − U(x)]1/2∫dx[E − U(x)]−1/2
,
with the integration in x going between the turning points of the motion xmin and xmax, i.e. in thedomain where E > U(x). The motion takes place with the positive momentum when moving fromxmin to xmax and with negative momentum on the way back. The expression in the enumeratiorof the second fraction is the action integral around the complete cycle,
I(E) =
∮p(x)dx,
91
with p(x) =√
2m[E − U(x)], the integral in the denominator is the derivative of this first integralw.r.t. to energy, ∂I
∂E, and is related to the corresponding frequency:(
∂I
∂E
)−1
=ω(E)
2π
(see action-angle coordinates in mechanics). Therefore
D(E) =γkBT
m
I(E)ω(E)
2π.
Noting that geq(E) = const · exp(−E/kBT ) we may rewrite this equation in the following form:
∂
∂tg(E, t) =
∂
∂E
γ
m
I(E)ω(E)
2π
[1 + kBT
∂
∂E
]g(E, t).
The main difference between the over- and the underdamped situation can be seen when con-sidering the example of the motion in a harmonic potential with a cutoff at E = Emax
29. For thiscase ω(I) = ω0 =
√k/m, where k is the elasticity constant of the spring, and I(E) = 2πE/ω0.
The problem of finding the mean first passage time from the initial energy E0 to Emax is given asa Homework.
Here we will see that the MFPT as a function of γ for small γ is proportional to γ−1, asituation opposite to the overdamped case where it is proportional to γ. Thus, the overall situationcorresponds to a non-monotonous γ-dependence: For initial energies below Emax the rate, asa function of γ, first grows and then starts to decay. The crossover between these two typesof behavior at moderate damping was discussed almost half a century later, in the works byMelnikov and Meshkov in 1986.
7 Simple bath model: Kac-Zwanzig bath
The Langevin equation describes the motion of a single particle (or maybe, if necessary, of a systemof few particles) in a bath. The particle-bath interaction leads to the continuous energy exchangebetween the particle and the bath, and is is separated into two parts. The first part, describedby the friction force, leads to the energy transfer from the particle to the bath, the second part,the random force, can work in both directions, but on the average leads to the energy transferfrom the bath to the particle by a mechanism akin to Fermi acceleration (see Homework 4.2).The description of these interactions is normally a very complex problem (we will see this whendiscussing a Boltzmann equation), but one can come up with a simple bath model, for which this isnot only possible, but also easy. This situation corresponds to a bath of harmonic oscillators. Themodel is often used when a reasonable explicit bath model is necessary in classical, or in quantumsetting, and is not as stupid as it may seem: If we consider e.g. the electron-phonon interaction atfinite temperature, the phonon bath, as described within the Debye model, is nothing else as the
29The simplification consisting the assumption that the maximum of the potential corresponds to a cusp at Emax,is not always realistic. Typically, the escape takes place through a saddle point of the potential. The frequency onthe trajectory passing through this saddle point vanishes, so that the diffusion coefficient 2πkBTγI/ω(I) formallydiverges. Thus, in the underdamped problem the behavior close to the separatrix is much less important than forthe overdamped case. Estimates show that our simple result is not too far from the reality, see N. van Kampen,Stochastic Processes in Physics and Chemistry, North Holland, Amsterdam, 1992
92
bath of harmonic oscillators. In what follows we consider this model (often called the Kac-Zwanzigbath30). Here we will consider the model in the classical case, in order to show how the Langevinequations (or their generalizations) appear from full description of the particle-bath interactions.
The Kac-Zwanzig model of a bath is non-generic, since the equations of motion of bath particlesin the classical case are integrable. Therefore the classical harmonic oscillator bath does not ther-malize by itself, and has to be initiated in a state of thermal equilibrium, after which initializationit is considered to represent a bona fide heat bath for a small canonical system (e.g. for a singleparticle in an external potential). On the other hand, this is exactly the integrability of the bath’sequations of motion which allows for rigorous derivation of the (generalized) Langevin equationswith no need for any additional assumptions or approximations.
Let us consider a probe particle whose motion is described by the Hamiltonian
HT =P 2
2M+ U(X) (112)
and the heat bath of N harmonic oscillators which we assume to have masses mj and springconstants Sj with the Hamiltonian
HB =N∑j=1
(p2j
2mj
+1
2Sjq
2j
). (113)
Here P and X are the momentum and the position of the distinguished particle, U(X) is theexternal potential acting on this particle, and pj and qj are the momenta and the coordinates ofthe bath particles.
Now we can switch the interaction of the bath with the test particle, so that the total Hamiltonfunction of the system is
H = HT +HB +Hint.
In different works this is done in several ways, some of them leading to the variants of the overalltheory which are quite non-transparent or awkward for numerical implementations and the inter-pretation of results. We here explicitly assume that the test particle is coupled to each of the bathoscillators’ particles by a spring with a spring constant sj:
Hint =1
2
∑j
sj(qj −X)2
so that the total Hamilton function reads
H =P 2
2M+ U(X) +
N∑j=1
(p2j
2mj
+1
2Sjq
2j +
1
2sj(qj −X)2
)
Now we denote Sj + sj = kj, ωj =√kj/mj, and κj = sj/kj = sj/(Sj + sj). According to this
definition 0 ≤ κj ≤ 1. The case with κj → 0 will be called the weak coupling case, and the case
30The model has more than two fathers. The important works on the issue are: G. Ford, M. Kac, and P. Mazur,Statistical mechanics of assemblies of coupled oscillators, J. Math. Phys. 6 504 (1965), G. Ford and M. Kac, On thequantum Langevin equation, J. Stat. Phys. 46 803 (1987), R. Zwanzig, Nonlinear generalized Langevin equations,J. Stat. Phys. 9 215 (1973), and R. Kupferman, Fractional Kinetics in Kac-Zwanzig Heat Bath Models, J. Stat.Phys. 114 291 (2004)
93
κj → 1 corresponds to the Galilean invariant model discussed by Kupferman (see the footnote30). This bath model will be called a mobile bath in what follows. In this notation the totalHamiltonian reads
H =P 2
2M+ U(X) +
1
2
[N∑j=1
kj(κj − κ2j)
]X2 +
N∑j=1
p2j
2mj
+1
2
N∑j=1
kj(qj − κjX)2.
The effective potential acting on the particle is
U(X) = U(x) +
[N∑j=1
kj(κj − κ2j)
]X2,
and differs from U(X) for all cases except for the mobile bath case with κj = 1. If the correctionto the external potential is undesirable (e.g. we want the bath to simulate the fluid system) thenthe mobile bath model is preferable. In this model, all bath particles interact with each other onlyvia the tagged one, and would be free otherwise (e.g. build an ideal gas). The cases with otherκj describe a viscoelastic medium (somehow attached to the laboratory table). In our furtherdiscussion we take κj = 1, the treatment of other cases is very similar (the only difference is that
some effective potential U(X) appears in the final equations instead of U(X)).The Hamiltonian equations of motion for our system read:
P = −U ′(X)−
(N∑j=1
kj
)X +
∑j
kjqj
and
pj = −∂H∂qj
= −kjqj + kjX. (114)
Taking into account that P = MX and that pj = mj qj and dividing the both parts of the secondequation by mj we obtain the final equations of motion
MX = −U ′(X)−
(N∑j=1
kj
)X +
∑j
kjqj
andqj = −ω2
j qj + ω2jX.
We start from this second equation. The general solution to this equation for given X(t) is:
qj(t) = qj(0) cosωjt+ qj(0)sinωjt
ωj+ ωj
∫ t
0
dt′X(t′) sinωj(t− t′). (115)
Performing integration by parts∫ t
0
dt′X(t′) sinωj(t− t′) = X(t′)cosωj(t− t′)
ωj
∣∣∣∣t0
−∫ t
0
dt′X(t′)cosωj(t− t′)
ωj,
we restore the explicit dependence on the initial position X(0):
qj(t) = [qj(0)−X(0)] cosωjt+ qj(0)sinωjt
ωj+X(t)−
∫ t
0
dt′X(t′) cosωj(t− t′).
94
We now introduce this solution into Eq.(114):
MX = −U ′(X(t))−
(N∑j=1
kj
)X(t) +
∑j
kjX(t)
−∑j
kj
∫ t
0
dt′X(t′) cosωj(t− t′)
+∑j
kj
[qj(0)−X(0)] cosωjt+ qj(0)
sinωjt
ωj
.
The second and third terms in the first line cancel. This equation has the structure of a Newtonequation with the friction term keeping the memory on the previous particle’s velocities:
MX = −U ′(X)−∫ t
0
dt′X(t′)K(t− t′) + F (t). (116)
We can interpret it as a generalized (non-Markovian) Langevin equation, if we say that the forceF (t) is fluctuating so strongly that it can be assumed random. Here K(t) is the friction memory-kernel
K(t) =∑j
mjω2j cosωjt
(note that, K(t) has a dimension of the elastic constant, namely [K(t)] = MT−2) and F (t) is the”noise” force,
F (t) =∑j
mjω2j
[(qj(0)−X(0)) cosωjt+ qj(0)
sinωjt
ωj
].
Now we fix X(0) and equilibrate the bath at temperature T by connecting it to an exter-nal thermostat (a harmonic bath, being integrable, does not equilibrate by itself). In such anequilibrated bath we get
〈qj(0)−X(0)〉 = 0, 〈pj(0)〉 = 0,
and, according to the equipartition theorem,
mω2j
⟨[qj(0)−X(0)]2
⟩= kBT, 〈p2
j(0)〉 = kBTm,
while all second cross-moments, 〈qipj〉 (with qj = qj(0)−X(0)) for all i, j, and 〈qiqj〉, 〈pipj〉 withj 6= j vanish. Now we evaluate the autocorrelation function of F (t):
〈F (t)F (t′)〉 =∑j
m2jω
4j
[⟨q2j (0)
⟩cosωjt cosωjt
′ +⟨p2j(0)
⟩ sinωjt sinωjt′
m2ω2j
]=
∑j
m2jω
4j
[kBT
mω2j
cosωjt cosωjt′ +
kBT
mω2j
sinωjt sinωjt′]
=
=∑j
mjω2jkBT cosωj (t− t′) = kBTK(t− t′). (117)
This is a corresponding fluctuation-dissipation theorem (of the second kind) for a non-MarkovianLangevin equation.
95
Let us first investigate the behavior of the system in the absence of external potential. Thegeneralized Langevin equation (GLE) in the absence of the external potential
MX = −∫ t
0
dt′X(t′)K(t− t′) + F (t)
is solved in the Laplace domain:
M [s2X(s)− sX0 − V0] = −K(s)[sX(s)−X0] + F (s)
(with X0 being the initial coordinate and V0 the initial velocity of the test particle) has a solution
X(s) =1
sX0 +
MV0
s[K(s) +Ms]+
F (s)
s[K(s) +Ms].
Defining the Green’s function of the GLE H(t) with the Laplace representation
H(s) =1
s[K(s) +Ms](118)
we may write the solution in the time domain as
X(t) = X0 +MV0H(t) +
∫ t
0
F (t′)H(t− t′)dt′.
Note that the dimension of H(t) is [H] = [T/M]. We moreover note that according to the initialvalue theorem
H(t = 0) = lims→∞
sH(s) = lims→∞
[K(s) +Ms]−1 = 0
for all cases when K(s) does not tend to −Ms for s → ∞. Since for t → 0 K(t) = 〈F 2(t)〉 > 0and K(t) in our model is essentially continuous (except in the limit of the white noise) its Laplacetransform for large s is non-negative, and the situation cannot be realized.
Now we start with X0 = 0 and calculate the mean squared displacement (MSD) of the particle:
〈[X −X0]2〉 =
⟨[MV0H(t) +
∫ t
0
F (t′)H(t− t′)dt′]2⟩
= 〈V 20 〉M2H2(t) (119)
+ 2
⟨V0MH(t)
∫ t
0
F (t′)H(t− t′)dt′⟩
+
⟨[∫ t
0
F (t′)H(t− t′)dt′]2⟩.
If we start from the thermalized conditions 〈MV 20 〉 = kBT then the first term reads MkBTH
2(t).If we start with zero velocity, it vanishes. The second term vanishes identically when we start atzero velocity, and the average vanishes in the thermalized case since V0 and F (t) are independent,and 〈V0〉 = 0. The third term now can be rewritten as follows:⟨[∫ t
0
F (t′)H(t− t′)dt′]2⟩
=
⟨∫ t
0
∫ t
0
F (t′)F (t′′)H(t− t′)H(t− t′′)dt′dt′′⟩
=
∫ t
0
∫ t
0
〈F (t′)F (t′′)〉H(t− t′)H(t− t′)dt′dt′′
96
Now we note that the correlation function of the noise is an even function of t′ − t′′ only and isgiven by kbTK(|t′ − t′′|). We now rewrite the last expression using this symmetry:⟨[∫ t
0
F (t′)H(t− t′)dt′]2⟩
= kBT
∫ t
0
∫ t
0
K(|t′ − t′′|)H(t− t′)H(t− t′′)dt′dt′′
= kBT
∫ t
0
∫ t
0
K(|ξ − η|)H(ξ)H(η)dξdη
= 2kBT
∫ t
0
dξH(ξ)
∫ ξ
0
K(ξ − η)H(η)dη.
The internal integral G(ξ) =∫ ξ
0K(ξ − η)H(η)dη is a convolution, and can be calculated in the
Laplace domain:
G(s) = K(s)H(s) =K(s)
s[K(s) +Ms]=
1
s− M
[K(s) +Ms]≡ 1
s−MsH(s).
In time domain this meansG(ξ) = 1−MH(ξ),
where the dot denotes the derivative. Now the whole integral reads⟨[∫ t
0
F (t′)H(t− t′)dt′]2⟩
= 2kBT
∫ t
0
dξH(ξ)[1−MH(ξ)
]= 2kBT
∫ t
0
H(ξ)dξ − kBTMH2(t).
Combining this with the first term in Eq.(119) (for thermalized initial velocity) we see that
〈R2(t)〉 ≡ 〈(X −X0)2〉 = 2kBT
∫ t
0
H(t′)dt′ (120)
Now we see the following: If H(t) tends to a constant for t large, the behavior of the MSDwill be diffusive: 〈R2(t)〉 ∝ t, and γ = limt→∞H(t) is essentially the usual friction coefficient.If H(t) grows with time, 〈R2(t)〉 will grow faster than linearly with time and the behavior willbe superdiffusive, and when H(t) decays with time, it will be subdiffusive. What case is realized,depends on the frequency spectrum and couplings in the system.
For normal diffusion (so-called Ohmic bath). The fact that H(t) tends to a constant value H∞for t → ∞ means in the Laplace domain that for s → 0 H(s) → H∞/s. According to Eq.(118),this would correspond to K(s) → K0 = const, and K(t) tend to a delta-function in the timedomain.
