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1 Commun. Theor. Phys. Beijing, China pp c Chinese Physical Society Vol. 49, No. 3, March 5, 008 Calculation of Entanglement Entropy for ContinuousVariable Entangled State Based on General TwoMode Boson Exponential Quadratic Operator in Fock Space DAI FangWen and MA Lei Department of Physics, East China Normal University, Shanghai 0006, China Received February 5, 007 Abstract We obtain an explicit formula to calculate the entanglement entropy of bipartite entangled state of general twomode boson exponential quadratic operator with continuous variables in Fock space. The simplicity and generality of our formula are shown by some examples. PACS numbers: Ud, p, Ta Key words: entanglement entropy, continuousvariable entangled states, linear quantum transformation Introduction 0 Σ B =, Continuous quantum variables have emerged as an alternative to discretelevel systems for performing quan 0 tum information processing tasks. Entanglement plays M M M c M Σ B = c d a central role in various aspects of quantum information processing because quantum teleportation, [] quan Thus, from Eq. one can conveniently obtain the c b c. 3 tum cryptography, [] and quantum dense coding, [3] etc., entanglement entropy of the states in Eq. only by calculating the eigenvalues of negative Hermitian matrix N. are all based on it. Thus, it is interesting to measure the entanglement of continuous variables systems. The von But in general cases, the form of a bipartite continuousvariable entangled state is not written as Eq. explicitly, Neumann entropy [4] of either partial trace of the density operator for any bipartite pure entangled state is considered as a good measurement of quantum entangle complicated to use Eq.. and then N is more implicitly. In certain cases, it is too ment. Based on the Schmidt decompositions, Parker et In this paper, we shall derive an explicit formula al. [5] have developed an elegant method to calculate the to calculate the entanglement entropy of any bipartite entanglement of the bipartite pure entangled states with continuousvariable entangled states in Fock space. This continuous variables by means of the integral eigenvalue paper is arranged as follows. In Sec., we shall obtain equations in the coordinatemomentum space. But in certain cases, it is somewhat harder to find such a Schmidt the general form of the bipartite continuousvariable entangled states, and then derive the explicit formula to calculate the entanglement entropy of the states. In Sec. 3, basis for the continuous variable systems. By virtue of the linear quantum transformation theory LQTT, [6] Lu et al. [3,4] have provided an alter by virtue of the explicit formula, we shall calculate the degree of entanglement of some bipartite continuousvariable entangled states. A brief conclusion will close the paper native method to calculate the entanglement entropy for in the last section. any bipartite pure entangled Gaussian state with continuous variables in Fock space. If the density operator of Explicit Formula for Calculating Entanglement Entropy of Bipartite Continuous the bipartite continuousvariable entangled state can be written as follows: { a ρ = A 0 : exp [a, a a Variable Entangled States M + a a, a M. General Form of Entangled State of Two a Mode Boson Exponential Quadratic Operator a ]} + a, a M : a where A 0 is a normalization factor, M i i =, are Hermitian matrices, and M = e f f e with e and f being two arbitrary complex numbers. Then the entanglement entropy of any bipartite pure entangled Gaussian state with continuous variables in Fock space can be calculated by the following formula, E = { ln det e N + tr [ N e N ]}, ln where a d N = lnm, M =, ac bd =, b c Let us consider the entangled state of a twomode boson operator with continuous variables as follows: ψ = U 00, 4 where U is a twomode boson exponential quadratic operator. If we denote Λ = a, ã a = a, a ã = a, a, 5 a i and a i i =, are twomode boson creation and annihilation operators, respectively. Thus, without any loss of generality, the ordinary form of U can be written as U = exp ΛNΣ Λ B, 6 where Σ B, N C 4 4, Σ B = 0 I I 0, I = 0 0, and N satisfies NΣ B = NΣ B. The project supported by the National Fundamental Research Program under Grant No. 006CB904 and National Natural Science Foundation of China under Grant No
