heat conduction in harmonic...
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Heat conduction in harmonic crystals
Abhishek DharInternational centre for theoretical sciences
TIFR, Bangalore
FPSP-XIII, LeuvenJune 16-29, 2013
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Outline
Introduction - why is heat conduction an interesting problem ?
Heat conduction in harmonic crystalsDerivation of a general formula for heat current in harmonic systems using the quantum Langevinequations approach.Applications of the heat current formula towards understanding heat conduction in disorderedharmonic crystals.Current fluctuations and Large deviation functions for heat conduction.
Discussion
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References
One dimensional harmonic chainsRieder, E. Lieb and J. L. Lebowitz (JMP, 1967)Casher and Lebowitz (JMP, 1971), O. Ajanki and F. Huveneers (CMP, 2011)Rubin and Greer (JMP, 1971), T. Verheggen (JMP, 1979)A. Dhar (PRL, 2001)S. Das and A. Dhar (EPJB, 2012)
Higher dimensionsA. Dhar and D. Roy (JSP, 2006)A. Kundu, D. Chaudhuri, D. Roy, A. Dhar, J.L. Lebowitz and H. Spohn (EPL, 2010)A. Kundu, D. Chaudhuri, D. Roy, A. Dhar, J.L. Lebowitz and H. Spohn (PRB, 2011)A. Dhar, K. Saito and P. Hanggi (PRE, 2012)
Current fluctuations and large deviationsP. Visco (JSM, 2006)K. Saito and A.Dhar (PRL, 2007)A. Kundu, A. Dhar and S. Sabhapandit (JSM, 2011)A. Dhar and K. Saito (PRL, 2011)H. Fogedby and A. Imparato (JSM, 2012)J.S. Wang, B. K. Agarwalla, and H. Li (2012)
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Heat CONDUCTION
Heat transfer through a body from HOT to COLD region.
T TL R
J
T(x)
x
TL
TR
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Fourier’s law
Fourier’s law of heat conduction
J = −κ∇T (x)
κ – thermal conductivity of the material.
Using Fourier’s law and the energy conservation equation
∂ε
∂t+∇J = 0
and, writing ∂ε/∂t = c∂T/∂t where c = ∂ε/∂T is the specific heat capacity,gives the heat DIFFUSION equation:
∂T∂t
=κ
c∇2T .
Thus Fourier’s law implies diffusive heat transfer.
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Proving Fourier’s law
Proving Fourier’s law from first principles (Newton’s equations of motion) is adifficult open problem in theoretical physics.
Review article:Fourier’s law: A challenge for theoristsBonetto, Rey-Bellet, Lebowitz (2000).
It seems there is no problem in modern physics for which there are on record as many false starts,and as many theories which overlook some essential feature , as in the problem of the thermalconductivity of nonconducting crystals.R. Peierls (1961)
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Theory of heat conduction
Equilibrium Systems
As far as equilibrium properties of a system are concerned there is a well definedprescription to obtain macroscopic properties starting from the microscopicHamiltonian – Statistical mechanics of Boltzmann and Gibbs. For a system inequilibrium at temperature T :
F = −kBT ln Z Z = Tr [ e−βH ]
Phase space density : ρ(r1, . . . , rN ,p1, . . . ,pN ) =e−βH
Z
Nonequilibrium Systems
Heat conduction is a nonequilibrium process and there is no equivalent statisticalmechanical theory for systems out of nonequilibrium.
We do not have a nonequilibrium free energy.
For a system in contact with two heat baths we not know what the appropriatephase space distribution is.
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Theories of heat transport
Kinetic theory
Peierl’s-Boltzmann theory
Green-Kubo linear response theory
Landauer theory – This is an open systems approach especially useful for “non-interacting”systems.
Nonequilibrium Green’s function formalism
Generalized Langevin equations formalism
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Kinetic theory
Heat conduction in a dilectric crystalline solid: Basic microscopic picture
. . . ..
.. .
..
.
. ..
.. ..
Ion Electron
Heat is carried by lattice vibrations i.e phonons.
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Kinetic theory for phonon gas
x
Simplest “derivation” of Fourier’s law.Heat carried by phonons which undergo collisions at time intervals ∼ τ - hence do random walks.Let ε(x) = be energy density at x and v = average velocity of particles. Then:
J =12
v [ε(x − vτ)− ε(x + vτ)] = −v2τ∂ε
∂T∂T∂x
= −v`K c∂T∂x
Therefore J = −κ∂T∂x
with κ = v`K c
where c = specific heat , `K = mean free path .
Can calculate thermal conductivity κ if we know average velocity, mean free path and specific heatof the phonons.
Note that this is not a proof since it assumes diffusive motion!But the heat carriers could be Levy walkers!!
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The set-up for a direct verification of Fourier’s law
.
L
T TL
R
J
Dynamics in bulk described by a Hamiltonian:
H =N∑`=1
p2`
2m`+ U(x` − x`−1) + V (x`)
where U(x) is the inter-particle interaction potential and V (x) an external pinning potential.
Boundary particles interact with heat baths. Example: Langevin baths.
m1x1 = −∂H/∂x1 − γx1 + (2γkBTL)1/2ηL
m`x` = −∂H/∂x` ` = 2, . . . ,N − 1
mN xN = −∂H/∂xN − γxN + (2γkBTR)1/2ηR
In nonequilibrium steady state compute temperature profile and current for different systemsizes.
kBT` = m`〈x2` 〉 and J` = 〈v`f`−1,`〉 ,
where f`,`−1 is the force on l th particle from (l − 1)th particle.
