232B Lecture NotesPath Integral IHitoshi Murayama

1 Feynmans Path Integral Formulation

Feynmans formulation of quantum mechanics using the so-called path inte-gral is arguably the most elegant. It can be stated in a single line:

xf , tf |xi, ti =Dx(t)eiS[x(t)]/h. (1)

The meaning of this equation is the following. If you want to know thequantum mechanical amplitude for a point particle at a position xi at timeti to reach a position xf at time tf , you integrate over all possible pathsconnecting the points with a weight factor given by the classical action foreach path. Hence the name path integral. This is it. Note that the positionkets form a complete set of basis, and knowing this amplitude for all x isenough information to tell you everything about the system. The expressionis generalized for more dimensions and more particles in a straightforwardmanner.

As we will see, this formulation is completely equivalent to the usualformulation of quantum mechanics. On the other hand, there are manyreasons why this expression is just beautiful.

First, the classical equation of motion comes out in a very simple way.If you take the limit h 0, the weight factor eiS/h oscillates very rapidly.Therefore, we expect that the main contribution to the path integral comesfrom paths that make the action stationary. This is nothing but the deriva-tion of EulerLagrange equation from the classical action. Therefore, theclassical trajectory dominates the path integral in the small h limit.

Second, we dont know what path the particle has chosen, even when weknow what the initial and final positions are. This is a natural generalizationof the two-slit experiment. Even if we know where the particle originates fromand where it hit on the screen, we dont know which slit the particle camethrough. The path integral is an infinite-slit experiment. Because you cantspecificy where the particle goes through, you sum them up.

1

Third, we gain intuition on what quantum fluctuation does. Around theclassical trajectory, a quantum particle explores the vicinity. The trajec-tory can deviate from the classical trajectory if the difference in the action isroughly within h. When a classical particle is confined in a potential well, aquantum particle can go on an excursion and see that there is a world outsidethe potential barrier. Then it can decide to tunnel through. If a classicalparticle is sitting at the top of a hill, it doesnt fall; but a quantum particlerealizes that the potential energy can go down with a little excursion, anddecides to fall.

Fourth, whenever we have an integral expression for a quantity, it is ofteneasier to come up with an approximation method to work it out, comparedto staring at a differential equation. Good examples are the perturbativeexpansion and the steepest descent method. One can also think of usefulchange of variables to simplify the problem. In fact, some techniques inquantum physics couldnt be thought of without the intuition from the pathintegral.

Fifth, the path integral can also be used to calculate partition functionsin statistical mechanics.

Sixth, the connection between the conservation laws and unitarity trans-formations become much clearer with the path integral. We will talk aboutit when we discuss symmetries.

There are many more but I stop here.Unfortunately, there is also a downside with the path integral. The ac-

tual calculation of a path integral is somewhat technical and awkward. Onemight even wonder if an integral over paths is mathematically well-defined.Mathematicians figured that it can actually be, but nonetheless it makes somepeople nervous. Reading off energy eigenvalues is also less transparent thanwith the Schrodinger equation.

Below, we first derive the path integral from the conventional quantummechanics. Then we show that the path integral can derive the conventionalSchrodinger equation back. After that we look at some examples and actualcalculations.

By the way, the original paper by Feynman on the path integral Rev.Mod. Phys. 20, 367-387 (1948) is quite readable for you, and I recommendit. Another highly recommended read is Feynmans Nobel Lecture. Youwill see that Feynman invented the path integral with the hope of replacingquantum field theory with particle quantum mechanics; he failed. But thepath integral survived and did mighty good in the way he didnt imagine.

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The derivation of path integral in this lecture note is based on J.J. Saku-rai, Modern Quantum Mechanics.

2 Physical Intuition

Take the two-slit experiment. Each time an electron hits the screen, there isno way to tell which slit the electron has gone through. After repeating thesame experiment many many times, a fringe pattern gradually appears on thescreen, proving that there is an interference between two waves, one from oneslit, the other from the other. We conclude that we need to sum amplitudesof these two waves that correspond to different paths of the electron. Nowimagine that you make more slits. There are now more paths, each of whichcontributing an amplitude. As you increase the number of slits, eventuallythe entire obstruction disappears. Yet it is clear that there are many pathsthat contribute to the final amplitude of the electron propagating to thescreen.

As we generalize this thought experiment further, we are led to concludethat the amplitude of a particle moving from a point xi to another pointxf consists of many components each of which corresponds to a particularpath that connects these two points. One such path is a classical trajectory.However, there are infinitely many other paths that are not possible classi-cally, yet contribute to the quantum mechanical amplitude. This argumentleads to the notion of a path integral , where you sum over all possible pathsconnecting the initial and final points to obtain the amplitude.

The question then is how you weight individual paths. One point is clear:the weight factor must be chosen such that the classical path is singled outin the limit h 0. The correct choice turns out to be eiS[x(t)]/h, whereS[x(t)] =

tfti dtL(x(t), x(t)) is the classical action for the path x(t) that

satisfies the boundary condition x(ti) = xi, x(tf ) = xf . In the limit h 0,the phase factor oscillates so rapidly that nearly all the paths would canceleach other out in the final amplitude. However, there is a path that makes theaction stationary, whose contribution is not canceled. This particular pathis nothing but the classical trajectory. This way, we see that the classicaltrajectory dominates the path integral in the h 0 limit.

As we increase h, the path becomes fuzzy. The classical trajectorystill dominates, but there are other paths close to it whose action is withinS ' h and contribute significantly to the amplitude. The particle does an

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excursion around the classical trajectory.

3 Propagator

The quantity

K(xf , tf ;xi, ti) = xf , tf |xi, ti (2)is called a propagator . It knows everything about how a wave function prop-agates in time, because

(xf , tf ) = xf , tf | =xf , tf |xi, tidxixi, ti|

=K(xf , tf ;xi, ti)(xi, ti)dxi. (3)

In other words, it is the Greens function for the Schrodinger equation.The propagator can also be written using energy eigenvalues and eigen-

states (if the Hamiltonian does not depend on time),

K(xf , tf ;xi, ti) = xf |eiH(tfti)/h|xi =n

xf |neiEn(tfti)/hn|xi

=n

eiEn(tfti)/hn(xf )n(xi). (4)

In particular, Fourier analyzing the propagator tells you all energy eigenval-ues, and each Fourier coefficients the wave functions of each energy eigen-states.

