On a Random Matrix Models of Quantum Relaxation

arXiv:0711.2356v1 [math-ph] 15 Nov 2007

Contemporary Mathematics

On a Random Matrix Models of Quantum Relaxation J. L. Lebowitz, A. Lytova, and L. Pastur Abstract. In paper [7] two of us (J.L. and L.P.) considered a matrix model for a two-level system interacting with a n × n reservoir and assuming that the interaction is modelled by a random matrix. We presented there a formula for the reduced density matrix in the limit n → ∞ as well as several its properties and asymptotic forms in various regimes. In this paper we give the proofs of assertions, announced in [7]. We present also a new fact about the model (see Theorem 2.1) as well as additional discussions of topics of [7]

1. Introduction The model considered in [7] can be viewed as a random matrix version of the spin-boson model, widely used in studies of open quantum systems (see e.g. review works [8, 12] and references therein). We mention here that one of the first models of this type, namely the model where the classical system is represented by a harmonic oscillator coupled linearly with the oscillator reservoir, was considered by N. Bogolyubov in 1945 [3], Chapter IV. We recall now the model, proposed and discussed in [7]. Let hn be a Hermitian (n) n × n matrix with eigenvalues Ej , j = 1, ..., n. We characterize the spectrum of hn by its normalized counting measure of eigenvalues Z ∞ n X (n) (n) (n) ν0 (dE) = 1, χ∆ (Ej ); (1.1) ν0 (∆) = n−1 −∞

j=1

(n)

where χ∆ is the indicator of an interval ∆ ⊂ R. We assume that ν0 converges weakly as n → ∞ to a limiting probability measure ν0 , i.e. that for any bounded and continuous function ϕ : R → R we have: Z ∞ Z ∞ Z ∞ (n) (1.2) lim ν0 (dE)ϕ(E) = ν0 (dE)ϕ(E), ν0 (dE) = 1. n→∞

−∞

−∞

−∞

Let wn be a Hermitian n × n random matrix, whose probability density is  2 (1.3) Q−1 n exp −Tr wn /2 , 1991 Mathematics Subject Classification. Primary , ; Secondary , Key words and phrases. Random matrix. Department of Physics of Rutgers University. c

0000 (copyright holder)

1

2

J. L. LEBOWITZ, A. LYTOVA, AND L. PASTUR

where Qn is the normalization constant. In other words, the entries wjk , 1 ≤ j ≤ k ≤ n of the matrix wn are independent Gaussian random variables with  1 





(1.4) hwjk i = 0, wjj 2 = 1, j, k = 1, ..., n, (ℜwjk )2 = (ℑwjk )2 = , j 6= k, 2

where the symbol h...i denotes here and below the expectation with respect to the distribution (1.3). This probability distribution is known as the Gaussian Unitary Ensemble (GUE) [9]. We define the Hamiltonian of our composite system S2,n as a random 2n × 2n matrix of the form H (n) = sσ z ⊗ 1n + 12 ⊗ hn + vσ x ⊗ wn /n1/2 ,

(1.5)

where 1l (l = 2, n) is the l × l unit matrix, σ z and σ x are the Pauli matrices     0 1 1 0 σx = , σz = , 1 0 0 −1 the symbol ⊗ denotes the tensor product, and s and v are positive parameters. The first term in (1.5) is the Hamiltonian of the two-level sytem S2 , the second term is the Hamiltonian of the n-level system (reservoir) Sn , and the third term is an interaction between them. Thus s determines the energy scale of the isolated small system (2s is its level spacing), and v plays the role of the coupling constant between S2 and Sn . We write the Hamiltonian H (n) in the form (n)

H (n) = H0

(1.6)

+ M (n) ,

where (n)

H0

= sσ z ⊗ 1n + 12 ⊗ hn ,

M (n) = vσ x ⊗ wn /n1/2 , (n)

and choose the basis in C2 ⊗Cn in which the matrix H0 (n)

(n)

(1.7) (H0 )αj,βk = λαj δαβ δjk ,

(n)

(n)

λαj = Ej

+ αs,

is diagonal:

α, β = ±,

j, k = 1, .., n.

Assume that at t = 0 the density matrix of the composite system S2,n is (n)

µ(n) m (Ek , 0) = ρ(0) ⊗ Pk ,

(1.8)

where ρ(0) is a 2 × 2 positive definite matrix of unit trace and Pk is the projection (n) on the state of energy Ek of the reservoir. Let µ(n) (t) be the density matrix of the composite system S2,n at time t, corresponding to the initial density matrix (n) µm (0) of (1.8): (1.9)

(n)

(n) µ(n) (t) = U (n) (−t)µ(n) (t), m (Ek , 0)U

U (n) (t) = eitH

(n)

.

Then the reduced density matrix of the small system is defined as (1.10)

(n)

(n)

ρbα,δ (Ek , t) =

n X j=1

(n)

µαj,δj (t),

α, δ = ±,

i.e. ρb(n) is obtained from the density matrix (1.9) of the whole composite (closed) system by tracing out the reservoir degrees of freedom. It follows from Theorem 2.1 below that the variance of the reduced density matrix vanishes as n → ∞, i.e.

ON A RANDOM MATRIX MODELS OF QUANTUM RELAXATION

3

that ρb(n) is selfaveraging. This allows us to confine ourselves to the study of the (n) mean reduced density matrix ρ(n) (Ek , t): n D E X X (n) (n) (n) (n) (n) (1.11) ρα,δ (Ek , t) = µαj,δj (t) = Tαβγδ (Ek , t)ρβ,γ (0), j=1

β,γ=±

where (1.12)

(n) (n) Tαβγδ (Ek , t)

=

n D X j=1

E (n) (n) Uαj,βk (−t) + Uγk,δj (t)

is the ”transfer” matrix, an analog of the influence functional by Feynman-Vernon [12]. (n) Notice that we can equally consider the factorized initial condition µc (β, 0) in which the microcanonical distribution Pk of the reservoir is replaced by its canonical (0) (n) distribution e−βHn /Z0 . We have evidently n X (n) (n) (n) (β, 0) = e−βEk µ(n) (1.13) µ(n) c m (Ek , 0)/Z0 . k=1

It is also easy to write the corresponding reduced density matrix. 2. Selfaveraging of Reduced Density Matrix (n)

Theorem 2.1. Let ρb(n) (Ek , t) be the reduced density matrix (1.10) of the composite system S2,n = S2 + Sn , given by (1.1) – (1.9). Then we have   8v 2 t2 (n) (n) (2.1) Var ρbα,δ (Ek , t) ≤ , α, δ = ±. n

To prove the theorem we use the following facts.