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Heat current and heat conductivity
L
T TL
R
J
Connect system of length L to heat baths with small temperature difference ∆T = TL − TR .We can measure the heat current J and this is always finite and it can be proven thatJ = G∆T , i.e that is the current response is linear.
Fourier’s law J = −κ∇T implies
J = κ∆TL
.
The size-dependence is difficult to prove.
We define a size-dependent conductivity
κL =J L∆T
.
We can directly measure this and ask whether limL→∞ κL exists and agrees with what isgiven by the Green-Kubo formula.
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Results so far
Direct studies (simulations and some exact results) in one and twodimensional systems find that Fourier’s law is in fact not valid. The thermalconductivity is not an intrinsic material property.
For anharmonic systems without disorder , κ diverges with system size L as:
κ ∼ L1/3 1D∼ Lα
′, log L 2D
∼ L0 3D
Lepri, Livi, Politi, Phys. Rep. 377 (2003),A.D, Advances in Physics, vol. 57 (2008).
Beijeren (PRL, 2012), Spohn (2013) - analytic understanding, connection to KPZ.A.D, K. Saito, B. Derrida (PRE, 2013) - Levy walk picture.
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Experiments
.
Breakdown of Fourier’s Law in Nanotube Thermal Conductors
C.W. Chang,1,2,* D. Okawa,1 H. Garcia,1 A. Majumdar,2,3,4 and A. Zettl1,2,4,+
1Department of Physics, University of California at Berkeley, California 94720, USA2Center of Integrated Nanomechanical Systems, University of California at Berkeley, California 94720, USA
3Departments of Mechanical Engineering and Materials Science and Engineering, University of California at Berkeley,
California 94720, USA4Materials Sciences Division, Lawrence Berkeley National Laboratory, Berkeley, California 94720, USA
(Received 11 March 2008; revised manuscript received 9 July 2008; published 15 August 2008)
We present experimental evidence that the room temperature thermal conductivity () of individual
multiwalled carbon and boron-nitride nanotubes does not obey Fourier’s empirical law of thermal
conduction. Because of isotopic disorder, ’s of carbon nanotubes and boron-nitride nanotubes show
different length dependence behavior. Moreover, for these systems we find that Fourier’s law is violated
even when the phonon mean free path is much shorter than the sample length.
DOI: 10.1103/PhysRevLett.101.075903 PACS numbers: 65.80.+n, 63.22.Gh, 73.63.Fg, 74.25.Kc
PRL 101, 075903 (2008) P HY S I CA L R EV I EW LE T T E R Sweek ending
15 AUGUST 2008
FIG. 3 (color online). (a) Normalized thermal resistance vs
normalized sample length for CNT sample 4 (solid black
circles), best fit assuming ¼ 0:6 (open blue stars), and best
fit assuming Fourier’s law (open red circles). (b) Normalized
thermal resistance vs normalized sample length for BNNT
sample 2 (solid black diamonds), best fit assuming ¼ 0:4
(open blue stars), and best fit assuming Fourier’s law (open
red circles).
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Experiments: graphene
.
Very large thermal conductivity.κ ∼ log(L) has been reported. [Balandin etal, Nat. Phys. (2011), Li etal (2012)]
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Experiments
.
+
Science 10 August 2007:
Vol. 317 no. 5839 pp. 787-790
DOI: 10.1126/science.1145220
REPORT
Ultrafast Flash Thermal Conductance of Molecular Chains
Zhaohui Wang1,*, Jeffrey A. Carter
1,*, Alexei Lagutchev
1,*, Yee Kan Koh
2, Nak-Hyun Seong
1,*,
David G. Cahill2,3
and Dana D. Dlott1,3,†
Author Affiliations
† To whom correspondence should be addressed. E-mail: [email protected]
* These authors contributed equally to this work.
ABSTRACT
At the level of individual molecules, familiar concepts of heat transport no longer apply. When large
amounts of heat are transported through a molecule, a crucial process in molecular electronic
devices, energy is carried by discrete molecular vibrational excitations. We studied heat transport
through self-assembled monolayers of long-chain hydrocarbon molecules anchored to a gold
substrate by ultrafast heating of the gold with a femtosecond laser pulse. When the heat reached the
methyl groups at the chain ends, a nonlinear coherent vibrational spectroscopy technique detected
the resulting thermally induced disorder. The flow of heat into the chains was limited by the interface
conductance. The leading edge of the heat burst traveled ballistically along the chains at a velocity of
1 kilometer per second. The molecular conductance per chain was 50 picowatts per kelvin.
Heat transport is central to the operation of mechanical and electronic machinery, but at the level of individual
Prev | Table of Contents | Next
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Experiments
.
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The Simplest models
Ordered harmonic Chain [Rieder, Lebowitz, Lieb (1967)].
TL TR
This case can be solved exactly and the nonequilibrium steady state distribution functioncomputed. One findsJ ∼ (TL − TR) unlike expected (TL − TR)/N from Fourier law.Flat Temperature profiles.
No local thermal equilibrium. Fourier’s law is not valid.