The propagator is a nice package that contains all dynamical informationabout a quantum system.

4 Derivation of the Path Integral

The basic point is that the propagator for a short interval is given by theclassical Lagrangian

x1, t+ t|x0, t = cei(L(t)t+O(t)2)/h, (5)Note that the expression here is in the Heisenberg picture. The base kets |x and the

operator x depend on time, while the state ket | doesnt.

4

where c is a normalization constant. This can be shown easily for a simpleHamiltonian

H =p2

2m+ V (x). (6)

The quantity we want is

x1, t+ t|x0, t = x1|eiHt/h|x0 =dpx1|pp|eiHt/h|x0. (7)

Because we are interested in the phase factor only at O(t), the last factorcan be estimated as

p|eiHt/h|x0= p|1 iHt/h+ (t)2|x0=

(1 i

h

p2

2mt i

hV (x0)t+O(t)

2

)eipx0/h

2pih

=12pih

expih

(px0 +

p2

2mt+ V (x0)t+O(t)

2

). (8)

Then the p-integral is a Fresnel integral

x1, t+ t|x0, t= dp

2piheipx1/hei(px0+

p2

2mt+V (x0)t+O(t)2)/h

=

m

2piihtexp

i

h

(m

2

(x1 x0)2t

V (x0)t+O(t)2). (9)

The quantity in the parentheses is nothing but the classical Lagrangian

times t by identifying x2 = (x1 x0)2/(t)2.Once Eq. (5) is shown, we use many time slices to obtain the propagator

for a finite time interval. Using the completeness relation many many times,

xf , tf |xi, ti =xf , tf |xN1, tN1dxN1xN1, tN1|xN2, tN2dxN2 dx2x2, t2|x1, t1dx1x1, t1|xi, ti. (10)

The expression is asymmetrical between x1 and x0, but the difference between V (x0)tand V (x1)t is approximately V t(x0 x1) ' V x(t)2 and hence of higher oder.

5

The time interval for each factor is t = (tf ti)/N . By taking the limitN , t is small enough that we can use the formula Eq. (5), and wefind

xf , tf |xi, ti = N1

i=1

dxieiN1

i=0L(ti)t/h, (11)

up to a normalization. (Here, t0 = ti.) In the limit N , the integralover positions at each time slice can be said to be an integral over all possiblepaths. The exponent becomes a time-integral of the Lagrangian, namely theaction for each path.

This completes the derivation of the path integral in quantum mechanics.As clear from the derivation, the overall normalization of the path integralis a tricky business.

A useful point to notice is that even matrix elements of operators can bewritten in terms of path integrals. For example,

xf , tf |x(t0)|xi, ti =dx(t0)xf , tf |x(t0), t0x(t0)x(t0), t0|xi, ti

=dx(t0)

tf>t>t0

Dx(t)eiS[x(t)]/hx(t0)t0>t>ti

Dx(t)eiS[x(t)]/h

=tf>t>ti

Dx(t)eiS[x(t)]/hx(t0), (12)

At the last step, we used the fact that an integral over all paths from xito x(t0), all paths from x(t0) to xf , further integrated over the intermediateposition x(t0) is the same as the integral over all paths from xi to xf . Thelast expression is literally an expectation value of the position in the form ofan integral. If we have multiple insertions, by following the same steps,

xf , tf |x(t2)x(t1)|xi, ti =tf>t>ti

Dx(t)eiS[x(t)]/hx(t2)x(t1). (13)

Here we assumed that t2 > t1 to be consistent with successive insertion ofpositions in the correct order. Therefore, expectation values in the pathintegral corresponds to matrix elements of operators with correct orderingin time. Such a product of operators is called timed-ordered Tx(t2)x(t1)defined by x(t2)x(t1) as long as t2 > t1, while by x(t1)x(t2) if t1 > t2.

In general,

xf , tf |TO1O2 On|xi, ti =tf>t>ti

Dx(t)O1O2 OneiS[x(t)]/h . (14)

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Another useful point is that EulerLagrange equation is obtained by thechange of variable x(t) x(t) + x(t) with xi = x(ti) and xf = x(tf ) heldfixed. A change of variable of course does not change the result of the integral,and we find

Dx(t)eiS[x(t)+x(t)]/h =Dx(t)eiS[x(t)]/h, (15)

and henceDx(t)eiS[x(t)+x(t)]/h

Dx(t)eiS[x(t)]/h =

Dx(t)eiS[x(t)]/h i

hS = 0. (16)

Recall (see Note on Classical Mechanics I)

S = S[x(t) + (t)] S[x(t)] = (L

x ddt

L

x

)x(t)dt. (17)

Therefore,

Dx(t)eiS[x(t)]/h i

h

(Lx ddt

L

x

)x(t)dt = 0. (18)

Because x(t) is an arbitrary change of variable, the expression must be zeroat all t independently,

Dx(t)eiS[x(t)]/h i

h

(L

x ddt

L

x

)= 0. (19)

Therefore, the EulerLagrange equation must hold as an expectation value,nothing but the Ehrenfests theorem.

5 Schrodinger Equation from Path Integral

Here we would like to see that the path integral contains all information weneed. In particular, we rederive Schrodinger equation from the path integral.

