On solving many-body Lindblad equation and quantum phase transition far from equilibrium

On solving many-body Lindblad equation and quantum phase transition far from equilibrium Department of Physics, FMF, University of Ljubljana, SLOVENIA MECO 35, Pont-a-Mousson 16.3.2010

Outline of the talk One dimensional quantum many body systems far from equilibrium: Paradigmatic models: open quantum spin chains with boundary openings On efficient computational methods: Analytical: Diagonalization of quantum Liouvillians in terms of Fock space of operators Numerical: Time-dependent DMRG in the Liouville space Idea: Non-equilibrium steady state may have similar computational complexity as the "ground state" of one-dimensional Hamiltonian models. On new exotic physics: Quantum phase transitions far-from equilibrium (example: anisotropic XY spin chain with transverse magnetic field) Finite size scaling offers surprising connection to wave resonators!

Many-body Lindblad equation The central equation we address is the Lindblad equation for the many-body density operator ρ(t): dρ dt = ˆLρ := i[h, ρ] + X µ 2L µρl µ {L µl µ, ρ} where H is a many-body (Hamiltonian) with local couplings, and L µ are Lindblad bath operators which act locally on degrees of freedom near each bath. The most general representation of infinitesimal generator of completely positive, Markovian dynamics (Lindblad 1976/Gorini, Kosakowski, Sudarshan 1976)

Solvable example: open XY quantum spin chains Consider magnetic and heat transport of a Heisenberg XY spin 1/2 chain, with arbitrary either homogeneous or positionally dependent (e.g. disordered) nearest neighbour interaction Xn 1 H = `Jx m σmσ x m+1 x + Jmσ y mσ y y m+1 + m=1 nx h mσm z (1) which is coupled to two thermal/magnetic baths at the ends of the chain, generated by two pairs of canonical Lindblad operators L 1 = 1 q Γ L 1 2 σ 1 L 3 = 1 q Γ R 1 2 σ n L 2 = 1 q Γ L 2 2 σ+ 1 L 4 = 1 q Γ R 2 2 σ+ n (2) where σ m ± = σm x ± iσm y and Γ L,R 1,2 are positive coupling constants related to bath temperatures/magnetizations. e.g. if spins were non-interacting the bath temperatures T L,R would be given with Γ L,R 2 /Γ L,R 1 = exp( 2h 1,n /T L,R ). m=1

Quantum phase transition far from equilibrium in XY chain TP & I. Pižorn, PRL 101, 105701 (2008) Jm x = (1 + γ)/2 Jm y = (1 γ)/2, h m = h C(j,k) = σj z σk z σj z σk z 1 Cres 10 4 10 5 1.5 h 0.9 hc h 0.3 hc h 10 6 10 7 ImΒ 1. 10 8 10 9 0.5 0 0 Γ 1 0 0.002 0.001 ReΒ 0 Π 2 1 0 1 2 Π 0 0.001 0.002 Φ ReΒ h c = 1 γ 2

Analytical solution: "Third quantization" TP, New J. Phys. 10, 043026 (2008) Consider a general solution of the Lindblad equation: dρ dt = ˆLρ := i[h, ρ] + X µ 2L µρl µ {L µl µ, ρ} for a general quadratic system of n fermions, or n qubits (spins 1/2) H = 2nX j,k=1 w j H jk w k = w H w L µ = 2nX j=1 l µ,j w j = l µ w where w j, j = 1, 2,..., 2n, are abstract Hermitian Majorana operators Two physical realizations: {w j,w k } = 2δ j,k j, k = 1,2,..., 2n canonical fermions c m, w 2m 1 = c m + c m,w 2m = i(c m c m),m = 1,..., n. spins 1/2 with canonical Pauli operators σ m, m = 1,..., n, Y w 2m 1 = σj x σ z m w2m = Y σy m m <m m <m σ z m

Fock space of operators Let us associate a Hilbert space structure x x to a linear 2 2n = 4 n dimensional space K of operators, with basis P α1,α 2,...,α 2n := w α 1 1 w α 2 2 w α 2n 2n α j {0, 1} orthogonal with respect to an inner product x y = tr x y K is just a usual Fock space with an unusual physical interpretation. Define a set of 2n adjoint annihilation linear maps ĉ j over K ĉ j P α = δ αj,1 w j P α and derive the actions of their Hermitian adjoints - the creation linear maps ĉ, P α ĉ j Pα = Pα ĉ j P α = δ α j,1 P α w j P α = δ αj,0 P α w j P α : ĉ j Pα = δα j,0 w jp α

