Nonequilibrium thermal entanglement for three-spin chain

0.28 MB·

by  N. Pumulo

Nonequilibrium thermal entanglement for three-spin
chain
N. Pumulo, I. Sinayskiy , F. Petruccione
∗
4
1 Quantum Research Group, School of Physics and National Institute for Theoretical
0 Physics, University of KwaZulu-Natal, Durban, 4001, South Africa
2

n
a
J

0
1 Abstract

]
The dynamics of a chain of three spins coupled at both ends to separate
h
p bosonic baths at different temperatures is studied. An exact analytical so-
-
t lution of the master equation in the Born-Markov approximation for the
n
a reduced density matrix of the chain is constructed. It is shown that for long
u
times the reduced density matrix converges to the non-equilibrium steady-
q
[ state. Dynamical and steady state properties of the concurrence between the

first and the last spin are studied.
1
v
Keywords: Non-equilibrium thermal entanglement, Spin chain,
2
0 Born-Markov approximation
3
2
.
1
1. Introduction
0
4
1 In the description of a real physical system effects of the environment
: play an important role [1]. Typically, the interaction with the surroundings
v
i destroys quantum correlations in the system. However, in some situations,
X
interaction with the environment can create extra quantum correlations in
r
a
the system [2]. Over the past few years the phenomenon of thermal entan-
glement has been extensively studied [3, 4, 5, 6]. When the quantum system
is in contact with a thermal reservoir, after a certain time the system relaxes
into its thermal state in a canonical distribution form and thermal entangle-
ment for such a scenario has been also widely studied in the past few years
[3]. A more general situation arises if the quantum system is coupled to two
∗
Corresponding author. Tel./Fax: [phone]/8090
e-mail address: [email protected]
Preprint submitted to Physics Letters A January 13, 2014
thermal reservoirs at different temperatures. To model the quantum analog
of the Fourier law one natural choice is to consider a chain of spins connected
at the ends to thermal reservoirs at different temperatures [7].
In a previous paper [4] thermal entanglement which is created in a two
spin system in contact with two heat reservoirs at different temperatures was
studied. Huang et al. [5] studied numerically a non-equilibrium three spin
system. The model investigated in this paper is slightly simpler than the one
studied by Huang et al. numerically [5]. We will assume, as Huang et al.
did, a chain of spins coupled by XX interaction. Huang et al. consider the
possibility of different strength of the spin-spin interaction between the first
and the second spin and the second and the third one. Here, we assume that
the strength of all spin-spin interactions is the same. This assumption will
allow us to construct an exact analytical expression for the density matrix of
the three spins system. We will show that the reduced density matrix of the
three spin chain is converging with time to the steady state.
The analytical solution for the three spin system is compared to the cor-
responding solution of a two spin model, that was studied previously [4].
Calculation of the concurrence between extreme spins in the three spin chain
and concurrence in the two spin chain, reveals that adding and intermediate
spin into the chain increases steady state concurrence for a certain range of
temperatures of the baths.
The paper is organized as follows. In Section 2 we introduce the model
of a chain of three spins coupled at both ends to bosonic baths at different
temperatures. The model is a natural generalization of the one studied in [4].
Forcompleteness wefollow[4]inderiving themasterequationforthereduced
density matrix of the spin system in the Born-Markov approximation. Thus,
the master equation has a structure similar to the one in [4]. In Section 3
we present the analytical solution and show the convergence of the obtained
solution to the non-equilibrium steady-state. Unlike in the two spin chain
case [4], in the three spin chain case the time dependence of the non-diagonal
elements has a more complicated form and we show analytically that all non-
diagonal elements vanish with time. Finally, in Section 4 we present results
and conclude.
2. The model
We consider a spin chain consisting of three spins. The first and the last