Let us first see that the momentum is given by a derivative. Startingfrom the path integral

xf , tf |xi, ti =Dx(t)eiS[x(t)]/h, (20)

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we shift the trajectory x(t) by a small amount x(t)+x(t) with the boundarycondition that xi is held fixed (x(ti) = 0) while xf is varied (x(tf ) 6= 0).Under this variation, the propagator changes by

xf + x(tf ), tf |xi, ti xf , tf |xi, ti = xfxf , tf |xi, tix(tf ). (21)

On the other hand, the path integral changes byDx(t)eiS[x(t)+x(t)]/h

Dx(t)eiS[x(t)]/h =

Dx(t)eiS[x(t)]/h iS

h. (22)

Recall that the action changes by (see Note on Classical Mechanics II)

S = S[x(t) + x(t)] S[x(t)]=

tftidt

(L

xx+

L

xx

)

=L

xx

xf

xi

+ tftidt

(L

x ddt

L

x

)x. (23)

The last terms vanishes because of the equation of motion (which holds asan expectation value, as we saw in the previous section), and we are left with

S =L

xx(tf ) = p(tf )x(tf ). (24)

By putting them together, and dropping x(tf ), we find

xfxf , tf |xi, ti =

Dx(t)eiS[x(t)]/h i

hp(tf ). (25)

This is precisely how the momentum operator is represented in the positionspace.

Now the Schrodinger equation can be derived by taking a variation withrespect to tf . Again recall (see Classical Mechanics II)

S

tf= H(tf ) (26)

after using the equation of motion. Therefore,

tfxf , tf |xi, ti =

Dx(t)eiS[x(t)]/hi

hH(tf ). (27)

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If

H =p2

2m+ V (x), (28)

the momentum can be rewritten using Eq. (25), and we recover the Schrodingerequation,

ih

txf , tf |xi, ti =

12m

(h

i

xf

)2+ V (xf )

xf , tf |xi, ti. (29)In other words, the path integral contains the same information as the

conventional formulation of the quantum mechanics.

6 Examples

6.1 Free Particle

Here we calculate the path integral for a free particle in one-dimension. Weneed to calculate

xf , tf |xi, ti =Dx(t)ei

m2x2dt/h, (30)

over all paths with the boundary condition x(ti) = xi, x(tf ) = xf .

6.1.1 The sloppy way

First, we do it somewhat sloppily, but we can obtain dependences on im-portant parameters nonetheless. We will come back to much more carefulcalculation with close attention to the overall normalization later on.

The classical path is

xc(t) = xi +xf xitf ti (t ti). (31)

We can write x(t) = xc(t) + x(t), where the left-over piece (a.k.a. quantumfluctuation) must vanish at the initial and the final time. Therefore, we canexpand x(t) in Fourier series

x(t) =n=1

an sinnpi

tf ti (t ti). (32)

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The integral over all paths can then be viewed as integrals over all an,Dx(t) = c

n=1

dan. (33)

The overall normalization factor c that depends on m and tf ti will be fixedlater. The action, on the other hand, can be calculated. The starting pointis

x = xc + x =xf xitf ti +

n=1

npi

tf tian cospin

tf ti (t ti). (34)

Because different modes are orthogonal upon t-integral, the action is

S =m

2

(xf xi)2tf ti +

m

2

n=1

1

2

(npi)2

tf tia2n. (35)

The first term is nothing but the classical action. Then the path integralreduces to an infinite collection of Fresnel integrals,

c

n=1

dan exp

[i

h

m

2

((xf xi)2tf ti +

n=1

1

2

(npi)2

tf tia2n

)]. (36)

We now obtain

xf , tf |xi, ti = cn=1

( ipih

m

2

1

2

(npi)2

tf ti

)1/2exp

[i

h

m

2

(xf xi)2tf ti

]. (37)

Therefore, the result is simply

xf , tf |xi, ti = c(tf ti) exp[i

h

m

2

(xf xi)2tf ti

]. (38)

The normalization constant c can depend only on the time interval tf ti,and is determined by the requirement that

dxxf , tf |x, tx, t|xi, ti = xf , tf |xi, ti. (39)And hence

c(tf t)c(t ti)2piih(tf t)(t ti)

m(tf ti) = c(tf ti). (40)

This argument does not eliminate the possible factor ei0(tfti), which would cor-respond to a zero point in the energy h0. Mahiko Suzuki pointed out this problem tome. Here we take the point of view that the normalization is fixed by comparison to theconventional method.

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We therefore find

c(t) =

m

2piiht, (41)

recovering Eq. (2.5.16) of Sakurai precisely.In order to obtain this result directly from the path integral, we should

have chosen the normalization of the measure to be

Dx(t) =

m

2piih(tf ti)n=1

m

2piih(tf ti)npi

2dan. (42)

6.1.2 The careful way with discretized time

Now we do it much more carefully with close attention to the overall normalization. Thepath integral for the time interval tf ti is divided up into N time slices each witht = (tf ti)/N ,

K =( m

2piiht

)N/2 N1n=1

dxneiS/h, (43)

where

S =12m

N1n=1

(xn+1 xn)2t

. (44)

The point is that this is nothing but a big Gaussian (to be more precise, Fresnel) integral,because the action is quadratic in the integration variables x1, , xN1. To make thispoint clear, we rewrite the action into the following form:

S =12m

t(x2N + x

20)

m

t(xN1, xN2, xN3, , x2, x1)

xN00...0x0

+

12m

t

(xN1, xN2, xN3, , x2, x1)

2 1 0 0 01 2 1 0 00 1 2 0 0...

......

. . ....

...0 0 0 2 10 0 0 1 2

xN1xN2xN3

...x2x1

(45)

Comapring to the identity Eq. (111), Nn=1

dxne 12xTAxxTy = (2pi)N/2(detA)1/2e+

12y

TA1y (46)

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we identify

A = ih

2mt

1 1/2 0 0 01/2 1 1/2 0 0

0 1/2 1 0 0...

......

. . ....

