The Nonequilibrium, Discrete Nonlinear Schroedinger...

2371-12

Advanced Workshop on Energy Transport in Low-Dimensional Systems: Achievements and Mysteries

Stefano LEPRI

15 - 24 October 2012

Istituto dei Sistemi Complessi ISC-CNR Firenze

Italy

The Nonequilibrium, Discrete Nonlinear Schroedinger Equation

The nonequilibrium, discrete nonlinear Schrodingerequation

Stefano Lepri

Istituto dei Sistemi Complessi ISC-CNR Firenze, Italy

Stefano Lepri (ISC-CNR) Nonequilibrium DNLS 1 / 39

Outline

The open, one-dimensional DNLS equation

iφn = Vnφn − φn+1 − φn−1 + αn|φn|2φn+ . . .

Part I : Finite-temperature coupled transport[S. Iubini, S.L., A. Politi, Phys.Rev E 86, 011108 (2012)]

Part II : Short driven chains, nonreciprocal transmission[S.L., G. Casati, Phys. Rev. Lett. 106, 164101 (2011)]


DNLS for layered photonic or phononic crystal

asymmetricNonlinear

Linear Linear

For linear propagation perpendicular to the layers:

cos k(d1 + d2) = cos(ωd1

c1) cos(

ωd2

c2) −

1

2(c1

c2+

c2

c1) sin(

ωd1

c1) sin(

ωd2

c2)


DNLS for layered nonlinear media

Thin layers d1 � d2: ”Kronig-Penney model”

Approximate dispersion for high-frequency bands:ω(k) = ω0 ± 2C cos kd (single band approx.)

Defective layers

Kerr nonlinearity

Rescale units, band center at ω = 0

Altogether:iφn = Vnφn − φn+1 − φn−1 + αn|φn|2φn

Conservation of energy and norm, no harmonics.[A. Kosevich, JETP (2001)]


DNLS for BEC in optical lattices

Tight-binding + semiclassical approximations → DNLS eq.[Franzosi, Livi, Oppo, Politi, Nonlinearity (2011)]


Part I

Finite-temperature transport


Equilibrium: Grand-canonical thermodynamics

Let φn = pn + iqn, the isolated systems has 2 integrals of motion (Vn = 0,αn = α).

H =α

4

N∑i=1

(p2

i + q2i

)2+

N−1∑i=1

(pipi+1 + qiqi+1)

A =N∑

i=1

(p2i + q2

i ) .

Statistical weight: exp[−β (H − μA)].Equilibrium states: identified by (μ, T ) or by the densities h = H/N ,a = A/N .


Phase diagram

0 1 2 3 4 5 6a

-1

0

1

2

3

4

5h

T=0T=co

nst.

T=∞

μ=co

nst.

T = 0: Ground state (for α > 0) φn =√

ae−iμt

h = −2a +α

2a2

T = ∞: random phases (almost uncoupled oscillators)

h = αa2

[Rasmussen et al, PRL 2001]Stefano Lepri (ISC-CNR) Nonequilibrium DNLS 8 / 39

The usual game ...

Put DNLS chain in contact with two thermostats at the edges:

T μ μTL RL R

Not trivial! for instance: ”naive” Langevin will not work!Dissipation must preserve the ground state.


Monte-Carlo heat baths

1 At random time intervals (distributed in [tmin, tmax]), let

p1 → p1 + δp; q1 → q1 + δq

δp and δq are i.i.d. random variables uniformly distributed in [−R, R].

2 If (ΔH − μLΔA) < 0 accept the move, otherwise accept withprobability

exp {−T−1L (ΔH − μLΔA)}

3 Evolve the Hamiltonian dynamics till the next collision


Moves for conservative Monte-Carlo heat baths

Norm conserving thermostat- Random change of the phase:

θ1 → θ1 + δθ mod(2π)

δθ i.i.d., uniform in [0, 2π]. The total norm A is conserved.

Energy conserving thermostat- Consider the local energy

h1 = |φ1|4 + 2|φ1||φ2| cos (θ1 − θ2) . (1)

Two steps:

1 |φ1| is randomly perturbed. As a result, both the local amplitude andthe local energy change.

2 Then, by inverting, Eq. (1), a value of θ1 that restores the initial energyis seeked. If no such solution exists, choose a new perturbation for |φ1|.