For subdiffusion, if R2(t) = D0tγ with γ < 1 we will get H(t) ∝ tγ−1, H(s) ' s−γ, and
K(s) ∝ sγ−1, and K(t) ∝ t−γ. The situation with superdiffusion in the Kac-Zwanzig model ismore subtle, due to the presence of the second term in the denominator of Eq.(118), and will benot discussed in the Lecture.
Depending on further needs, we can make specific assumptions about the distribution of fre-quencies and masses: thus, we can take all frequencies the same, and masses different (c.f. Ein-stein’s model of a solid), all masses the same and frequencies different (Zwanzig’s model), etc. Nowwe assume that the mass of the bath particle is a function of the oscillator’s frequency.
97
According to Eq.(117), we have
K(t) =∑j
mjω2j cos(ωjt) '
∫ ∞0
m(ω)ω2g(ω) cos(ωt)dω,
where g(ω) is the density of frequencies in the system. Now, just for the sake of simplicity, let usopt for the Zwanzig’s model. In this case m are constant, and to obtain K(t) = δ(t) we have totake g(ω) ∝ ω−2. To simulate subdiffusion, we will need to take g(ω) ∝ ωγ−1.
8 The Boltzmann equation
Our approaches so far were based on phenomenological description of systems of high physicalinterest, but hardly gave the connections between the phenomenological coefficients and the mi-croscopic properties of the system, like intermolecular interaction potentials. This task is not fullysolved (except in simulations), but considerable progress was achieved in many important cases.In this chapter we consider the simplest situation pertinent to rarefied gases.
The kinetic equation for the distribution function of velocities and coordinates of molecules ina gas at low density (particle density in the phase space) f(x,v, t) was proposed by Boltzmannin 1872. This equation was a great step forward in statistical mechanics since it explained how thecorresponding distribution function relaxes to its equilibrium (Boltzmann-Maxwell distribution)form, it gave the basis of understanding how the equations of hydrodynamics (e.g. the Navier-Stokes equation) emerge from the kinetic theory, and showed the way of how the correspondingkinetic coefficients (e.g. viscosity or heat conductivity) can be calculated when the intermolecularpotentials are known. The Boltzmann equation corresponds to a level of description intermediatebetween the full mechanical description of a multiparticle system via its full Liouville dynamics,and the stochastic description concentrating on the behavior of a probe particle and replacingthe whole complex interaction by noise. It is also a very interesting example of a self-consistentapproach, which was proved useful in many domains of nonequilibrium statistical physics, wheredifferent kinetic equations were derived along the similar lines of argumentation.
The Boltzmann equation in its “native” form is only applicable to gases of molecules interactingvia short-range forces at sufficiently low densities. However, many conclusions of the followingdiscussion still hold for higher densities and more complex types of interactions. In many casesphenomenological models based on the Boltzmann-like approaches, giving rise to different “kineticmodels” which give a correct quantitative description. For these ones read Chapter 5 of Zwanzig’sbook.
8.1 The BBGKY-Hierarchy
In order to get a flavor of the Boltzmann’s approach let us first consider the full mechanicaldescription of a system of many interacting particles, leading to an infinite (in thermodynam-ical limit) hierarchy of equations for the distribution functions, the BBGKY-Hierarchy (Born-Bogolyubov-Green-Kirkwood-Yvon). The N classical (i.e. distinguishable) particles, whichwe for simplicity take to be identical, move according to the Hamiltonian dynamics, and the evo-lution of their distribution function
fN(q1, ..., q3N , p1, ..., p3N , t) ≡ f(x1...xN)
98
(in what follows, all canonical variables for a particle i will be considered as comprising a vectorxi = (pi,qi)) is given by the Liouville equation
∂f
∂t= H, f =
3N∑n=1
(∂H
∂qn
∂f
∂pn− ∂H
∂pn
∂f
∂qn
)≡ Lf,
where H is the Hamilton function of the system, and L is the Liouville-operator (Liouvillian).The Hamilton function of the system consists of the single-particle part and of the part describinginteractions:
H = H1 +H2 =N∑i=1
(p2i
2m+ Uext(qi)
)+∑i<j
Uint(qi,qj),
where Uext(qi) is the external potential, and Uint(qi,qj) is the interaction potential between the twoparticles (each interacting particle pair is counted once). Therefore, the corresponding Liouvillianalso consists of the two parts, which can be expresses as the sum over the one- and two-particleselementary Liouville operators:
∂f
∂t= H, f =
N∑i=1
L0i f +
∑i<j
L′ijf
where L0 is a single-particle Liouvillian (having the same form for all particles)
L0i =
∑n
(∂Uext(ri)
∂qn
∂f
∂pn− pnm
∂f
∂qn
),
(here ri is the position of particle i and qn and pn are the canonical coordinates and momenta ofparticle i) and L′ij describes the interactions,
L′ij =∑n,k
(∂Uint(rij)
∂qn
∂f
∂pn+∂Uint(rij)
∂qk
∂f
∂pk
),
where rij is the distance between the particles i and j, and qk and pk are the canonical coordinatesand momenta of particle j.
If the total distribution function f is known, the distribution functions for smaller groups ofparticles (single particles, pairs, triples, etc.) can be obtained by building marginal distributions
fs(x1, ...,xs) =N !
(N − s)!
∫dxs+1...dxNfN(x1, ...,xN).
The time evolution of fs(x1, ...,xs) is given by
∂
∂tfs(x1, ...,xs) =
N !
(N − s)!
∫dxs+1...dxN
∂
∂tfN(x1, ...,xN)
=N !
(N − s)!
∫dxs+1...dxN
(N∑i=1
L0i fN +
∑i<j
L′ijfN
)
=N !
(N − s)!
N∑i=1
∫dxs+1...dxN L
0i fN
+N !
(N − s)!∑i<j
∫dxs+1...dxN L
′ijfN .
99
Now we have to distinguish between the variables with indices i and j belonging to the group(1, ..., s) and to the group (s+ 1, ..., N), since the first ones are the free variables, and the secondones are the dumb variables of integration. Let us first consider the first term in the final expression,where we separate the corresponding variables:
N !
(N − s)!
N∑i=1
∫dxs+1...dxN L
0i fN =
N !
(N − s)!
s∑i=1
L0i
∫dxs+1...dxNfN
+N !
(N − s)!
N∑j=s+1
∫dxs+1...dxN L
0jfN
=s∑i=1
L0i fs(x1, ...,xs) +
N !
(N − s)!(N − s)
∫dxs+1...dxN L
0s+1fN ,
where we have noted that the second sum is essentially a sum of N − s equal integrals obtainedfrom each other by renumbering the variables. Now we show that the second integral vanishes. Tosee so we first note that f has to vanish both for qi → ±∞ and for pi → ±∞, since it is integrable.We now consider first the integration in the s+ 1-st variable:∫
dxs+1L0s+1f =
∫dqs+1,αdps+1,α
(∂Uext
∂qs+1,α
∂f
∂ps+1,α
− ps+1,α
m
∂f
∂qs+1,α
)where α = 1, 2, 3, and see (e.g. for the first term) that∫
dqdp∂Uext
∂q
∂f
∂p=
∫ ∞−∞
dq∂Uext(q)
∂q
∫ ∞−∞
∂f
∂pdp = 0.
The second term vanishes on the same reasons (for a system of a finite volume). The termcontaining single-particle Liouvillians of particles 1, ..., s is simply
∑si=1 L
0i fs(x1, ...,xs). Now let
us consider the interaction term,
N !
(N − s)!
N∑i<j
∫dxs+1...dxN L
′ijfN .
Here we have to consider three cases: a) both indices belong to the group (s+1, ..., N), b) i belongsto the group (1, ..., s) and j belongs to the group (s+1, ..., N) of indices, and c) both i and j belongto the group (1, ..., s).
a) In this case one integrates over the momenta of both particles i and j and the correspondingintegrals vanish, like in the previous case. Therefore no contribution to the sum arises.
b) Now we use the fact that f is symmetric in all its arguments (except for t!) and thereforethe corresponding integral, like in the previous case, is the sum of N − s equal integrals, obtained
100
from each other by renaming the corresponding integration variable, i.e.
N !
(N − s)!
s∑i=1
n∑j=s+1
∫dxs+1...dxN L
′ijfN
=N !
(N − s)!(N − s)
s∑i=1
∫dxs+1...dxN L
′i,s+1fN
=s∑i=1
∫dxs+1L
′i,s+1
N !
(N − s− 1)!
∫dxs+2...dxNfN
=s∑i=1
∫dxs+1L
′i,s+1fs+1(x1, ...,xs+1).
c) Here no transformations can be made, and the corresponding contribution is
s∑j=2
j∑i=1
L′ijfs.
Thus, the total evolution of an s-particle distribution function fs is given by the following equations
∂
∂tfs =
s∑i=1
L0i fs +
s∑j=2
j∑i=1
L′ijfs
+s∑i=1
∫dxs+1L
′i,s+1fs+1(x1, ...,xs+1). (121)
The first term describes the free motion of s particles, the second one their interaction within thegroup and the third one the interaction with the particles not included into the group. Note thatthis term depends on the next, s+ 1-particles, distribution functions. Therefore Eqs.(121) are notclosed but build a hierarchy of N equations (in thermodynamical limit we have N → ∞). Theirusefulness is quite limited until effective truncation schemes are proposed.
8.2 Single-particle distribution function
Let us now concentrate on the first equation of the hierarchy, the one for f1(q,p, t). This equationdoes not contain the second term (the group consists only of one particle) and reads formally
∂
∂tf1(q,p, t) = L0
1f1(q,p, t) +
∫ ∫dq2dp2L
′12f2(q1,p1,q2,p2, t).
Now we move the first term in the r.h.s of this equation to its l.h.s. and use the explicit expressionfor the one- and two-particle Liouvillians to put this equation into the form
∂
∂tf1(q,p, t) +
∑α
F extα
∂f1
∂pα+∑α
pαm
∂f1
∂qα
=
∫ ∫dq2dp2
∑α
(F 12α
∂f2
∂p2,α
− F 12α
∂f2
∂p1,α
), (122)
101
where F extα = −∂Uext/∂qα and F 12
α = −F 21α = −∂Uint(rij)/∂qα, and f2 is the function of q1,p1,q2,p2
and t as above. The right hand side of this equation is called the collision integral (Stossintegral)Istoss; it is a functional of f2 and has a “gain – loss” structure typical for master equations, withthe corresponding rates determined by interaction forces and by f2. In the case of non-interactingparticles (F12 = 0) it is absent, and the left hand side gives us the free evolution of the distribu-tion function described by a one-particle Liouvillian. It is convenient to pass from momenta tovelocities of the particles, and the rewrite this equation as31
∂
∂tf1(q,v, t) +
Fext
m∇vf1(q,v, t) + v∇qf1(q,v, t) = Istoss. (123)
From here on we suppress the index 1 and denote the single-particle distribution function by f .Consider the situation in thermodynamical equilibrium, when the single-particle distribution
function is given by the time-independent Maxwell-Boltzmann distribution
f(q,v, t) = feq(q,v) ∝ exp
(−U
ext(q)
kT
)· exp
(−mv2
2kT
).
we get that the l.h.s. of our equation vanishes identically: for each coordinate (here x)
0 +1
m
∂U
∂x
mvxkT
f − vx∂U
∂x
1
kTf = 0.
This means that in equilibrium the collision integral must vanish identically.The Boltzmann equation follows under assumption that the collision integral can be represented
as a non-linear functional of f1 which leads to a closed integro-differential equation for this function.
8.2.1 Kinetic models
The collision integral has a meaning of the rate of the change of the distribution function due tointeractions. This fact allows for formulation of a class of phenomenological kinetic models whichdescribe equilibration. The simplest kinetic model (relaxation time model) arises when assuming
Istoss = −ν(f − feq)
where the inverse τ = ν−1 of ν is the characteristic relaxation time of the distribution. Such amodel is quite adequate for many phenomena (but not for gases at low densities where it leadsto wrong coefficients in hydrodynamic equations). A famous Drude model of electric conductivitycan be cooked down to such a relaxation time model.
Let us assume that our system of charge carriers can be modeled as a classical gas of chargedparticles (not a good model for metals, where it has to be slightly altered to take into account thefermionic nature of charge carriers, but quite OK for electrolytic solutions), stays homogeneous ateach time (so that the terms with spatial derivatives vanish) and consider its response to a weakexternal electric field E. The kinetic equation in this case reads:
∂f
∂t+
e
mE∇vf = −ν(f(v, t)− feq(v)).
31 Note that the l.h.s. of this equation resembles the one of the Klein-Kramers equation, in which the collisionintegral is represented by some differential operator. Indeed, many equations we discussed before can be consideredas special (linearized) approximations to our general scheme, i.e. they belong to the class of kinetic models asdiscussed below.
102
Let us look for the stationary solution of this equation describing the constant current density inour system. The time derivative in the l.h.s. vanishes. Let us denote f(v) − feq(v) = δf andassume it to some extent small. We thus get
e
mE∇vfeq +
e
mE∇vδf = −νδf,
and solve it iteratively (looking on it “from the right to the left”). In the lowest order we get
δf = −1
ν
e
mE∇vfeq,
in the next order
δf = −1
ν
e
mE∇vfeq +
1
ν2
e2
m2E2∇2
vfeq,
etc. We see that the liner response to the field corresponds to the first order in the expansion.Therefore, in the first order in the field we get
f(v) = feq(v)− 1
ν
e
mE∇vfeq(v).
Since feq(v) is given by the Maxwell distribution
feq(v) = C · exp
(−mv2
2kT
)we have
∇vfeq(v) = −mv
kTfeq(v)
and
f(v) =
(1 +
1
ν
e
kTEv
)feq(v).
The electric current is connected with the mean velocity of the charge carriers
j = ne〈v〉 = ne
∫vf(v)dv = ne
∫vfeq(v)dv +
n
ν
e2
kT
∫(Ev)vfeq(v)dv.
The first integral vanishes due to the rotational symmetry of feq(v), and the second one can beeasily calculated in the Cartesian coordinates:
ji =n
ν
e2
kT
∑j
Ej
∫vivjfeq(v)dv.
Again, due to symmetry of feq(v),∫vivjfeq(v)dv = 〈v2
i 〉δij = (kT/m)δij, and we get finally
jj = n e2
νmEjδij, i.e., the Ohm’s law
j = σE
with σ = ne2/νm.
103
8.3 The Boltzmann equation
The form of our Eq.(123) corresponds to
d
dtf = Istoss
with d/dt denoting the total derivative, ∂/∂t − L0 with the single-particle Liouvillian expressedthrough the velocity (not momentum). The main assumption of the Boltzmann’s treatment is thatIstoss can be expressed as a functional of the same function f(x,p, t) (the index 1 is now omitted,and the vector of coordinates is denoted by x):
d
dtf = If.