...0 0 0 1 1/20 0 0 1/2 1

(47)

This is nothing but the matrix KN1 in Eq. (111) with a = 1/2 up to a factor of ih 2mt .One problem is that, for a = 1/2, + = = 1/2 and hence Eq. (116) is singular.Fortunately, it can be rewritten as

detKN1 =N+ N+ =

N1+ +

N2+ + + ++N2 +N1 = N

(12

)N1. (48)

Therefore,

detA =(

2miht

)N1N

12N1

= N( miht

)N1. (49)

Therefore, the prefactor of the path integral is( m2piiht

)N/2(2pi)(N1)/2(detA)1/2

=( miht

)N/2(2pi)1/2N1/2

( miht

)(N1)/2=( m

2piihNt

)1/2=(

m

2piih(tf ti))1/2

. (50)

The exponent is given by

i

h

12im

ht(x2N + x

20) +

12

( mt

)2(xN , 0, 0, , 0, x0)A1

xN00...0x0

(51)

We need only (1, 1), (N 1, N 1), (1, N 1), and (N 1, 1) components of A1.

(A1)1,1 = (A1)N1,N1 =iht2m

detKN2detKN1

=ihtm

N 1N

. (52)

On the other hand,

(A1)1,N1 = (A1)N1,1

12

=iht2m

1detKN1

(1)N2det

1/2 1 1/2 0

0 1/2 1 0...

......

. . ....

0 0 0 1/2

=iht2m

2N1

N

12N2

=ihtNm

. (53)

We hence find the exponent

i

h

12m

t(x2N + x

20) +

12

( mt

)2(xN , 0, 0, , 0, x0)A1

xN00...0x0

=

i

h

12m

t(x2N + x

20) +

12

(im

ht

)2(ihtm

N 1N

(x2N + x20) + 2

ihtNm

xNx0

)=

i

h

12m

t1N

(x2N + x20)

12

im

hNt2xNx0

=i

h

m

2(xN x0)2

Nt

=i

h

m

2(xf xi)2tf ti . (54)

Putting the prefactor Eq. (50) and the exponent Eq. (81) together, we find the propagator

K =(

m

2piih(tf ti))1/2

eihm2

(xfxi)2tfti . (55)

We recovered Eq. (2.5.16) of Sakurai without any handwaving this time!

6.2 Harmonic Oscillator

Now that we have fixed the normalization of the path integral, we calculatethe path integral for a harmonic oscillator in one-dimension. We need tocalculate

xf , tf |xi, ti =Dx(t)ei

(m2 x2 12m2x2)dt/h, (56)

over all paths with the boundary condition x(ti) = xi, x(tf ) = xf . Theclassical path is

xc(t) = xisin(tf t)sin(tf ti) + xf

sin(t ti)sin(tf ti) . (57)

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The action along the classical path is

Sc = tfti

(m

2x2 1

2m2x2

)dt =

1

2m

(x2i + x2f ) cos(tf ti) 2xixf

sin(tf ti) .(58)

The quantum fluctuation around the classical path contributes as theO(h) correction to the amplitude, relative to the leading piece eiSc/h. Weexpand the quantum fluctuation in Fourier series as in the case of free particle.Noting that the action is stationary with respect to the variation of x(t)around xc(t), there is no linear piece in an. Because different modes areorthogonal upon t-integral, the action is

S = Sc +n=1

m

2

((npi)2

tf ti 2(tf ti)

)1

2a2n. (59)

Therefore the path integral is an infinite number of Fresnel integrals over anusing the measure in Eq. (42)

xf , tf |xi, ti = eiSc/h

m

2piih(tf ti)

n=1

m

2piih(tf ti)npi

2

dan exp

[i

h

m

2

((npi)2

tf ti 2(tf ti)

)1

2a2n

]

= eiSc/h

m

2piih(tf ti)n=1

1 ((tf ti)npi

)21/2 . (60)Now we resort to the following infinite product representation of the sinefunction

n=1

(1 x

2

n2

)=

sin pix

pix. (61)

We find

xf , tf |xi, ti = eiSc/h

m

2piih(tf ti)

(tf ti)sin(tf ti)

= eiSc/h

m

2piih sin(tf ti) (62)

This agrees with the result from the more conventional method as given inSakurai Eq. (2.5.18).

One can obtain exactly the same result using the discretized time asshown in the Appendix.

14

7 Partition Function

The path integral is useful also in statistical mechanics to calculate partitionfunctions. Starting from a conventional definition of a partition function

Z =n

eEn , (63)

where = 1/(kBT ), we rewrite it in the following way using completenessrelations in both energy eigenstates and position eigenstates.

Z =n

n|eH |n =dxn

n|xx|eH |n

=dxn

x|eH |nn|x =dxx|eH |x. (64)

The operator eH is the same as eiHt/h except the analytic continuationt i = ih. Therefore, the partition function can be written as a pathintegral for all closed paths, i.e., paths with the same beginning and endpoints, over a time interval ih. For a single particle in the potentialV (x), it is then

Z =Dx() exp

1h

h0

d

m2

(x

)2+ V (x)

. (65)The position satisfies the periodic boundary condition x(h) = x(0).

For example, in the case of a harmonic oscillator, we can use the resultin the previous section

xf , tf |xi, ti=

m

2piih sin(tf ti) exp[i

h

1

2m

(x2i + x2f ) cos(tf ti) 2xixf

sin(tf ti)],

(66)

and take xf = xi = x, tf ti = ih

x|eH |x =

m

2pih sinh hexp

[1h

1

2mx2

2 cosh h 2sinh h

]. (67)

15

Then integrate over x,

Z =dxx|eH |x

=

m

2pih sinh h

pih

2

m

sinh h

2 cosh h 2

=

1

4 sinh2 h/2

=1

2 sinh h/2

=eh/2

1 eh=

n=0

e(n+12

)h. (68)

Of course, we could have redone the path integral to obtain the same result.The only difference is that the classical path you expand the action aroundis now given by sinh rather than sin.