This can be understood since there are no mechanism for scattering of phonons and heattransport is purely ballistic. In real systems we expect scattering in various ways:
Phonon-phonon interactions: In oscillator chain take anharmonic springs ( e.gFermi-Pasta-Ulam chain ).
Disorder: In harmonic systems make masses random.—WE WILL DISCUSS THIS INDETAIL.
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Heat conduction in disordered harmonic crystals
This is best understood using the Landauer approach.
Open systems approach to transport. Write coupled Hamiltonian for system and two baths.Evolve initial state ρ = ρL(TL)⊗ ρR(TR)⊗ ρS and obtain the nonequilibrium steady statedensity matrix.
Exact results for non-interacting systems ( both phononic and electronic systems withquadratic Hamiltonians).
Current in terms of transmission coefficients.
Results can be derived using:Langevin equations approachKeldysh Green functions techniquesPath integral approach
Simple derivation is the Langevin equations approach which I will derive in this lecture.
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Landauer formula for heat current
TL TR
J =kB(TL − TR)
4π
∫ ∞0
dωT (ω) ,
where T (ω) = 4γ2ω2|G1N |2 , (Phonon transmission)
with the matrix G = [−ω2M + Φ− Σ]−1
M = mass matrix ,Φ = force matrix
Σ is a “self-energy” correction from baths.
Quantum mechanical case:
J =1
2π
∫ ∞0
dω ~ ω T (ω) [ f (TL)− f (TR) ]
where f (ω) = (eβ~ω − 1)−1.
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Harmonic chains
The Hamiltonian is
H =N∑`=1
(p2`
2m`+
kox2`
2
)+
N∑l=0
k(x`+1 − x`)2
2.
Boundary conditionsFixed BC: x0 = xN+1 = 0.
To drive a heat current in this system, one needs to connect it to heat reservoirs. Various modelsof baths exist and here we discuss Langevin type baths.
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The classical Langevin equation
Langevin baths: Add extra terms in the equation of motion of the particles in contact with baths.In the simplest form, the additional forces from the heat bath consist of a dissipative term, and astochastic term.For a single oscillator connected to a heat bath, the Langevin equations of motion are
x =pm, p = −kx −
γ
mp + η(t)
where the noise term given by η is taken to be Gaussian, with zero mean, and related to thedissipation coefficient γ by the fluctuation dissipation relation
〈η(t)η(t ′)〉 = 2kBTγ(.t − t ′) .
The Langevin dynamics ensures that the system reaches thermal equilibrium in the long time limit.
More general Langevin baths where the noise is correlated can also be used an example beingthe Rubin model. Quantum description also possible.
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Quantum Langevin baths- Rubin model
Heat Bath
mm
k
System
ko
o
The bath is itself a semi-infinite harmonic oscillator chain with HamiltonianHb =
∑∞`=1 P2
` /2m +∑∞`=0 k(X` − X`+1)2/2
The bath is initially assumed to be in thermal equilibrium at temperature T and then coupled, attime t = −∞, to the system. System-bath coupling is of the form −k x X1.
Starting with Heisenberg equations of motion for x , p it can be shown that the effective equation ofmotion of the particle is a generalized Langevin equation of the form
x = p/mo , p = −kox +
∫ t
−∞dt ′Σ(t − t ′)x(t ′) + η(t) ,
where the Kernel Σ(t) is given by:
Σ(ω) =
∫ ∞0
dtΣ(t)eiωt = k1−mω2/2k + iω(m/k)1/2[1−mω2/(4k)]1/2 = keiq ,
and q is defined through cos(q) = 1−mω2/2k .
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Quantum baths
The noise is Gaussian and the correlations are now given by:
〈η(ω)η(ω′)〉 =~Γ(ω)
π[1 + f (ω,T )] δ(ω + ω′),
where η(ω) = (1/2π)∫∞−∞ dtηL(t)eiωt , Γ(ω) = Im[Σ(ω)], f (ω,T ) = 1
e~ω/kB T−1.
Alternative form:
12〈 η(ω)η(ω′) + η(ω′)η(ω) 〉 =
~Γ(ω)
2πcoth
~ω2kBT
δ(ω + ω′) .
Ohmic bath Σ(ω) = iγω. High temperature classical limit gives white noise.
Note Γ(ω) is non-zero only within bath bandwidth — infinite for Ohmic baths.
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Definition of Current
To define the local energy current inside the wire we first define the local energy densityassociated with the l th particle (or energy at the lattice site l) as follows:
εl =p2
l2ml
+ V (xl ) +12
[ U(xl−1 − xl ) + U(xl − xl+1) ] .
Taking a time derivative of these equations, and after some straightforward manipulations, we getthe discrete continuity equations given by:
εl = −jl+1,l + jl,l−1 where jl,l−1 =12
(xl−1 + xl )fl,l−1 ,
and fl,l−1 = −∂xl U(xl − xl−1) is the force that the (l − 1)th particle exerts on the l th particle.Hence jl+1,l is the energy current from site l to l + 1.In the steady state using the fact that 〈dU(xl−1 − xl )/dt〉 = 0 it follows that 〈xl−1fl,l−1〉 = 〈xl fl,l−1〉and hence:
〈jl,l−1〉 = 〈(xl + xl−1)fl,l−1〉/2 = 〈xl fl,l−1〉 ,
and this has the simple interpretation as the average rate at which the (l − 1)th particle does workon the l th particle.Similarly it can be shown that the current flowing from reservoirs into the system is given by
jL = 〈x1fLB〉 , jR = 〈xN fRB〉 ,
for the two baths respectively.