Representing quantum Liouvillean in 3rd quantization picture The key is to reallize now that quatum Liouville oprator ˆL of a general quadratic system can be represented as a quadratic form in a-fermi maps ĉ j,ĉ j. This we do in two steps, decomposing the Liouvillean as ˆL = ˆL 0 + ˆL bath where ˆL 0 is the unitary part ˆL 0ρ := i[h, ρ] and ˆL bath = P ˆL µ µ is a sum of dissipative bath operators ˆL µρ := 2L µρl µ {L µl µ, ρ}

Step 1: Unitary part Let us define a Lie derivative on K: ad A B := [A, B].

Step 1: Unitary part Let us define a Lie derivative on K: ad A B := [A, B]. Then: ad w j w k P α = w j w k P α P αw j w k = = 2(δ αj,1δ αk,0 + δ αj,0δ αk,1) w j w k P α = = 2(ĉ j ĉ k + ĉ j ĉk) P α = 2(ĉ j ĉk ĉ kĉj) P α

Step 1: Unitary part Let us define a Lie derivative on K: ad A B := [A, B]. Then: ad w j w k P α = w j w k P α P αw j w k = = 2(δ αj,1δ αk,0 + δ αj,0δ αk,1) w j w k P α = = 2(ĉ j ĉ k + ĉ j ĉk) P α = 2(ĉ j ĉk ĉ kĉj) P α Extending this relation by linearity to an arbitrary element of K, it follows that for an arbitrary quadratic Hamiltonian its Lie derivative has a very similar quadratic form in a-fermi maps ˆL 0 = iad H = 4i 2nX j,k=1 ĉ j H jkĉ k = 4iĉ H ĉ

Step 2: Dissipative part ˆL µρ = 2L µρl µ {L µl µ, ρ} = 2nX j,k=1 where we write ˆL j,k ρ := 2w j ρw k w k w j ρ ρw k w j. l µ,j l µ,k ˆL j,k ρ

Step 2: Dissipative part ˆL µρ = 2L µρl µ {L µl µ, ρ} = 2nX j,k=1 where we write ˆL j,k ρ := 2w j ρw k w k w j ρ ρw k w j. l µ,j l µ,k ˆL j,k ρ Straighforward calculation shows that, if we restrict ˆL j,k to the subspace of K with even number a fermions, i.e. subspace spanned by P α with even α := α 1 +... + α 2n, then ˆL j,k = 4ĉ j ĉ k 2ĉ j ĉk 2ĉ kĉj

Liouvillean in 3rd quantization: result Putting 1 and 2 together we find the original problem formulated as a quadratic problem in 2n a-fermions d dt ρ = ˆLρ namely ˆL = 2ĉ (2iH + M + M T ) ĉ + 2ĉ (M M T )ĉ where M = M is a complex Hermitian matrix parametrizing the Lindblad operators M jk = P µ l µ,jl µ,k. The question which remains is whether a linear canonical transformation to a new set of canonical a-fermi maps, we shall call them normal master modes ˆb j, so that ˆL becomes diagonal quadratic form in ˆb j?

Reduction to normal master modes: Step 1 Following Lieb, Schultz and Mattis (1961), generalizing to non-hermitean case:

Reduction to normal master modes: Step 1 Following Lieb, Schultz and Mattis (1961), generalizing to non-hermitean case: Define 4n adjoint Hermitian Majorana maps â r = â r, r = 1,2,..., 4n: â 2j 1 = 1 2 (ĉ j + ĉ j ) â 2j = i 2 (ĉ j ĉ j ) satisfying the anti-commutation relations {â r,â s} = δ r,s in terms of which the Liouvillean can be rewritten as ˆL = â A â A 0ˆ½ (3) where A 0 = 2 P 2n j=1 M jj = 2trM and A antisymmetric complex 4n 4n matrix A 2j 1,2k 1 = 2iH jk M jk + M kj A 2j 1,2k = 2iM kj A 2j,2k 1 = 2iM jk A 2j,2k = 2iH jk + M jk M kj

Reduction to normal master modes: Step 2 Lemma: Diagonalizable complex anti-symmetrix matrix can be written as 0 1 0 β 1 0 0 β 1 0 0 0 A = V T BV where B = 0 0 0 β 2 B @ 0 0 β 2 0 C A....... where the set of 2n c-numbers ordered as Reβ 1 Reβ 2... β 2n 0 shall be called rapidities, and eigenvector matrix V satisfies 0 1 0 1 0 0 1 0 0 0 VV T = J = 0 0 0 1 B @ 0 0 1 0 C A.......