Let us apply the path-integral representation of the partition function toa simple harmonic oscillator. The expression is

Z =Dx() exp 1

h

h0

d

m2

(x

)2+m

22x2

. (69)We write all possible paths x() in Fourier series as

x() = a0 +n=1

an cos2pin

h +

n=1

bn sin2pin

h, (70)

satisfying the periodic boundary condition x(h) = x(0). Then the exponentof the path integral becomes

1h

h0

d

m2

(x

)2+m

22x2

= m

2

a20 + 12n=1

(2pinh

)2+ 2

(a2n + b2n) . (71)

16

The integration over all possible paths can be done by integrating over theFourier coefficients an and bn. Therefore the partition function is

Z = cda0

n=1

dandbn expm

2

2a20 + 12n=1

(2pinh

)2+ 2

(a2n + b2n) .

(72)This is an infinite collection of Gaussian integrals and becomes

Z = c

2pi

m2

n=1

4pi

m

(2pinh

)2+ 2

1 . (73)Here, c is a normalization constant which can depend on because of nor-malization of Fourier modes, but not on Lagrangian parameters such as m or. Now we use the infinite product representation of the hyperbolic function

n=1

(1 +

x2

n2

)=

sinh pix

pix. (74)

The partition function Eq. (82) can be rewritten as

Z = c

2pi

m2

n=1

h2

pimn2

1 + ( h2pin

)21

= c2

sinh h/2

= ceh/2

1 eh , (75)

where c is an overall constant, which does not depend on , is not importantwhen evaluating various thermally averages quantities. If you drop c, thisis exactly the partition function for the harmonic oscillator, including thezero-point energy.

8 Schrodinger Field

We now apply the path integral to field theory. As a starter, we computethe partition function of the non-relativistic identical bosons. This calcula-tion shows non-trivially that the quantized Schrodinger field theory indeed

17

contains multi-body states automatically and is also useful from a practicalpoint of view. It is also meant to be a demonstration that path integral infield theory makes sense.

The path integral in the particle quantum mechanics is given by thesummation over all possible paths in the configuration space xi(t). In the fieldtheory, (~x) is the canonical coordinate, and it can follow various paths(~x, t). Therefore, the path integral in field theory is a summation over allpossible field configurations (~x, t). This discussion defines the path integral

D(~x, t)D(~x, t)eiS/h, (76)

where the action is that of the field theory, such as

S =d~xdt

(ih +

h2

2m

)(77)

in the free case.To calculate the partition function, we go to the imaginary time t = i ,

Z =D(~x, )D(~x, ) exp

[1h

h0

dd~x

(h +

h22m

)].

(78)We impose the periodic boundary condition (~x, t + h) = (~x) for =1/kT .

Because of the periodic boundary condition in and also in space due tothe box normalization, we can expand in Fourier series both in the imaginarytime as well as in space,

(~x, ) =1

L3/2~p,n

z~p,nei~p~x/he2piin/h (79)

and simiarly for

(~x, ) =1

L3/2~p,n

z~p,nei~p~x/he2piin/h. (80)

Then the exponent in the partition function Eq. (97) is

1h

h0

dd~x

(h +

h22m

)=

~p,n

(2piin+

~p2

2m

)z~p,nz~p,n.

(81)

18

Therefore, the path integral is simply a product of many many Gaussianintegrals

Z =~p,n

dz~p,ndz~p,ne

(2piin+~p2/2m)z~p,n

z~p,n =~p,n

pi

2piin+ ~p2/2m. (82)

Now we use the infinite product representation of the hyperbolic functions

n=1

(1 +

x2

n2

)=

sinhpix

pixn=1

(1 +

x2

(2n 1)2)

= coshpix

2. (83)

We use the first identity here. The partition function Eq. (82) can be rewrit-ten as

Z =~p

pi

~p2/2m

n=1

pi

(2pin)2 + (~p2/2m)2

=~p

pi

~p2/2m

n=1

pi

(2pin)2

(1 +

(~p2/2m/2pi)2

n2

)1

=~p

( n=1

pi

(2pin)2

)pi

~p2/2m

~p2/4m

sinh ~p2/4m

= c~p

2

sinh ~p2/4m

= c~p

e~p2/4m

1 e~p2/2m . (84)

where c is an (infinite) overall constant, which is not important when evaluat-ing various thermally averages quantities. This is a grand partition functionwhich contains summation over all possible number of particles but with nochemical potential. One small difference from the conventional calculation isthat the path integral automatically includes the zero-point energy ~p2/4m,which results in the exponential factor in the numerator. The reason whythere is this zero-point energy is because the Fock space is a collection of(infinite number of) harmonic oscillators and each harmonic oscillator hasthe zero-point energy. But the zero-point energy does not lead to any phys-ical consequences (while the zero-point fluctuation does) and we can always

19

shift the energy of the system by an infinite constant~p ~p

2/4m to removethe zero-point energy in the expression.

The inclusion of the chemical potential is obvious. Because the numberoperator is N =

d~x(~x)(~x) and the grand canonical emsenble is summed

with a factor of e(EN), the path integral Eq. (97) is modified to

Z =D(~x, )D(~x, )

exp

[1h

h0

dd~x

(h +

h22m

)]

. (85)

The result of the path integral is then

Z = c~p

e(~p2/2m)/2

1 e(~p2/2m) . (86)

Note that Z = e with = pV = F N for the grand partitionfunction. Namely,

= kT~p

lne(~p

2/2m)/2

1 e(~p2/2m)

=~p

[1

2

(~p2

2m

)+ kT ln(1 e(~p2/2m))

]. (87)

The thermodynamic quantities are related by

F = U TS, (88) = F N, (89)

and have the following differentials

dU = TdS pdV + dN, (90)dF = SdT pdV + dN, (91)d = SdT pdV Nd. (92)

Therefore, the number is given by

N =

T,V

= ~p

[1

2+e(~p2/2m)1 e(~p2/2m)

]=~p

[1

2+

1

e(~p2/2m) 1].

(93)

20

Apart from the zero-point piece 1/2, this is the familiar expression for theoccupation number for Bose-Einstein distribution.

The thermal average energy U is given by the Legendre transform U = + TS +N with S = /T |V,. We obtain

U =~p

~p2

2m

(1

2+

1

e(~p2/2m) 1). (94)

This must be a famililar expression to you, again except the zero-point energyterm that can be dropped without changing physical content.