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Solving the linear Langevin equations for steady stateproperties-simple example
Consider single harmonic oscillator coupled to a single Langevin-type heat bath.
x = p/m , p = −kx − γpm
+ η(t) , 〈η(t)η(t ′)〉 = 2γkBTδ(t − t ′) .
Equivalently the system is described by the Fokker-Planck equation for the phase-spaceprobability distribution P(x , p, t)
∂P(x , p, t)∂t
= −∂
∂xpm
P +∂
∂pkxP +
[∂
∂pγpm
P + γkBT∂2P∂p2
].
First two terms on r.h.s — Hamiltonian evolutionSecond two terms — heat bath dynamics.In steady state ∂PSS/∂t = 0. Here steady state is the equilibrium distribution:
P(x , p) =e−β[
p22m + kx2
2
]Z
, where Z =
∫dx∫
dp e−β[
p22m + kx2
2
].
From PSS(x , p) we can then obtain steady state properties, for example〈x2〉 = kBT/k , 〈p2〉 = mkBT .
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Method 2: Fourier transforms
Define the Fourier transforms x(t) =∫∞−∞ dωe−iωt x(ω), η(t) =
∫∞−∞ dωe−iωt η(ω).
We then get
x(ω) = G+(ω)η(ω), where G+(ω) =1
−mω2 + k − iωγ.
Noise correlations
〈η(ω)η(ω′)〉 =γkBTπ
(.ω + ω′) .
Hence
〈x2(t)〉 =
∫dω∫
dω′ ei(ω+ω′)t 〈x(ω)x(ω′)〉
=
∫dω∫
dω′ ei(ω+ω′)t G+(ω)G+(ω′)〈η(ω)η(ω′)〉
=γkBTπ
∫ ∞−∞
dω G+(ω) G+(−ω) =kBT
k.
Similarly
〈u2(t)〉 =γkBTπ
∫ ∞−∞
dω ω2 G+(ω) G+(−ω) =kBTm
.
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Steady state current
Classical problem with white noise.
Gaussian steady state measure. Let q = (x1, . . . , xN , p1, . . . , pN )T . Then
PSS(q) ∼ e−qC−1q/2 ,
where C is the correlation matrix with elements 〈xi xj 〉, 〈xi pj 〉 and 〈pi pj 〉.
Two ways to obtain the correlation matrix C.
Method 1– Write the Fokker-Planck equation:
∂P(q, t)∂t
= LP(q, t) ,
Solve LPSS = 0 .
Linear equation for C which can be explicitly solved for ordered chainNote that current and temperature profile are both given by two-point correlations.(Rieder,Lebowitz, Lieb, 1967).
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Simpler and more general approach- Langevin equations
m1x1 = −k1x1 − k(x1 − x2) +
∫ ∞−∞
dt ′Σ+1 (t − t ′)x1(t ′) + η1(t)
m`x` = −kox` − k(2x` − x`−1 − x`+1) 1 < ` < N
mN xN = −kNx1 − k(xN − xN−1) +
∫ ∞−∞
dt ′Σ+N (t − t ′)xN (t ′) + ηN (t) .
Noise correlations are
〈 ηL(ω) ηTL (ω′) 〉 =
~ΓL(ω)
π[ 1 + f (ω,TL) ] δ(ω + ω′),
〈 ηR(ω) ηTR (ω′) 〉 =
~ΓR(ω)
π[ 1 + f (ω, TR) ] δ(ω + ω′) .
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Solution by Fourier transforms
As for the single particle case, the cheapest way to obtain steady state properties is by Fouriertransforming the equations of motion.We get
−m1ω2x1(ω) = −k0x1 − k(x1 − x2) + Σ+
1 (ω)x1(ω) + η1(ω)
−m`ω2x`(ω) = −k0xl − k(2x` − x`+1 − x`+1) 1 < ` < N
−mNω2xN (ω) = −k0xN − k(xN − xN−1) + Σ+
N (ω)xN (ω) + ηN (ω)
Useful to write in matrix form —[−Mω2 + Φ− Σ+
L (ω)− Σ+R (ω)
]X(ω) = ηL(ω) + ηR(ω) ,
where X T = x1, x2, ...xN, ηTL = η1, 0, 0, ..., ηT
R = 0, 0, 0, ...ηN .
Σ+L , Σ+
R are N × N matrices whose only non-vanishing elements are[Σ+
L ]11 = Σ+1 (ω), [Σ+
R ]NN = Σ+N (ω).
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Thus we get the following solution of the equations of motion:
X(ω) = G+(ω) [ ηL(ω) + ηR(ω) ] , where G+(ω) =1
−M ω2 + Φ− Σ+L (ω)− Σ+
R (ω).
Recall single particle solution:
x(ω) = G+(ω) η(ω), where G+(ω) =1
−m ω2 + k − iγω.
Noise correlations are now
〈 ηL(ω) ηTL (ω′) 〉 =
~ΓL(ω)
π[ 1 + f (ω,TL) ] δ(ω + ω′),
〈 ηR(ω) ηTR (ω′) 〉 =
~ΓR(ω)
π[ 1 + f (ω, TR) ] δ(ω + ω′) .
where Γ(ω) = Im[Σ+(ω)].
