Degradation of entanglement in moving frames

0.12 MB·English

by  Shahpoor Moradi

Degradation of entanglement in moving frames
Shahpoor Moradi1,
∗
2
1
0 January 4, 2012
2

n
a
J 1 Department of Physics, Razi University, Kermanshah, IRAN

2

Abstract
]
h
p The distillability of bipartite entangled state as seen by moving observers has been
t- investigated. It is found that the same initial entanglement for a state parameter α and
n its normalized partner √1 α2 will be degraded as seen by moving observer. It is shown
a
−
u that in the ultra relativistic limit, the state does not have distillable entanglement for any
q α.
[

1 Relationship between special relativity and quantum information theory is discussed by many
v
5 authors [1]. Peres et al [2] investigated the relativistic properties of spin entropy for a single,
0 free particle of spin 1/2. They show that the usual definition of quantum entropy has no
5 −
invariant meaning in special relativity. The reason is that, under a Lorentz boost, the spin
0
. undergoes a Wigner rotation whose direction and magnitude depend on the momentum of
1
0 the particle. Even if the initial state is a direct product of a function of momentum and a
2
function of spin, the transformed state is not a direct product. Lamata et al[3] define weak
1
: and strong criteria for relativistic isoentangled and isodistillable states to characterize relative
v
i and invariant behavior of entanglement and distillability. In this letter, we choose a generic
X
state as the initial entangled state and we will try to show that the entanglement is degraded
r
a as seen by the relativistically observer. This help us to understand the relationship between
special relativity and quantum information theory. The initial nonmaximal entangled state is
Φ = αΨ(a)(p )Ψ(b)(p )+√1 α2Ψ(a)(p )Ψ(b)(p ), (1)
| i 1 a 1 b − 2 a 2 b
where α is some number that satisfies α (0,1). Here p and p are the corresponding
a b
| | ∈
momentums vectors of particles A and B and
g(p )
Ψ(a)(p ) = g(p ) 0 = 0,p = a , (2)
1 a a | i | ai 0 !
0
Ψ(a)(p ) = g(p ) 1 = 0,p = . (3)
2 a a | i | bi g(pa) !
∗e-mail: [email protected]
1
For simplicity assume that they are sufficiently well localized around momenta p, Under the
Lorentz transformation the states (2) and (3) transformed as [3]
b (p) cos(θ /2)
Λ[Ψ (p)] = 1 = p g(p) = cosθ θ /2) 0,p +sin(θ /2) 1,p , (4)
1 b2(p) ! sin(θp/2) ! ( p | i p | i
b (p) sin(θ /2)
Λ[Ψ (p)] = − 2 = − p g(p) = sin(θ /2) 0,p +cos(θ /2) 1,p , (5)
2 b1(p) ! cos(θp/2) ! − p | i p | i
where θ is Wigner angle satisfies the relation
p
sinhξsinhδ
tanθ = , (6)
p
coshξ +coshδ
here coshξ = (1 β2)−1/2 where β is boost speed and coshδ = p /m. Now under Lorentz
0
−
transformation the state transformed to (after tracing over momentum eigenket p ),
| i
Φ = αcos2(θ /2)+√1 α2sin2(θ /2) 00 +
Λ p p
| i − | i
(cid:16) (cid:17)
sin(θ /2)cos(θ /2)(α √1 α2)( 01 + 10 )+
p p
− − | i | i
αsin2(θ /2)+√1 α2cos2(θ /2) 11 . (7)
p p
− | i
(cid:16) (cid:17)
A very popular measure for the quantification of bipartite quantum correlations is the concur-