9 Non-relativistic Fermions

In the case of fermions, their fields are supposed to satisfy canonical anti-commutation relation. In the classical limit, they become anti-commutingnumbers which are called Grassmann odd. Clearly, a Grassmann-oddclassical Fermi field is not an observable as no measurement would yield aGrassmann-odd number as a result. However, this issue is not completelyacademic because the path integral requires the classical integration variable:we need to integrate over Grassmann-odd Fermi field (~x, t).

What is the definition of an integral over a Grassmann-odd number? Ido not go into details, but rather give you a consistent definition for our use:the integration is the same as the differentiation! To make it explicit whatI mean, let me take one Grassmann-odd number (no ~x, t dependence).Since it anti-commutes with itself, {, } = 0, we find 2 = 0. Namely, anyhigher powers than one give vanishing results. Therefore, any function of can always be expanded as

f() = f(0) + f (0). (95)

The definition of the Grassmann integral isdf() = f (0). (96)

One could have tried to take f(0) instead as the definition of the integralrather than f (0); however, it would violate the shift invariance of the integra-tion variable = +. We would like to retain this property to be anal-ogous to the bosonic integral which satisfies

dxf(x) =

dxf(x + a).

21

To see this, let us imagine defining the integral to pick up the piece f(0) in-stead of f (0). Then

df() = f(0) by definition. However,

df( + )

for a constant would bedf( + ) =

d(f(0) + ( + )f (0)). Since

this (wrong) definition would pick all independent pieces, the result wouldbe f(0) + f (0), which is different from the result without the shift. On theother hand, picking up the piece proportional to yields f (0) in both cases.

The partition function is given by the same expression as the bosonic case

Z =D(~x, )D(~x, ) exp

[1h

h0

dd~x

(h +

h22m

)].

(97)However, there is another peculiarity with Fermi field: we need to take theanti-periodic boundary condition (h) = (0) instead of the periodicboundary condition. This can be shown by carefully working out the traceTreH in terms of fermionic path integral (see the appendix). We do notgo into the detail of the derivation here, but proceed with the anti-periodicboundary condition. Fourier expansion is then modified

(~x, ) =1

L3/2~p,n

z~p,nei~p~x/hepii(2n1)/h. (98)

The partition function is

Z =~p,n

dz~p,ndz~p,ne

(pii(2n1)+(~p2/2m))z~p,n

z~p,n

=~p,n

[pii(2n 1) +

(~p2

2m

)], (99)

where we used the rule for the Grassmann integrals after Taylor expandingthe exponential up to order z~p,nz~p,n for each ~p, n. This time we use theformula

n=1

(1 +

x2

(2n 1)2)

= coshpix

2, (100)

and find

Z = c~p

cosh

2

(~p2

2m

)= c

~p

(e(~p

2/2m) + 1)e(~p

2/2m)/2. (101)

22

The energy expectation value is hence

E =

lnZ

=~p

~p2

2m

[1

2+

1

e(~p2/2m) + 1

], (102)

which must be familiar to you except the zero-point energy contribution. Itis interesting to note that the zero-point energy is negative for fermions.

A Path Integrals in Phase Space

As we saw above, the normalization of the path integral is a somewhat tricky issue. Thesource of the problem was when we did p integral in Eq. (9), that produced a singularprefactor

m/2piiht. On the other hand, if we do not do the p integral, and stop at

x1, t+ t|x0, t =

dp

2piheipx1/hei(px0+

p2

2mt+V (x0)t+O(t)2)/h

=

dp

2pihei(p

x1x0t p

2

2mV (x0))t, (103)

we do not obtain any singular prefactor except 1/2pih for every momentum integral. Thepath integral can then be written as

xf , tf |xi, ti =Dx(t)Dp(t)eiS[x(t),p(t)]/h, (104)

where the action is given in the phase space

S[x(t), p(t)] = tfti

(pxH(x, p)) dt. (105)

The expression (104) is an integral over all paths in the phase space (x, p). It is understoodthat there is one more p integral than the x integral (albeit both infinite) because of thenumber of times the completeness relations are inserted.

Using the path integral in the phase space, we can work out the precise normalizationof path integrals up to numerical constants that depend on t, but not on m, or any otherparameters in the Lagrangian. For instance, for the harmonic oscillator, we expand

x(t) = xc(t) +

2

tf tin=1

xn sinnpi

tf ti (t ti), (106)

p(t) = pc(t) +

1

tf ti p0 +

2tf ti

n=1

pn cosnpi

tf ti (t ti). (107)

If you have both bosons and fermions with the same mass, the zero-point enegiesprecisely cancel. This is what happens with supersymmetry.

23

The action is then

S = Sc p20

2m 1

2

n=1

(pn, xn)

(1m

npitfti

npitfti m

2

)(pnxn

), (108)

where Sc is the action for the classical solution Eq. (58). Using the integration volumeDx(t)Dp(t) = c(tf ti)

dp02pih

n=1

dxndpn2pih

, (109)

we find the path integral to be

xf , tf |xi, ti = eiSc/hc(tf ti) 12pih(

2pimhi

)1/2n=1

12pih

(2pimhi

)1/2(2pihim

12 (npi/(tf ti))2

)1/2. (110)

Up to an overall constant c(tfti) that depends only on tfti, both m and dependencesare obtained correctly.

B A Few Useful Identities

To carry out the integral, a few identities would be useful. For N 1-dimensional columnvectors x, y, and a symmetric matrix A, N

n=1

dxne 12xTAxxTy = (2pi)N/2(detA)1/2e+

12y

TA1y (111)

It is easy to prove this equation first without the linear term y by diagonalizatin the matrixA = OTDO, where O is a rotation matrix and D = diag(1, , N ) has N eigenvalues.Under a rotation, the measure

dxn is invariant because the Jacobian is detO = 1.