Reduction to normal master modes: result Using the previous decomposition of A we rewrite the Liouvillean as ˆL = â V T BVâ A 0ˆ½ = (Vâ) B(Vâ) A 0ˆ½

Reduction to normal master modes: result Using the previous decomposition of A we rewrite the Liouvillean as ˆL = â V T BVâ A 0ˆ½ = (Vâ) B(Vâ) A 0ˆ½ Let us define the Normal-master-mode (NMM) maps ˆb := (ˆb 1, ˆb 1, ˆb 2, ˆb 2,..., ˆb 2n, ˆb 2n) := Vâ which, due to orthogonality of matrix V satisfy canonical relations {ˆb j, ˆb k } = 0 {ˆb j, ˆb k} = δ j,k {ˆb j, ˆb k} = 0 ˆb j could be interpreted as annihilation map and ˆb j as a creation map of jth NMM, but we should note that ˆb j is in general not the Hermitian adjoint of ˆb j.

Non-equilibrium steady state: Liouvillean vacuum

Non-equilibrium steady state: Liouvillean vacuum From ˆL = 2ĉ (2iH+M+M T ) ĉ +2ĉ (M M T ) ĉ it follows immediately that 1 = P (0,0...,0) is left-annihilated by ˆL, 1 ˆL = 0.

Non-equilibrium steady state: Liouvillean vacuum From ˆL = 2ĉ (2iH+M+M T ) ĉ +2ĉ (M M T ) ĉ it follows immediately that 1 = P (0,0...,0) is left-annihilated by ˆL, 1 ˆL = 0. So we have just shown that 0 is always an eigenvalue of ˆL +, hence there should also exist the corresponding right eigenvector NESS, normalized as 1 NESS = tr ρ NESS = 1, which represents density matrix of non-equilibrium steady state ˆL NESS = 0

Complete excitation spectrum of quantum Liouvillean The complete excitation spectrum and the corresponding left/right eigenvectors of the Liouvillean are given in terms of a sequence of 2n binary integers (NMM occupation numbers) ν = (ν 1, ν 2,..., ν 2n), ν j {0, 1}, namely λ ν = 2 Θ L ν ˆL = λ ν Θ L ν ˆL Θ R ν = λ ν Θ R ν 2nX j=1 β j ν j Θ L ν = 1 ˆb ν 2n 2n ˆb ν 2 ˆb ν 1 2 1 Θ R ν = ˆb ν 1 1 ˆb ν 2 2 ˆb ν 2n 2n NESS where by construction, left and right eigenvectors satisfy the bi-orthonormality relation Θ L ν ΘR ν = δ ν,ν.

NESS expectation values of physical observables Assume that all the rapidities are strictly away from the real line Reβ j > 0. Then: 1 NESS is unique. 2 Any initial density matrix relaxes to NESS with an exponential rate = 2min Reβ j. 3 Then the expectation value of any quadratic observable w j w k in a (unique) NESS can be explicitly computed as where w j w k NESS = δ j,k + 1 ĉ j ĉ k NESS = δ j,k + 2 = δ j,k 1 π 2nX m=1 Z G(ω) = (A iω½) 1 V 2m,2j 1 V 2m 1,2k 1 dω G 2j 1,2k 1 (ω)

Coming back to XY chain 0 1 B L h 1R R 1 0 0. R T 1 h 2R R.. 2 0 A = 0 R T 2 h 3R., A 0 = Γ L + + Γ R +, B @....... C Rn 1 A 0 0 R T n 1 B R h nr where B L := B Γ L +,Γ L, BR := B Γ R +,ΓR, ΓL,R ± R m := := ΓL,R 2 ± Γ L,R 1, and 0 1 0 1 0 0 Jm y 0 0 0 1 0 B 0 0 0 Jm y C @ Jm x 0 0 0 A, R := B 0 0 0 1 C @ 1 0 0 0A, 0 Jm x 0 0 0 1 0 0 0 i 0 2 Γ+ 1 i 2 Γ 1 2 Γ B Γ+,Γ := B i 2 Γ+ 0 1 2 Γ i 2 Γ @ i 2 Γ 1 2 Γ 0 C i 2 Γ+ A 1 2 Γ i 2 Γ i 2 Γ+ 0