Proposition 2.2. (Poincare-Nash inequality). If a random Gaussian vector X = {ξj }pj=1 satisfies conditions hξj i = 0, hξj ξ¯k i = Cjk , j, k = 1, .., p, and functions Φ1,2 : Rp → C have bounded partial derivatives, then (2.2)

Cov{Φ1 , Φ2 } := h(Φ1 , Φ2 )i − hΦ1 ihΦ2 i 1

where

1

≤ h(C∇Φ1 , ∇Φ1 )i 2 h(C∇Φ2 , ∇Φ2 )i 2 , (C∇Φ, ∇Φ) =

p X

¯ k. Cjk (∇Φ)j (∇Φ)

j,k=1

For the proof of the inequality see e.g. [2], Theorem 1.6.4. Proposition 2.3. (Duhamel formula). If M1 , M2 are n × n-matrices, then Z t (M1 +M2 )t M1 t (2.3) e =e + eM1 (t−s) M2 e(M1 +M2 )s ds. 0

The proof is elementary. Notice, that Duhamel formula allows us to obtain the derivative of the matrix U (n) (t) with respect to the entry wlm , l, m = 1, ..., n of the matrix wn in (1.5): Z tX (n) ∂Uαj,βk (t) iv (n) (n) (2.4) = √ U (t − s)U−κm,βk (s)ds. ∂wlm n 0 κ=± αj,κl

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J. L. LEBOWITZ, A. LYTOVA, AND L. PASTUR

Proof. (of the Theorem 2.1). By using the Poincare-Nash inequality (2.2) with n X (n) (n) Uγk,δj (t)Uαj,βk (−t), X = {wlm }nl,m=1 , C = 1n2 , Φ1 (X) = Φ2 (X) = j=1

differentiation formula (2.4), and then Schwartz inequality, we obtain (2.5) Var

X j

 (n) (n) Uγk,δj (t)Uαj,βk (−t)  Z v 2 X X t X (n) (n) (n) Uγk,κl (t − s)U−κm,δj (s)ds Uαj,βk (−t) ≤ n 0 κ l,m j 2  Z tX X (n) (n) (n) Uγk,δj (t) Uαj,κl (−t + s)U−κm,βk (−s)ds + 0

j

κ

2 Z t X (n) 2 X X (n) 2v 2 t (n) U ds ≤ (t − s) U (s)U (−t) γk,κl −κm,δj αj,βk n 0 κ,l κ,m j 2  Z t X X X (n) 2 (n) (n) + Uαj,κl (−t + s)Uγk,δj (t) U−κm,βk (−s) ds , 0

κ,l

κ,m

j

and here and below all the sums over the Latin indices will be from 1 to n, and the sum over the Greek indices will be over ±. Notice that X (n) 2 X (n) 2 U U (2.6) (−s) = (t − s) = 1, −κm,βk

γk,κl

κ,m

and

(2.7)

κ,l

2  X X (n) X X (n) (n) (n) = (t − s) U (−t + s)U U (−t + s)U (t) αj1 ,κl κl,αj2 αj,κl γk,δj κ,l

j1 ,j2

j

×

κ,l

(n) (n) Uγk,δj1 (t)Uδj2 ,γk (−t)

Hence, we have by (2.5), (2.6), and (2.7):  X 4v 2 t2 (n) (n) , Uγk,δj (t)Uαj,βk (−t) ≤ Var n j

=

X (n) U

2 ≤ 1.

γk,δj1 (t)

j1

α, δ = ±.

Now, taking into account (1.10) and the fact that ρ(0) is a 2 × 2 positive definite matrix and of unit trace, we obtain (2.1).  3. Equilibrium Properties We begin by considering the equilibrium (time independent) microcanonical density matrix of the composite system S2,n : . (3.1) Ω(λ) = δ(λ − H (n) ) Tr δ(λ − H (n) ) . Following a standard prescription of statistical mechanics, we will replace the Dirac delta-function in (3.1) by the function (2ε)−1 χε , where χε is the indicator of the interval (−ε, ε), and ε ≪ λ. Then the reduced microcanonical density matrix, i.e.

ON A RANDOM MATRIX MODELS OF QUANTUM RELAXATION

5

the microcanonical density matrix of S2,n , traced with respect to the states of Sn , is the 2 × 2 matrix of the form ω (n) (λ) = P

(3.2)

ν (n) (λ)

(n)

δ=± ν δ,δ (λ)

where

−1 ν (n) αγ (λ) = (2εn)

(3.3)

n X j=1

,

χε (λ − H (n) )αj,γj .

The corresponding canonical distribution of the composite system is  (3.4) e−βHn Tr e−βHn , and the reduced distribution of the small system is R ∞ −βλ (n) ν (dλ) −∞ e (3.5) , R∞ P −βλ ν (n) (dλ) δ=± −∞ e δ,δ where (cf (3.3)) (3.6)

(n) ν (n) (∆) = {ναγ (∆)}α,γ=± ,

(n) ναγ (∆) = n−1

n X

χ∆ (H (n) )αj,γj ,

j=1

and χ∆ is the indicator of an interval ∆ of the spectral axis. Theorem 3.1. Consider the 2 × 2 matrix measure ν (n) of (3.6). Then (i) there exists non-random diagonal 2 × 2 matrix measure such that the weak convergence:

ν = {να δαγ }α,γ=± lim ν (n) = ν

(3.7)

n→∞

holds with probability 1; (ii) if

Z

∞

να (dλ) , ℑz 6= 0, −∞ λ − z is the Stieltjes transform of να , and ν0 is defined by (1.2), then the pair fα (z), α = ± is a unique solution of the system of two coupled functional equations Z ∞ ν0 (dE) (3.8) fα (z) = , α=± 2 −∞ E + sα − z − v f−α (z) fα (z) =

in the class of functions analytic for ℑz 6= 0, and satisfying the condition ℑfα (z) · ℑz > 0, ℑz 6= 0; (iii) nonnegative measures να , α = ± have the unit total mass, να (R) = 1, and if the measure ν0 of (1.2) is absolute continuous and sup ν0′ (λ) < ∞, then να , λ∈R

α = ± are also absolute continuous, and we have (3.9)

να′ (λ) ≤ sup ν0′ (µ); µ∈R

(iv) for any λ ∈ R with probability 1 there exists the limit of the reduced microcanonical distribution (3.10)

lim ω (n) = ω,

n→∞

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J. L. LEBOWITZ, A. LYTOVA, AND L. PASTUR

where ω(λ) = P

(3.11) and (3.12)

ν αγ = δα,γ ν α ,

ν(λ) , δ=± ν δ,δ (λ) −1

ν α (λ) = (2ε)

Z

λ+ε

να (dµ),

λ−ε

analogous formulas are also valid for the limits of the reduced canonical distribution (3.5). Remark 3.2. The limiting measures να , α = ± can be found from their Stieltjes transforms fα , α = ± via the inversion formula [1]: Z −1 (3.13) να (△) = π lim ℑfα (λ + iτ )dλ. τ →0

△

To prove the theorem we need the following auxiliary fact. Proposition 3.3. Let Φ be a C1 function of n × n hermitian matrix, bounded together with its derivatives. Then we have for the GUE matrix wn of (1.3): 

∂Φ(wn ) 

= Φ(wn )wkj . (3.14) ∂wjk

The proof of proposition follows from (1.3)–(1.4) and the integration by parts formula. Proof. (of the Theorem 3.1). Denote G(n) (z) = (H (n) − z)−1 , ℑz 6= 0

(3.15)

the resolvent of (1.5) and set

(n) gαγ (z) = n−1

(3.16)

n X

(n)

Gαj,γj (z).

j=1

(n)

It follows from the spectral theorem for Hermitian matrices that gαγ is the Stieltjes (n) transform of ναγ and in view of the one-to-one correspondence between measures and their Stieltjes transforms (see [1], Section 59) to prove the weak convergence (n) (3.7) with probability 1 it suffices to prove that with probability 1 gαγ converges to δαγ fα uniformly on a compact set of C\R. Denote (3.17)

n X 

(n)

(n)  (n) Gαj,γj (z) . (z) := gαγ (z) = n−1 fαγ j=1

For further purposes it is convenient to start by considering the functions n X 

(n) −1 Uαj,γj (t) , (3.18) u(n) (t) = n αγ j=1

(n)

where the matrix U (t) is defined in (1.9). By the spectral theorem for Hermitian (n) (n) (n) matrices uαγ is the Fourier transfrom of ναγ and fαγ (z) is the generalized Fourier (n) transform (see e.g. [11]) of uαγ (t): Z ∞ (n) −1 (3.19) fαγ (z) = i e−izt u(n) ℑz < 0. αγ (t)dt, 0

ON A RANDOM MATRIX MODELS OF QUANTUM RELAXATION

7

 Notice that the matrix U (n) (t) is diagonal with respect to the Latin indices. Indeed, since wn in (1.5) is the GUE random matrix whose probability law (1.3) is unitary invariant, we have for any unitary n × n-matrix U :




(n) exp{itH (n) } = exp{it H0 + vn−1/2 σ x ⊗ U w(n) U ∗ } .