Therefore, after rotation of the integration variables, the integral is

Nn=1

dxne

nx2n/2 =

Nn=1

2pin

= (2pi)N/2(Nn=1

n)1/2 (112)

The product of eigenvalues is nothing but the determinant of the matrix detA = detOTdetDdetO =Nn=1 n, and hence N

n=1

dxne 12xTAx = (2pi)N/2(detA)1/2. (113)

In the presence of the linear term, all we need to do is to complete the square in theexponent,

12xTAx xTy = 1

2(xT yTA1)A(xA1y) + 1

2yTA1y, (114)

24

and shift x to eliminate A1y, and diagonalize A. This proves the identity Eq. (111).For a N N matrix of the form

KN =

1 a 0 0 0a 1 a 0 00 a 1 0 0...

......

. . ....

...0 0 0 1 a0 0 0 a 1

, (115)

one can show

detKN =N+1+ N+1+ , =

12

(1

1 4a2). (116)

This is done by focusing on the top two-by-two block. The determinant that picksthe (1, 1) element 1 comes with the determinant of remaining (N 1) (N 1) block,namely detKN1. If you pick the (1, 2) element a, it is necessary to also pick the (2, 1)element which is also a and the rest comes from the remaining (N 2) (N 2) block,namely detKN2. Therefore, we find a recursion relation

detKN = detKN1 a2detKN2. (117)

We rewrite this recursion relation as(detKN

detKN1

)=(

1 a21 0

)(detKN1detKN2

). (118)

The initial conditions are detK1 = 1, detK2 = 1 a2. Using the recursion relationbackwards, it is useful to define detK0 = 1. We diagonalize the matrix(

1 a21 0

)=(+ 1 1

)(+ 00

)(1 1 +

)1

+ . (119)

Hence,(detKN

detKN1

)=

(+ 1 1

)(+ 00

)N1( 1 1 +

)1

+

(detK1detK0

)=

1+

(+ 1 1

)(N1+ 0

0 N1

)(1 1 + +

)=

1+

(N+

N

N1+ N1

)(+

)=

1+

(N+1+ N+1N+ N

). (120)

25

C Harmonic Oscillator with Discretized Time

We perform path integral for a harmonic oscillator again using discretized time. This way,we can pay closer attention to the normalization by going through the same steps as for afree particle using discretized time slices. The action is

S =12m

N1n=1

(xn+1 xn)2t

12m2

N1n=1

x2n 12m2

(12x20 +

12x2N

). (121)

The last term is there to correctly acccount for all N 1 time slices for the potentialenergy term V dt, but to retain the symmetry between the initial and final positions,we took their average. Once again, this is nothing but a big Fresnel (complex Gaussian)integral:

S =12m

t(x2N + x

20)

m

t(xN1, xN2, xN3, , x2, x1)

xN00...0x0

+

12m

t

(xN1, xN2, xN3, , x2, x1)

2 1 0 0 01 2 1 0 00 1 2 0 0...

......

. . ....

...0 0 0 2 10 0 0 1 2

xN1xN2xN3

...x2x1

12m2(xN1, xN2, xN3, , x2, x1)

xN1xN2xN3

...x2x1

t 1

4m2(x2N + x

20)t

(122)

Comapring to the identity Eq. (111), Nn=1

dxne 12xTAxxTy = (2pi)N/2(detA)1/2e+

12y

TA1y (123)

26

we identify

A =i

hm2tIN1 i

h

2mt

1 1/2 0 0 01/2 1 1/2 0 0

0 1/2 1 0 0...

......

. . ....

...0 0 0 1 1/20 0 0 1/2 1

(124)

This is nothing but the matrix KN1 in Eq. (111) up to an overall factor of

ih

2mt

+i

hm2t =

m

iht(2 2(t)2) (125)

with

a =12

2m/t2mt m2t

=1

2 2(t)2 . (126)

Using Eq. (116),

=12

(1

1 4

(2 2(t)2)2)

(127)

Therefore,

detA =( miht

)N1(2 2(t)2)N1

N+ N+ . (128)

Therefore, the prefactor of the path integral is( m2piiht

)N/2(2pi)(N1)/2(detA)1/2

=

m

2piiht(2 2(t)2)(N1)/2(N+ N

+

)1/2. (129)

Now we use the identity

limN

(1 +

x

N

)N= ex. (130)

We find in the limit of t = (tf ti)/N and N ,

=12

(1

1 4

(2 2(t)2)2)

=12

(1 it) +O(N3), (131)

and hence

N+ N+

12N

ei(tfti) ei(tfti)it

+O(N1)

=N

2N1sin(tf ti)(tf ti) +O(N

1). (132)

27

In the same limit,(2 2(t)2)(N1)/2 = 2(N1)/2(1 1

22(t)2

)(N1)/2= 2(N1)/2 +O(N1). (133)

Therefore, the prefactor Eq. (129) ism

2piiht2(N1)/2

(N

2N1sin(tf ti)(tf ti)

)1/2=

m

2piih sin(tf ti) . (134)

The exponent is given by

i

h

12im

ht(x2N + x

20)

i

h

14m2(x2N + x

20) +

12

(im

ht

)2(xN , 0, 0, , 0, x0)A1

xN00...0x0

(135)

We need only (1, 1), (N 1, N 1), (1, N 1), and (N 1, 1) components of A1.(A1)1,1 = (A1)N1,N1

=ihtm

(2 2(t)2)1 detKN2detKN1

=ihtm

(2 2(t)2)1N1+ N1N+ N

(136)

On the other hand,

(A1)1,N1 = (A1)N1,1

=ihtm

(2 2(t)2)1 1detKN1

(1)N2det

a 1 a 00 a 1 0...

......

. . ....