Transverse Ising chain First we discuss a simple special case of homogeneous transverse Ising chain, say with numerical values J x m 1.5, J y m 0, h m 1.0 and take non-equilibrium bath couplings Γ L 1 = 1,Γ L 2 = 0.6, Γ R 1 = 1, Γ R 2 = 0.3

Transverse Ising chain First we discuss a simple special case of homogeneous transverse Ising chain, say with numerical values ImΒ 2 J x m 1.5, J y m 0, h m 1.0 and take non-equilibrium bath couplings Γ L 1 = 1,Γ L 2 = 0.6, Γ R 1 = 1, Γ R 2 = 0.3 1 0 1 2 ImΒ 0.1 0.2 0.3 0.4 ReΒ Rapidity spectrum for different n: (note analytic results for n = with colored symbols!) 2 1 0 1 0.1 0.2 0.3 0.4 ReΒ 2 ImΒ 2 1 0 1 0.1 0.2 0.3 0.4 ReΒ 2

Spectral gap of Liouvillean The rate of relaxation to NESS is given by the spectral gap of ˆL.

Spectral gap of Liouvillean The rate of relaxation to NESS is given by the spectral gap of ˆL. We find explicit analytical result: = C(J x,h, Γ L,R 1,2 ) n 3

Spectral gap of Liouvillean The rate of relaxation to NESS is given by the spectral gap of ˆL. We find explicit analytical result: Comparing to numerics: 15 = C(J x,h, Γ L,R 1,2 ) n 3 10 n 3 5 Log n 3 C 1 0 1 2 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 Log n 0 0 50 100 150 200 250 300 350 n

Complete spectrum of Liouvillean for n = 6 15 10 5 ImΛ 0 5 10 15 5 4 3 2 1 0 ReΛ

Spin-spin correlations in NESS (XY chain) Saturation vs. exponential decay & power law critical scaling at the critical point. ln C r 5 10 15 20 a 0 25 10 20 30 30 r n ln C r Α ln n h h c 35 40 n 160 320 640 35 0.0 0.2 0.4 0.6 0.8 40 0 1 2 3 4 5 6 40 0 20 40 60 80 100 120 140 ln r r 5 10 15 20 25 30 b h h c lnξ 2.5 2.0 1.5 1.0 0.5 0.0 8 7 6 5 4 ln h

Fluctuation of spin-spin correlation in NESS and "wave resonators" Near QPT: Scaling variable z = (h c h)n 2

Fluctuation of spin-spin correlation in NESS and "wave resonators" Near QPT: Scaling variable z = (h c h)n 2 Scaling ansatz: C 2j+α,2k+β = Ψ α,β (x = j/n, y = k/n,z)

Fluctuation of spin-spin correlation in NESS and "wave resonators" Near QPT: Scaling variable z = (h c h)n 2 Scaling ansatz: C 2j+α,2k+β = Ψ α,β (x = j/n, y = k/n,z) Certain combination Ψ(x,y) = ( / x + / y)(ψ 0,0 (x,y) + Ψ 1,1 (x,y)) obeys Helmoltz equation!!! «2 x + 2 2 y + 4z Ψ = octopole antenna sources 2

Operator space entanglement entropy of NESS (XY chain) Von Neumann entropy of a bipartition of NESS as an element of a Fock space Drastically different behaviour than for entanglement entropy of ground states of 1D critical/non-critical models! S 25 20 15 3.0 2.5 2.0 1.5 1.0 0.5 0.0 0 10 20 30 40 n ln slope 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7 6 5 4 3 2 ln h c h 10 h 0.5 h 0.74 5 h 0.7498 0 h 0.75 0 200 400 600 800 1000 1200 n

Summary Method for exact solution of many-body open quantum systems has been outlined. Quantum phase transition far from equilibrium discovered - experimental tests desirable! Possible new realizations of wave/quantum chaos!?