Microscopic expressions for T and μ

For nonseparable Hamiltonians kinetic temperature is not simply 〈p2〉!1

T=

∂S∂H

,μ

T= −∂S

∂A,

where S is the thermodynamic entropy.[Franzosi, PRE 2011] For a system with two conserved quantities C1, C2

∂S∂C1

=

⟨W‖ξ‖∇C1 · ξ

∇ ·(

ξ

‖ξ‖W

)⟩mic

where

ξ =∇C1

‖∇C1‖− (∇C1 · ∇C2)∇C2

‖∇C1‖‖∇C2‖2

W 2 =2N∑

j,k=1

j<k

[∂C1

∂xj

∂C2

∂xk− ∂C1

∂xk

∂C2

∂xj

]2

,

and x2j = qj , x2j+1 = pj .Stefano Lepri (ISC-CNR) Nonequilibrium DNLS 12 / 39

Microscopic expressions for T and μ

Setting C1 = H and C2 = A: expression for T

Setting C1 = A and C2 = H: expression for μ

Both expressions are (ugly and) nonlocal (involve severalneighbouring pn and qn)

In practice: time-average expressions on short subchains around site nto obtain local values Tn and μn.

Check in equilibrium conditions TL = TR, μL = μR


Equilibration

0 1 2 3 4 5 6a

-1

0

1

2

3

4

5h

T=0T=1

T=∞

Computation of the isochemicals μ = 0, μ = 1 and μ = 2


Microscopic Currents

The expressions for the local energy- and particle-fluxes are derived in theusual way from the continuity equations for norm and energy densities,respectively

ja(n) = 2 (pn+1qn − pnqn+1)

jh(n) = − (pnpn−1 + qnqn−1)

Steady state : (ja(n) = ja and jh(n) = jh). Moreover it is also checkedthat ja and jh are respectively equal to the average energy and normexchanged per unit time with the reservoirs.


Linear irreversible thermodynamics

For small applied gradients:

ja = −Laad(βμ)

dy+ Lah

dβ

dy(2)

jh = −Lhad(βμ)

dy+ Lhh

dβ

dy

where we have introduced the continuous variable y = i/N ,L is the symmetric, positive definite, 2 × 2 Onsager matrix.detL = LaaLhh − L2

ha > 0.In energy-density representation the thermodynamic forces are ∇(−βμ)and ∇μ.


Thermodiffusion

The particle (σ) and thermal (κ) conductivities

σ = βLaa; κ = β2 detL

Laa.

”Seebeck coefficient“ (ja = 0)

S = β

(Lha

Laa− μ

),

Figure of merit

ZT =σS2T

κ=

(Lha − μLaa)2

detL;


Transport

102

103

N

10-2

10-1

ja,h

10-3

10-2

10-1

ja,h

(a)

(b)

(a) High-temperature regime TL = 2, TR = 4, μ = 0(b) Low-temperature regime TL = 0.3, TR = 0.7, μ = 1.5


Linear response: Onsager coefficients

1 2 3-1 0

1 2

0

400

800

Laa

Tμ

1 2 3-1 0

1 2 0

800

1600

Lah

Tμ

1 2 3-1 0

1 2 0

800

1600

Lha

Tμ

1 2 3-1 0

1 2

0 800

1600 2400 3200

Lhh

Tμ

N = 500; ΔT = 0.1, Δμ = 0.05


Linear response: Seebeck coefficient

(a)

0.5 1 1.5

2 2.5

3

T

-0.5 0 0.5 1 1.5 2

μ

0

0.1

0.2

0.3S

-0.1-0.05 0 0.05 0.1 0.15 0.2 0.25 0.3

S = 0 for Lha/Laa = μ


Linear response

0 0.5 1y

1.1

1.2

1.3

1.4

T(y)μ(y)

0 0.5 1y

0

0.5

1

atot

= 1.5atot

= 4

(b)

TL = 1, TR = 1.5; norm-conserving thermostats


Nonlinear regimes

Nonmonotonous profiles:

0

1

2

3T(y)μ(y)

0 0.2 0.4 0.6 0.8 1y

0

2

4

6

a(y)h(y)

(a)

(b)

N = 3200 sites and TL = TR = 1, μL = 0, μR = 2


Nonlinear regimes

0 1 2 3 4 5 6a

-1

0

1

2

3

4

5h

T=0T=1

T=∞

N = 200, 800, 3200, (a(y), h(y)) “pushed” away from the T = 1isothermal.


Profile reconstruction

1 Rewrite constitutive equations as(ja

jh

)= A(μ, T )

d

dy

(μT

)

A is expressed in terms of L, T and μ (e.g. A11 = −Laa/T ).

2 If A−1 exists

d

dy

(μT

)= A

−1(μ, T )

(ja

jh

)3 Compute A in the linear response regime

4 Integrate the two ”nonautonomous” linear differential equations withthe numerical values (ja, jh) to reconstruct the profile.