0 0 0 a

=ihtm

(2 2(t)2)1 + N+ N

aN2

ihtm

212N1

N

(tf ti)sin(tf ti)

12N2

=ihtNm

(tf ti)sin(tf ti) . (137)

We hence find the exponent

i

h

(12im

ht 1

4m2

)(x2N + x

20) +

12

( mt

)2(xN , 0, 0, , 0, x0)A1

xN00...0x0

28

=i

h

(12im

ht 1

4m2

)(x2N + x

20) +

12

(im

ht

)2ihtm(

(2 2(t)2)1N1+ N1N+ N

(x2N + x20) +

2N

(tf ti)sin(tf ti)xNx0

)(138)

Taking the limit N , the second term is obviously regular, and the last term in theparentheses is also:

12

(im

ht

)2ihtm

2N

(tf ti)sin(tf ti)xNx0 =

i

h

m

sin(tf ti)xNx0. (139)

The other terms are singular and we need to treat them carefully.

i

h

12m

t(x2N + x

20)

12im

ht(2 2(t)2)1

N1+ N1N+ N

(x2N + x20)

=im

2ht

(1 (2 2(t)2)1

N1+ N1N+ N

)(x2N + x

20). (140)

We need O(N1) term in the parentheses. The factor (2 2(t)2)1 = 21 + O(N2)is easy. Using Eq. (131), the next factor is

N1+ N1N+ N

=2N

2N1(1 + it)N1 (1 it)N1 +O(N2)

(1 + it)N (1 it)N +O(N2) (141)

Therefore, dropping O(N2) corrections consistently,

1 (2 2(t)2)1N1+ N1N+ N

=(1 + it)N (1 it)N (1 + it)N1 + (1 it)N1

(1 + it)N (1 it)N

= it(1 + it)N1 + (1 it)N1

(1 + it)N (1 it)N

= i(tf ti)

N

2 cos(tf ti)2i sin(tf ti) =

1N

(tf ti) cos(tf ti)sin(tf ti) . (142)

Collecting all the terms, the exponent is

i

hSc = i

h

m

sin(tf ti)xNx0 +im

2ht1N

(tf ti) cos(tf ti)sin(tf ti) (x

2N + x

20)

=i

h

m

2(x2N + x

20) cos(tf ti) 2xNx0

sin(tf ti) (143)

which is the same as the classical action Eq. (58).

29

D Anti-periodic Boundary Condition

Here is the derivation why fermion fields must satisfy the anti-periodic boundary condition.It turns out it is a somewhat long story.

Let us start with a Lagrangian

L = ih h. (144)(t) depends only on time but not on space. The classical EulerLagrange equation is

ih = 0, (145)with an obvious solution

(t) = (0)eit. (146)

To see what this Lagrangian describes quantum mechanically, note that the can be re-garded as the canonical coordinate, while its canonically conjugate momentum is L/ =i. Therefore, we find the canonical anti-commutation relation

{,} = 1. (147)This defines a two-dimensional Hilbert space spanned by

|0, |1 = |0, (148)where the state |0 is defined by the requirement |0 = 0. The Hamiltonian of the systemis

H = h( 1

2

). (149)

(The zero-point energy was added to be consistent with the result from the path integral.)In other words, this is a two-state system with energies 12 h.

Now we need to find the decomposition of unity analogous to |xdxx| = 1 to be

used to derive the path integral for . Note that the conventional path integral wasderived using the position eigenstates x|q = q|q. Likewise, we need the eigenstates ofthe operator . It can be obtained in analogy to coherent states

| = |, (150)where the coherent state | is defined by

| = |0 |1. (151)Note that is a Grassman number, and anti-commutes with the annihilation operator even though it is not an operator. That guarantees that

| = |0 |1 = |1 = |0 = |, (152)where we used 2 = {, }/2 = 0 at the very last step. This coherent state, however, isnot normalized, because

| = 1 + = e, (153)

30

where the last expression is more convenient for later purposes. Now it is easy to see thatthe following decomposition of unity holds:

1 =dd|e|. (154)

Starting from the r.h.s., we can check thatdd|e| =

dd(|0 |1)e(0| 1|)

=dd (|00|+ |11|)

= |00|+ |11|. (155)Here and below, we stick to the convention that the integrals are done from right to left,so that

dd =

d = 1.

We need another quantity, |eiHt/h|, analogous to x|eiHt/h|x for small t forthe ordinary path integral as in Eq. (9). This can be calculated easily

|eiHt/h| = |eit| = |eit| = eite. (156)for small t.

Now we can keep inserting the decomosition of unity between the initial and the finalstates.

(t)|eiHt/h|(0)= lim

N

N |eiHt/h|N1eN1N1dN1dN1N1|eiHt/h|N2

eN2N2dN2dN2 e

11d1d11|eiHt/h|0

= limN

ei

NN1e

NN1e

N1N1dN1dN1e

iN1N2eN1N2

eN2N2dN2dN2 e

11d1d1e

i10e10

=d(t)d(t) exp

[i

h

dt(ih h)

]. (157)

Note that (t) = N and (0) = 0 are not integrated over. This defines the path integral.Now comes the question of the trace. The trace of an operator O is defined by

TrO = 0|O|0+ 1|O|1. (158)The point is to rewrite it in terms of an integral. It is easy to show that

dd|O|e =dd(0| 1|)O(|0 |1)e

=dd(0|O|0+ 1|O|1)

= 0|O|0 1|O|1. (159)

31

This is off by a sign for the second term for the trace we wanted to calculate. We canavoid this if we had taken not | but rather |:

dd|O|e = 0|O|0+ 1|O|1. (160)

This is the origin of the anti-periodic boundary condition.We finally calculate N |eH |0 in terms of a path integral, and to obtain the trace

from this result, we do the following integral

Z =d(t)d(t) exp

[ 1h

h0

d((ih) + h)]

(161)

with the understanding that (t), (t) follow the anti-periodic boundary conditions.

32

Feynman's Path Integral FormulationPhysical IntuitionPropagatorDerivation of the Path IntegralSchrdinger Equation from Path IntegralExamplesFree ParticleThe sloppy wayThe careful way with discretized time

Harmonic Oscillator

Partition FunctionSchrdinger FieldNon-relativistic FermionsPath Integrals in Phase SpaceA Few Useful IdentitiesHarmonic Oscillator with Discretized TimeAnti-periodic Boundary Condition