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Green-Kubo Formula for Open Systems

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Green-Kubo Formula for Open Systems Green-Kubo formula for open systems Abhishek Dhar Anupam Kundu (CEA, Grenoble) Onuttom Narayan (UC, Santa Cruz) Keiji Saito (Tokyo University) Raman Research Institute SNBose, Kolkata, India November 2010 (RRI) July 2010 1 / 31 Outline Introduction to the problem. Green-Kubo formula for thermal conductivity. Anomalous heat conduction in low-dimensional systems. Conductance for open systems: Landauer formula, Fluctuation theorem. Exact linear response formula for conductance. Frequency dependent conductance. Summary. Classical treatment. (RRI) July 2010 2 / 31 Response to small temperature difference J T TL R For small ∆T = TL − TR and system size L : ∆T Fourier0s law implies : j ∼ κ L The thermal conductivity κ is expected to be an intrinsic material property. (RRI) July 2010 3 / 31 Computing κ j L κ = ∆T Computing κ is difficult since this is a nonequilibrium problem. In equilibrium we know that e−βH(x,v) Prob[x, v] = P (x, v) = 0 Z Hence we can find the expectation value of any physical observable, say A(x, v). Thus Z Z hAi = dx dv A(x, v) P0(x, v) . For a system with one end at T and the other end at T + ∆T , in general, one does not know Prob(x, v) even in the nonequilibrium steady state. Hence difficult to find j = hj(x, v)i. (RRI) July 2010 4 / 31 Green-Kubo linear response theory For small deviations from equilibrium caused by small changes of Hamiltonian, one can do perturbation theory. Start with equilibrium phase space distribution P0(x, v) corresponding to Hamiltonian H. In this case hJi = 0. Add perturbation (e.g electric field): st X HT = H + Ve V = −e φ(xi ) ∂P Solve eqn. of motion ∂t = −{P, HT } to linear order in V to find Pneq = Peq + δP . Calculate hji = R dx R dv j(x, v) δP . This gives a formula for the conductivity in terms of equilibrium current autocorrelation functions. (RRI) July 2010 5 / 31 Green-Kubo formula for thermal conductivity 1 Z τ κ = lim lim dt h J(t) J(0) i , 2 d τ→∞ L→∞ kBT L 0 where J(t) = R j(x, t)dx, and j(x, t) is the heat flux density in x -direction. L is the linear size of the system. However..... For heat conduction derivation is more difficult since current is response to boundary driving forces which cannot be included in the Hamiltonian. Requires assumption of local thermal equilibrium. Luttinger’s mechanical derivation: assumes local thermal equilibium and relation between responses to applied field and temperature gradient. Limit of infinite system size necessary. Order of limits important. (RRI) July 2010 6 / 31 Heat conduction in finite systems Clearly the usual Green-Kubo formula cannot be directly used for: (i) small systems (e.g single molecules) and (ii) low-dimensional systems with anomalous transport. Here, instead of conductivity, one is interested in the actual current through the system or its conductance defined by: J G = . ∆T (RRI) July 2010 7 / 31 Anomalous transport Fourier’s law is not generally valid in low-dimensional systems . κ depends on system size L. Necessary and sufficient conditions for validity of Fourier’s law ? Role of anharmonicity, disorder and dimensionality. Lepri, Livi, Politi, Phys. Rep. (2003) . Dhar , Adv. Phys. (2008) . (RRI) July 2010 8 / 31 One-dimensional systems with nonintegrable interactions Momentum conserving system: FPU - model N 2 N+1 " 2 3 4 # X p X (ql − ql−1) (ql − ql−1) (ql − ql−1) H = l + k + k + λ . 2m 2 2 3 3 4 l=1 l=1 Momentum non-conserving system: φ4 - model N " 2 2 # 2 N 4 X pl ql X (ql − ql−1) X ql H = + ko + k + λ . 