Solution in time is
X(t) =
∫ ∞−∞
dω X(ω)e−iωt .
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Computing steady state current
We now calculate the heat current and the local kinetic energies (temperature) in the steady state.
These are two point correlation functions.
The steady state current can be found by evaluating the current on any bond. To get Landauerform — evaluate the boundary current which is
J = 〈 x1(t)[η1(t) +
∫ ∞−∞
dt ′ Σ+1 (t − t ′) x1(t ′)
]〉 .
Using matrix notation thi is
J =
⟨ [X T (t) ηL(t) +
∫ ∞−∞
dt ′X T (t) Σ+L (t − t ′) X(t ′)
] ⟩=− i
∫ ∞−∞
dω∫ ∞−∞
dω′ e−i(ω+ω′)t ω⟨ [
X T (ω) ηL(ω′) + X T (ω) Σ+L (ω′)X(ω′)
] ⟩=− i
∫ ∞−∞
dω∫ ∞−∞
dω′ e−i(ω+ω′)t ω⟨
Tr[ηL(ω′) X T (ω)
]+ Tr
[Σ+
L (ω′) X(ω′) X T (ω)] ⟩
.
Now use the solution X = G+(ω) [ηL + ηR ]. We get
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Computing the current
J = −i∫ ∞−∞
dω∫ ∞−∞
dω′ e−i(ω+ω′)t ω(
Tr[ ⟨
ηL(ω′) ( ηTL (ω) + ηT
R (ω) )⟩
G+(ω)]
+ Tr[
Σ+L (ω′) G+(ω′)
⟨( ηL(ω′) + ηR(ω′) ) ( ηT
L (ω) + ηTR (ω) )
⟩G+(ω)
] ),
where we used G+T = G+.
Now collect terms in J which depend only on TR —-
JR = −i∫ ∞−∞
dω∫ ∞−∞
dω′ e−i(ω+ω′)t ω Tr[
Σ+L (ω′) G+(ω′)
⟨ηR(ω′) ηT
R (ω)⟩
G+(ω)].
Using the noise correlations we then get:
JR = −iπ
∫ ∞−∞
dω ~ω Tr [ Σ+L (−ω) G−(ω) ΓR(ω) G+(ω) ] f (ω,TR) ,
where we have defined G+(ω) = G−(−ω).We also used 1 + f (−ω) = −f (ω) and ΓR(ω) = −Γ(ω).
Now take the complex conjugate of this.
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Computing the current
JR = −iπ
∫ ∞−∞
dω ~ω Tr [ Σ+L (−ω) G−(ω) ΓR(ω) G+(ω) ] f (ω,TR) ,
Using (Tr [A])∗ = Tr [A†])
J∗R =iπ
∫ ∞−∞
dω ~ω Tr [ G−(ω) ΓR(ω) G+(ω) Σ−L (−ω) ] f (ω,TR) .
Using the cyclic property of trace we get
J∗R =iπ
∫ ∞−∞
dω ~ω Tr [ Σ−L (−ω) G−(ω) ΓR(ω) G+(ω) ] f (ω,TR) .
Hence taking the real part (JR + J∗R)/2 and noting that ΓL(−ω) = Im[Σ+L (−ω)] = −ΓL(ω) gives
JR = −1π
∫ ∞−∞
dω ~ω Tr[
G+(ω) ΓL(ω) G−(ω) ΓR(ω)]
f (ω,TR) .
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Computing the current
Contribution of the left reservoir to the current must have a similar form (since for TL = TR wemust have J = 0 — hence
J =1π
∫ ∞−∞
dω ~ω Tr[
G+(ω) ΓL(ω) G−(ω) ΓR(ω)]
[ f (ω,TL)− f (ω,TR) ]
=1
4π
∫ ∞−∞
dω ~ω T (ω) [f (ω,TL)− f (ω,TR)]
where T (ω) = 4Tr [G+(ω) ΓL(ω) G−(ω) ΓR(ω) ]
is the phonon-transmission coefficient.
Linear response — ∆T = TL − TR << T , where T = (TL + TR)/2
J =∆T4π
∫ ∞−∞
dω T (ω)~ω∂f (ω,T )
∂T.
Classical limit —- ~ω/kBT → 0.
J =kB ∆T
4π
∫ ∞−∞
dω T (ω) .
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Local kinetic energy
The average kinetic energy at a site is given by
ki = [MK ]ii ,
where the correlation matrix K is given by
K = 〈 X X T 〉
=
∫ ∞−∞
dωω
π[ G+(ω)ΓL(ω)G−(ω)~ω(1 + f (ω,TL)) + G+(ω)ΓR(ω)G−(ω)~ω(1 + f (ω,TR)) ]
=
∫ ∞−∞
dωω
π[ G+(ω)ΓL(ω)G−(ω)
~ω2
coth(~ω
2kBTL) + G+(ω)ΓR(ω)G−(ω)
~ω2
coth(~ω
2kBTR) ]
where the last line is easily obtained after writing K = (K + K∗)/2 since K is a real matrix.
Classical limit —-
K = 〈X X T 〉
=kBTL
π
∫ ∞−∞
dω ω G+(ω)ΓL(ω)G−(ω) +kBTR
π
∫ ∞−∞
dω ω G+(ω)ΓR(ω)G−(ω) .