2 No. 3 Calculation of Entanglement Entropy for ContinuousVariable Entangled State Based on General TwoMode 59 Denoting M = e N = A D B C, A, B, C, and D are four complex matrices, U and M thus satisfy [6] By using Eq. 8, we immediately have UΛU = ΛM, 7 MΣ B M = Σ B. 8 Ã B BA = 0, Ã C BD =, 9 If C exists, by using Eq. 9, we can rewrite matrix M as I C M = e N D C 0 I 0 = 0 I 0 C BC. 0 I Therefore, from LQTT, the action of U is equivalent to the product of three LQT operators, i.e., U = U + U 0 U, where U +, U 0, and U satisfy the following expressions, I C U + ΛU+ D = ΛM + = Λ, 0 I C U 0 ΛU0 0 = ΛM 0 = Λ, 0 C I 0 U ΛU = ΛM = Λ BC, I By virtue of LQTT, we can easily obtain 0 C D N + = lnm + =, 0 0 lnc 0 N 0 = lnm 0 =, N = lnm = 0 ln C, BC 0 U + = exp ΛN +Σ B Λ = exp a C Dã, U 0 = exp ΛN 0Σ B Λ = detc / : exp [ a C a ] :, U = exp ΛN Σ B Λ = exp ãbc a. 4 U +, U 0, and U only contain the terms of a i a j, a i a j, and a i a j, respectively, where i, j =,. By noticing expa i a j 00 = exp a i a j 00 = 00, 5 the bipartite continuousvariable entangled state 4 can be rewritten as ψ = U 00 = U + U 0 U 00 = det C / exp a C Dã From the above discussions, we obtain the general form of the entangled state of any twomode boson exponential quadratic operator 4 as the following form, ψ = det C / expαa + βa + γa a 00, 7 where α, β, and γ are determined by C D.. Explicit Formula of Calculating Entanglement Entropy It is easy to see that the entanglement entropy, i.e., the degree of entanglement of the entangled state 7 between a and a, is determined by α, β, and γ, but which parameter is more predominant? Are α and β really symmetrical? If γ = 0, obviously, the state 7 is a separable state, and its entanglement entropy should be zero. When γ 0, the density operator of Eq. 7 can be written as ρ =detc expαa + βa + γa a exp α a + β a + γ a a. 8 By noticing the normal ordering form of the twomode vacuum state projector = : e a a a a : 9 after substituting Eq. 9 into Eq. 8, we have, { α ρ =det C : exp a [, a α so we get α M = α a a β + a, a β a a γ 0 + a, a 0 γ a ]} : 0 β γ 0, M = β, M = 0 γ. By using Eq. 3, we find the explicit form of M as follows: M = γ 4 4 α 4 β 4 αβγ + α β γ α 4α β + β γ γ. α 4α β + βγ 4 β Then, substituting it into Eq., we obtain the explicit formula for calculating the entanglement entropy for the continuous variables entangled states 7 as follows: L 4 L + L E = log 4 L L 4L log 4 L + L log 4 L L 4 + log 4 + log L, 3 where the logarithm is defined in the complex domain and L = γ + γ 4 α 4 β 4 αβγ γ + α β γ. 4 From Eq. 4, we can see that α and β are symmetrical and exchangeable for calculating the entanglement entropy. Obviously, it is true because the modes a and a are symmetrical in Fock space. a
3 59 DAI FangWen and MA Lei Vol. 49 When 0 L <, equation 3 can be simplified as E = + L L + i 4 L L arctan log L. 5 When L >, equation 3 can be simplified as E = L + L log L + L 4 log L. 6 With Eqs. 5 and 6, one can easily have E L and E L +, so we have E L=. 7 Equation 7 shows that if, the entangled state 7 will be maximally entangled. The entanglement entropy is plotted as a function of L in Fig., and figure shows the entanglement entropy as a function of α and γ β is fixed to be 0.5. Fig. Entanglement entropy as a function of L E when L. Fig. Entanglement entropy as a function of α and γ β is fixed to be Some Applications 3. Calculating Entanglement Entropy of Common TwoMode Squeezed Vacuum State Now, let us consider twomode squeezed vacuum state ξ = exp ξ a a ξa a 00, where ξ = r e iθ is the squeezing parameter, by virtue of LQTT, [6] we can obtain ξ = sech rexp a a eiθ tanhr 00, where α = 0, β = 0, γ = e iθ tanhr, according to Eq. 4, we have By using Eq. 6, we can easily obtain L = tanh r + coth r >. E = cosh r log cosh r sinh r log sinh r. This result has been obtained by other authors through different methods, [5,68] but our formula is more explicit. figure 3a shows that the amount of entanglement entropy of twomode squeezed vacuum state is approximately linear against the amount of squeezing parameter r. 3. Calculating Entanglement Entropy of TwoMode OneSided Squeezed Vacuum State Reference [5] has constructed twomode onesided squeezed vacuum state { λ S = exp a + a 4[ a + ]} [ a 00 = sech / λ exp where α = β = tanhλ/4, γ = tanhλ/, so tanhλ 4 a + ] a 00, L = 4 coth λ > and then by Eq. 6 we obtain E = coshλ log + cosh λ cosh λ log sinh λ log cosh λ.