The collision integral consists of two terms with opposite signs, the gain and the loss term: If =Gf − Lf. The further simplifications we make are:
• Only binary collisions (this is why the theory only works for dilute gas)
• Molecular chaos (the velocities of colliding molecules are independent)
• The forces acting during the collision are short ranged, therefore the time during which themolecules interact in negligible
• The collisions are elastic (reversible)
The third and the fourth assumptions may be relaxed in more complicated theories, but the firsttwo are crucial for the whole approach.
Figure 7: The geometry of the collision in a center-of-mass system of two particles. Shown are allparameters used in what follows.
For the sake of simplicity we consider only central collisions (see Fig. 7). At each collision, dueto energy and momentum conservation,
v21 + v2
2 = v′21 + v′
22
v1 + v2 = v′1 + v′2,
104
which give us the total of four equations for six components of velocities. Here v1 and v2 arethe velocities of the two particles before the collision, and v′1 and v′2 are their velocities after thecollision. Note that the conservation laws imply that in the system of the center of mass of thetwo particles they simply interchange their velocities, which will allow us for interchanging theintegrations variables, with d3v1d
3v2 = d3v1d3v′1, a trick which will be repeatedly used in what
follows.Two parameters are free. These will be the collision parameter b and the angle ϕ. The
dependence on ϕ is essentially absent due to circular symmetry. The number of particles withvelocities v1 and v2 colliding during some short time interval ∆t within the volume ∆3x is
∆N(v1,v2, b) = 2πb∆b|v1 − v2|f(x,v1, t)f(x,v2, t)∆3x∆3v1∆3v2∆t. (124)
Here we assumed the the distribution functions are normalized in such a way that∫ ∫
f(x,v, t)d3xd3v =N i.e. that the integral over the velocities defines the spacial particle density, n(x, t) =
∫f(x,v, t)d3v.
The x-arguments of both f -functions are taken equal since the interaction radius is assumed smallso that the gas can be assumed to be homogeneous on the typical scales of the molecular interaction.
Therefore the loss term (the number of collisions of particles with velocity within some ∆3v1
domain per unit time per unit volume) is
Lf =
∫∆N(v1,v2, b)
∆3x∆3v1∆tdv2 =
∫dv2
∫db 2πb|v1 − v2|f(x,v1, t)f(x,v2, t).
The gain term corresponds to the number of collisions during the time ∆t in volume element ∆3xwhich lead to the final velocity value around v1. Due to reversibility of the collision, the numberof such particles is given by Eq.(124) with interchanged v1,2 and v′1,2. Therefore
∆N(v′1,v′2, b) = 2πb∆b|v′1 − v′2|f(x,v′1, t)f(x,v′2, t)∆
3x∆3v′1∆3v′2∆t,
where v′1(v1,v2) and v′2(v1,v2) are the functions of v1 and v2. Therefore, the total equation reads
df
dt=
∫dv2
∫db 2πb|v1 − v2| [f(x,v′1, t)f(x,v′2, t)− f(x,v1, t)f(x,v2, t)] . (125)
Here we used in addition that due to the phase volume conservation dv1dv2 = dv′1dv′2. The
equation can be rewritten using the differential scattering cross-section for the collision
dσ
dΩ=b(χ)
sinχ
∣∣∣∣db(χ)
dχ
∣∣∣∣ :
it now reads
df
dt=
∫dv2
∫dΩ
dσ
dΩ|v1 − v2| [f(x,v′1, t)f(x,v′2, t)− f(x,v1, t)f(x,v2, t)] ,
one of the standard forms of the Boltzmann equation.
8.3.1 Maxwell’s transport equations
Let us now assume that the form of the collision integral is known, and that we are (in principle)able to solve the Boltzmann equation. Let us now consider some physical property of the system
105
(e.g. density, or energy) depending on the coordinates and velocities of single particles, which canbe obtained as a mean value of a single-particle property ψ(r,v, t):
〈ψ(r, t)〉 =
∫ψ(r,v, t)f(r,v, t)dv∫
f(r,v, t)dv.
Since the integral in denominator is the spacial density, we can rewrite this expression as
n(r, t)〈ψ(r, t)〉 =
∫ψ(r,v, t)f(r,v, t)dv.
Now we multiply both parts of the Boltzmann equation by ψ(r,v, t) and integrate over the veloc-ities. Let us first perform this procedure for the left hand side of the equation:∫
ψ(r,v, t)
[∂f
∂t+ v∇rf +
F ext(r)
m∇vf
]dv
=
∫ [∂
∂t(ψf) + v∇r(ψf) +
F ext(r)
m∇v(ψf)
]dv
−∫f
[∂
∂tψ + v∇rψ +
F ext(r)
m∇vψ
]dv
Now, the integral of the last term in the second line vanishes:
F ext(r)
m
∫∇v(ψf)dv = 0
since it can be reduced to a surface integral and has to vanish since f(r,v, t) vanishes fast for|v| → ∞.
Therefore the whole expression on the left hand side reads:
∂
∂tn〈ψ〉+∇rn〈ψv〉 −
∫f
[∂
∂tψ + v∇rψ +
F ext(r)
m∇vψ
]dv. (126)
Let us now turn to the collision integral on the right hand side of the equation, and denote it asI(ψ) =
∫Ifψ(r,v, t)dv. The equation
∂
∂tn〈ψ〉+∇rn〈ψv〉 −
∫f
[∂
∂tψ + v∇rψ +
F ext(r)
m∇vψ
]dv = I(ψ)
are called the Maxwell’s transport equation (for ψ).Using our expression for the collision integral as used in Eq.(125) (where I renamed the velocity
variables as v1 = v, v2 = u) v′1 = v′ and v′2 = u′ we write
I(ψ) = 2π
∫d3vd3udbbψ(r,v, t)|v − u| [f(r,v′, t)f(r,u′, t)− f(r,v, t)f(r,u, t)]
(note that v′ and u′ are the functions of v, u and of the collision parameter b). Now we note thatduring the elastic collisions d3vd3u = d3v′d3u′ and obtain three other equivalent forms of I(ψ) bychanging or renaming the variables of integration:
I(ψ) = 2π
∫d3ud3vdbbψ(r,u, t)|u− v| [f(r,u′, t)f(r,v′, t)− f(r,u, t)f(r,v, t)]
106
I(ψ) = 2π
∫d3v′d3u′dbbψ(r,v′, t)|v′ − u′| [f(r,v, t)f(r,u, t)− f(r,v′, t)f(r,u′, t)]
I(ψ) = 2π
∫d3u′d3v′dbbψ(r,u′, t)|u′ − v′| [f(r,u, t)f(r,v, t)− f(r,u′, t)f(r,v′, t)] .
Now we call the integration variables in each integral v and v′ and denote the other pair of variablescorrespondingly, and afterwords sum up all the corresponding expressions (and divide the resultby four). We get:
I(ψ) =π
2
∫d3v1d
3v′db b|v − u| [ψ(r,v, t) + ψ(r,u, t)− ψ(r,v′, t)
−ψ(r,u′, t)] [f(r,v′, t)f(r,u′, t)− f(r,v, t)f(r,u, t)] . (127)
This is the most convenient form of the corresponding integral.
8.3.2 Collision invariants
The first square bracket in the last equation can be considered as the difference in the total valueof ψ for the two particles before and after collision. If this sum before and after the collision is thesame, i.e.
ψ(r,v, t) + ψ(r,u, t) = ψ(r,v′, t) + ψ(r,u′, t)
the integral I(ψ) vanishes, and the transport equation reads
∂
∂tn〈ψ〉+∇rn〈ψv〉 =
∫f
[∂
∂tψ + v∇rψ +
F ext(r)
m∇vψ
]dv. (128)
This is for example true for the mass of the particles (which does not depend on coordinates,velocities and on the time). In this case the right hand side of Eq.(128) vanishes:
∂
∂tnm+∇rnm〈v〉 = 0,
a continuity equation which can be rewritten in its usual form
∂
∂tρ+∇rρ〈v〉 = 0,
for the density.Another example is a component of the momentum mvi. We get
∂
∂tn〈mvi〉+∇rn〈mviv〉 =
∫f
[∂
∂tmvi + v∇rmvi +
F ext(r)
m∇vmvi
]dv.
Note that the component of the momentum mvj does not depend on time and on coordinatesexplicitly, so that the first two terms in square brackets vanish, and therefore the equation takes aform
∂
∂tρ〈vi〉+
∂
∂xjρ〈vivj〉 = Fin
The second term contains the momentum flux density (i.e. stress tensor). The comparison ofthe stress tensor with the corresponding mean can be used for deriving kinetic coefficients (like
107
viscosity). We will fulfill this program within the simpler framework of the relaxation time kineticmodel.
The equation for the kinetic energy reads:
∂
∂tn
⟨mv2
2
⟩+∇rn
⟨mv2
2v
⟩=
∫f
[∂
∂t
mv2
2+ v∇r
mv2
2+F ext(r)
m∇v
mv2
2
]dv,
and translates to the form∂
∂tρ〈v2〉
2+∇rρ
〈v2〉v2
= Fnv.
The first term in the left hand side is dominated by the kinetic energy of the molecular motion
and is therefore connected with the temperature, ρ 〈v2〉2≈ 3
2nkBT , the second term is the heat flow
density, and the term in the right hand side is the total power.Comparing these equation with phenomenological hydrodynamic description of Sec. 2.5.3 we
see that these are special cases of the corresponding equations for density, momentum and en-ergy transport. More complicated situations (e.g. reactions) can be considered when turning tocollisions which may change the particle nature, and inelastic collisions.
We note that in order to calculate the averages we need to know the corresponding distributionfunction, and calculating them within the genuine Boltzmann schema is not easy. However, tounderstand what happens, it is sufficient to discuss a kinetic model like the one we consideredabove. This will be done in Sec. 8.4. Before going into this detail let us however discuss anotherimportant consequence of the Boltzmann equation.
8.3.3 The H -Theorem
According to statistical physics, the (statistical, surrogate-) entropy of a state with distributionfunction f is
S = S0 − kB∫f ln fd3rd3v.
This entropy has to reach its maximum in an equilibrium state. The Boltzmann’s H-function wassimply H =
∫f ln fd3rd3v, therefore the name. This function reaches its minimum in equilibrium.
Let us consider the time evolution of this surrogate entropy:
dS
dt= −kB
∫∂
∂t(f ln f) d3rd3v = −kB
∫∂f
∂tln fd3rd3v − kB
∫∂f
∂td3rd3v. (129)
The last term vanishes since∫fd3rd3v = N = const. According to the equation Eq.(123),
∂f
∂t= −v∇rf(r,v, t)− Fext
m∇vf(r,v, t) + Istossf.
We use this expression in our previous one (Eq. (129)), and evaluate the first integral
kB
∫ln f
[v∇rf(r,v, t) +
Fext
m∇vf(r,v, t)
]d3rd3v.
It consists of two similar terms, with the gradients in r and in v, each of them including the scalarproduct of a vector and a gradient. We consider them component-wise. For example, the firstterm contains (before integrating over velocities and coordinates x and y) the contribution∫
vx ln f∂
∂xfdx = vx
∫ln f
∂
∂xfdx = vx f ln f |∞−∞ −
∫f
1
f
∂f
∂xdx = 0
108
vanishes (where we performed partial integration and noted that f(x, ..., t)→ 0 for x→ ±∞ dueto integrability). The same is true for all other contributions. Thus, the free motion of moleculesdoes not change the entropy of the system, and
dS
dt= −kB
∫ln fIstossfd3rd3v.
The left hand side is the integral of the type described by Eq.(127). Let us put it down in anexplicit way:
dS
dt= −kB
π
2
∫d3v1d
3v′db b|v − u| [ln f(r,v, t) + ln f(r,u, t)− ln f(r,v′, t)
− ln f(r,u′, t)] [f(r,v′, t)f(r,u′, t)− f(r,v, t)f(r,u, t)]
= kBπ
2
∫d3v1d
3v′db b|v − u| [ln f(r,v′, t) + ln f(r,u′, t)− ln f(r,v, t)
− ln f(r,u, t)] [f(r,v′, t)f(r,u′, t)− f(r,v, t)f(r,u, t)]
(in the last two lines we ordered the logarithms of f -functions in the same way the correspondingfunctions are ordered in the last square bracket, the minus sign is gone!). We note that
ln f(r,v′, t) + ln f(r,u′, t)− ln f(r,v, t)− ln f(r,u, t) = lnf(r,v′, t)f(r,u′, t)
f(r,v, t)f(r,u, t)
We now denote the f -functions of the corresponding arguments by f1, f2, f3 and f4, and b|v− u|as a function g, and note that g is non-negative. The integral on the right hand side has then theform ∫
d3v1d3v′db g ln
(f1f2
f3f4
)(f1f2 − f3f4).
Note that the integrand is non-negative, since g ≥ 0 and
(x− y) ln
(x
y
)≥ 0
for all x > 0 and y > 0: when x ≥ y, one has x− y ≥ 0 and ln(x/y) ≥ 0, and the product of thesetwo functions is non-negative; when x ≤ y, one has x− y ≤ 0 and ln(x/y) ≤ 0, and the product isnon-negative as well. Therefore
dS
dt≥ 0.
Note that the non-negativity holds not only for the total entropy but also for local entropy pro-duction (before the volume integration), so that
∂s
∂t≥ 0
where s is the entropy density. The entropy production is totally due to collisions, but the entropydensity in each volume of the system does not have to grow since the entropy is also transported.The entropy is produced as long as the equilibrium state is not reached.
109
8.4 Relaxation time approximation for hydrodynamics: The BGKequation
In the case of relaxation to total equilibrium or small deviations from such in a spatially homoge-neous system we used in Sec. 8.2.1 the relaxation time approximation:
∂f
∂t+
Fext
m∇vf = −ν(f(v, t)− feq(v)).
For complicated fluid motions one can assume that for a small fluid volume moving with determin-istic velocity V(x, t) one can define the local equilibrium distribution function floc(x,v, t), whichalso depends on the local density and on the local temperature. Now we generalize our approach tosuch case, and assume that Istoss in Eq.(123) is given by Istoss = −ν(f − floc). This approximationof the collision integral
Istoss = −ν(f − floc)
is called the Bhatnagar-Gross-Krook (BGK) model (1954). What we do further followsessentially the Grad’s approach to the solution of the Boltzmann equation (1949) within theapproximation done.