rence [4]. This quantity can be defined
C(ρ) = max 0,λ λ λ λ . (8)
1 2 3 4
{ − − − }
with λ being the square roots of the eigenvalues of ρ (σ σ )ρ∗ (σ σ ) where the asterisk
i AB y⊗ y AB y⊗ y
0 i
denotes complex conjugation and and σ = − . Now the concurrence for this state is
y i 0
!
C( Φ ) = 2α√1 α2. (9)
Λ
| i −
Which isthe Lorentz invariant concurrence. To bemore precise oneshould take wave packets in
momentum space, with Gaussian momentum distributions g(p) = π−3/4w−3/2exp( p 2/2w2).
−| |
If we trace the momentum degrees of freedom we obtain the usual entangled state φ =
| i
α 00 +√1 α2 11 . The general density matrix for two particle systems with momentums p
a
| i − | i
and p is given by
b
ρ = C Ψ (p ) Ψ (p )[Ψ (p′) Ψ (p′)]†. (10)
Φ ijkl i a ⊗ j b l a ⊗ m b
ijkl=1,2
X
For state (1) the coefficients C are
ijkl
C = α2, C = 1 α2,
1111 2222
−
C = C = α√1 α2 (11)
1122 2211
−
2
For obtaining the Lorentz transformation of (10), we need the relativistic properties of spin
entropy for a single, free particle of spin 1/2. The quantum state of a spin-1 particle can be
− 2
written in the momentum representation as follows
a (p)
Ψ(p) = 1 , (12)
a (p)
2 !
where
( a (p) 2 + a (p) 2)dp = 1. (13)
1 2
| | | |
Z
The density matrix corresponding to state (6) is
a (p′)a (p′′)∗ a (p′)a (p′′)∗
ρ(p′,p′′) = 1 1 1 2 . (14)
a (p′)a (p′′)∗ a (p′)a (p′′)∗
1 2 2 2 !
By setting p′ = p′′ = p and integrating over p we obtain the reduced density matrix for spin
1 1+n n in
σ = z x − y , (15)
2 nx +iny 1 nz !
−
where the Bloch vector n is given by
n = ( a (p) 2 a (p) 2)dp = 1, (16)
z 1 2
| | −| |
Z
n in = a (p)a (p)∗dp. (17)
x y 1 2
−
Z
Now under Lorentz boost density matrix (10) transformed into
Λρ Λ† = C Λ(p )Ψ (p ) Λ(p )Ψ (p )
Φ ijklmn a i a b j b
⊗
ijkl=1,2
X
[Λ(p )Ψ (p′) Λ(p )Ψ (p′)]†. (18)
× a l a ⊗ b m b
The reduced density matrix for spin is obtained by setting p = p′, p = p′ and tracing
a a b b
over momentum
τ = Tr [Λρ Λ†] = C Tr Λ(p )Ψ (p )[Λ(p )Ψ (p )]†
papb Φ ijkl pa a i a a k a
ijkXl=1,2 n o
Tr Λ(p )Ψ (p )[Λ(p )Ψ (p )]† . (19)
⊗ pb b j b b l b
n o
To leading order w/m 1 we have
≪
2
w ξ
n = n 1 tanh , n = n 0, (20)
z x y
≈ − 2m 2! ≈
It can be appreciated in Eq. (19) that the expression is decomposable in the sum of the tensor
productsof2 2spinblocks, eachcorresponding toeachparticle. Wecompute nowthedifferent
×
blocks, corresponding to the four possible tensor products of the states (3) and (4):
1 1+n 0
Tr Λ(p)Ψ (p)[Λ(p)Ψ (p)]† = , (21)
p 1 l 2 0 1 n !
n o −
3
1 1 n 0
Tr Λ(p)Ψ (p)[Λ(p)Ψ (p)]† = − , (22)
p 2 2 2 0 1+n !
n o
1 0 1+n
Tr Λ(p)Ψ (p)[Λ(p)Ψ (p)]† = , (23)
p 1 2 2 (1 n) 0 !
n o − −
1 0 (1 n)
Tr Λ(p)Ψ (p)[Λ(p)Ψ (p)]† = − − . (24)
p 2 1 2 1+n 0 !
n o
With the help of Eqs . (21)-(24), it is possible to compute the effects of the Lorentz trans-
formation, associated with a boost in the x direction, on any density matrix of two spin-1/2
particles with factorized Gaussian momentum distributions. In particular density matrix (19)
is reduced to
4α2n+(1 n)2 0 0 2α√1 α2(1+n2)
− −
1 0 1 n2 2α√1 α2(1 n2) 0
 