spin are coupled to separate bosonic baths at different temperatures. The
2
total Hamiltonian of the system is given by
Hˆ = Hˆ +Hˆ +Hˆ +Hˆ +Hˆ , (1)
S B1 B3 SB1 SB3
where Hˆ is the Hamiltonian of the spin subsystem,
S
3 2
ǫ
Hˆ = σˆz +κ σˆ+σˆ +σˆ σˆ+ . (2)
S 2 i i i−+1 i− i+1
i=1 i=1
X X(cid:0) (cid:1)
In the Hamiltonian Hˆ the constant ǫ denotes the energy level of the spins
S
and κ the strength of the spin-spin interaction.
The Hamiltonians Hˆ of the reservoirs coupled to the first spin (j = 1)
Bj
and the last spin (j = 3) are given by
HˆBj = ωn,jˆb†n,jˆbn,j. (3)
n
X
The interaction between the spin subsystem and the bosonic baths is de-
scribed by
HˆSBj = σˆj+ gn(j)ˆbn,j +σˆj− gn(j)∗ˆb†n,j. (4)
n n
X X
Of course, σˆj±,σˆjz are the well-known Pauli matrices and ˆb†n,j and ˆbn,j denote
bosonic creation and annihilation operators. The constants ω and g(j)
n,j n
denote frequencies of the bosonic modes and amplitudes of the transitions
due to spin-boson interaction, respectively. In this paper units are chosen
such that ~ = k = 1.
B
In the Born-Markov approximation the dynamics of the reduced density
matrix ρˆ of the spin subsystem is described by [1]:
dρˆ
= i[Hˆ ,ρˆ]+ (ρˆ)+ (ρˆ), (5)
S 1 3
dt − L L
with dissipators (j = 1 and j = 3)
Lj(ρˆ) ≡ Jµ(j,ν)(ωj,ν){[Vˆj,µ,[Vˆj†,ν,ρˆ]] (6)
µ,ν
X
−(1−eβjωj,ν)[Vˆj,µ,Vˆj†,νρˆ]}.
The spectral density is given by
Jµ(j,ν)(ωj,ν) = ∞dseiωj,νshe−isHˆBjfˆj†,νeisHˆBjfˆj,µij. (7)
Z0
3
In the derivation of the master equation Eq. (5) it is assumed that the
interaction Hamiltonian Eq. (4) can be represented in the following way,
HˆSBj = Vˆj†,µfˆj,µ +Vˆj,µfˆj†,µ, (8)
µ
X
where the operators Vˆ and fˆ describe transitions in the spin and bath
j,µ j,ν
subsystems, respectively. Note, that the index j identifies the bath (j = 1
and j = 3) and the index µ,ν the number of the transition. As usual (cf. [4,
6, 8, 9]), the transition operators Vˆ originate from the decomposition of the
j,µ
interaction Hamiltonian Hˆ into eigenoperators of the system Hamiltonian
SBj
Hˆ and satisfy the following condition,
S
[Hˆ ,Vˆ ] = ω Vˆ . (9)
S j,µ j,µ j,µ
−
To construct the explicit form of the transition operators we need to find
the eigenvectors and eigenvalues of the Hamiltonian of the spin system Hˆ .
S
After straightforward calculations we get a set of eigenvectors λ , where
i
| i
(i = 1,..,8):
λ = 0,0,0 , (10)
1
| i | i
0,0,1 1,0,0
λ = | i−| i, (11)
2
| i √2
0,1,1 1,1,0
λ = | i−| i, (12)
3
| i √2
λ = 1,1,1 , (13)
4
| i | i
1,0,0 √2 0,1,0 + 0,0,1
λ = | i− | i | i, (14)
5
| i 2
1,1,0 √2 1,0,1 + 0,1,1
λ = | i− | i | i, (15)
6
| i 2
1,0,0 +√2 0,1,0 + 0,0,1
λ = | i | i | i, (16)
7
| i 2
1,1,0 +√2 1,0,1 + 0,1,1
λ = | i | i | i, (17)
8
| i 2
with corresponding eigenvalues λ , where (i = 1,..,8):
i
3ǫ
λ = λ = , (18)
1 4
− − 2
4
ǫ
λ = λ = , (19)
2 3
− −2
ǫ
λ = λ = √2κ, (20)
5 8
− −2 −
ǫ
λ = λ = √2κ. (21)
6 7
− 2 −
Following [8] it is easy to see that in the case of three spins the operators
Vˆ have the form:
j,µ
1
Vˆ = ( λ λ + λ λ λ λ + λ λ ), (22)
1,1 1 2 3 4 5 6 7 8
√2 −| ih | | ih |−| ih | | ih |
1
Vˆ = ( λ λ λ λ λ λ + λ λ ), (23)
1,2 1 5 2 6 7 3 8 4
2 | ih |−| ih |−| ih | | ih |
1
Vˆ = ( λ λ + λ λ + λ λ + λ λ ), (24)
1,3 1 7 2 8 5 3 6 4
2 | ih | | ih | | ih | | ih |
1
Vˆ = ( λ λ λ λ λ λ + λ λ ), (25)
3,1 1 2 3 4 5 6 7 8
√2 | ih |−| ih |−| ih | | ih |
1
Vˆ = ( λ λ + λ λ + λ λ + λ λ ), (26)
3,2 1 5 2 6 7 3 8 4
2 | ih | | ih | | ih | | ih |
1
Vˆ = ( λ λ λ λ λ λ + λ λ ). (27)
3,3 1 7 2 8 5 3 6 4
2 | ih |−| ih |−| ih | | ih |
The frequencies of the transitions are given by:
ω ω ω = ǫ, (28)
1,1 3,1 1
≡ ≡
ω ω ω = ǫ √2κ, (29)
1,2 3,2 2
≡ ≡ −
ω ω ω = ǫ+√2κ. (30)
1,3 3,3 3
≡ ≡
In following [4] we choose the bosonic bath as the infinite set of harmonic
oscillators and the coupling constants to be frequency independent, so that
J(j)(ω ) = γ n (ω ), where n (ω ) is the Bose distribution, n (ω ) = (eβjων
ν j j ν j ν j ν
−
1) 1. The dissipators of the master equation Eq. (5) take the following
− j
L
form,
3
1
Lj(ρˆ) = γj(nj(ωi)+1) Vˆj,iρˆVˆj†,i − 2[Vˆj†,iVˆj,i,ρˆ]+ (31)
i=1 (cid:18) (cid:19)
X
1
+γjnj(ωi) Vˆj†,iρˆVˆj,i − 2[Vˆj,iVˆj†,i,ρˆ]+ ,
(cid:18) (cid:19)
5
where [A,B] AB +BA denotes the anticommutator.
+
In the basis≡of eigenvectors of the Hamiltonian Hˆ the equation for the
S
diagonal elements of the density matrix ρˆ can be written as:
ρ (t) ρ (t)
11 11
d ρ (t) ρ (t)
 22.  = B 22. , (32)
dt .. ..
   