Eq.(123) now reads:
∂
∂tf(x,v, t) +
Fext
m∇vf(x,v, t) + v∇xf(x,v, t) = −ν(f(x,v, t)− floc(x,v, t)). (130)
First let us introduce some new notation. Again let f(v,x, t) be the distribution function, andn(x, t) =
∫f(v,x, t)dv the particles’ number density. Then
ρ(x, t) = m
∫f(v,x, t)dv
is the density of the gas, and
ρ(x, t)〈v(x)〉 = m
∫vf(v,x, t)dv
is the density of momentum.Our macroscopic velocity V(x, t) is nothing else than the mean velocity 〈v〉. The deviation of
the molecular velocity v from its mean value will be denoted by
C(x, t) = v −V(x, t).
The density of the internal energy is
ρ(x, t)e(x, t) =m
2
∫|C(x, t)|2f(v,x, t)dv
so that e(x, t) = 〈C2/2〉 (note that the internal energy of a rarified gas is simply the mean kineticenergy of molecules in a comoving coordinate frame).
The local equilibrium distribution function floc corresponds to the Maxwell distribution ofmolecular velocities around the deterministic velocity V(r) under the given local temperatureT (x, t) which doesn’t have to be the same in different parts of the system:
floc(v,x, t) = n(x, t)
(m
2πkBT (x, t)
)3/2
exp
[− m
2kBT (x, t)(v −V(x, t))2
].
110
This is properly normalized and defined such that the mean values of density, mean molecularvelocity and its square deviations from the deterministic value are the same when averaging overthe true f and over floc.
In the absence of the volume forces Fext Eq.(130) reads
∂f
∂t+ v∇xf = −ν(f − floc). (131)
We note that since the local equilibrium is defined so that the means of the hydrodynamic fieldsover the true distribution and over the local equilibrium are the same, 〈v〉 = 〈v〉loc = V and〈|C(x, t)|2〉 = 〈|C(x, t)|2〉loc, the approximation preserves the structure of the collision invariants,so that for m, p and e the right hand side of the kinetic model vanishes, as is also true for theBoltzmann equation. As before we define the mean as
〈ψ(r, t)〉 =
∫ψ(r,v, t)f(r,v, t)dv∫
f(r,v, t)dv.
Now we multiply the both parts of the BGK equation by ψ (as we did it for the Boltzmannequation), assume that ψ is not explicitly time-dependent, and integrate:
∂
∂tnψ =
∫ψ∂f
∂tdv =
∫ψ[−v∇f − ν(f − floc)]dv.
Now we rewrite this equation in the form
∂
∂tnψ = −∇
∫ψvfdv +
∫(v∇ψ)fdv − ν
∫ψ(f − floc)dv. (132)
(as we did this for the Boltzmann equation, Eq.(126)). We moreover note that the last termvanishes for the collision invariants.
For the mass density there is no difference with the standard discussion, and we get the conti-nuity equation
∂
∂tρ+∇rρ〈v〉 = 0.
Now let us consider the transport equation for the momentum. Using ψ = mvα we get
∂ρVα∂t
+∇∫vαρvfdv = 0.
The second integral in Eq.(132) vanishes while v (and thus mvα) is an independent variable (fromx). We note that the summation in the gradient operator goes over the index of the cartesiancomponent of v. In the tensor form this would look like
∂ρV
∂t= −∇
∫ρvvfdv
where vv is an outer product of v with itself, i.e. a tensor with components vαvβ.Now we note that v = V + C and that 〈Cα〉 = 0, and rewrite the last equation as
∂ρV
∂t= −∇ρVV −∇σ
111
where σ is the stress tensor defined as
σα,β = ρ
∫CαCβfdv.
Using ψ = mv2/2 we obtain the equation for the energy balance
∂
∂t
(ρe+
1
2ρV 2
)= −∇
∫vv2
2fdv.
Here we again can go from v to V and C, and separate the terms:
vαv2 = (Vα + Cα)(V2 + 2VC + C2)
= VαV2 + 2VαVC + VαC
2 + CαV2 + 2CαVC + CαC
2.
Comparing the ensuing form with previously defined quantities we may write
∂
∂t
(ρe+
1
2ρV 2
)= −∇
[V
(ρe+
1
2ρV 2
)+ σV + q
],
where the last term q, a vector with components
qα =1
2
∫CαC
2fdv
is associated with the heat flux. Now we have the hydrodynamic equations for a single-componentfluid, but do not know what the stress tensor and the heat flux are, i.e. we don’t yet have theNavier-Stokes equation and the Fourier law because we do not know the values of the kineticcoefficients.
First we note that the trace of the stress tensor is∑α
σαα = ρ〈C2〉 = 3nkBT = 3P
where P is the local pressure (we note that our assumption is that the local equilibrium functionperfectly reproduces the distribution of the collision invariants). The viscous part of the stresstensor (i.e. its trace-free part) is then
σ′αβ = σαβ − Pδαβ = ρ
(〈CαCβ〉 −
1
3〈C2〉δαβ
)vanishes under local equilibrium. Therefore the viscous stress is governed by the deviation fromthis local equilibrium. It doesn’t have to vanish in general since the product CαCβ is not a collisioninvariant.
Now some complications get evident. Let us consider our general equation of motion for ψ,Eq.(132), and take ψ = vv (a tensor-valued function). Our equation now reads
∂
∂tρ〈vv〉 = −∇ρ〈vvv〉 − ν[〈vv〉 − 〈vv〉loc].
Analog for the three-component construct ψ = vvv we get
∂
∂tρ〈vvv〉 = −∇ρ〈vvvv〉 − ν[〈vvv〉 − 〈vvv〉loc].
112
Therefore the equations do not build a closed system and need some closure. We will assume thatthe higher moments relax very fast to their local equilibrium value, at least when ν is large, andthen do not change. Thus we consider the quasi-stationary situation. The higher moment in thecorresponding equation will be approximated by its local equilibrium value. Thus, for the secondmoment we take
−∇ρ〈vvv〉loc − ν[〈vv〉 − 〈vv〉loc] = 0.
Analogous equation ensues for the third one.Separating v into V and C we get (using our previous definitions) that
ρ〈vαvβ〉 = ρVαVβ + Pδαβ + σ′αβ,
so thatρ〈vαvβ〉 − ρ〈vαvβ〉loc = σ′αβ.
For the third moment we get
ρ〈vαvβvγ〉 = ρVαVβVγ + ρVα〈CβCγ〉+ ρVβ〈CαCγ〉+ ρVγ〈CαCβ〉+ ρ〈CαCβCγ〉
The local equilibrium value of the third moment is
ρ〈vαvβvγ〉loc = ρVαVβVγ +ρkBT
m(Vαδβγ + Vβδαγ + Vγδαβ) .
Now we consider the simplest situations. Let us first turn to viscosity and assume that thepressure, temperature and density are uniform in space and do not change with time. Therefore,since ∂ρ/∂t = 0 and ∇ρ = 0 we get ∇V = 0 from the continuity equation. We assume themacroscopic motion to be slow compared to the thermal velocity
√kBT/m (which is of the order
of the velocity of sound), and therefore neglect the nonlinear term ρVαVβVγ in the equation for thethird moment and keep only the linear ones. The equation for the second moment reduces to
0 = −ρkBTm
[∂Vα∂xβ
+∂Vβ∂xα
]− νσ′αβ.
from which
σ′αβ = −nkBTν
[∂Vα∂xβ
+∂Vβ∂xα
].
We can compare this with the standard form of the Navier-Stokes stress tensor (Caution: Thetensor σ differs from the tensor τ defined in Eq.(25) by the sign. I will change the notation later.)
σαβ = −η[∂Vα∂xβ
+∂Vβ∂xα− 2
3δαβ∇V
]− δαβη1∇V.
(with η and η1 being the coefficients of the shear and the volume viscosity, respectively), where thelast term in the brackets and the last term of the whole expression describing the volume viscosityvanish under approximations done (since ∇V = 0). Since fluid has constant density everywhere,it behaves as an incompressible one. The shear viscosity coefficient is
η =nkBT
ν.
113
In gases it is independent on the density since the collision rate governing the relaxation rate ν isitself proportional to it (this lays however outside of the model), a result first obtained by Maxwellin 1860.
The same procedure for slightly more complicated quantities can be used for the heat flow.Now we assume that the density and pressure are constant, and that the macroscopic velocityvanishes everywhere. In this case v = C. The heat flux is
q =ρ
2〈v(v2)〉.
Our transport equation for q now reads
∂q
∂t= 0 = −∇ρ
2〈vvv2〉 − νq.
The fourth moment is replaced by the local equilibrium value:
〈vαvβvγvγ〉 = 〈vαvβ〉loc〈vγvγ〉loc + 2〈vαvγ〉loc〈vβvγ〉loc
In the local equilibrium the cross moments vanish, and after summation over γ we get
〈vvv2〉loc = 5
(kBT
m
)2
I
where I is the unit matrix. The heat flux than is
q = −∇ 5
2νρ
(kBT
m
)2
.
Now we remember that the pressure was assumed constant everywhere, so that P = n(x)kBT (x)is independent on the position, and that n = ρ/m. Therefore we can rewrite this expression asfollows:
q = − 5
2ν
PkBm∇T = − 5
2ν
nk2BT
m∇T.
This has the form of the Fourier’s law. The thermal conductivity is
κ =5nk2
BT
2mν.
The ratio of thermal conductivity to the viscosity has the universal value
λ
η=
5
2
k
m.
If we had started with the full Boltzmann equation, made similar assumptions about thesystems, about the separation between the macroscopic variables and fluctuations, and abouthomogeneity (but linearized the collision integral close to equilibrium correctly) we would haveobtained
κ
η=
15
4
k
m,
which is larger than the BGK result by a factor of 3/2. What is however more important, wewould get a reasonable approximation for the collision rate and the relaxation speed, which in theBGK theory is given by the coefficient ν which is introduced ad hoc.
114
9 Kinetics of quantum systems
The presentation in this section mostly follows the Zwanzig’s book.
9.1 The Golden Rule
The Golden Rule gives the rate of the transition between states n and m in presence of the externalperturbation in the first order of the perturbation theory. Although we have considered this inthe quantum mechanics course (where I typically do this for time-dependent perturbations to passthen to the emission/absorption of light by atoms and to the theory of the photoelectric effect),there is a subtle reason to shortly repeat the derivation to grasp the difference between the time-dependent and the time-independent perturbation. I will perform the derivation in the simplestform of the time-dependent perturbation theory.
The assumptions done in the derivation of the Golden Rule are:
• Perturbation is weak
• Only slow processes are discussed.
We start from the Schrodinger equation
∂
∂tΨ(t) = − i
~[H + V (t)]Ψ(t),
where H is the Hamiltonian of the unperturbed system and V (t) is the perturbation. We expandthe states over the basis of the eigenstates of the unperturbed Hamiltonian, H|j〉 = Ej|j〉:
Ψ(t) =∑j
aj(t)|j〉.
In our basis the Schrodinger equation reads:
∂
∂taj(t) = − i
~Ejaj(t)−
i
~∑k
Vjk(t)ak(t) (133)
with Vjk = 〈j|V |k〉. We start from a system prepared in state |n〉: Ψ(0) = |n〉, so that aj(0) = δjn,and integrate Eq.(133) over time:
aj(t) = e−iEjt/~δjn −i
~∑k
∫ t
0
dt′e−iEj(t−t′)/~Vjk(t
′)ak(t′).
To get the result in the first order of the perturbation theory (i.e. in the first order in V ) we takeak to be equal to its unperturbed value ak(t) = e−iEkt/~δkn. We get:
aj(t) = e−iEjt/~δjn −i
~∑k
∫ t
0
dt′e−iEj(t−t′)/~Vjk(t
′)e−iEkt′/~δkn +O(V 2).
The terms of higher order in V are now neglected. The amplitude of the state m 6= n after theaction of the perturbation is
am(t) = − i~
∫ t
0
dt′e−iEm(t−t′)/~Vmn(t′)e−iEnt′/~ (134)
This expression can be integrated over t′.There are two important situations to be considered: i) the constant perturbation V (t) = V ,
and ii) the perturbation depending on time periodically, say, V (t) = V cos(ωt).
115
Case i) Integrating Eq.(134) we obtain
am(t) = −1
~e−iEmt/~Vmn
eiωmnt − 1
ωmn
with ωmn = (Em − En)/~. The probability to find the system in state m after time t is given by
Pm(t) = |am(t)|2 =1
~2|Vmn|2
|eiωmnt − 1|2
ω2mn
=1
~2|Vmn|2
4 sin2(ωmnt/2)
ω2mn
.
The function
f(t, ωmn) =4 sin2(ωmnt/2)
ω2mn
is (for t ω−1mn) strongly peaked around ωmn = 0, the height of the peak is t2, and the total area∫∞
−∞ f(t, ω)dω is 2πt. Thus, for t→∞
f(t, ωmn) = 2πtδ(ωmn) = 2π~tδ(Em − En).
This defines the transition rate
wn→m =Pm(t)
t=
2π
~|Vnm|2δ(Em − En). (135)
In this case the rate essentially follows from the normal formula for the first order of perturbationtheory for degenerate states, where the perturbation causes the mixing of the degenerate states andremoves the degeneracy by slightly shifting their energies. We note that for the finite perturbationstrength Pm(t) = wn→mt with wn→m given by Eq.(135) diverges at longer times, while we knowthat it can never exceed unity. This is due to the breakdown of the first order approximation inour perturbation theory. It is possible to show, that the whole series would essentially converge,and that our approximation works only in the intermediate time domain, where the width of the“δ-function” in ω (of the order of 1/t) is already small, but the first order perturbation is stillapplicable: t ~/|Vmn|. This upper bound is large for weak perturbations, but is never infinite.Master equations allow for overcoming this upper bound for finite perturbations.
Case ii) In a harmonic field
V (t) = Veiωt + e−iωt
2.
Integrating the equation (134) with respect to time we get
am(t) = − 1
2~e−iEmt/~Vmn
(ei(ωmn+ω)t − 1
ωmn + ω+ei(ωmn−ω)t − 1
ωmn − ω
),
and
Pm(t) = |am(t)|2 =1
4~2|Vmn|2
∣∣∣∣ei(ωmn+ω)t − 1
ωmn + ω+ei(ωmn−ω)t − 1
ωmn − ω
∣∣∣∣2 .The function
f(t, ωmn, ω) =1
4
∣∣∣∣ei(ωmn+ω)t − 1
ωmn + ω+ei(ωmn−ω)t − 1
ωmn − ω
∣∣∣∣2
116
has two peaks around ωmn = ω and ωmn = −ω which are well separated if t ω−1. For t ω−1
we may approximate
f(t, ...) =πt
2[δ(ωmn + ω) + δ(ωmn − ω)]
and get
wn→m =π
2~|Vnm|2[δ(Em − En + ~ω) + δ(Em − En − ~ω)]. (136)
Note that the transition to Eq.(135) for ω → 0 is discontinuous: there is a jump by a factorof 4. The reason is the nature of the approximation and the constructive interference. While theEq.(136) can be used, say, for the calculation of the optical absorption coefficient, Eq.(135) is thefirst step to deriving the quantum mechanical master equations.