In particularly, for any diagonal unitary matrix U = {eiϕj δjk }nj,k=1 with distinct ϕj ∈ [0, 2π), j = 1, ..., n we obtain



 (exp{itH (n) })αj,βk = ei(ϕj −ϕk ) (exp{itH (n)})αj,βk .

This implies

(n)  (n) (3.20) Uαj,βk (t) = Uαβ,j (t)δjk ,

Hence we can write (3.18) as

 (n) Uαβ,j (t) = exp{itH (n) }αj,βj . n

u(n) αγ (t) =

(3.21)

1 X (n) (t). U n j=1 αγ,j

It follows now from (2.3), (2.4), (1.6), and (3.14) that Z t

i(t−s)H (n) (n) itH (n) 

itH (n)  (n)
itH0 0 (3.22) e + i ds e M e = e αj,βj αj,βj αj,βj 0 Z t X

 (n) (n) iv (n) = eitλαj δαβ + √ wjm U−αm,βj (s) ds ei(t−s)λαj n 0 m Z s XX 2 Z t

 (n) (n) v (n) (n) ds ei(t−s)λαj dτ = eitλαj δαβ − U−αm,νm (s − τ )U−νj,βj (τ ) . n 0 0 m ν Hence taking into account (3.20) and (3.21), we obtain: Z t (n) (n) (n) (3.23) Uαβ,j (t) = ds ei(t−s)λαj rαβ,j (s) 0 Z t Z s X (n) (n) (n) (n) + eitλαj δαβ − v 2 ds ei(t−s)λαj dτ u−αν (s − τ )U−νβ,j (τ ), 0

where (n)

rαβ,j (s) = −

v2 n

Z

s

0

dτ

0

ν

X X ◦ (n)  (n) U −αm,νm (s − τ )U−νj,βj (τ ) , m

ν

◦

(U = U − hU i).

By using Schwartz inequality and inequality (3.34) below we have that (n) Cs3 r (3.24) αβ,j (s) ≤ 3/2 . n Here and below we use the notation C for all positive quantities that do not depend on n, t, z and indexes. (n) (n) We will use the notations Gj (z), f (n) (z) and Rj (z) for the generalized Fourier transforms (see (3.19)) of the 2 × 2-matrices (n)

(3.25) Uj

(n)

(n)

(n)

(n)

= {Uαβ,j (t)}αβ=± , u(n) = {uαβ (t)}αβ=± , rj (t) = {rαβ,j (t)}αβ=± .

We have from the spectral theorem, (3.18), (3.20), and (3.15) (cf (3.19)) Z ∞  (n) (n) −1 (3.26) Gj (z) = i e−izt Uj (t)dt = hGαj,βj(z) i α,β=± , ℑz < 0. 0

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J. L. LEBOWITZ, A. LYTOVA, AND L. PASTUR

This, (3.17), and (3.23) lead to the matrix relation: (n)

(Ej

(n)

(n)

+ sσ z − z − v 2 σ x f (n) (z)σ x )Gj (z) = 1 + Rj (z).

Since the resolvent G(n) (z) possesses the property ℑzℑ(G(n) (z)x, x) ≥ 0, ∀x ∈ C2n , the matrix f (n) (z) possesses the same property ∀ξ ∈ C2 , and (n)

ℑ((Ej +sσ z −z−v 2 σ x f (n) (z)σ x )ξ, ξ) = −ℑz||ξ||2 −v 2 (f (n) (z)σ x ξ, σ x ξ) ≥ −ℑz||ξ||2 . (n)

Thus the matrix Ej (3.27)

+ sσ z − z − v 2 σ x f (n) (z)σ x is invertible, its inverse (n)

(n)

f (n) (Ej , z) = (Ej

admits the bound

+ sσ z − z − v 2 σ x f (n) (z)σ x )−1 (n)

||f (n) (Ej , z)|| ≤ |ℑz|−1 ,

(3.28) (n)

and equation for Gj (z) takes the form (n)

(n)

(n)

(n)

Gj (z) = f (n) (Ej , z) + f (n) (Ej , z)Rj (z). P Applying to the equation the operation n−1 j we obtain in view of (1.1) Z ∞ n X (n) (n) (n) f (n) (Ej , z)Rj (z). ν0 (dE)f (n) (E, z) + n−1 (3.30) f (n) (z) = (3.29)

−∞

(n)

Since the resolvent G we have the bound

j=1

(z) is analytic if ℑz 6= 0 and bounded from above by |ℑz|−1 , ||f (n) (z)|| ≤ |ℑz|−1 ,

(3.31)

implying that the sequence {f (n) (z)}n≥1 consists of functions, analytic and uniformly bounded in n and in z by η0−1 if |ℑz| ≥ η0 > 0. Hence there exists analytic 2 × 2 matrix function f (z), ℑz 6= 0, such that ||f (z)|| ≤ |ℑz|−1 , and an infinite subsequence {f (nk ) (z)}k≥1 that converges to f (z) uniformly on any compact set of C \ R. This and estimates (3.24), (3.28) allow us to pass to the limit nk → ∞ in (3.30) and obtain that the limit of any converging subsequence of the sequence {f (n) (z)}n≥1 satisfies the matrix functional equation Z ∞ (3.32) f (z) = ν0 (dE)f (E, z), −∞

where (3.33)

f (E, z) = (E + sσ z − z − v 2 σ x f (z)σ x )−1 ,

||f (E, z)|| ≤ |ℑz|−1 .

The equation is uniquely soluble in the class of 2 × 2 matrix functions, analytic for ℑz 6= 0, and such that ℑzℑ(f (z)ξ, ξ) ≥ 0, ∀ξ ∈ C2 . Indeed, for any two solutions f1 , f2 of the class, and g = f1 − f2 we have Z ∞ −1 2 x  v σ g(z)σ x ν0 (dE) E + sσ z − z−v 2 σ x f1 (z)σ x g(z) = −∞

 −1 × E + sσ z − z − v 2 σ x f2 (z)σ x ,

and by (1.2), (3.33) we obtain inequality ||g(z)|| ≤ v 2 |ℑz|−2 ||g(z)|| from which it follows that g(z) = 0 for vℑz < 1, hence for any ℑz 6= 0 by analyticity. The solution of (3.32) is diagonal, fαβ = fα δαβ , and pair fα , α = ± satisfies system (3.8). This follows from the unique solvability of (3.8). We can rewrite (3.8) in