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Note: The expressions for current and kinetic energy density have been written using a somewhatformal matrix notation. The reason for doing so is that in this form they generalize immediately tothe case when the wire is not one dimensional but any harmonic system. In the next section weagain go back to our one-dimensional chain and will write more explicit expressions.
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Comments on the Landauer formula
T is the transmission probability of phonons from the left to right reservoir (Or thetransmission probability of phonons from right to left).
Same derivation works for arbitrary harmonic networks in any dimension (assuming a uniquesteady state exists).
Derivation can also be done for electronic systems described by tight-binding typeHamiltonian where one can obtain Landauer-type expressions for both the heat and particlecurrent.Results for electron—-Landauer conductance —–ADD
Interacting problems
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Current in terms of phonon transmission
Consider classical case with Σ1(ω) = ΣN (ω) = Σ(ω).
J =∆T2π
∫ ∞0
dωT (ω) ,
whereT (ω) = 4Γ2(ω)|G1N |2
and
kG = k [−ω2M + Φ− ΣL − ΣR ]−1 = Z−1
where Z =
a1 − Σ′(ω) −1 0 ... 0−1 a2 −1 0 ... 0... ... ... ... ...0 ... 0 −1 aN−1 −10 ... 0 −1 aN − Σ′(ω)
,
a` = (2k −m`ω2)/k , Σ′ = Σ/k .
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For the two heat bath models (a) Casher-Lebowitz and (b) Rubin-Greer,
Σ(ω) = −iγω model(a)
Σ(ω) = k1−mω2/2k − iω(m/k)1/2[1−mω2/(4k)]1/2 = k eiq model(b),
G+ =1k
Z−1
G+1N =
1k
Z−11N =
1k
1∆N
where ∆N = Det[Z ] .
Let us define ∆l,m to be the determinant of the submatrix of Z beginning with the l th row andcolumn and ending with the mth row and column.Also let Dl,m be defined to be the determinant of the submatrix of [−ω2M + Φ]/k beginning withthe l th row and column and ending with the mth row and column.
Using recursion relations such as ∆1,N = (a1 − Σ′)∆2,N −∆3,N it is easy to see that
∆N (ω) = D1,N − Σ′(D2,N + D1,N−1) + Σ′2D2,N−1
= (1 − Σ′)
(D1,N −D1,N−1D2,N −D2,N−1
)(1
Σ′
)
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Product of random matrices
Again using relations such as D1,N = a1D2,N − D3,N we can write
D =
(D1,N −D1,N−1D2,N −D2,N−1
)= T1T2....TN
where Tl =
(2−mlω
2/k −11 0
)Thus finally we find that G1N and thus the transmission T (ω) can be expressed in terms of aproduct of random matrices.
∆N = [kG1N ]−1 = (1 − Σ′) TN TN−1...T1
(1
Σ′
)where Tl =
(2−mlω
2 −11 0
)
Generalization to higher dimensions (Nd lattice):T (ω) can be expressed in terms of a product of 2Nd−1 × 2Nd−1 random matrices.
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Ordered case
In this case it is easy to show that
D1,N = sin(N + 1)q/ sin q , ω2 = 2 (k/m) [1− cos(q)] .
Rubin-Greer model: Σ = keiq , therefore Γ = k sin(q) and is non-zero only for ω within the bathspectrum. It then follows that T = 1 as expected. Hence
J =kB∆T
2π
∫ ωm
01 =
kB∆Tωm
2π.
Casher-Lebowitz model: In this case T 6= 1 but for large N, only phonons within the spectral bandof system transmit, and the integral over ω can be converted to one over q. We get
J =2γ2kB∆T
k2π
∫ π
0dq|
dωdq|
ω2q
|∆N |2.
The integrand has the form g1(q)/(1 + g2(q) sin Nq) and we use the result
limN→∞
∫ π
0dq
g1(q)
1 + g2(q) sin Nq=
∫ π
0dq
g1(q)
[1− g22 (q)]1/2
,
to get in the limit N →∞ the Rieder-Lebowitz-Lieb result
J =kkB∆T
2γ
[1 +
ν
2−ν
2
√1 +
4ν
]where ν =
mkγ2
.
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Disordered Harmonic systems: Results in 1D
In this case we use the following reult:Furstenberg’s theorem for product of random matrices —- D1,N ∼ eN/`(ω)
(This is infact a proof of Anderson localization).This then implies: T (ω) ∼ e−N/`(ω) with `(ω) ∼ 1/ω2 for ω → 0 .
Hence frequencies ω <∼ N−1/2 “do not see” the randomness and can carry current.
These are ballistic modes which give a current J ∼∫ N−1/2
0 T (ω)dω.
Form of T (ω) (at small ω) depends on boundary conditions and from our earlier result for orderedcase we get:
Casher-Lebowitz model: T (ω) ∼ ω2 J ∼ 1/L3/2
Rubin-Greer model: T (ω) ∼ ω0 J ∼ 1/L1/2
If all sites are pinned then low frequency modes are cut off. Hence we get:
J ∼ e−L/` .
[Ford and Allen (1966), Matsuda, Ishi (1968), Rubin, Greer, Casher, Lebowitz (1972),Verheggen(1979), Dhar (2001), Ojanki and Huveneers (2012)]
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Higher dimensional systems
What about systems in two three dimensions ?Is Fourier’s law valid ?Heat conduction in a classical harmonic crystal with isotopic massdisorder.