τ = − − − − .
4 0 2α√1 α2(1 n2) 1 n2 0
 − − − − 
 2α√1 α2(1+n2) 0 0 4α2n+(1+n)2 
 − − 
 (25) 
We can apply now the positive partial transpose criterion [5] to know whether this state is
entangled and distillable. The partial transpose criterion provides a sufficient condition for the
existence of entanglement in this case: if at least one eigenvalue of the partial transpose is
negative, the density matrix is entangled; but a state with positive partial transpose can still
be entangled. It is the well-known bound or nondistillable entanglement [6]. Partial transpose
of density matrix (25) yields
4α2n+(1 n)2 0 0 2α√1 α2(1 n2)
− − − −
1 0 1 n2 2α√1 α2(1+n2) 0
τT =  − − .
4 0 2α√1 α2(1+n2) 1 n2 0
 − − 
 2α√1 α2(1 n2) 0 0 4α2n+(1+n)2 
 − − − − 
 
It is possible diagonalize τT and get it’s eigenvalues
1 1
λ = (1 n2)+ α√1 α2(1+n2), (26)
1
4 − 2 −
1 1
λ = (1 n2) α√1 α2(1+n2), (27)
2
4 − − 2 −
1 1
λ = (1+n2)+ n2 +α2(1 α2)(n4 6n2 +1), (28)
3
4 2 − −
q
1 1
λ = (1+n2) n2 +α2(1 α2)(n4 6n2 +1). (29)
4
4 − 2 − −
q
For 0 < n,α < 1 eigenvalues λ , λ and λ are always positive. The eigenvalue λ is negative
1 3 4 2
for α√1 α2 > R where
− 1 n2
R = − . (30)
2(1+n2)
4
In this range the logarithmic negativity takes the form
1
N = log (1+n2) 1+2α√1 α2 . (31)
2 2 −
(cid:26) (cid:16) (cid:17)(cid:27)
In ultra relativistic limit n 0: N log 1 +α√1 α2 , then the state does not have
→ → 2 2 −
distillableentanglement foranyα. Fortherestnframen = 1: No= log 1+2α√1 α2 . Inthe
2 −
range0 < α < 1/√2thelargerα,thestrongertheinitialentanglement;nbutintherangeo1/√2 <
α < 1 the larger α, the weaker the initial entanglement. For finite velocity, the monotonic
decrease of N with increasing boost speed for different α means that the entanglement of the
initial state is lost due to Wigner rotation. From Fig.1 it is found that the entanglement in
moving frame, for α and it’s normalized partner √1 α2, will be degraded as boost speed
−
increases. Here we calculate the concurrence which is defined as follows
C = 2 detρ = 2 detρ . (32)
A B
q q
For density matrix (25) we have
α2n+(1 n)/2 0
ρ = ρ = − , (33)
A B 0 α2n+(1+n)/2
!
−
then
1
C = (1 n+2α2n)(1+n 2α2n). (34)
2 − −
q
In non relativistic limit (n = 1) we have: C = 2α√1 α2 and in ultrarelativistic limit (n = 0):
−
C = 1/2. It is interesting that for maximally entangled state as α = 1/√2 for all values of n
concurrence is 1.
References
[1] A. Peres and D. R. Terno, Rev. Mod. Phys. 76, 93 (2004)(References therein)
[2] A. Peres, P. F. Scudo, and D. R. Terno, Phys. Rev. Lett. 88, 230402 (2002).
[3] L. Lamata, M. A. Martin-Delgado, and E. Solano, Phys. Rev.Lett. 97, 250502 (2006).
[4] . Hill, W. K. Wootters, Phys. Rev. Lett. 78, 5022 (1997).
[5] A. Peres, Phys. Rev. Lett. 77, 1413 (1996).
[6] G. Vidal and R. F. Werner, Phys. Rev. A 65, 032314 (2002)
5
1.0
0.5
0.0
−0.5
0.0
−1.0
0.25
1.0 0.5
0.75 alpha
0.5 0.75
0.25 1.0
n 0.0
Figure 1: Plot of negativity versus n and α
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