9.2 Quantum mechanical Master equations
Pauli master equation. We consider a QM system with the Hamiltonian
H = H0 + λV
with a time-independent perturbation. The unperturbed system has the eigenstates |m〉 and theeigenvalues Em:
H0|n〉 = En|n〉.
Now we include the diagonal matrix elements (causing the shifts of the state energies) of V intoEm, and consider the purely non-diagonal perturbation which mixes the states (the states |n〉 areno more the eigenstates of H0), i.e. introduces the transitions between them. Thus we considerV to contain only non-diagonal matrix elements. We perform a single-time measurement and askfor the probability to find the system in the state n. Note that this probability is essentially thediagonal matrix element of the density matrix ρnn(t).
Now we put down the gain-loss equation for the probability of the occupation of a state, i.e.the Master equation:
d
dtPn(t) =
∑m
Wm→nPm(t)−∑m
Wn→mPn(t)
which is the Pauli Master equation. The transition rates are given by the Golden Rule, Eq.(135),Wm→n = wm→n, since the perturbation is time-independent. This is the microcanonical equation,since all states involved in transitions have to have an approximately the same energy. Thetransition rates given by Eq.(135) are symmetric in states
Wn→m = Wm→n,
which is called the property of microscopic reversibility. At microcanonical equilibrium all P eqm
will have the same value of P eqm = 1/g, where g is the degeneracy of the state with energy En, in
accordance with the standard microcanonical assumption.
Heat bath master equation. Now we take the Hamiltonian to be
H = Hs + Hb + λV
117
with Hs being the Hamiltonian of the system disconnected from the bath, Hb being the Hamiltonianof the bath alone, and λV is the interaction between them, whose strength is governed by the(small) perturbation parameter λ. We have:
Hs|n〉 = En|n〉Hb|α〉 = εn|α〉.
The unperturbed states are considered in the product basis:
(Hs + Hb)|nα〉 = (En + εα)|nα〉.
The whole system is isolated and therefore microcanonical, the total energy is E = En + εα and ishighly degenerated because the spectrum of the bath is assumed quasi-continuous, the perturbationis small and causes transitions between states. We put down the Pauli Master equation:
d
dtPnα(t) =
∑mβ
Wmβ→nαPmβ(t)−∑mβ
Wnα→mβPnα(t). (137)
The Golden rule rates are
Wnα→mβ =2π
~λ2|〈nα|V |mβ〉|2δ(Em + εβ − En − εα). (138)
Now we assume (without proof) that the bath stays in equilibrium no matter what the state ofthe system is (the proof is typically done assuming a special model of the bath). Then Pnα(t)factorizes into Pn(t) and ρα, where ρα is the probability to find the bath in its state α, and thisone is time-independent: Pnα(t) = Pn(t)ρα. We assume this form in Eq.(137) and sum over all α:
d
dtPn(t)
∑α
ρα =∑α,β
∑m
Wmβ→nαPm(t)ρβ −∑α,β
∑m
Wnα→mβPn(t)ρα. (139)
The sum over α in the l.h.s. is unity due to normalization of ρα. Now we denote
Wm→n =∑α,β
Wmβ→nαρβ
Wn→m =∑α,β
Wnα→mβρα
and get the standard formal master equation for Pn:
d
dtPn(t) =
∑m
Wm→nPm(t)−∑m
Wn→mPn(t).
The rates now are no more symmetric. However we know thatρα = Z−1 exp(−εα/kBT ) and ρβ = Z−1 exp(−εβ/kBT ) with Z being the partition function ofthe bath. Moreover, according to Eq.(138) only the transitions between the states with Em−En =εα − εβ are possible, due to the presence of the δ-function. Therefore one may write
ρβ = ρα exp[(εα − εβ)/kBT ] = ρα exp[(Em − En)/kBT ].
Now, for n and m fixed we have
Wm→n
Wn→m=
∑α,βWmβ→nαρβ∑α,βWnα→mβρα
=
∑α,βWmβ→nαραe
Em−EnkBT∑
α,β
∑mWnα→mβρα
= eEm−EnkBT .
The corresponding rates satisfy the detailed-balance condition.
118
Bilinear coupling. For simplicity we will assume that the coupling between the system and thebath is of the form
λV = F (system)G(bath),
i.e. is a bilinear form in operators involving only the system’s and the bath’s variables. Oneexample is the case of NMR, with F acting on the nuclear spin, and G describing the randommagnetic fields generated by the random currents in the bath. Now we can calculate the transitionrates. The matrix elements of the perturbation now are:
Vnα,mβ = FnmGαβ.
We put down the rates as given by Eq.(138), sum them over α and β, and get for the thermallyactivated rate
wn→m =2π
~|Fnm|2
∑α,β
δ(Em + εβ − En − εα)|Gαβ|2ρβ.
The properties of the sum in this expression can be connected with the fundamental correlationproperties of the bath.
To see this, we proceed as follows:
• We replace the delta-function via its integral representation δ(x) = 12π
∫∞−∞ e
−ixkdk, denotek = t/~ and write
δ((Em + εβ − En − εα) =1
2π~
∫ ∞−∞
ei(En−Em)t
~ eiεαt~ e−i
εβt
~ dt.
We denote ωnm = (En − Em)/~: this is the frequency of the transition.
• Noting that |Gαβ|2 = GβαGαβ we rewrite our expression in the following form:
wn→m =1
~2|Fnm|2
∫ ∞−∞
dteiωnmt∑α,β
Gβαei εαt~ Gαβe
−iεβt
~ ρβ.
• We recognize the Heisenberg representation of the time-dependent operator G(t):
[G(t)]αβ = eiεαt~ Gαβe
−iεβt
~ .
In this Heisenberg representation Gβα = [G(0)]∗αβ.
• We sum over α and get ∑α
eiεαt~ Gαβe
−iεβt
~ Gβα = [G(0)G(t)]ββ.
• Now we can sum over β and obtain∑β
[G(0)G(t)]ββρβ = 〈G(0)G(t)〉eq.
The final result reads:
Wn→m =1
~2|Fmn|2
∫ ∞−∞
dteiωnmt〈G(0)G(t)〉eq.
The transition rate is proportional to the Fourier component of the bath’s correlation function atthe frequency of transition.
119
A simple example. A typical model of the heat bath is the phonon bath of Einstein’s orDebye’s models or any other assembly of harmonic oscillators (i.e. a Kac-Zwanzig bath, see Sec.7). This is a pet model of quantum kinetics, leading, for a two-level system (say, nuclear spin), toa so-called “spin-boson” model. The spin-boson model will be considered later. Let us considera quantum mechanical system coupled weakly to an oscillator bath. For the sake of simplicity allbath oscillators masses are assumed to be the same.
We first repeat our calculation of Sec. 7 slightly changing the notation: we assume that
H = Hs(p, x) +∑j
(p2j
2m+m
2ω2j
(qj −
γjω2j
x
)2).
Now the equations of motion are
dx
dt=
p
M,
dp
dt= −U ′(x) +
∑j
γj
(qj −
γjω2j
x
)dqjdt
=pjm,
dpjdt
= −mω2j qj +mγjx
(U ′(x) is the force). These equations can be treated either as classical equations of motion or asHeisenberg equations of motion for operators p and x.
Now we write:Mx = p = −U ′(x)−
∑j
γjqj (140)
and
qj =pjm
= −ω2j qj − γjx. (141)
The corresponding system of equations can be solved analytically, like in Sec. 7. Let us start fromthe second equation, and assume that x(t) is some known function of t. The general solution ofEq.(141) is:
qj(t) = qj(0) cosωjt+ pj(0)sinωjt
ωj− γj
∫ t
0
dt′x(t′)sinωj(t− t′)
ωj. (142)
Neglecting back-coupling as described by the third term in Eq.(142) we get
qj(t) = qj(0) cosωjt+pj(0)
m
sinωjt
ωj.
Now we can multiply both parts of this equation with qi(0) and average over the equilibriumdistribution. We get:
〈G(0)G(t)〉eq =∑j
(γ2j cosωjt〈q2
j 〉eq + γ2j
sinωjt
mωj〈pjqj〉eq
). (143)
The only difference in the quantum case is that the whole expression has to be considered as anthe operator, and that the second term in equilibrium does not vanish due to non-commutativity.Now we assume all γj to be the same, γj = γ. Note that the dimension of γ is [M/T2].
The difference to the classical case is the second term due non-vanishing contribution of non-commuting variables pj and qj. The term 〈pjqj〉eq vanishes in the classical case: 〈pjqj〉eq = 0, and
120
it’s non-zero value in the quantum case is solely due to the fact of non-commutativity of pj andqj. Therefore one can assume that 〈pjqj〉eq ∼ [p, q] ∼ ~ in the quantum case. In the classicalapproximation 〈qiqj〉eq = (kBT/mω
2j )δij, and this term definitely dominates if kBT ~ωj, i.e. at
high temperatures. The quantum effects at high temperatures only correspond to freezing out thecontributions of high-frequency oscillators. The calculations will be given as a Homework.
In what follows we will use the classical approximation:
〈G(0)G(t)〉eq = kBTγ2
m
∑j
cosωjt
ω2j
.
Instead of summation we can integrate over the density D(ω) of the oscillator modes,
〈G(0)G(t)〉eq = kBTγ2
m
∫ ∞0
cosωt
ω2D(ω)dω.
For the Debye model in three dimensions (so-called “Ohmic bath”) D(ω) = 32π2cC3ω
2 (with Cbeing the velocity of sound and c = N/V being the atomic density) and therefore 〈G(0)G(t)〉eq =
kBT3γ2
2mπ2cC3 δ(t), In this case
Wn→m =λkBT
~2|Fnm|2.
where λ is a constant comprising the prefactor in D(E) and γ2.If our system is a harmonic oscillator with mass M and frequency Ω, then the matrix elements
squared, |Fnm|2 = |〈n|x|m〉|2, differ from zero only for m = n ± 1 and are equal to |Fnm|2 =~n/2MΩ. The rate of the transition form n to n− 1 is
Wn→n−1 =kBTζ
2MΩn
where in our model
ζ =1
~3γ2
2mπ2cC3.
is a constant of the dimension of the inverse length. The transitions to the states of higher energyfulfill the detailed-balance condition
Wn−1→n = e− ~ΩkBTWn→n−1.
The resulting heat bath master equation for the harmonic oscillator reads
dPndt
= kne−θPn−1 + (n+ 1)Pn+1 − [n+ (n+ 1)e−θ]Pn
where k = kBTζ/2MΩ is the rate constant and θ = ~Ω/kBT . We note that this master equationgives us the possibility to translate the Golden rule description to longer times. From this masterequation the equation for the evolution of the mean energy of the oscillatior follows (Homework10).
A similar situation arises when considering the inelastic energy transfer between the molecularharmonic oscillator and the gas of inert molecules (bath). At the end, it only differs in the valueof the prefactor (Landau and Teller, 1936).
Returning to the role of the back-coupling we note that all calculation of Sec. 7 can beconsidered as fully quantum mechanical, if the equations of motion are considered as the Heisenberg
121
equations of motion for operators. Therefore the equation of motion for x is still the generalizedLangevin equation
Mx = −U ′(x)−∫ t
0
dt′x(t′)K(t− t′) + F (t),
where the coordinate and the force are now considered as operators. The only difference betweenthe quantum and the classical case is the fluctuation-dissipation relation between 〈F (t)F (t′)〉 andK(t). For the case when U(x) is harmonic, this equation can be formally solved. For non-harmonicpotentials the solution is tricky. Even in the Ohmic case there are in general no such instrumentsas Fokker-Planck equation for quantum Langevin equations.
9.3 Electron transfer kinetics
As an example of the system where its back-action on the bath plays a crucial role let us nowconsider the electron transfer between two quantum states. Let us assume we have a quantumanalog of the Kramers problem, and the electron can tunnel between the two minima of somepotential with two potential wells (there is no practical need in quantum description of any otherreactions at “normal” temperatures, maybe except for proton transfer at very low ones). Now oursystem is described by the Hamiltonian
Hs =p2
2m+ U(x).
The model of the bath will be taken as in our previous one:
H = Hs(p, x) +∑j
(p2j
2m+m
2ω2j
(qj −
γjω2j
x
)2).
We assume that the good starting basis for the problem is a system with two localized states |L〉in the left and |R〉 in the right well (all other states are considered to be well separated in energy,and neglected altogether). The states |L〉 and |R〉 can be taken as the harmonic oscillator groundstates for the quadratic approximation of the corresponding minima of the potentials. Thereforethe system’s Hamiltonian can be represented as a two by two matrix
Hs =
(EL VV ∗ ER
).
Here V are the matrix elements V = 〈L|Hs|R〉 describes interaction (tunneling) between |L〉 and|R〉. We assume that the non-diagonal matrix elements are small (tunneling is weak) and that theinteraction with the bath does not change them (this is not always true and leads to a different,very interesting class of problems both in quantum mechanical and in classical setting). The bathHamiltonian has two different forms, depending on whether the system is in the |L〉 or in the |R〉state (this is exactly the back action we were talking about). Assuming that the coordinate of theparticle in the corresponding state is xL or xR and neglecting the uncertainty of this coordinatewe may write
HL,R =∑j
(p2j
2m+m
2ω2j
(qj −
γjω2j
xL,R
)2). (144)
122
The total Hamiltonian in this approximation reads
H =
(EL + HL V
V ∗ ER + HR
).
This operator acts on a wave function being a two-component vector (“spin”) with componentsdepending on the coordinates of all oscillators, and is called a “spin-boson” Hamiltonian (excitationof a harmonic oscillator = boson). The states of the system are easy to classify. In the absence oftunneling (V = 0) we have a spin state (1, 0) for |L〉 and (0, 1) for |R〉, and the states |µL〉 and|νR〉 for HL and HR, respectively:
HL|µL〉 = εµL|µL〉,HR|νR〉 = ενR|νR〉.
The transitions between these unperturbed states are caused by the non-diagonal perturbationoperator
V =
(0 VV ∗ 0
).
According to the Golden rule
WµL→νR =2π
~|〈µL|V |νR〉|2δ(EL + ενL − ER − εµR)
=2π
~|V |2|〈µL|νR〉|2δ(EL + ενL − ER − εµR)
(note that the eigenstates |µL〉 and |νR〉 are the eigenstates of two different Hamiltonians, HL andHR, and are not orthonormal. Moreover, since HL and HR have different systems of eigenfunctions,which can be easily checked by explicitly writing them down, they do not commute). Summingover all final oscillator states and averaging over the equilibrium distribution ρµL of the initialstates we get
WL→R =2π
~|V |2
∑µ
ρµL∑ν
|〈µL|νR〉|2δ(EL + ενL − ER − εµR).