ON A RANDOM MATRIX MODELS OF QUANTUM RELAXATION

9

the form fα (z) = fα0 (z + v 2 f−α (z)), where fα0 (z) is the Stieltjes transform of the unit non-negative measure ν0α (E) = ν0 (E − αs). Since fα0 (z) possesses the property limη→∞ η|fα0 (iη)| = 1 and |ℑ(z + v 2 f−α (z))| ≥ |ℑz|, then limη→∞ η|fα (iη)| = 1 and fα (z), α = ± are Stieltjes transforms of the unit non-negative measures να (λ) (3.13) (see [1], Section 59). In addition, the Tchebyshev inequality and bound (3.35) below imply that for any ε > 0 (n) (n) P{|fαγ (z) − gαγ (z)| > ε} ≤

2v 2 1 (n) Var{gαγ (z)} ≤ 2 2 . 2 ε ε n |ℑz|4

Hence the series ∞ X

n=1

(n) (n) P{|fαγ (z) − gαγ (z)| > ε}

converges for any ε > 0, |ℑz| ≥ η0 > 0, and by the Borel-Cantelli lemma we have for (n) any fixed z, |ℑz| ≥ η0 > 0, limn→∞ gαγ (z) = fαγ (z) with probability 1. With the same probability this limiting relation is valid for all points of an infinite countable sequence {zj }j≥1 , |ℑzj | ≥ η0 > 0, possessing an accumulation point. Hence on (n) any compact of C \ R with probability 1 limn→∞ gαγ (z) = fαγ (z), and we have the weak convergence (3.7) with the formulas (3.1)-(3.8). Let us prove assertion (ii) of theorem. It follows from the (3.8) Z ∞ ℑ(z + v 2 f−α (z))dE ℑfα (z) ≤ sup ν0′ (µ) 2 2 2 2 µ∈R −∞ (E + sα − ℜ(z + v f−α (z))) + (ℑ(z + v f−α (z))) = πsup ν0′ (µ). µ∈R

We have now by (3.13) να (△) ≤ |△| sup ν0′ (µ). µ∈R

This implies (ii). To prove (iii) we notice that by (ii) measures να , α = ± are continuous. Thus we can pass to the limit n → ∞ in (3.3), written as (n) ν (n) α (λ) = να ([λ + ε, λ − ε)).

This and (3.2) imply (3.10)-(3.11).



Lemma 3.4. Under the conditions of Theorem 3.1 (3.34) (3.35)

v 2 t2 , n v2 (n) , Var{Gαj,γk (z)} ≤ n|ℑz|4 (n)

Var{Uαj,γk (t)} ≤

v 2 t2 , n2 2v 2 (n) Var{gαγ (z)} ≤ 2 , n |ℑz|4

Var{u(n) αγ (t)} ≤

ℑz 6= 0.

Proof. Acting as in the case of Theorem 2.1 we obtain (3.34). The differentiation formula for the resolvent (n)

∂Gαj,βk (z) (n)

∂wlm

iv X (n) (n) = √ G (z)G−κm,βk (z), n κ αj,κl

10

J. L. LEBOWITZ, A. LYTOVA, AND L. PASTUR

following from the resolvent identity, together with Poincare-Nash inequality (2.2) imply the first inequality in (3.35): 2  v 2 X X (n) (n) (n) Gαj,κl (z)G−κm,βk (z) Var{Gαj,γk (z)} ≤ n κ l,m

2 X

v n

≤

l,κ

X (n) 2  G(n) (z) 2 G ≤ αj,κl −κm,βk (z) m,κ

v2 , n|ℑz|4

ℑz > 0.

The second inequality in (3.35) can be proved by a similar argument.



We will return now to functions of variable find the n → ∞ limits of the P t and (n) (n) sequences {Uj (t)}n≥1 and {u(n) (t) = n−1 nj=1 Uj (t)}n≥1 . (n)

Theorem 3.5. Consider the 2 × 2 matrices {Uj (t)}n≥1 and {u(n) (t)}n≥1 , (n)

defined in (3.25) and (3.21) and choose a subsequence {Ejn } that converges to a given E of the support of ν0 of (1.2). Then there exist the limits (3.36)

(n)

n→∞

n→∞

where i (3.37) UE (t) = 2π

Z

e

L

izt

i lim f (E, z)dz := 2π N →∞

(3.38)

||UE (t)|| = 1

and (3.39)

u(t) = lim u(n) (t),

UE (t) = lim Ujn (t),

i u(t) = 2π

Z

dz e

L

izt

Z

N −iη

eizt f (E, z)dz,

−N −iη

∀η > 0,

∀t ≥ 0,

Z

∞

ν0 (dE)f (E, z)

−∞

with f (E, z) defined in (3.33): (3.40) fαβ (E, z) = fα (E, z)δαβ ,

fα (E, z) = (Eα −z−v 2 f−α (z))−1 , Eα = E+αs.

Proof. It follows from (3.26), (3.29), and the inversion formula for the generelized Fourier transform [11] that Z t (n) (n) (n) (n) −1 (n) (3.41) Ujn (t) = Q (Ejn , t) + i Q(n) (Ejn , t − s)rjn (s)ds, 0

where

Z i (n) (n) (3.42) Q(n) (Ejn , t) = eizt f (n) (Ejn , z)dz = UE (t) 2π L Z   i (n) (n) eizt f (n) (Ejn , z) E − Ejn + v 2 σ x (f (n) (z) − f (z))σ x f (E, z)dz. + 2π L

The resolvent identity yields  1 1  z (n) (n) (n) f (n) (Ejn , z) = − sσ + (E − Ejn ) + v 2 σ x f (n) (z)σ x f (n) (Ejn , z), + z−E z−E and we have for sufficiently big η = |ℑz|: (3.43)

(n)

||f (n) (Ejn , z)|| ≤

2 , |z − E|

||f (E, z)|| ≤

2 . |z − E|

ON A RANDOM MATRIX MODELS OF QUANTUM RELAXATION

11

This together with (3.31) allow us to pass to the limit under the integral in the r.h.s. of (3.42) and to show that it vanishes as n → ∞. Moreover, we conclude that integral in the r.h.s. of (3.42) is bounded uniformly in n and ∀t ≥ 0. The uniform boundedness of the matrix UE (t) follows from the equalities (UE )αβ (t) = (UE )αα (t)δαβ and Z v 2 f−α (z) i . dz eitz (UE )αα (t) = eiEα t + 2π L (Eα − z − v 2 f−α (z))(Eα − z) (n)

Hence Q(n) (Ejn , t) converges to UE (t) as n → ∞ and is uniformly bounded in n (n)

and t. This together with (3.41), (3.24) and equality ||Uj (t)|| = 1, ∀t ≥ 0 give us (3.37) and (3.38). To prove (3.39) notice first that we have from (3.30) Z Z ∞ i (n) dz eizt (3.44) ν0 (dE)f (E, z) u(n) (t) = 2π L −∞ Z Z ∞   i (n) + dz eizt ν0 (dE) f (n) (E, z) − f (E, z) 2π L −∞ Z t n X (n) (n) Q(n) (Ej , t − s)rj (s)ds. + i−1 n−1 0

j=1

We integrate by parts with respect to z in the first integral to obtain in view of (3.40) Z ∞ Z ′ 1 + v 2 f−α (z) 1 (n) . ν0 (dE) dz eizt − 2 2πt −∞ (Eα − z − v f−α (z))2 L It follows from (3.32)-(3.33) and (1.2) that ||fα (z)|| = |z|−1 (1 + o(1)), |z| → ∞ and ||fα′ (z)|| ≤ |ℑz|−2 . Thus the integral with respect to z is bounded and continuous (n) function of E. This and the weak convergence ν0 to ν0 (see (1.2)) yield the convergence of the first term of the r.h.s. of (3.44) to the r.h.s. of (3.39). Furthermore, by using (3.43), (3.31) and (1.1) we obtain Z ∞ Z ||f (n) (z + E) − f (z + E)|| (n) ν0 (dE) |dz| |z|2 L −∞ Z  Z dx (n) (n) ≤C + max ||f (y − iη) − f (y − iη)|| . ν0 (dE) + 2 2 |y|≤A+T |x|≥A x + η |E|≥T