Look at L dependence of J in nonequilibrium steady state. (J ∼ 1/L ? )
Look at phonon transmission:
Effect of localization and boundary conditions.
A. Kundu, D.Chaudhuri, A. Dhar, D. Roy, J.L. Lebowitz and H. Spohn,EPL 90, 40001 (2010), PRB 81, 064301 (2010).
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Heat conduction in mass disordered harmonic crystal
H =∑
x
mx
2q2
x +∑x,e
k2
(qx − qx+e)2 +∑
x
ko
2q2
x
qx: scalar displacement, masses mx random.ko = 0: Unpinned. ko > 0: Pinned.
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Kinetic theory.
In the harmonic approximation the solid can be thought of as a gas of phonons withfrequencies corresponding to the vibrational spectrum.
The phonons do a random walk. A phonon of frequency ω has a mean free path `K (ω).
Thermal conductivity given by:
κ =
∫ ωD
0dω D(ω) ρ(ω) c(ω,T )
D(ω) = v`K (ω) = phonon diffusivity ,v = phonon velocity ,ρ = density of states ,c = specific heat capacity ,ωD = Debye frequency .
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Kinetic theory
Rayleigh scattering of phonons gives a mean free path `K (ω) ∼ 1/ωd+1.
For low frequency phonons `K>∼ L , where L is the size of system (ballistic phonons).
ρ(ω) ∼ ωd−1. Hence
κ =
∫ ωD
1/Ldω ρ(ω) v`K (ω)
∼ L1/(d+1)
1D : κ ∼ L1/2 , 2D : κ ∼ L1/3 , 3D : κ ∼ L1/4.
Thus kinetic theory predicts a divergence of the conductivity in all dimensions.For a pinned crystal it predicts a finite conductivity.
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Problems with kinetic theory
Is a perturbative result valid in the limit of weak disorder. Rayleigh scattering result for `K (ω)is obtained for single scattering of sound waves.
Effect of Anderson localization is not accounted for since this requires one to considermultiple scattering of phonons.
The kinetic theory result can also be derived directly from the Green-Kubo formula and `K (ω)can be computed using normal mode eigenvalues and eigenstates.
Green-Kubo probably correct for normal transport.Probably WRONG for anomalous transport.
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Anderson localization
Consider harmonic Hamiltonian:
H =∑
x
12
mxq2x +
∑x,x′
qxφxx′q′x .
x denotes point on a d-dimensional hypercube of size Nd .φ - Force-matrix .
Equation for qth normnal mode, q = 1, 2, ...,Nd . [Closed system]
mx Ω2q aq(x) =
∑x′φx,x′ aq(x′) .
Normal mode frequency: Ωq .Normal mode wave-function: aq(x).
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Character of normal modes
20
40
60
20
40
60
-1-0.5
00.51
20
40
60
Extended periodic mode(Ballistic)
20
40
60
20
40
60
-5
0
5
20
40
60
Extended random mode(Diffusive)
20
40
60
20
40
60
02.55
7.510
20
40
60
Localized mode(Non-conducting)
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Predictions from Renormalization group theory
S. John, H. Sompolinsky, and M.J. Stephen
For unpinned system:
In 1D all modes with ω > 1/N1/2 are localized.
In 2D all modes with ω > [log(N)]−1/2 are localized.
In 3D there is a finite band of frequencies of non-localized states between0 to ωL
c , where ωLc is independent of system-size.
System size dependence of current ??Nature of non-localized states ??Ballistic OR Diffusive OR .....
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Nature of phonon modes
Localization theory tells us that low frequency modes are extended.
At low frequencies the effect of disorder is weak and so we expect kinetic theory to be accurate.
Setting `K (ω) = N gives us the frequency scale ωKc below which modes are ballistic. Since
`K (ω) = 1/ωd+1 this gives:
1D: ωKc ∼ N−1/2
2D: ωKc ∼ N−1/3
3D: ωKc ∼ N−1/4
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Nature of phonon modes
Kinetic theory and localization theory gives us two frequency cut-offs: ωKc and ωL
c :
ω ωLc
Kc
Ballistic LocalizedDiffusive
ω
1D : ωKc ∼ ωL
c ∼ N−1/2
2D : ωKc ∼ N−1/3, ωL
c ∼ [ln N]−1/2
3D : ωKc ∼ N−1/4, ωL
c ∼ N0
Try to estimate the N dependence of J.
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Heuristic theory
Nonequilibrium heat current is given by the Landauer formula:
J =∆T2π
∫ ∞0
dω T (ω) ,
where T (ω) is the transmission coefficient of phonons.
Phonons can be classified as:
Ballistic: Plane wave like states with T (ω) independent of system size.
Diffusive: Extended disordered states with T (ω) decaying as 1/N.
Localized states: T (ω) ∼ e−N/`L .
Find contribution to J of the different types of modes and estimate the asymptotic sizedependence.
Ballistic contribution to current depends on boundary conditions.
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Boundary conditions
Free BC: T (ω) ∼ ωd−1
ω → 0
Fixed BC: T (ω) ∼ ωd+1
ω → 0
Pinned system: No low frequencyballistic modes
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Predictions of heuristic theory
1D unpinned fixed BC: J ∼ N−3/2
1D unpinned free BC : J ∼ N−1/2
1D pinned: J ∼ exp(−bN)
These have been proved rigorously.