Now, as we already did it before, we use the integral representation of the δ-function and separatethe terms:
WL→R =1
~2|V |2
∫ ∞−∞
dtei(EL−ER)t/~∑µ
ρµLeiεµLt/~
∑ν
|〈µL|νR〉|2e−iενLt/~.
The sum over ν is now rewritten as∑ν
|〈µL|νR〉|2e−iενLt/~ =∑ν
〈µL|νR〉e−iενLt/~〈νR|µL〉.
Now we see that this is simply a representation of
〈µL|e−iHRt/~|µL〉
123
in the system of eigenfunctions of HR. Now we move the eiεLt/~ inside the scalar product andrewrite the whole double sum as∑
µ
ρµLeiεµLt/~
∑ν
|〈µL|νR〉|2e−iενLt/~ =∑µ
ρµL〈µL|eiHLt/~e−iHRt/~|µL〉.
Therefore the whole expression for the transition rate reads:
WL→R =1
~2|V |2
∫ ∞−∞
dtei(EL−ER)t/~〈eiHLt/~e−iHRt/~〉L,
where the average of the corresponding operator is taken over the equilibrium distribution for bathin the case when the particle in the left well.
Even this model in general is still too complex. The reason is that HL and HR do not commute,and therefore eiHLt/~e−iHRt/~ 6= ei(HL−HR)t/~ (we know that eA+B = eBeAe−[A,B]/2).
As before, the sake of simplicity, we treat the bath oscillators as classical ones. In the classicallimit the operators do commute, and
〈eiHLt/~e−iHRt/~〉L = 〈ei(HL−HR)t/~〉L
(note that the commutator [HL, HR] can be expressed via commutators of [pj, qj] multiplied bypowers of pi and qj, and therefore is proportional to ~, the corresponding correction disappearswhen ~→ 0). The difference HL −HR is
HL −HR = −∑j
γjqj(xL − xR) +K
2(x2
L − x2r)
with K =∑
j(γ2j /ω
2j ). This difference is linear in the bath oscillator’s coordinate. The last term
is simply a constant which will be denoted by λ in what follows.Now we may take the average:
〈ei(HL−HR)t/~〉L = eitK(x2L−x
2r)/2~
⟨e−it
∑j γjqj(xL−xR)/~⟩
L.
In the classical bath we have the distribution of qi following the Boltzmann distribution, i.e. theGaussian distribution with the mean value
〈qi〉L =γiω2i
xL
and variance ⟨(qi − 〈qi〉)2
⟩L
=kBT
ω2i
,
as evident from the form of the second term in Eq.(144) (remember that the mass of the bathparticles is unity), and different coordinates are the independent random variables. Therefore thecorresponding mean is the product of characteristic functions of these Gaussians taken at the valueof k = −tγj(xl − xr)/~. Finally we get:
〈ei(HL−HR)t/~〉L = eitK(x2L−x
2r)/2~e−(2~)−2t2(xL−xR)2kBT
∑j(γ
2j /ω
2j )
≡ eitK(x2L−x
2r)/2~e−t
2(xL−xR)2kBTK/2~2
.
124
The quantity
λ = 〈HL −HR〉L =K
2(xL − xR)2
having the dimension of the energy (which gets clear when restoring the mass of the oscillatorparticles in the final expression) is the change in the bath’s energy under the transition from xLto xR and is called the Marcus’s reorganization energy. Therefore the transition rate is:
WL→R =|V |2
~2
∫ ∞−∞
eit(EL−ER)/~e−itλ/~e−λkBTt2/~2
.
Performing integration over t we get the Marcus’s result
WL→R =|V 2|~
√π√
λkBTe− 1kBT
(ER−EL+λ)2
4λ ,
Nobel prize in chemistry 1992. Note that this is not of Kramers form. First, there is a dependenceof the rate both on the energy of the initial and of the final state. Instead of the height of thebarrier there is a new energetic parameter which governs the rate, namely the reorganization energyλ. The height of the barrier comes in the tunneling matrix element (essentially, in an exponentialmanner, as always in tunneling problems), but the properties of the barrier are weighted with thetemperature differently than in the Kramers expression, since the effective activation energy in theArrhenius-dependence differs. All in all, thermally assisted tunneling does not look like overcomingthe classical barrier.
9.4 The density matrix approaches
9.4.1 The density matrix and the quantum Liouville equation.
Let us consider a quantum system whose states are described in terms of the eigenfunctions of itsHamiltonian,
Hψi = Eiψi
orH|i〉 = Ei|i〉.
The wave functions ψi build a full orthonormal system of functions. A quantum mechanicalmean value of some observable (operator) A in state i is
〈A〉i = 〈i|A|i〉 =
∫dqψ∗j (q)Aψ(q).
Now let us imagine an ensemble of quantum systems prepared in different initial states i withprobability ρi. For a microcanonical ensemble the initial state energies are taken in a narrowenergy interval, for a canonical one the probability that a system in thermal equilibrium is in itsstate i is given by
ρeqi = Z−1 exp(−Ei/kBT ),
where Z is the partition function Z =∑
i exp(−Ei/kBT ). We consider ρ = diag(ρ1, ρ2, ...) as adiagonal matrix, and call this the (canonical) density matrix in energy representation. The meanover many experiments with the ensemble will be
〈〈A〉〉 =∑i
ρi〈A〉i,
125
for example the equilibrium mean is then given by
〈A〉eq =∑i
ρeqi 〈A〉i.
In what follows we will consider only 〈A〉eq as an example, but our approach works for any doublemean over the initial conditions. To use this expression we need to work in the representation inwhich the Hamiltonian is diagonal (i.e. in the energy representation), which is not always the mostconvenient one. Let us use any other representation, i.e. reexpand the set of ψi over any other fullorthonormal set of functions φj. We have
|ψi〉 =∑k
Sik|φk〉,
with Sik = 〈φk|ψi〉. Since both sets are considered as orthonormal, the transformation matrixS is unitary, so that S−1 = S+, i.e S−1
ij = S∗ji. Under this transformation the Hamiltonian istransformed into
H → SHS+,
which is represented by a matrix with elements Hjk =∑
m SmjEmS∗mk. Both, the Hamiltonian,
and the density matrix are non-diagonal in this representation, and the last one has the elements
ρeqjk =
∑m
SmjρeqmS∗mk.
Any dynamical quantity (operator A) has a representation
Ajk = 〈φj|A|φk〉,and the average for a single energy quantum state is
〈A〉i = 〈ψi|A|ψi〉 =∑k
(S+)ik〈φk|A|∑n
Sin|φn〉 =∑m,n
S∗imAmnSin.
Now we get the equilibrium average
〈A〉eq =∑i
ρeqi
∑m,n
S∗imAmnSin
=∑m,n
Amn∑i
ρeqi S∗imSin.
Now we rename m to n and n to m:
〈A〉eq =∑m,n
(∑i
Simρeqi S∗in
)Anm.
The expression in brackets is the equilibrium density matrix, so that
〈A〉eq =∑m,n
ρeqmnAnm = TrρeqA.
Since we will often switch between the classical and quantum descriptions, we note the followingclassical - quantum correspondence:
classical quantum∫dXA(X)f(X) ↔ TrρA
(X are the phase variables which describe the system; the same equation states true when passingto time-dependent observables or densities).
126
Quantum Liouville operator. Now we turn to dynamics. The evolution of the state is givenby
i~∂
∂tψ = Hψ.
If the Hamiltonian is time-independent, the standard time evolution operator follows:
|ψ(t)〉 = e−itH/~|ψ(t)〉.
The averages get to be time-dependent:
〈A(t)〉 = 〈ψ(t)|A|ψ(t)〉 = 〈e−itH/~ψ(0)|A|e−itH/~ψ(0)〉= 〈ψ(0)|eitH/~Ae−itH/~|ψ(0)〉.
This can be considered as an average of the operator in the Heisenberg representation:
〈A(t)〉 = 〈ψ(0)|A(t)|ψ(0)〉.
The operator A(t) in the Heisenberg representation is obtained from the time-independent one inthe Schrodiner representation (equal to A(0)) via
A(t) = eitH/~A(0)e−itH/~. (145)
The initial condition can be posed at any time, so that
A(t+ ∆t) = ei∆tH/~A(t)e−i∆tH/~.
Taking this ∆t small we get
A(t+ ∆t) = A(t) + ∆ti
~
[HA(t)− A(t)H
]+O(∆t2)
= A(t) + ∆ti
~
[H, A(t)
]+O(∆t2).
Therefore∂
∂tA(t) =
i
~
[H, A(t)
]≡ LA.
The commutator in the r.h.s. defines the quantum Liouville operator L (superoperator acting onthe operator A):
L· = i
~
[H, ·
]In its “matrix form” L is a tetradic, i.e. four-index construct:
(LA)jk =∑m,n
Ljk,mnAmn
with
Ljk,mn =i
~(Hjmδnk − δjmHnk)
as following from the definition of the commutator. The formal solution of the equation
∂
∂tA(t) = LA(t).
127
for the time-independent L (i.e. for the time-independent Hamiltonian) is given by
A(t) = etLA(0).
Using Eq.(145) it is easy to find the tetradic representation of etL:
(etL)jk,mn = (eitH/~)jm(e−itH/~)nk.
Now we can calculate the average of 〈A(t)〉 over the initial conditions when starting at equi-librium. What is essentially averaged is A(0): 〈A(0)〉 = TrρeqA = TrAρeq (the trace is invariantunder the cyclic permutations). We can proceed formally or start from the energy representationand then change to a arbitrary one. In any case we get:
〈A(t)〉eq = TreiLtA(0)ρeq
= TreitH/~A(0)e−itH/~ρeq
= TrA(0)e−itH/~ρeqeitH/~
(where we again used the invariance of trace under cyclic permutations). Now we obtained theSchrodinger representation of the same average:
〈A(t)〉eq = TrAρ(t) :
the operator A does not change in time, but the density matrix does:
ρ(t) = e−itH/~ρeqeitH/~.
From this equation the equation of motion for ρ(t) can easily be read out:
∂
∂tρ(t) = − i
~
[H, ρ(t)
].
This quantum Liouville (or von Neumann) equation differs from the Heisenberg equations fortime-dependent operators by the opposite sign of time evolution.
We note that the averages in the Heisenberg and in the Schrodinger pictures are the same:
TrρeqA(t) = TrρeqeLtA(0) = Trρ(t)A(0) = Tre−LtρeqA(0). (146)
This means that the operator eLt is anti-self-adjoint.Note that all scheme developed is valid only for free evolution of an isolated system prepared
at equilibrium. It is not valid for time-dependent perturbations where time-ordering of the expo-nentials has to be taken into account. However this scheme is enough for our purposes when weconsider a system prepared in thermal equilibrium and follow is further evolution.
9.4.2 Two level systems in the bath.
Simple example: dephasing. As the first example let us consider the simplest model, a spin-boson model for a two-level system, in the eigenbasis of the levels, and assume that the interactionwith the bath shifts the level (essentially, in our model only the higher one) but does not induce
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the transitions. The main difference with the Marcus model is the absence of the non-diagonalelements:
H =
(HB 00 ~ω0 +HB + ~VB
).
(~VB here plays the role of the difference between bath energies in the left and the right state inour previous example, the notation is chosen in a way that allows to drop out ~ at further stepsof calculations). The back-action of the system on the bath is neglected. The density matrix hasa form
ρ =
(ρ11 ρ12
ρ21 ρ22
),
where the elements of this matrix still depend on the bath variables. For a quantum bath ρij areoperators comprising the bath’s particles coordinates and momenta. For a classical bath they arethe functions thereof. The spectral shape of the line is defined by the correpation function of thetransition dipole moments of the transition, i.e. of the Hermitean operator
µ =
(0 µ12
µ21 0
),
A dipol moment averaged over the bath variables comprizing a vector B is
µ(t) =
∫dB(µ12ρ21 + µ21ρ12).
(in the quantum case we take the trace over the bath variables instead of integrating over them).The partial traces (or integrals) of the matrix elements of ρij over the bath variables will be denotedby cij:
cij
∫ ∫dBρ12(t).
The equation of motion of the density matrix
∂
∂tρ(t) = − i
~
[H, ρ(t)
]now separates in four independent equations (this is only due to the absence of non-diagonalelements in H). Writing out the commutator we get
∂
∂tρ11 = − i
~[HB, ρ11],
∂
∂tρ22 = − i
~[HB, ρ22],
∂
∂tρ12 = − i
~[HB, ρ12] + i(ω0 + V )ρ12,
∂
∂tρ21 = − i
~[HB, ρ21]− i(ω0 + V )ρ21.
Due to the absence of back-coupling of the system to the bath, if the bath was prepared inequilibrium at the beginning it stays in equilibrium all the time. According to the Liouvilletheorem, the Hamiltonian evolution of the bath doesn’t change the equilibrium distribution, thusLBf = 0. The right hand sides of the first two equations, and the terms with commutators in the
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last two ones (the terms of the type LBρ) vanish. As initial condition we assume that the bathwas prepared in equilibrium. As the initial condition we will assume ρ12(t = 0,B) = f(B), so thatc12(0) = 1. The dipole time correlation function 〈c(0)c(t)〉 is then proportional to Re c12(t).
Let us concentrate on the equation for ρ12. We consider our bath to be classical, and changethe commutator for a classical Liouvillian, i.e. for a Poisson bracket, but it doesn’t play any roleup to the end, except for the fact that integrals are changed for the traces. Our equation reads:
∂
∂tρ12 = −LBρ12 + i(ω0 + V )ρ12. (147)
Now let us introduce the new variable g(t,B) so that
ρ12(t) = e−LBteiω0tg(t,B).
For this function our equation (147) reads:
∂
∂tg(t,B) = iV (t,B)g(t,B),
where V is now time-dependent and is given by
V (t,B) = eLBtV (B) = V (B(t)).
The solution for g(t,B) now reads
g(t,B) = exp
(i
∫ t
0
dt′V (t′,B)
)g(0,B).
The initial value of g is f(B) and therefore we get32
c12(t) = eiω0t
∫dB exp
(i
∫ t
0
dt′V (t′,B)
)f(B).
32 Essentially, we have to discuss what happens with e−LBt, since
ρ12(t) = eiω0te−LBt exp
(i
∫ t
0
dt′V (t′,B)
)f(B).