For any ε > 0 choosing consequently A = A(ε), T = T (ε, A), N0 = N0 (ε, A, T ), and taking in account (1.2), and convergence f (n) (z) to f (z) on any compact set in C \ R, we obtain that the second term of the r.h.s. of (3.44) vanishes as n → ∞. At last (3.24) yields for the third term of the r.h.s. of (3.44): Z Z t Cs3 1 X || dz eiz(t−s) f (n) (Ej , z)|| ds n n j L 0   Z t Z Z ∞ Cs3 (n) ≤ ds ν0 (dE) ||UE (t − s)|| + dz||f (n) (E, z) − f (E, z)|| , n −∞ 0 L and taking into account (3.38), (3.43), and (3.31) we conclude that the term also vanishes as n → ∞ uniformly in t, varying on a compact interval. 

12

J. L. LEBOWITZ, A. LYTOVA, AND L. PASTUR

4. Time Evolution We will prove now the main general result of [7], a formula for the limit as n → ∞ of the expectation (1.11) of the reduced density matrix (1.10) of our model formula (4.7) of [7]. Theorem 4.1. Consider the model of composite system, defined by (1.1)-(1.9). (n) Choose a subsequence {Ekn } of eigenvalues of hn of (1.5) that converges to a certain E ∈ supp ν0 . Then we have for the limit as n → ∞ of the expectation (1.11) of the reduced density matrix (1.10) uniformly in t varying on a finite interval: Z Z 1 (n) (n) dz (4.1) ρα,δ (E, t) := lim ρα,δ (Ekn , t) = dz1 eit(z2 −z1 ) 2 n→∞ (2π)2 L2 L1 ×

fα (E, z1 )fδ (E, z2 )ρα,δ (0) + v 2 f−α (E, z1 )f−δ (E, z2 )fα,δ (z1 , z2 )ρ−α,−δ (0) , 1 − v 4 fα,δ (z1 , z2 )f−α,−δ (z1 , z2 )

where L1 = (−∞ + iη1 , ∞ + iη1 ), L2 = (−∞ − iη2 , ∞ − iη2 ), η1 > 0, η2 > 0; Z ∞ ν0 (dE)fβ (E, z1 )fγ (E, z2 ) (4.2) fβ,γ (z1, z2 ) = −∞

with fα (E, z) defined in (3.40). Proof. In view of (1.11) it suffices to prove the following expression for the average transfer matrix (1.12): (n)

(4.3)

(n)

Tαβγδ (E, t) := lim Tαβγδ (Ekn , t) n→∞ Z Z 1 = dz dz1 eit(z2 −z1 ) Teαβγδ (E, z1 , z2 ), 2 (2π)2 L2 L1

where the ”two-point” functions Teαβγδ (E, z1 , z2 ) are analytic in z1 and in z2 outside the real axis and have the form (4.4) Teαβγδ (E, z1 , z2 ) = fβ (E, z1 )fγ (E, z2 ) ×(δα,β δγ,δ + v 2 δ−α,β δ−γ,δ f−β,−γ (z1 , z2 ))[1 − v 4 fβ,γ (z1 , z2 )f−β,−γ (z1 , z2 )]−1 .

Acting as in the proof of the Theorem 3.1, i.e., by using the Duhamel formula (2.3), (3.14) and differentiation formula (2.4) we obtain the relation (cf (3.22)) n D E X (n) (n) (n) (n) (n) (n) Uαj,βk (−t1 )Uγk,δj (t2 ) = eit2 λγkn δγδ Uαβ,kn (−t1 ) (4.5) Tαβγδ (Ekn , t1 , t2 ) := 2

Z

+ v2

Z

−v

j=1

t2

ds e

i(t2 −s)λγkn

Z

(n) i(t2 −s)λγkn

Z

(n)

0

+

dτ

0

t2

ds e

0

0

Z

s

t2

ds e

0

(n) i(t2 −s)λγkn

X κ

t1

dτ

X κ

(n)

(n)

(n)

Tα,β,−κ,δ (Ekn , t1 , τ )u−γκ (s − τ ) (n)

(n)

U−κβ,kn (−τ )Kα,κ,−γ,δ (t1 − τ, s)

(n)

rαβγδ (t1 , s),

where (n)

(4.6) Kαβγδ (t1 , t2 ) =

n n E 1 X X D (n) (n) Uαj,βm (−t1 )Uγm,δj (t2 ) , n m=1 j=1

(n)

|Kαβγδ (t1 , t2 )| ≤ 1,

ON A RANDOM MATRIX MODELS OF QUANTUM RELAXATION

13

and (n)

(4.7)

rαβγδ (t1 , t2 )  Z t2 X X ◦ (n)  v2 X (n) (n) U −γm,κm (t2 − τ )U−κkn ,δj (τ )Uαj,βkn (−t1 ) dτ − = n j 0 m κ  Z t1 X X ◦ (n)  (n) (n) U −κkn ,βkn (−τ )U−γm,δj (t2 )Uαj,κm (τ − t1 ) . dτ + 0

m

κ

It follows from Schwartz and Poincare-Nash inequalities and estimate (3.34) that (cf (3.24)) (n)

|rαβγδ (t1 , t2 )| = O(n−1/2 ).

(4.8)

(n) (n) Let Teαβγδ (E, z1 , z2 ), ℑz1 > 0, ℑz2 < 0 be generalized Fourier transform of Tαβγδ (E, t1 , t2 ):  Z ∞  Z ∞ 1 (n) (n) Teαβγδ (E, z1 , z2 ) = i dt1 eiz1 t1 dt2 e−iz2 t2 Tαβγδ (E, t1 , t2 ) , i 0 0

so that

(n)

Tαβγδ (E, t1 , t2 ) =

1 (2π)2

Z

L2

dz2

Z

L1

(n)

dz1 ei(t2 z2 −t1 z1 ) Teαβγδ (E, z1 , z2 ),

where L1 = (−∞ + iη1 , ∞ + iη1 ), L2 = (−∞ − iη2 , ∞ − iη2 ), η1 > 0, η2 > 0. In view of relations (n)

(n)

Uαj,βk (−t1 ) = Uαk,βj (t1 ), we have i

Z

∞

0

(n)

(n)

Gαj,βk (z1 ) = Gαk,βj (z1 )

(n)

(n)

dt1 eiz1 t1 Uαj,βk (−t1 ) = Gαj,βk (z1 )

and (4.5) yields (4.9)

(n) (n) Teαβγδ (Ekn , z1 , z2 )

=

1

(n)

 X (n) (n) (n) (n) Gαβ,kn (z1 )δγδ + v 2 f−γκ (z2 )Teαβ−κδ (Ekn , z1 , z2 )