3D unpinned fixed BC: J ∼ N−1
3D unpinned free BC : J ∼ N−3/4
3D pinned: J ∼ N−1
2D unpinned fixed BC: J ∼ N−1(ln N)−1/2
2D unpinned free BC : J ∼ N−2/3
2D pinned: J ∼ exp(−bN)
We now check these predictions numerically.
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Disordered harmonic systems
Equations of motion:
mxqx = −∑
e(qx − qx−e)− koqx +fx from heat baths
mx chosen from a random distribution.
fx is force from heat baths and acts only on boundary particles. We choose white noise Langevinbaths:
fx = −γqx + (2γT )1/2ηx(t).〈ηx(t)ηx′ (t ′) = δ(t − t ′)δx,x′
Periodic boundary conditions in transverse directions.
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Measured quantities
In the nonequilibrium steady state we measure the energy current and temperature profile. Theseare given by the following time-averages:
J =∑x′
q1,x′ q2,x′
Tx =∑x′
mx,x′ q2x,x′
We use two approaches:1. Numerical evaluation of phonon transmission and use the Landauer formula to find J.2. Nonequilibrium Simulations.
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Current in terms of phonon transmission
J =∆T2π
∫ ∞0
dωT (ω) ,
whereT (ω) = 4γ2ω2|G1N |2
and
G = [−ω2M + Φ− ΣL − ΣR ]−1
=
a1 − iγω −1 0 ... 0−1 a2 −I 0 ... 0... ... ... ... ...0 ... 0 −1 aN−1 −10 ... 0 −1 aN − iγω
−1
,
al = 2 + ko −mlω2.
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Product of random matrices
Transmission function T (ω) can be expressed in terms of a product of randommatrices.
G1N = (1 − iγ) TN TN−1...T1
(1iγ
)where Tl =
(2−mlω
2 −11 0
)
Generalization to higher dimensions (Nd lattice):T (ω) can be expressed in terms of a product of2Nd−1 × 2Nd−1 random matrices (Anupam Kundu).
We implement this numerically for d = 2,3.
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Numerical and simulation results
Open system. We look at:
• Disorder averaged transmission: T (ω) = [T (ω)]
Nd−1
• Current: = J ∼∫∞
0 dω T (ω).
Closed system without heat baths.For the normal modes find:
• Density of states: ρ(ω) = 1Nd
∑q δ(ω − Ωq)
• Inverse participation ratio: P−1 =∑
x a4x
(∑
x a2x)2 .
• Diffusivity D(ω) from Green-Kubo.
IPR is large (∼ 1) for localized states, small (∼ N−d ) for extended states.
Binary mass distribution: m1 = 1 + ∆, m2 = 1−∆ (∆ = 0.8.
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J-versus-N plots: 3D
8 16 32 64N
0.001
0.01
JFree BCFixed BCPinned
N-0.75
N-1
(b)
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J-versus-N plots: 2D
4 16 64 256 1024N
0.001
0.01
0.1J
Free BCFixed BC
4 16 641e-06
1e-05
0.0001
0.001
N-0.75
N-0.6
N-3.7
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Comparision with numerical results
3D
Heuristic prediction (Numerical result)Unpinned fixed BC: J ∼ N−1 [∼ N−0.75]
Unpinned free BC: J ∼ N−3/4 [∼ N−0.75]
Pinned: J ∼ N−1 [∼ N−1]
2D
Heuristic prediction [Numerical result]Unpinned fixed BC: J ∼ N−1(ln N)−1/2 [∼ N−0.75]
Unpinned free BC: J ∼ N−2/3 [ ∼ N−0.6]
Pinned: J ∼ exp(−bN) [∼ exp(−bN) ]
Look at Phonon transmission functions.
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Phonon transmission: pinned 3D
2 2.4 2.8 3.20
0.08
0.16
0.24
0.32
N T
(ω)
8 8.4 8.8 9.2 9.6 10 10.4
N=8N=16N=32
ω
(i)
2 2.4 2.8 3.20
0.5
1
1.5
ρ(ω
)
DisorderedOrdered
8 8.4 8.8 9.2 9.6 10 10.4ω
(ii)
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Transmission: Unpinned 3D
0 1 2 3 4 5 6 7 8ω
1
10
100
1000
N3P
-1
N=8N=16
0 1 2 3 4 5 6 7 80
0.2
0.4
0.6
0.8
1
ρ(ω
)
Ordered LatticeDisordered Lattice
(iii)
(iv)
0 1 2 3 4 5 6 7 8ω
0
0.4
0.8N
T(ω
)
N=8N=16N=32
0 1 2 3 4 5 6 7 80
0.04
0.08
T(ω
)
N=8N=16N=32
(i)
(ii)
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Conclusions
Unpinned case: Results sensitive to boundary conditions. Heuristic arguments suggest thatFourier’s law valid for generic boundary conditions (fixed).Numerical verification presumably requires larger system sizes.
Pinned system: First numerical verification of Fourier’s law in a 3D system.
Pinned system: Direct verification of a 2D heat insulator.
Landauer and Green-Kubo formulas.N T (ω)/2π ? = ? D(ω) ρ(ω) (Proof)Our results suggest this is true for normal transport but not true for anomalous tranport.
Future questions
Exact results.
Improving numerical methods and study larger system sizes.
What happens in real systems ? Effect of anharmonicity, other degrees of freedom.
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