The classical integration over the bath variables corresponds to the building of the trace over quantum states ofthe bath, all functions in our expression are essentially operators in the bath variables. By integration we trace outthese bath variables:
c12(t) = eiω0tTre−LBt exp
(i
∫ t
0
dt′V (t′,B)
)f(B),
where f(B) is the equilibrium density matrix of the bath states. Here we have a construct C = Tre−LBt(Af),
with A = A(t). We write the quantum Liouvillian as two exponentials of the Hamiltonian: Tre−iHt/~(Af)eiHt/~.Now we have a trace of a product of four operators (don’t forget that the Liouvillian was not an operator, buta superoperator, i.e. the tetradic construct) and can use the fact that it is invariant under cyclic permutations:
Tre−iHt/~(Af)eiHt/~ = Tr(Af)eiHt/~e−iHt/~ = Tr(Af). Then we use the classical-quantum correspondence. Alter-native: Write explicitly the classical Liouvillian, use chain rules for differentiations, and perform partial integrations.
The construct has some other interesting properties: for example, we can insert a unit operator between A andf , and get
Tr(e−iHt/~AeiHt/~)(e−iHt/~f eiHt/~) = Tr(e−LBtA)(e−LBtf).
Now we note that the equilibrium density matrix of the bath f is invariant under the Liouvillian evolution (because
it is diagonal in the energy representation): e−LBtf = f . Now we have the expression C = Tr(e−LBtA)f and use
the anti-self-adjointness of the exponential of the Liouvillian, i.e. make the cyclic permutation C = Tr(e−LBtA)f =
Tre−iHt/~AeiHt/~f = TrAeiHt/~f e−iHt/~ = TrAeLBtf . Now we again use the invariance of f under the time
evolution, and see that eLBt drops out: we have C = TrA(t)f . Essentially, we have a relative freedom to move theexponential of the Liouvillian inside the trace, and will use this possibility repeatedly.
130
Now we return to our harmonic oscillator bath (Kac-Zwanzig model) and consider V as abilinear combination of the particle’s coordinate and the bath oscillator’s coordinates. We moreoverassume that the distribution of the bath coordinates is Gaussian at any time, and has a zeromean. Now we can perform the averaging by noting that what we have here is nothing else asthe characteristic function of the distribution of
∫ t0dt′V (t′,B), where V is assumed to be a linear
functional in bath’s coordinates, and therefore also has a Gaussian distribution. This is essentiallya trick we used many times in discussing the Brownian motion:∫
dB exp
(i
∫ t
0
dt′V (t′,B)
)f(B) = exp
[−1
2
∫ t
0
dt′∫ t
0
dt′′〈V (t′)V (t′′)〉].
Here is the explanation once again (for one dimension): if x(t) is a Gaussian process with zero meanand correlation function C(t), then the integral sum for I(t) =
∫ t0x(t′)dt′ is a sum of Gaussian
random variables, and therefore a Gaussian random variable again. Its distribution is fully givenby its mean (which is zero anyhow) and its dispersion which is
〈I2(t)〉 =
⟨∫ t
0
∫ t
0
dt′dt′′x(t′)x(t′′)
⟩=
∫ t
0
∫ t
0
dt′dt′′ 〈x(t′)x(t′′)〉 .
The characteristic function of a symmetric Gaussian is exp(−12〈I2(t)〉). The correlation function
of the linear bath variables is exactly the memory kernel K(t′ − t′′) we have already encounteredin our discussion above. Therefore
c12(t) = eiω0t exp
[−1
2
∫ t
0
dt′∫ t
0
dt′′K(t′ − t′′)]. (148)
For the Ohmic bath K(t′ − t′′) = γδ(t− t′), so that
c12(t) = eiω0t−γt, (149)
for the exponentially decaying memory kernel K(t) = γe−t/τ the integral is the one which we havealready encountered in the discussion of the Taylor-Kubo formula in Sec. 4.3, and the result is
c12(t) = exp[iω0t− γ
(t− τ + e−t/τ
)],
shpowing an exponential decay at longer times. Without contact to the bath c12 oscillates; theeffect of the bath introduces the decay of c12, the dephasing or “decoherence”.
The decay will be different for other memory kernels K(t), but the fact of decay of non-diagonalelements of the density matrix averaged over the bath variables (i.e. of the non-diagonal matrixelements of the reduced density matrix characterizing the system only) and of the propertiesdepending thereon stays true. We note that for non-ohmic bath with integrable memory kernel Kwe can approximate
∫∞0K(t′)dt′ by some effective value γ, as we did it for Brownian motion.
We note that we have chosen a bath model which allows for an exact solution. Generally wecannot do so, and have to rely on approximations. One of them corresponds to the local equilibriumassumption, and is as follows.
We return to our equation for ρ12
∂
∂tρ12 = −LBρ12 + i(ω0 + V )ρ12
131
and integrate over the bath variables. The term with the bath’s Liouvillian vanishes, as discussedbefore. We therefore get
∂
∂tc12 = iω0c12 + i
∫dBV (B)ρ12(t,B).
Now we return back to the equation for ρ12 and integrate it formally as an initial condition problemwith ρ12(t = 0,B) = f(B). We get
ρ12(t) = eiω0tf(B) +
∫ t
0
dt′e(iω0−LB)(t−t′)iV (B)ρ(t′,B)
(the values of bath variables in V (B) have to be taken at time t′). Now we insert this formalsolution into the integral in the previous equation. After this insertion, the integral containingfirst term in ρ12(t),
i
∫dBV (B)eiω0tf(B)
drops our, since the first moment of V in equilibrium is zero. What stays is
∂
∂tc12 = iω0c12 + i
∫dBV (B)
∫ t
0
dt′e(iω0−LB)(t−t′)iV (B)ρ12(t′,B)
This integral is formally of the second order in V . The equation is however not closed, the equationfor c12 still contains ρ12(t) which depends on the bath variables. As a closure the local equilibriumapproximation may be done:
ρ12(t) ≈ c12(t)f(B).
Now we get a closed equation
∂
∂tc12 = iω0c12 + i
∫dBV (B)
∫ t
0
dt′e(iω0−LB)(t−t′)iV (B)c12(t′)f(B).
Now we note that the exponential e−LB(t−t′) being applied to V (B) produces V taken for the stateof the bath at time (t′ − t). This exponential does not act on the variables of the system (i.e. onc12). Its action on f(B) does not lead to any changes as long as integration (or taking the trace)over the bath states is considered, as we have discussed it in the previous paragraph. Interchangingthe sequence of integrating over the bath variables and over the time we get for the integral overbath variables ∫
dBV (B)e−LB(t−t′)V (B)f(B) = K(t− t′),
with K being the correlation function of the bath. We thus get
∂
∂tc12 = iω0c12 −
∫ t
0
dt′eiω0(t−t′)K(t− t′)c12(t′).
The local equilibrium approximation can be justified a-posteriori when putting down the higherorder perturbation, i.e. formally reiterating the equation once again, and seeing that the differencebetween ρ12(t) and c12(t)f(B) is of the third order in V .
For the Ohmic bath, when K is a δ-function, the result will be the same as in in the previousdiscussion, namely Eq.(149). For non-ohmic bath we can make a Markovian approximation. Onehas however to note that the main behavior of c12 is the oscillation at frequency ω0. Therefore the
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variable e−iω0t′c12(t′) is slow, i.e. hardly changes at the time scales of the order of ω−10 . If we assume
that K decays considerably at the time scale ω−10 , we can first the integral over t′ neglecting this
dependence, and then change the variable of integration and move its upper boundary to infinity.We then get:
∂
∂tc12 = iω0c12 −
[∫ t
0
dt′′K(t′′)
]c12(t),
which redefines the value of γ.
Bloch equations. Initially the problem of a dynamics of a two-level system in the contact withthe bath was posed for a nuclear spin in a magnetic field that fluctuates due to random currentsin the bath, but is equally applicable to any two-level approximation (say, optical transition). Westay within a spin-1/2 picture. This will lead to a slightly different notation than in the previousparagraphs.
The angular momentum σ = (σx, σy, σz) is expressed via Pauli matrices
σx = σ1 =
(0 11 0
), σy = σ2 =
(0 −ii 0
), σz = σ3 =
(1 00 −1
).
Here we note that the three Pauli matrices and the matrix I form a full basis, over which any 2×2matrix M can be expanded33:
M = m0I +m1σx +m2σy +m3σz. (150)
We remind thatσiσj = δij I + i
∑k
εijkσk
where εijk = 1 if ijk is a cyclic permutation of 123, εijk = −1 if ijk is a cyclic permutation of 321,
and εijk = 0 otherwise. Note that Trσj = 0, and therefore Trσiσj = 2δij. The matrix I can beconsidered as σ0, and Trσ0σj = 2δ0j. Therefore the coefficient mj in Eq.(150) is
mj =1
2TrσjM. (151)
The nucleus interacts with a constant magnetic field in z-direction, causing level shifts of±~ω0/2 of the corresponding levels, and the additional fluctuation field ~V/2 (the factor ~/2 isadded to cancel later!). The full two-level Hamiltonian is now
H = H0 + H1
with
H0 =
(HB − ~ω0
20
0 HB + ~ω0
2
)= HB I −
~ω0
2σz,
with I being a unit matrix, and
H1 = −~2
(Vxσx + Vyσy + Vzσz) = −~2Vσ
33 For those who don’t believe: consider the elements of the matrix as a string with four entries:(M11,M22,M12,M21), i.e. as a vector in a four-dimensional space. Note that the vectors (1, 1, 0, 0) (correspondingto a unit matrix), (0, 0, 1, 1), (0, 0,−i, i) and (1,−1, 0, 0) are linearly independent.
133
(note that compared to the previous discussion both levels are shifted down by −~ω0/2). Theequilibrium density matrix is
ρeq(B) =1
Ze−H/kBT with Z =
∫dBTre−H/kBT .
The time-dependent density matrix is ρ(t,B), and we will be interested in the deviations from theequilibrium, as described by the matrix δρ = ρ− ρeq. As in our previous example the bath will beconsidered classical.
Now we express the deviation of the density matrix from equilibrium via Pauli matrices:
δρ(t,B) =1
2
[m0(t,B)I +m1(t,B)σx +m2(t,B)σy +m3(t,B)σz
](here 1/2 is added to simplify the coefficients in the following equations). The coefficients mj(t,B)can be found by taking the trace
mj(t,B) = Tr(σjδρ(t,B))
as it follows from Eq.(151).The mean magnetization is obtained by integrating over the bath variables. The deviation of
the mean magnetization from its equilibrium value is
〈δσj; t〉 = 〈σj; t〉 − 〈σj〉eq =
∫mj(t,B)dB.
Since the equilibrium density matrix satisfies the equation LBρeq = 0, the deviation δρ satisfiesthe same Liouville equatuion as ρ(t), namely
∂
∂tδρ = − i
~[H, δρ] = −LBδρ+
i
2[ω0σz, δρ] +
i
2[Vσ, δρ],
where LB is the Liouvillian of the bath assumed classical. Now we considermj(t,B) = Tr(σjδρ(t,B)).
Note that σi commute with LB.Let us first consider the zeroth component. Due to Trσ0σj = δ0j, the contributions of the two
last terms on the right hand side of the last equation vanish. Therefore the equation for m0 reads
∂
∂tm0 = −LBm0, (152)
and decouples. For other components we get the result by explicitly writing e.g.:
∂
∂tmx = Tr
(σj∂
∂tδρ(t,B)
)= −LBmx +
iω0
2Tr (σx[σz, δρ]) +
i
2[Tr (σx[Vxσx, δρ])
+ Tr (σx[Vyσy, δρ]) + Tr (σx[Vzσz, δρ])] .
Now we note that
Trσi[σj, A] = Tr(σiσjA− σiAσj
)= TrAσiσj − TrAσjσi =
Tr(Aσiσj − Aσjσi
)= TrA[σi, σj] = 2iεijkTrAσk.
134
(The last expression in the first line is obtained by cyclic permutations, in the last expression inthe second line the commutation properties of Pauli matrices are used). The equation for mx thusreads:
∂
∂tmx = −LBmx + ω0my − Vymz + Vzmy.
The two other equations are
∂
∂tmy = −LBmy − ω0mx − Vzmx + Vxmz
and∂
∂tmz = −LBmz − Vxmy + Vymx.
All three equations for mx,my and mz can be put down as a single equation for the vector m =(mx,my,mz):
∂
∂tm = −LBm +Rm +Wm
with two matrices
R =
0 ω0 0−ω0 0 0
0 0 0
, W =
0 Vz −Vy−Vz 0 VxVy −Vx 0
.
The only difference with our previous case is that now the corresponding evolution equation is onefor the vector. All the methods of our discussion in the previous section can be used. Here −LBand R describe the evolution of the unperturbed system, and W describes the perturbation. Theformal solution of the inhomogeneous equation with inhomogeneity Wm reads:
m(t) = e−LBteRtm(0) +
∫ t
0
dt′e−LB(t−t′)eR(t−t′)Wm(t′). (153)
Now we start from a specific initial condition
δρ(0,B) = f(B)1
2
∑i=1,2,3
〈δσi; 0〉σi
which fixes the system state at the beginning on the average. We assume that this averagecorresponds to some valid spin state, i.e. that m0(0) = 0. According to Eq.(152) this valueof m0 stays constant during the following evolution. The initial density matrix decouples
mi(0,B) = f(B)〈δσi; 0〉.
From here on we can use both approaches described in the previous section: use our pet oscillatorbat, or rely on the local equilibrium approximation. We use here the second way.
We take our evolution equation
∂
∂tδρ = −LBδρ+
i
2[ω0σz, δρ] +
i
2[Vσ, δρ],
project it onto the Pauli matrices, and integrate over the bath variables. At the end we get:
∂
∂t〈δσj; t〉 = R〈δσj; t〉+
∫dBW (B)m(t,B).
135
Now we insert our formal solution, Eq.(153), for m(t). We require that the bath average offluctuating field vanishes (otherwise we should include this in ω0):
∫dBVj(B)f(B) = 0. Therefore
the mean of the matrix W is zero, and the term containing initial conditions drops out. Theremaining part is
∂
∂t〈δσj; t〉 = R〈δσj; t〉+
∫dBW
∫ t
0
dt′e−LB(t−t′)eR(t−t′)Wm(t′)
This equation is still exact. Now we use the local equilibrium approximation
mj(t, B) = 〈δσj; t〉f(B)
and get∂
∂t〈δσj; t〉 = R〈δσj; t〉+
∫ t
0
dt′K(t− t′)〈δσj; t′〉 (154)
with
K(t) =
∫dBW (0)e−LBteRtW (0)f(B).
Now we use our trace trick which allows for moving e−LBt all the way to the left, and get
K(t) =
∫dB[e−LBtW (0)]eRtW (0)f(B) = 〈W (t)eRtW (0)〉.