Profile reconstruction

0

20

40

dT/d

y

0 0.2 0.4 0.6 0.8 1y

0

10

20

30

dμ/d

y

(a)

(b)

N = 250, N = 1000 (dot)


Part II

Driven DNLS: Nonreciprocal transmission


A DNLS chain embedded in a linear lattice

iφn = Vnφn − φn+1 − φn−1 + αn|φn|2φn

Infinite lattice , Vn, αn = 0 only for 1 ≤ n ≤ NIntegrate out the φn for n ≤ 0 and n > N :

φ0(t) = F0(t) − i

∫ t

0G(t − s)φ1(s)ds

φN+1(t) = FN+1(t) − i

∫ t

0G(t − s)φN (s)ds

Memory term G(t) = J1(2t)/t.From Hamiltonian problem to driven, dissipative


Motivation: the (ideal) “wave diode”

ωA,

A,ω

To violate the reciprocity theorem (without breaking time-reversal) bothasymmetry and nonlinearity are necessary !


Transmission problem

Stationary DNLS, φn = ψne−iωt, Vn = 0 and αn = for 1 ≤ n ≤ N

ωψn = Vnψn − ψn+1 − ψn−1 + αn|ψn|2ψn

ψ1 . . . ψN

Teikn

Re−ikn

R0eikn

ω = −2 cos k, 0 ≤ k ≤ π


Transmission problem

Look for complex solutions such that:

ψn =

{R0e

ikn + Re−ikn n ≤ 1

Teikn n ≥ N

ψn complex, current J = 2|T |2 sin k

Non-mirror symmetric couplings: Vn = VN−n+1 and/or αn = αN−n+1

Convention: k < 0 is for (Vn, αn) −→ (VN−n+1, αN−n+1) (“flippedsample”)

For αn = 0: reciprocity for any Vn


Reduction to nonlinear map

Let un = ψn and vn = ψn+1. Back iterating from uN = T exp(ikN),vN = T exp(ik(N + 1))

un−1 = −vn + (Vn − ω + αn|un|2)un, vn−1 = un

Map is area preserving.For given T and k

R0 =exp(−ik)u0 − v0

exp(−ik) − exp(ik), R =

exp(ik)u0 − v0

exp(ik) − exp(−ik)

Transmission coefficient

t(k, |T |2) =|T |2|R0|2


The simplest case: the dimer N = 2

For k > 0:

t =

∣∣∣∣ eik − e−ik

1 + (ν − eik)(eik − δ)

∣∣∣∣2

δ = V2 − ω + α2T2, ν = V1 − ω + α1T

2[1 − 2δ cos k + δ2].

For k < 0: exchange the subscripts 1 and 2Symmetric case (V1,2 = V0, α1,2 = α): two nonlinear resonances

V0 + αT 2 = 0 (V0 < 0)

V0 + αT 2 = ω (V0 < ω)


The dimer N = 2: transmission curves

0 1 2 3 4 5

|R0|2

0

0.5

1

wav

e tr

ansm

issi

on c

oeff

icie

nt t

k=+π/2k=-π/2

I

II

k = π/2, αn = 1, V1,2 = V0(1 ± ε) V0 = −2.5 ε = 0.05.


Stability

-10 -5 0 5 10lattice index n

-2

-1

0

1

2I

-10 -5 0 5 10lattice index n

-2

-1

0

1

2II

-2 -1 0 1 2Re λ

-0.4

-0.2

0

0.2

0.4

0.6

Im λ

-2 -1 0 1 2Re λ

-0.4

-0.2

0

0.2

0.4

0.6

Im λ

|R0|2 = 2 tI = 0.99 and tII = 0.30


Oscillatory instability

-100

-50

0

50

100 0 10

20 30

40 50

2

4|φn|2

n

t

|φn|2

Radiation and birth of localized mode (with frequency outside of thephonon band) on a plane-wave background.


Quasiperiodic solution

-10 -5 0 5 10 15lattice site n

-2

0

2

4am

plitu

de


Wavepacket transmission

Numerical simulation on a finite lattice |n| < M

iφn = Vnφn − φn+1 − φn−1 + αn|φn|2φn

Initial condition: Gaussian

φn(0) = I exp

[−(n − n0)

2

w2+ ik0n

]

Transmission coefficient (for n0 < 0)

tp =

∑n>N |φn(tfin)|2∑

n<0 |φn(0)|2


Wavepacket transmission

50

100

150

200

-400 -200 0 200 400

t

n

-400 -200 0 200 400

n

0 0.5 1 1.5 2 2.5 3 3.5 4


Summary

1 Steady coupled transport

� Monte Carlo thermostats� Normal transport, except at very low T� Nonmonotonous energy and density profiles� S changes sign increasing the interaction

2 Driven chains: nonreciprocal transmission

� Simple modeling of “wave diode”� Nonlinear resonances and multistability� Oscillatory instabilities� Nonreciprocal wavepacket trasmission


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