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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 concurrence [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 transformation, 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 6