λγkn − z2 κ  X (n) (n) e (n) + v2 (z , z ) + r e (z , z ) G−κβ,kn (z1 )K α,κ,−γ,δ 1 2 αβγδ 1 2 . κ

e (n) and re(n) are generalized Fourier transforms of K (n) and r(n) of (4.6) and Here K e (n) (z1 , z2 ) (4.7) respectively, and as it follows from (4.6) the absolute values of K αβγδ −1 −1 are bounded uniformly in n by |ℑz1 | |ℑz2 | . e (n) , Se(n) , To write (4.9) in the matrix form for any fixed pair α, β we denote K αβ αβ (n) (n) (n) e )γδ = K e e −(n) are re the 2 × 2-matrices, which entries are (K etc., and K αβ

αβ

αβγδ

e −(n) )γδ = K e (n) 2 × 2-matrices with the entries (K αβ α,β,−γ,δ , γ, δ = ±, so

αβ

(n) (n) (n) (n) (n) rαβ (z1 , z2 ) + f (n) (Ekn , z2 ) Teαβ (Ekn , z1 , z2 ) = f (n) (Ekn , z2 )e   X (n) (n) 2 −(n) e × Gαβ,kn (z1 )12 + v G−κβ,kn (z1 )Kακ (z1 , z2 ) . κ

14

J. L. LEBOWITZ, A. LYTOVA, AND L. PASTUR (n)

Plugging expression (3.29) for Gkn (z2 ) we obtain (4.10)

(n) (n) e(n) (E (n) , z1 , z2 ) Teαβ (Ekn , z1 , z2 ) = R αβ kn   X (n) (n) (n) (n) (n) 2 (n) −(n) e + f (Ekn , z2 ) fαβ (Ekn , z1 )12 + v f−κβ (Ekn , z1 )Kακ (z1 , z2 ) , κ

e(n) (E (n) , z1 , z2 ) is a 2 × 2-matrix, and according to (3.24), (4.8), where reminder R αβ kn (n) e (n) (z1 , z2 ) and f (n) (Em and uniform boundedness of K , z), we have (4.11)

(n)

(n)

e (E , z1 , z2 )|| = 0, lim ||R αβ kn

n→∞

Applying the operation n (4.12)

Pn −1

e (n) (z1 , z2 ) = v 2 K αβ +

Z

∞

−∞

m=1

XZ κ

∞

−∞

lim ||n−1

n→∞

n X

m=1

(n) e (n) (Em R , z1 , z2 )|| = 0. αβ

to (4.10) with kn = m we obtain: (n) (n) −(n) e ακ ν0 (dE)f−κβ (E, z1 )f (n) (E, z2 )K (z1 , z2 )

(n) (n) ν0 (dE)fαβ (E, z1 )f (n) (E, z2 )

+ n−1

X m

e(n) (E (n) , z1 , z2 ). R m αβ

e αβγδ (z1 , z2 ) = This implies that for any fixed α, β, γ, δ the limiting values K (n) e e limn→0 Kαβγδ (z1 , z2 ) and Kα,−β,−γ,δ (z1 , z2 ) satisfy the system of linear equations   e αβγδ (z1 , z2 ) = fβγ (z1 , z2 ) δαβ δγδ + v 2 K e α,−β,−γ,δ (z1 , z2 ) , K

where fβγ (z1 , z2 ) are defined in (4.2). Solving this system we obtain   f−β,−γ (z1 , z2 ) δα,−β δ−γ,δ + v 2 fβ,γ (z1 , z2 )δα,β δγ,δ e (4.13) Kα,−β,−γ,δ (z1 , z2 ) = . 1 − v 4 fβ,γ (z1 , z2 )f−β,−γ (z1 , z2 )

Now we return to the variables t1 , t2 . It follows from (4.10) and (4.11) that Z Z 1 dz2 (4.14) Tαβγδ (E, t1 , t2 ) = lim dz1 ei(t2 z2 −t1 z1 ) n→∞ (2π)2 L L1 2  (n) (n) (n) (n) × fαβ (Ekn , z1 )fγδ (Ekn , z2 )  X (n) (n) (n) (n) 2 (n) e +v f−κβ (Ekn , z1 )fγν (Ekn , z2 )Kα,κ,−ν,δ (z1 , z2 ) , κ,ν

and we have to prove the equality: Z Z 1 dz (4.15) Tαβγδ (E, t1 , t2 ) = dz1 ei(t2 z2 −t1 z1 ) fβ (E, z1 )fγ (E, z2 ) 2 (2π)2 L2 L1   e α,−β,−γ,δ (z1 , z2 ) , × δαβ δγδ + v 2 K

which together with (4.13) yields (4.3). Notice that for any fixed non-real z1 , z2 the integrand of (4.14) tends to integrand of (4.15), but it has no an integrable majorant. e (n) Because of this fact we replace K α,κ,−ν,δ (z1 , z2 ) in (4.14) by the corresponding entry

ON A RANDOM MATRIX MODELS OF QUANTUM RELAXATION

15

of the r.h.s. matrix of (4.12) to obtain Tαβγδ (E, t1 , t2 ) = (UE )ββ (−t1 )(UE )γγ (t2 )δαβ δγδ Z v2 X (n) (n) (n) (n) (n) (Ekn , z2 ) ν0 (dµ)dz2 dz1 ei(t2 z2 −t1 z1 ) f−κβ (Ekn , z1 )fγν + lim n→∞ (2π)2 κ,ν X (n)   (n) (n) (n) e α,κ1 ,−ν1 ,δ (z1 , z2 ) . (µ, z2 )K f (µ, z1 )f × f (µ, z1 )f (µ, z2 ) + v 2 ακ

−κ1 κ

−νδ

−νν1

κ1 ,ν1

R R (n) R ∞ (n) R Here we denote ν0 (dµ)dz2 dz1 = −∞ ν0 (dµ) L2 dz2 L1 dz1 . Now it remains to prove that the following expressions Z   (n) (n) (n) ν0 (dµ)dz2 dz1 f (n) (Ekn , z1 ) − f (E, z1 ) f (n) (Ekn , z2 )f (n) (µ, z1 )f (n) (µ, z2 ), Z   (n) ν0 (dµ)dz2 dz1 f (E, z1 )f (E, z2 ) f (n) (µ, z1 ) − f (µ, z1 ) f (n) (µ, z2 ), Z   (n) ν0 (dµ)dz2 dz1 f (E, z1 )f (E, z2 )f (µ, z1 )f (µ, z2 ) K (n) (z1 , z2 ) − K(z1 , z2 ) , Z Z Z ∞   (n) dz1 f (E, z1 )f (E, z2 )f (µ, z1 )f (µ, z2 )K(z1 , z2 ), dz2 ν0 − ν0 (dµ) −∞

L1

L2

(n)

where K (n) (z1 , z2 ) = Kαβγδ (z1 , z2 ), vanish as n → ∞. Since ||f (n) (µ, z1 )|| ≤ η1−1 and there exist ηi , i = 1, 2 such that ||f (n) (µ, z2 )|| ≤ (n) 2|z2 − µ|−1 , ||f (n) (Ekn , zi )|| ≤ 2|zi |−1 , |ℑzi | ≥ ηi , then the norm of the first expression is bounded by Z Z Z
∞ (n) |dz2 | 1 1 (n) 2 (n) v ||f (z1 ) − f (z1 )|| + |Ekn − E| . ν0 (dµ) |dz1 | 2 η1 L1 |z1 | −∞ L2 |z2 ||z2 − µ| We have by Schwartz inequality 1/2 Z ∞ Z Z ∞ |dz2 | π dx dx ≤ = . 2 + η2 2 + η2 |z ||z − µ| x (x − µ) η 2 2 2 L2 −∞ 2 −∞ 2 This and the uniform convergence of f (n) to f on a compact set of C \ R imply that the first expression vanishes as n → ∞. Treating similarly the remaining three expressions we prove that they tends to zero as n → ∞.  5. Van-Hove Limit In this section we study the limiting case, where the coupling constant v of the system-reservoir interaction tends to zero, the time t tends to infinity while the transition rate, given by first order perturbation in the interaction, is kept fixed [4, 5, 6, 10]. In terms of (4.3) this corresponds to making simultaneously the limits (5.1)

v → 0,

t → ∞,

τ = tv 2

fixed

after the limit n → ∞, i.e., in formula (4.1). We note that this limit as well as several other important topics of the small system-reservoir dynamics were considered by N.N. Bogolubov in 1945 [3] in the context of classical oscillator interacting linearly with the oscillator reservoir.