 ρ (t)   ρ (t) 
88 88
   
   
where B is a 8 8 matrix of constant coefficients. The fact that the equation
×
for the diagonal elements decouples from the non-diagonal ones is a conse-
quence of the diagonal form of the semigroup generator, Eq. (31), and the
diagonal form of the product of the transition operators Vˆj,iVˆj†,i. There are
28 non-diagonal elements. The dynamical equations for them can be divided
in three groups. The first group consists of 4 non-diagonal elements the time
dependence of which is trivial:
ρ (t) = ρ (0)esi,jt, (33)
i,j i,j
where s is a constant. The second group consists of 6 couples of non-
i,j
diagonal elements which satisfy the following system of equations:
d ρ (t) ρ (t)
i,j = Mp i,j , (34)
dt ρ (t) 2 ρ (t)
k,l k,l
(cid:18) (cid:19) (cid:18) (cid:19)
where Mp is a 2 2 matrix with constant coefficients. The third group
2 ×
consists of 3 quadruples of non-diagonal elements, each of them satisfies the
following system of differential equations:
ρ (t) ρ (t)
i1,j1 i1,j1
ddt  ...  = M4p ... , (35)
ρ (t) ρ (t)
i4,j4 i4,j4
   
   
where Mp is a 4 4 matrix of constant coefficients.
4 ×
6
3. Analytical Solution
The exact analytical solution for the diagonal elements in the basis of
eigenvectors λ of the Hamiltonian Hˆ has the following form:
i S
{| i}
ρ (t) ρ (0)
11 11
ρ (t) ρ (0)
 22..  = RJ(t)R−1 22.. , (36)
. .
   