4 No. 3 Calculation of Entanglement Entropy for ContinuousVariable Entangled State Based on General TwoMode 593 Figure 3b shows that the amount of entanglement entropy of twomode onesided squeezed vacuum state is also approximately linear against the amount of squeezing parameter λ, but is about only half of amount of entanglement entropy of the twomode squeezed vacuum state with the squeezing parameter λ = r. Fig. 3 a Entanglement entropy of twomode squeezed vacuum state as a function of squeezing parameter r. b The entanglement entropy of twomode onesided squeezed vacuum state as a function of squeezing parameter λ. 3.3 Calculating the Entanglement Entropy of Entangled States Produced by a Beam Splitter with Squeezed States Inputs When the two input fields are squeezed, the output state from a beam splitter is [6] B θ, φs ζ S ζ 00, 8 where B θ, φ is the beam splitter operator, [ θ B θ, φ = exp a a e iφ a a e iφ]. t = cosθ/ and r = sinθ/ are the amplitude reflection and transmission coefficients. The beam splitter gives the phase difference between the reflected and transmitted fields. S ζ is the singlemode squeezed operator S ζ = exp ζ a ζa, ζ = s e iϕ is the squeezing parameter. By virtue of LQTT, [6] we can obtain B θ, φs ζ S ζ 00 = coshs coshs exp αa + βa + γa a 00, where α = [ e iϕ cos θ tanhs + e iϕ+φ sin θ ] tanhs, β = [ e iϕ φ sin θ tanhs + e iϕ cos θ ] tanhs, γ = sinθ e iϕ φ tanhs e iϕ+φ tanhs. 9 Thus, substituting Eq. 9 into Eq. 4, the entanglement entropy of the state 8 can be calculated explicitly. The entanglement entropy of the state 8 is plotted in Fig. 4 against the squeezing parameter s and reflection coefficient for s = 0.5. The relative phase φ = 0 in Fig. 4a and φ = π/ in Fig. 4b. These two figures have been obtained in Ref. [6], but it did not give the explicit form of the entanglement entropy. It is necessary to point out that in Ref. [6], natural logarithm is used to calculate the entanglement entropy, but in this paper, the logarithm with base is used. 3.4 Calculate the Entanglement Entropy of the Thermal Vacuum State of a Free Boson Now, we investigate the entanglement entropy of the thermal vacuum state of a free boson, [8], 0 β = e βω / exp e βω/ a ã 00. Substituting α = β = 0, γ = e βω/ into Eq. 4, we have and then, by using Eq. 6, we can easily get L = e βω + e βω >, E = log e βω e βω e βω log e βω.
5 594 DAI FangWen and MA Lei Vol. 49 Reference 8] obtained the same result by virtue of the technique of IWOP, [5] but we get the result more simply. Fig. 4 Entanglement entropy of the beamsplitter output field. The squeezing parameter for one squeezed input is fixed to S = 0.5 while the squeezing parameter for other squeezed state is varied from S = 0 to. The transitivity is R R r. The beam splitter gives phase difference φ = 0 a and φ = π/ b between reflected and transmitted fields. 4 Conclusions In Fock space, by virtue of LQTT, we show that the entangled state of any twomode boson exponential quadratic operator ψ = U 00 can be rewritten as ψ = detc / expαa + βa + γa a 00, and we obtain the explicit formula to calculate the entanglement entropy of the state. The simplicity and generality of our formula are shown by several entangled states of continuous variables. These states include the common twomode squeezed vacuum state, the twomode onesided squeezed vacuum state, the output states produced by a beam splitter with squeezed states inputs and the thermal vacuum state of a free single boson. It is shown that LQTT and the explicit formula we obtained are power tools for investigating the degree of entanglement for entangled states of twomode boson system with exponential quadratic operator. Acknowledgments We thank Dr. JinMing Liu for beneficial discussions. References [] C.H. Bennett, G. Brassard, et al., Phys. Rev. Lett [] K. Ekert, Phys. Rev. Lett [3] C.H. Bennett and S.J. Wiesner, Phys. Rev. Lett [4] C.H. Bennett, H.J. Herbert, S. Popescu, and B. Schumacher, Phys. Rev. A [5] S. Parker, S. Bose, and M.B. Plenio, Phys. Rev. A [6] Y.D. Zhang and Z. Tang, J. Math. Phys [7] Y.D. Zhang and Z. Tang, Nuovo Cimento B [8] L. Ma and Y.D. Zhang, Nuovo Cimento B [9] Y.D. Zhang and Z. Tang, Commun. Theor. Phys. Beijing, China [0] Y.D. Zhang, L. Ma, X.B. Wang, et al., Commun. Theor. Phys. Beijing, China [] S.X. Yu and Y.D. Zhang, Commun. Theor. Phys. Beijing, China [] L. Ma, Doctoral Dissertation, University of Science and Technology of China 995. [3] H.X. Lu, Z.B. Chen, J.W. Pan, and Y.D. Zhang, LANL eprint quantph/ [4] H.X. Lu, Doctoral Dissertation, University of Science and Technology of China 003. [5] H.Y. Fan, Int. J. Mod. Phys. B [6] M.S. Kim, W. Son, V. Buzek, and P.L. Knight, Phys. Rev. A [7] S.J. van Enk, Phys. Rev. A [8] X.T. Liang, Commun. Theor. Phys. Beijing, China
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