16

J. L. LEBOWITZ, A. LYTOVA, AND L. PASTUR

Theorem 5.1. Let the Fourier transform νb0 (u) of the density ν0′ of the measure ν0 in (1.2) be absolutely integrable function: Z ∞ (5.2) |νb0 (u)|du = c0 < ∞, −∞

(5.3)

νb0 (u) =

Z

∞

−∞

e−iuE ν0′ (E)dE.

Then the diagonal entries of the limiting reduced density matrix in (4.1) in the van Hove limit are  ′ ν ′ (E − 2αs) ν0 (E) (5.4) ρα,α (0) + 0 ρ−α,−α (0) ρvH (E, τ ) = 2π α,α Γα (E) Γ−α (E)  ν ′ (E + 2αs) ν ′ (E − 2αs) + e−τ Γα (E) 0 ρα,α (0) − e−τ Γ−α(E) 0 ρ−α,−α (0) Γα (E) Γ−α (E) where Γα (E) = 2π [ν0′ (E) + ν0′ (E + 2αs)] ;

(5.5)

and the off-diagonal entries are (5.6)

−2αsit iτ (f0 (E+2αs+i0)−f0 (E−2αs−i0)) ρvH e , α,−α (E, τ ) = ρα,−α (0)e

where f0 is the Stiltjes transform of ν0′ : Z ∞ ′ ν0 (E)dE , f0 (z) = −∞ E − z

ℑz 6= 0.

Lemma 5.2. In conditions (5.2), (5.3) of the Theorem 5.1 next statements for the functions fα (z), α = ± are valid: (i) supℑz≥0 |fα (z)| ≤ c0 , (ii) limv→0 π1 ℑfα (λ + i0) = ν0′ (λ − αs), λ ∈ R. Proof. Estimate (i) follows from the representation of the functions fα (z), ℑz > 0 in the form Z ∞ 2 (5.7) fα (z) = i eiu(−αs+z+v f−α (z)) νb0 (u)du 0

and condition ℑzℑfα (z) ≥ 0. It also follows from (5.2), (5.7) that Z ∞ (5.8) lim fα (z) = f0 (z − αs) = i eiu(z−αs) νb0 (u)du, ℑz ≥ 0. v→0

0

Hence (5.9)

1 1 lim ℑfα (λ + i0) = ℜ v→0 π π

Z

∞

0

eiu(−αs+λ) νb0 (u)du = ν0′ (λ − αs).

Proof. (of the Theorem 5.1). By using equalities (see (4.2)) fα,α (z1 , z2 ) =

δfα , δz + v 2 δf−α

δz = z1 − z2 ,

δfα = fα (z1 ) − fα (z2 ),



ON A RANDOM MATRIX MODELS OF QUANTUM RELAXATION

17

and by using analyticity of the integrand of (4.1) in z1 and in z2 , we can write the following representation for the diagonal entries of the limiting reduced density matrix (5.10)

Z Z ρα,α (0) dz2 dz1 e−itδz fα (E, z1 )fα (E, z2 ) ρα,α (E, t) = (2π)2 Lv2 Lv 1 Z Z ρα,α (0) v 4 δfα δf−α fα (E, z1 )fα (E, z2 ) + dz dz1 e−itδz 2 2 (2π) δz(δz + v 2 δfα + v 2 δf−α ) Lv Lv 2 1 Z Z 2 2 ρ−α,−α (0) −itδz v δfα (δz + v δfα )f−α (E, z1 )f−α (E, z2 ) dz dz e + 2 1 (2π)2 δz(δz + v 2 δfα + v 2 δf−α ) Lv Lv 2 1 = I1v (E, t) + I2v (E, t) + I3v (E, t),

where Lv1 = {z1 : ℑz1 = v 2 η1 }, Lv2 = {z2 : ℑz2 = −v 2 η2 }, η1 and η2 are arbitrarily chosen positive constants. To compute the limit (5.1) of I1v (E, t) we change variables to ζj = v −2 (zj −Eα ), j = 1, 2, and by Lemma 5.2 we have Z Z eiτ ζ2 e−iτ ζ1 ρα,α (0) dζ2 dζ 1 2 2 (2π) ζ1 + f−α (Eα + v ζ1 ) L2 ζ2 + f−α (Eα + v 2 ζ2 ) L1 Z Z eiτ ζ2 e−iτ ζ1 ρα,α (0) + o(1). dζ2 dζ1 = 2 (2π) ζ1 + f0 (E + 2αs + i0) L2 ζ2 + f0 (E + 2αs − i0) L1

I1v (E, t) =

Computing last integrals by residues and applying equality f0 (λ + i0) − f0 (λ − i0) = 2πiν0′ (λ)

(5.11) we obtain (5.12)

′

vH− lim I1v (E, t) = ρα,α (0)e−2πτ ν0 (E+2αs) ,

where the symbol ”vH-lim” denotes the double limit (5.1). Changing variables in I3v (E, t) to ζ2 = v −2 (z2 − E−α ) ∈ L2 = {ζ : ℑζ = −η2 }, ζ1 = v −2 (z1 − z2 ) ∈ L1 = {ζ : ℑζ = η1 + η2 } yields (5.13)

Z Z ρ−α,−α (0) δfα (ζ1 + δfα ) dζ dζ1 e−iτ ζ1 2 (2π)2 ζ (ζ 1 1 + δfα + δ−α ) L2 L1 1 1 × · . ζ2 + fα (E−α + v 2 ζ2 ) ζ1 + ζ2 + fα (E−α + v 2 (ζ1 + ζ2 ))

I3v (E, t) =

It follows from (5.2) that the absolute value of integrand of (5.14) is bounded from above by c c p p = p 2 , 2 |ζ1 ||ζ2 ||ζ1 + ζ2 | λ1 + (η1 + η2 )2 λ2 + η22 (λ1 + λ2 )2 + η12