 ρ (t)   ρ (0) 
88 88
   
   
where R is the nondegenerate matrix
A+C+ C+ A+C+ C+ A+ 1 A+ 1
A−C− −C− A−C− −C− −A− −A−
C+ C+ C+ C+ 1 1 1 1
 C− C− C− C− − − − − 
A+B− B− A+ 1 A+B− B− A+ 1
A−B+ −B+ −A− A−B+ −B+ −A−
 B− B− 1 1 B− B− 1 1 
R =  A+BB−+C+ BB−+C+ A−+C+ −C+ AB++B− BB−+ −A+ −1 ,
 AB−B−+CC+− −BB−+CC+− −A−CC+− CC−+ −AB−B−+ BB+− A1− −1 
 B+C− B+C− −C− −C− −B+ −B+ 
 A+ 1 A+ 1 A+ 1 A+ 1 
 A− − A− − A− − A− − 
 1 1 1 1 1 1 1 1 
 
 (37)
and J(t) is a diagonal matrix given by
J(t) = diag 1,e At,e Bt/2,e (A+B/2)t,e Ct/2,e (A+C/2)t,e (B+C)t/2,e (A+B/2+C/2)t ,
− − − − − − −
(38)
(cid:0) (cid:1)
where the coefficients A ,B ,C are
± ± ±
A = J(1)( ω )+J(3)( ω ), A = A+ +A , (39)
± 1 1 −
± ±
B = J(1)( ω )+J(3)( ω ), B = B+ +B , (40)
± 2 2 −
± ±
C = J(1)( ω )+J(3)( ω ), C = C+ +C . (41)
± 3 3 −
± ±
The time dependence of the non-diagonal elements of the first group has the
form:
ρ1,4(t) = ρ1,4(0)e−t(A2+B+4C−i3ǫ), (42)
ρ2,3(t) = ρ2,3(0)e−t(A2+B+4C−iǫ), (43)
ρ5,8(t) = ρ5,8(0)e−t(A2+B+4C−iǫ−i2√2κ), (44)
7
ρ6,7(t) = ρ6,7(0)e−t(A2+B+4C+i3ǫ−i2√2κ). (45)
The first couple of the non-diagonal elements of the second group (Eq. (34),
p = 1) satisfy the following equation:
d ρ (t) ρ (t)
2,5 = M1 2,5 , (46)
dt ρ (t) 2 ρ (t)
8,3 8,3
(cid:18) (cid:19) (cid:18) (cid:19)
where
A B 1 C C+
M1 = i√2κ + − − . (47)
2 −2 − 4 − 2 C− C+
(cid:18) (cid:19) (cid:18) − (cid:19)
All the other couples of non-diagonal elements from the second group satisfy
equations of a similar kind. The explicit form of the matrices Mi (i = 2,..,6)
2
can be found in the Appendix. The solution for the corresponding couple
of non-diagonal elements can be constructed with the help of the following
formula:
α 0 0 β
exp t∆+t − t = (48)
0 β ± α 0
(cid:20) (cid:18) − (cid:19) (cid:18) (cid:19)(cid:21)
e∆t β +αe−(α+β)t 0 e∆t 1−e−(α+β)t 0 β .
α+β 0 α+βe (α+β)t ± α+β α 0
(cid:18) − (cid:19) (cid:0) (cid:1) (cid:18) (cid:19)
In the case α > 0, β > 0 and Re[∆] < 0 it is obvious that
α 0 0 β
lim exp t∆+t − t = 0. (49)
t 0 β ± α 0
→∞ (cid:20) (cid:18) − (cid:19) (cid:18) (cid:19)(cid:21)
This means that all non-diagonal elements of the second group will vanish
at asymptotic times.
The first quadruple of the non-diagonal elements of the third group (Eq.
(35), p = 1) satisfies the following system of equations:
ρ (t) ρ (t)
1,2 1,2
d ρ (t) ρ (t)
 3,4  = M1 3,4 , (50)
dt ρ (t) 4 ρ (t)
5,6 5,6
 ρ (t)   ρ (t) 
 7,8   7,8 
   