Now we need some matrix algebra. We get:
eRt =
cosω0t sinω0t 0− sinω0t cosω0t 0
0 0 1
.
The resulting matrix K(t) contains many cross products of different Vi. Assuming bilinear couplingto a Gaussian bath, we may only take the diagonal ones to be nonzero:
〈Vi(t)Vj(0)〉 = kj(t)δij.
In this case the correlation function is
K(t) =
ky(t) + kz(t) cosω0t kz(t) sinω0t 0−kz(t) sinω0t kx(t) + kz(t) cosω0t 0
0 0 (kx(t) + ky(t))) cosω0t
.
This memory kernel is time-dependent and gives rise to a non-markovian behavior. The integralequation Eq.(154) is in general hard to solve except for the case when ki(t) are delta-functions.Due to symmetry, one can now take kx(t) = ky(t) = k⊥(t), and try to find a correct Markovianapproximation for the case when this is not the case. To do so we introduce a new slow variableg(t) so that
〈δσj; t〉 = eRtg(t)
(this eliminates the main frequency, as we have done this in our previous example), and substitutethis in Eq.(154):
∂
∂teRtg(t) = ReRtg(t) +
∫ t
0
dt′K(t− t′)eRt′g(t′).
136
This gives us the equation for g(t):
∂
∂tg = e−Rt
∫ t
0
dt′K(t− t′)eRt′g(t′) =
∫ t
0
dt′′e−RtK(t′′)eR(t−t′′)g(t− t′′).
Assuming the the decay of K(t′′) is the fastest change in the integrand, we can replace g(t − t′′)by g(t) and extend the upper boundary of integration to infinity:
∂
∂tg ' e−Rt
[∫ ∞0
dt′′K(t′′)e−Rt′′]eRtg(t).
Now we substitute this in the expression for 〈δσj; t〉 and get the equation for this:
∂
∂t〈δσj; t〉 = R〈δσj; t〉+
[∫ ∞0
dt′′K(t′′)e−Rt′′]〈δσj; t〉.
Performing matrix algebra and integration we get the equations for the components of the spin:
∂
∂t〈σx〉 = ω1〈σy〉 −
1
T2
〈σx〉,
∂
∂t〈σy〉 = −ω1〈σx〉 −
1
T2
〈σy〉,
∂
∂t〈σz〉 = − 1
T1
(〈σz〉 − 〈σz〉eq)
(the equilibrium values 〈σx〉eq = 〈σy〉eq = 0). These are the Bloch equations (1946). Theparameters are:
ω1 = ω0 −∫ ∞
0
dtk⊥(t) sinω0t
1
T1
= 2
∫ ∞0
dtk⊥(t) cosω0t
1
T2
=
∫ ∞0
dt (kz(t) + k⊥(t) cosω0t) .
The first two equations can be rewritten as equations for 〈σx + iσy〉,∂
∂t〈σx + iσy〉 = −
(iω1 +
1
T2
)〈σx + iσy〉
and its complex conjugate. These can be solved exactly and give us e.g. the correlation function
〈σx(t)σx(0)〉 = cos(ω1t)e−t/T2 .
If the system is subject to the action of a weak external oscillating field with frequency ω, this,according to the Golden rule, will cause transitions between the states with the rates proportionalto ∫ t
0
dt cos(ωt)〈σx(t)σx(0)〉 =
∫ t
0
dt cos(ωt) cos(ω1t)e−t/T2 ,
i.e. will form a couple of Lorentzian lines centered at ±ω1 (one corresponds to absorption, anotherto emission) of the total shape
W ∝ 1
1 + (ω − ω1)2T 22
+1
1 + (ω + ω1)2T 22
,
where T−12 defines the line widths. ω1 defines the resonant frequencies, and T−1
1 the rate ofspontaneous transitions, defining the equilibration.
137
9.5 Master equation from density matrix approach.
Now we will use the methods which we used for two-level systems for deriving the (Pauli) masterequation, first in the second order in perturbation, and then discuss how a generalized masterequation may be derived, which is formally valid in all orders. Let
H = H0 + V ,
with V being the perturbation causing the transitions between the eigenstates of H0. We work inthe representation in which H0 is diagonal, because we want to describe the transitions betweenthe eigenstates of H0. We assume that V only causes the transitions, i.e. that its diagonal elementsvanish (otherwise they should be included in H0). The equation for the density matrix reads
d
dtρ(t) = − i
~[H, ρ(t)] = − i
~
([H0, ρ(t)] + [V , ρ(t)]
)= −(L0 + LV )ρ(t).
Since H0 is diagonal, the tetradic form for L0 is
(L0)jk,lm =i
~(Ej − Ek)δjlδkm,
and LV has a general tetradic form
(LV )jk,lm =i
~(Vjlδkm − δjlVmk).
We note that diagonal elements of L0 vanish, and that
(LV )jj,ll = 0. (155)
Moreover, ∑l
(LV )jk,ll =i
~∑l
(Vjlδkl − δjlVlk) = Vjk − Vjk = 0.
The tetradic form of the Liouville equation is
d
dtρjk(t) = −
∑lm
Ljk,lmρlm(t).
In the Master equation, we are only interested in the evolution of the diagonal elements ρjj.Now we take the Liouville equation
d
dtρ(t) = −L0ρ(t)− LV ρ(t),
consider the second term in the r.h.s. as an inhomogeneity, integrate it once:
ρ(t) = e−L0tρ(0)−∫ t
0
dt′e−L0(t−t′)LV ρ(t′),
and substitute this back into the Liouville equation:
d
dtρ(t) = −L0ρ(t)− LV e−L0tρ(0) + LV
∫ t
0
dt′e−L0(t−t′)LV ρ(t′).
138
This equation is still exact, but now prepared to get the second order perturbation result out ofit.
We want to obtain the equation for diagonal elements only, while the Liouville equation mixesdiagonal and non-diagonal elements:
d
dtρjj(t) = −(L0ρ(t))jj − (LV e
−L0tρ(0))jj +
(LV
∫ t
0
dt′e−L0(t−t′)LV ρ(t′)
)jj
.
We assume that ρjk(0) was prepared diagonal (for example as in a heat bath, or in a mi-crocanonical case, or simply by putting a system into a well-defined state i). This is called theassumption (or approximation) of the initial random phase. The first term vanishes because the
diagonal elements of L0 are zero. Then e−L0tρ(0) stays diagonal, and the second term in the r.h.s.drops our since (LV )jj,ll = 0. We get:
d
dtρjj(t) =
∑kl
(LV
∫ t
0
dt′e−L0(t−t′)LV
)jj,kl
ρkl(t′).
Now we assume that the density matrix is at all times dominated by its diagonal part, and replaceρkl by ρkkδkl (this assumption will be discussed later in more detail).
If we assume so, we simplify the summation, and, changing the time integration variable tos = t− t′, write
d
dtρjj(t) =
∑k
∫ t
0
ds(LV e
−L0sLV
)jj,kk
ρkk(t− s).
We denote (LV e
−L0sLV
)jj,kk
= Kjj,kk(s)
and getd
dtρjj(t) =
∫ t
0
ds∑k
Kjj,kk(s)ρjj(t− s).
The memory kernel satisfies the sum rule∑k
Kjj,kk(s) = 0.
To see this we note that(LV e
−L0sLV
)jj,kk
=∑k
∑cd
(LV e−L0s)jj,cd(LV )cd,kk =
∑cd
(LV e−L0s)jj,cd
∑k
(LV )cd,kk
and remember that∑
k(LV )cd,kk = 0. Using this sum rume we write:
d
dtρjj(t) =
∫ t
0
ds∑k 6=j
Kjj,kk(s)[ρkk(t− s)− ρjj(s)],
a form called a generalized master equation (due to the additional time-integration). We now needto evaluate Wk→j(s) = Kjj,kk(s) for k 6= j.
Kjj,kk(s) =∑abcd
(LV )jj,ab(esL0)ab,cd(LV )cdkk
=∑abcd
(i
~
)2
(Vjaδjb − Vbjδja)eis(Eb−Ea)/~δacδbd(Vckδdk − Vkdδck)
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Now one write this out explicitly, and gets at the end for k 6= j
Kjj,kk(s) =1
~2|Vjk|22 cos(ωjks)
with ωjk = (Ej − Ek)/~.Now we can make the Markovian approximation. The rate of change of the diagonal elements
is of the order of |V |2, and is slow. Therefore, if K(s) is strongly oscillating on this time scale wecan use the approximation∫ t
0
dsK(s)(ρkk(t− s)− ρjj(t− s)) ≈(∫ ∞
0
K(s)ds
)(ρkk(t)− ρjj(t)).
Now we getd
dtρjj(t) =
∑k 6=j
2
~2|Vjk|2
sinωjkt
ωjk(ρkk(t)− ρjj(t)).
Now we take t large and see that sinωjkt/ωjk behaves like a delta-function (of “width” 1/t) andwith integral over ω being π. We get the Pauli master equation with the rates given by the Goldenrule:
d
dtρjj(t) =
∑k 6=j
2π
~|Vjk|2δ(~ωjk)(ρkk(t)− ρjj(t)).
From this the heat bath Master equation follows by the procedure we have discussed before.
The projection operator approach. In our previous calculations we simply assumed that thedensity matrix is diagonal-dominated. Now we have to show this. On this way we present a generalmethod (which we wouldn’t have time to discuss in detail): The projection operator formalism(Nakajima, 1958, Zwanzig, 1960).
The diagonal part of any matrix is obtained by the action of a tetradic operator
Pjk,lm = δjkδjlδkm :
(PA)jk =∑
kl δjkδjlδkmAlm = δjkAjj. This is a projection operator since P 2 = P : A diagonal partof a diagonal matrix is the matrix itself. We use this operator to separate the diagonal and thenon-diagonal parts of the density matrix.
The master equation only involves the diagonal part, but the Liouville equation mixes thediagonal and the non-diagonal parts of the density matrix.
We separate the density matrix in it diagonal and off-diagonal parts:
ρ = ρd + ρod, ρd = P ρ, ρod = (I − P )ρ.
Acting on the both sides of the Liouville equation by P and I − P respectively, we get:
∂
∂tρd = −PLρd − PLρod
∂
∂tρod = −(I − P )Lρd − (I − P )Lρod
The first equation is essentially the one we have already considered.
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The following properties of Liouville operators involved will be of importance. Explicitly writingout PL0 we see that
PL0 = 0. (156)
To see this note that since H0 is diagonal, (L0)ab,mn = (Haaδamδbn − δamHbbδnb)
(PL0)jk,mn = δjk∑ab
δjaδkb(Haaδamδbn − δamHbbδnb)
= δjk(Hjjδjmδkn − δjmHkkδkn)
= δjk(Hjjδjmδkn −Hjjδjmδkn) = 0.
By a similar calculation we can show that L0P = 0 as well. We moreover note that the result ofaction of L0 on any diagonal matrix M vanishes:
L0M = 0. (157)
To see this we make the calculation
(L0M)ab =∑mn
(Haaδamδbn − δamHbbδnb)Mmmδmn
=∑m
(Haaδamδbm − δamHbbδmb)Mmm
= (Haaδba −Hbbδba)Maa = 0.
Similarly, ML0 = 0. We moreover note that for a diagonal matrix M it holds, that
PLM = 0. (158)
To see this we make a straightforward calculation:
(PLM)jk = δjk∑lm
δjlδkm∑p
Llm,ppMpp
=i
~δjk
i
~∑lm
δjlδkm∑p
(Hlpδmp − δlpHmp)Mpp
=i
~δjk∑p
(Hjpδkp − δjpHkp)Mpp =i
~δjk(HjkMkk −HkjMjj)
=i
~δjk(HjjMjj −HjjMjj) = 0.
Due to this last property the first term on the r.h.s. of the equation for ρd vanishes, and theequation reads:
∂
∂tρd = −PLρod.
Due to the first property L can be replaced by LV in this equation.Instead of finding ρod explicitly, in our previous discussion we have incorporated it by the
iteration procedure. Now we will find it formally solving the second equation. The formal solutionof the second equation (when considering the term with ρd as an inhomogeneity) is:
ρod(t) = e−(I−P )Ltρod(0)−∫ t
0
dt′e−(I−P )L(t−t′)(I − P )Lρd(t′).
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As before, we consider the initial conditions where the density matrix is diagonal, and thereforethe first term drops out. Now we can substitute the solution into the first equation and get
∂
∂tρd = PL
∫ t
0
dt′e−(I−P )L(t−t′)(I − P )Lρd(t′).
This equation is still exact, and is a closed equation for the diagonal part of the density matrix. Wenote that since PL0 = 0 (Eq.(156)), the combination PL outside of the integral can be changed forPLV . Now we note that Lρd = L0ρ
d + LV ρd. The expression L0ρ
d vanishes according to Eq.(157).Therefore the leftmost and the rightmost Liouville operators in our expression can be changed forLV . Now we change the time variable of integration to s = t− t′ and write:
d
dtρjj(t) =
∑k
∫ t
0
ds(LV e−(I−P )Ls(I − P )LV )jj,kkρkk(t− s)
=∑k
∫ t
0
dsKjj,kk(s)ρkk(t− s).
The new memory kernel K = LV e−(I−P )Ls(I − P )LV = LV e
−(I−P )(L0+LV )s(I − P )LV satisfies thesame sum rule as the old one, because the expression for it ends with LV . It is formally secondorder in LV but essentially contains all orders, as it gets clear when expanding the exponential.This gives us the possibility to build the theory which is formally exact up to any order, or toformulate a perturbation approach. Our aim here is modest, and this is to validate the assumptionmade in the previous discussion.
We note that the operator LV is itself first order in V and therefore expanding the exponentialwill give us the next (third) order in V :
Kjj,kk = (LV e−(I−P )L0s(I − P )LV )jj,kk +O(V 3).
Now we note that since PL0 = 0 so that the operator in the exponent can be changed for −L0s:
Kjj,kk = (LV e−L0s(I − P )LV )jj,kk +O(V 3).
Now we consider the operator standing between the two LV , namely e−L0s(I − P ). Formallyexpanding the exponential, we get the weighted sum of L0...L0(I−P ) and remember that L0P = 0
(and that I(I − P ) = I − P ). Therefore the operator e−L0s(I − P ) = e−L0s − P . On the otherhand
(PLV )ab,kk =∑ij
(P )ab,ij(LV )ij,kk =∑ij
δabδaiδbj(LV )ij,kk = δab(LV )ab,kk.
These martix elements are identically zero since if a 6= b the Kronecker delta vanisches, and ifa = b, then (PLV )aa,kk = 0 by virtue of Eq. (155). Therefore PLv = 0 and
Kjj,kk = (LV e−L0sLV )jj,kk +O(V 3),
i.e. the correction to what we obtained before is of the higher order in perturbation V .
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