18

J. L. LEBOWITZ, A. LYTOVA, AND L. PASTUR

where c > 0 does not depend on v, λj = ℜζj , j = 1, 2. Now Schwartz inequality yields for any B > 0 Z Z ∞ dλ2 dλ1 ∞ p p 2 λ 1 λ2 + 1 (λ1 + λ2 )2 + 1 −∞ B Z Z ∞ dλ2 dλ1 ∞ q q =2 λ1 0 B (λ2 − λ21 )2 + 1 (λ2 + λ21 )2 + 1  21 Z ∞ Z dλ2 dλ1 ∞ ≤2 λ1 0 ((λ2 − λ21 )2 + 1)((λ2 + λ21 )2 + 1) 14 B Z ∞  12 Z dλ1 ∞ dλ2 × < ∞. λ1 0 ((λ2 + λ21 )2 + 1) 43 B This allows us to pass to limit in integral (5.13) by using (5.8) and ( 5.11): vH− lim I3v (E, t) Z Z ∞ ′ ′ ρ−α,−α (0) ∞ −iτ λ1 2πiν0 (E − 2αs)(λ1 + 2πiν0 (E − 2αs)) = dλ dλ e 2 1 (2π)2 λ1 (λ1 + 2πi(ν0′ (E) + ν0′ (E − 2αs)) −∞ −∞ 1 1 × · . λ2 + f0 (E − 2αs − i0) λ1 + λ2 + f0 (E − 2αs + i0)

Here integration path in λ1 encircles zero from above. Computing last integrals by residues we have   ′ ′ ν0 (E − 2αs) −τ Γ−α (E) ν0 (E − 2αs) v . −e (5.14) vH− lim I3 (E, t) = 2πρ−α,−α (0) Γ−α (E) Γ−α (E) Treating similarly the term I2v in the r.h.s. of (5.10) we obtain (5.15)

vH− lim I2v (E, t)  ′  ′ ν (E) ν ′ (E + 2αs) = 2πρα,α (0) 0 − ρα,α (0)e−2πτ ν0 (E+2αs) . + e−τ Γα (E) 0 Γα (E) Γα (E)

Now the assertion (5.4) of theorem follows from the (5.10), (5.12), (5.14) and (5.15). Consider now the off-diagonal entry of (4.1): Z Z ρα,−α (0) dz (5.16) ρα,−α (E, t) = dz1 e−itδz fα (E, z1 )f−α (E, z2 ) 2 v (2π)2 Lv L 2 1  Z Z 4 v f (z , z2 )f−α,α (z1 , z2 )fα (E, z1 )f−α (E, z2 ) α,−α 1 −itδz dz1 e dz2 + 1 − v 4 fα,−α (z1 , z2 )f−α,α (z1 , z2 ) Lv Lv 1 2 Z Z v 2 fα,−α (z1 , z2 )f−α (E, z1 )fα (E, z2 ) ρα,−α (0) + dz2 dz1 e−itδz 2 (2π) 1 − v 4 fα,−α (z1 , z2 )f−α,α (z1 , z2 ) Lv Lv 2 1 =

 ρα,−α (0) v ρα,−α (0)  v I1 (E, t) + I2v (E, t) + I (E, t). 2 (2π) (2π)2 3

To find the limit of I1v (E, t) of (5.16) we change variables to ζ1 = v −2 (z1 − Eα ), ζ2 = v −2 (z2 − E−α ). This yields (5.17)

I1v (E, t) = exp(−2αsit) J1v ,

ON A RANDOM MATRIX MODELS OF QUANTUM RELAXATION

where J1v =

Z

dζ2

L2

eiτ ζ2 ζ2 + fα (E−α + v 2 ζ2 )

Z

dζ1 L1

19

e−iτ ζ1 , ζ1 + f−α (Eα + v 2 ζ1 )

and by (5.8) (5.18) vH− lim J1v (E, t) =

Z

L2

eiτ ζ2 dζ2 ζ2 + f0 (E − 2αs − i0)

Z

L1

e−iτ ζ1 dζ1 ζ1 + f0 (E + 2αs + i0)

2

= (2π) exp{iτ ((f0 (E + 2αs + i0) − f0 (E − 2αs − i0))}.

We have similarly:

I2v = exp(−2αsit) J2v ,

(5.19) J2v =

Z

L1

dζ1

Z

L2

dζ2 e−iτ δζ

v 4 fα,−α f−α,α 1 − v 4 fα,−α f−α,α 1 1 · . × ζ1 + f−α (Eα + v 2 ζ1 ) ζ2 + fα (E−α + v 2 ζ2 )

Here we denote fβ,γ = fβ,γ (Eα + v 2 ζ1 , E−α + v 2 ζ2 ) (see (4.2)). By using the relations: δf−α,α δfα,−α , f−α,α = , fα,−α = 2 v (δζ + δf−α,α ) −4αs + v 2 (δζ + δfα,−α ) where δfβ,γ = fβ (Eα + v 2 ζ1 ) − fγ (E−α + v 2 ζ2 ), we obtain Z Z v 2 f−α,α δfα,−α 1 dζ2 dζ1 e−iτ δζ (5.20) J2v = · 4 1 − v fα,−α f−α,α ζ1 − ζ2 + δf−α,α L2 L1 1 1 × · . 2 ζ1 + f−α (Eα + v ζ1 ) ζ2 + fα (E−α + v 2 ζ2 )

Notice that |fα,−α | ≤ Hence

2c0 , + η2 )

v 2 (η1

α = ±.



2 2c0 1 |1 − v fα,−α fα,−α | ≥ 1 − > , if ηj ≥ 2c0 , (η1 + η2 ) 2 and integrand of (5.20) is uniformly bounded from above by integrable function C(|ζ1 − ζ2 ||ζ1 ||ζ2 |)−1 . This allows us to pass to the limit in the integral in (5.20) and obtain that 4

(5.21)

vH− lim J2v (E, t) = 0.

Treating similarly the term I3v (E, t) of (5.16) we obtain (5.22)

vH− lim I3v (E, t) = 0.

Now assertion (5.6) of the theorem following from (5.16)-(5.22).



According to (5.6) the off-diagonal entry of the reduced density matrix in the van Hove limit does not vanish but just oscillates as const·e−2iαst . The exponential that determines these fast oscillations (recall that t → ∞) is the same as in the zero coupling (S2 -isolated) limit of our model (1.5), where the reduced density matrix is ραβ (Ek , t) |v2 =0 = e−its(α−β) ραβ (0),

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J. L. LEBOWITZ, A. LYTOVA, AND L. PASTUR

hence is again const·e−2iαst if α 6= β, (α = −β). In the case where the two-level system models a continuous quantum mechanical degree of freedom associated with a potential with two wells (see e.g. [8] for examples and discussion), the above oscillation reflects the phase coherence between the quantum mechanical amplitudes for being in the left and right wells, a pure quantum mechanical effect. In this case our result means that an environment, modeled by a random matrix, does not destroy the quantum mechanical coherence, at least in the weak coupling regime corresponding to the van Hove limit. However, from the statistical mechanics point of view the absence of decay, moreover, fast oscillations, of the off-diagonal entries of the reduced density matrix seems not too natural. In this connection it worth noting that the fast (”microscopical”) oscillating behaviour of ραβ , α 6= β can be converted into a decaying behaviour by several modification of our initial setting. One of them is to assume that the spacing 2s of our two-level system is random and continuously distributed, although concentrated around a certain 2s0 . In other words, it is necessary to assume that the two-level system is the subject of a certain (even small) noise. Another modification is to replace the van Hove limit (5.23)

lim ρ(E, t) |v2 =τ /t

t→∞

by (5.24)

lim

t→∞,△t→∞

(2△t)−1

Z

t+△t

t−△t

ρ(E, t′) |v2 =τ /t′ dt′.

If △t = t, we just replace the limit t → ∞ by the Cesaro limit (time average limit), a rather often used procedure in statistical mechanics. However the off-diagonal entry vanishes even for t → ∞, bat △t/t → 0, although with a smaller rate of decay. One can view this as an assumption on a sufficiently large (macroscopic) measurent time: s−1