where M1
4
A
M1 = +iǫ+T1, (51)
4 −2 4
8
where T1 is the following matrix
4
B C 0 G+ H+
− −
− − −
1 0 B+ C+ H G
T1 =  − − − − −  (52)
4 2 G H+ B+ C 0
− −
− − −
 H G+ 0 B C+ 
− −
 − − − 
 
and the constants are
F = J(1)( ω ) J(3)( ω ), (53)
± 1 1
± − ±
G = J(1)( ω ) J(3)( ω ), (54)
± 2 2
± − ±
H = J(1)( ω ) J(3)( ω ). (55)
± 3 3
± − ±
The solution of the above system of equations has a complicated form. In-
stead of presenting it we will show that the real part of the maximum eigen-
values of the matrix M1 is strictly negative:
4
Re λ (M1) < 0. (56)
Max 4
This means that for long times(cid:2)the solutio(cid:3)n for this set of the non-diagonal
elements will converge tozero. Inordertoprove that, weneedfirst toanalyze
the eigenvalues of the matrix T1. After some straightforward calculation one
4
can see that:
1
λ = ( B C B C ), (57)
1,..,4 G H
4 − − ±| ± |
where the constant B is
G
B = (B+ B )2 +4G+G (58)
G − −
−
p
and the constant C is
H
C = (C+ C )2 +4H+H . (59)
H − −
−
p
Taking into account that B, C, B , C > 0 the maximum eigenvalue of the
G H
matrix T1 is:
4
1
λ = ( B C +B +C ). (60)
Max G H
4 − −
It is easy to see that the difference B B is non negative:
G
−
B B B2 B2 (61)
≥ G ⇔ ≥ G
9
(B+ +B )2 (B+ B )2 +4G+G (62)
− − −
⇔ ≥ −
B+B G+G 0. (63)
− −
⇔ − ≥
Recalling the explicit expression for B and G it follows:
± ±
B+B G+G = (64)
− −
−
2J(1)(ω )J(3)( ω )+2J(3)(ω )J(1)( ω ) 0.
2 2 2 2
− − ≥
In a similar way one can show that C C 0. This means that
H
− ≥
A 1
Re λ (M1) = (B +C B C ) < 0, (65)
Max 2 −2 − 4 − G − H
i.e., the maxim(cid:2)um value (cid:3)of the real part of the eigenvalues of the matrix
M1 is negative and with time the corresponding non-diagonal elements will
4
converge to zero. A similar proof holds for the other non-diagonal elements
from the third group. An explicit form of the corresponding matrices can be
found in the Appendix. More detailed calculations can be found in [10].
From the above discussion it follows that all non-diagonal elements con-
verge to zero and a stationary state for the reduced density matrix is given
by the long time limit of the diagonal elements, i.e.,
A+B+C+
A B+C+
−
 
A+B C
− −
1  A B C 
lim ρ (t) =  − − − . (66)
t ii ABC  A+B−C+ 
→∞  
 A B C+ 
− −
 
 A+B+C 
−
 
 A B+C 
− −
 
 
As an example of the dynamics of the system, we consider an initial W
3
| i
state for the three spin chain, i.e.,
1
W = ( 1,0,0 + 0,1,0 + 0,0,1 ). (67)
3
| i √3 | i | i | i
The only non-zero non-diagonal elements will be λ ρˆλ , λ ρˆλ and
5 7 6 8
h | | i h | | i
their transpositions,
λ ρˆλ = λ ρˆλ (68)
5 7 7 5 ∗
h | | i h | | i
A+ +A e At
= e (B+C)t/4+i2√2κt − − ,
−
6A
10