2m 2 2 2 4 l=1 l=1,N+1 l=1 Momentum conserving: κ ∼ L1/3 Momentum nonconserving (pinned case): κ ∼ L0 Disorder + Anharmonicity: κ ∼ Lα. Computing α is a difficult problem. Diverging conductivity leads to a problem in use of the Green-Kubo formula. (RRI) July 2010 9 / 31 Conductance of finite systems: earlier results Landauer-like formula for thermal conductance. Can be derived using: (i) nonequilibrium Green’s function (Keldysh formalism) , (i) Langevin equations of motion approach, (iii) Green-Kubo formula by including infinite reservoirs. Most useful for non-interacting (e.g. harmonic) systems. Steady state fluctuation theorem of Gallavotti-Cohen implies a Green-Kubo-like formula. Valid for finite OPEN systems. No general proof exists. (RRI) July 2010 10 / 31 Landauer-like formula for thermal conductance TL TR Exact expression for nonequilibrium heat current [“Landauer-like” formula for phonons.] In classical case: I k Z ∞ G = = B dωT (ω) , ∆T 2π 0 where T (ω) is the phonon transmission function. [Casher and Lebowitz (1971), Rubin and Greer (1971), Dhar and Roy (2006)]. In quantum case: 1 Z ∞ I = dω ~ωT (ω)[f ( ω, TL) − f (ω, TR )] . 2π 0 (RRI) July 2010 11 / 31 Steady state fluctuation theorem Cohen-Gallavotti SSFT: Let ∆β = 1/TR − 1/TL. Consider rate of entropy production over time τ. Z τ Q Q = dt j(t) , s = (∆β) 0 τ P(s) = exp[sτ] τ → ∞ P(−s) Let Z(λ) = hexp[−λQ]i ∼ exp[g(λ)τ]. SSFT implies the symmetry relation: g(λ) = g(∆β − λ). Expanding both sides and comparing coefficients gives: h j i 1 Z ∞ G = lim ∆T = hj(0)j(t)i . 2 T ∆T →0 ∆T kBT 0 (Gallavotti, Lebowitz-Spohn, Andrieux-Gaspard): (RRI) July 2010 12 / 31 New results For J = R dxj(x): h J i 1 Z ∞ lim ∆T = hJ(0)J(t)i . 2 T ∆T →0 ∆T kBT L 0 Correlation is evaluated not with Hamiltonian dynamics but with the dynamics of the open system. Heat bath dynamics either deterministic or stochastic. No need of infinite size limit. In this work: A general derivation of the open-system conductance formula, derived by perturbation theory, and without invoking the fluctuation theorem. Generalization of the result to the case of current response to time-dependent temperature perturbations. (RRI) July 2010 13 / 31 Outline of the derivation Hamiltonian system in contact with heat baths: stochastic Markovian dynamics. Write Fokker-Planck equation for phase space distribution P(x, v, t). Set P = P0 + p and solve for p −βH perturbatively to O(∆T ).[ P0 = e /Z]. Use p to compute hJi∆T . This gives the response in terms of the correlator hJ(0)Jfp(t)i where Jfp is a boundary current operator. Relate hJ(0)Jfp(t)i to hJ(0)Jb(t)i where Jb is an instantaneous boundary current operator. Use continuity equations to relate hJ(0)Jb(t)i to hJ(0)J(t)i. (RRI) July 2010 14 / 31 One dimensional lattice Hamiltonian: Langevin reservoirs j 1, L −jNR, m i TL TR j i+ 1 , i N " 2 # N−1 X ml vl X H = + V (x ) + U(x − x + ) 2 l l l 1 l=1 l=1 L L m1v˙1 = f1 − γ v1 + η ml v˙l = fl l = 2, 3, ...N − 1 R R mN v˙N = fN − γ vN + η . fl = −∂H/∂xl , L L 0 L 0 R 0 0 hη (t)η (t )i = 2γ kBTLδ(t − t ) , hηR (t)η (t )i = 2γR kBTR δ(t − t ) . (RRI) July 2010 15 / 31 Definitions Local energy: l (t) Continuity equation : ∂l (t)/∂t = jl,l−1 − jl+1,l + δl,1j1,L + δl,N jN,R . This defines the local current operators: 1 j , − = (v − + v ) f , − l l 1 2 l 1 l l l 1 j1,L = v1 (−γLv1 + ηL) jN,R = vN (−γR vN + ηR ) P Total current: J = l=1,N−1 jl+1,l (RRI) July 2010 16 / 31 Fokker-Planck equation ∂P(x, v, t) = LˆH P(x, v, t) + LˆBP(x, v, t) . ∂t N X „ ∂ ∂ f « LˆH = − + v + l ∂x l ∂v m l=1 l l l L „ « R „ « ˆB γ ∂ kBTL ∂ γ ∂ kBTR ∂ L = v1 + + vN + . m1 ∂v1 m1 ∂v1 mN ∂vN mN ∂vN 1 Let T = 2 (TL + TR ) and ∆T = (TL − TR ). (RRI) July 2010 17 / 31 Fokker-Planck equation ∂P(x, v, t) = LPˆ (x, v, t) + Lˆ∆T P(x, v, t) . ∂t L „ « R „ « ˆ ˆH γ ∂ kBT ∂ γ ∂ kBT ∂ L = L + v1 + + vN + , m1 ∂v1 m1 ∂v1 mN ∂vN mN ∂vN ! k ∆T γL ∂2 γR ∂2 Lˆ∆T (v) = B − . 2 2 2 2 2 m1 ∂v1 mN ∂vN At time t = −∞ system is in equilibrium: P = P0 = exp(−βH)/Z. Solve Fokker-Planck equation perturbatively upto order ∆T . (RRI) July 2010 18 / 31 Solution of Fokker-Planck equation Let P(x, v, t) = P0(x, v) + p(x, v, t). Z t 0 Lˆ(t−t0) ˆ∆T p(x, v, t) = dt e L (v) P0(x, v) −∞ Z t 0 Lˆ(t−t0) = ∆β dt e Jfp(v) P0(x, v) , −∞ L » – R » – γ 2 kBT γ 2 kBT with Jfp(v) = − v1 − + vN − . 2 m1 2 mN Note that: 2 • hj1,Li = hv1(−γLv1 + ηL)i = −γL (v1 − kBTL/m1). −1ˆ∆T ˆ 1 ∂P ˜ • Jfp = (∆β P0) L P0 = ∆β . P ∂t P=P0 (RRI) July 2010 19 / 31 Finding hJi∆T The expectation value of total current is given by: Z hJi∆T = dxdv J p(x, v) Z ∞ Z Ltˆ = ∆β dt dxdv J e Jfp P0 0 Z ∞ = ∆β dt hJ(t)Jfp(0)iT 0 Z ∞ = −∆β dt hJ(0)Jfp(t)iT (using detailed balance principle) . 0 Jfp does not have any obvious physical interpretation.
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