Entanglement property of the Werner state in accelerated frames

Abstract

We study the entanglement property of a free Dirac field in a Werner state as seen by two relatively accelerated parties. We study the concurrence, negativity, mutual information and $\pi$-tangle of the tripartite system. We show how these entanglement properties depend on both the free parameter F, which is a real parameter called fidelity, and the acceleration parameter r. The degree of entanglement is degraded by the Unruh effect, but we notice that the Werner state always remains entangled even in the acceleration limit, and thus, it can become a good candidate to quantum teleportation in uniform acceleration frame. We notice that the entropy $S(\rho _{A\, \mathrm{I}\, \mathrm{II}})$ decreases with the free parameter F, and also $S(\rho _{A\, \mathrm{I}\, \mathrm{II}})$, $S(\rho _{A})$ and $S(\rho _{\mathrm{I}\, \mathrm{II}})$ are independent of the acceleration parameter r. The von Neumann entropy is not a good entanglement measure any more for this mixed state. We verify that the Werner state in a noninertial frame obeys the Coffman–Kundu–Wootters (CKW) monogamous inequality and find that two useful relations for the concurrence and negativity.

Appendix A: Explicit expressions for \(1-2\) and \(1-1\) tangles

In this Appendix, we are going to write out explicitly the analytical expressions for these $1-1$ and $1-2$ tangles as follows:

$$\begin{aligned} N_{\mathrm{A}({\mathrm{I}})}= & {} \frac{1}{24} \left( 16 F \cos ^2(r)\right. \nonumber \\&\left. +\sqrt{8 (8 F-5) (8 F+1) \cos (2 r)+256 F (2 F-1)+18 \cos (4 r)+86}-2 (\cos (2 r)+7)\right) ,\nonumber \\ \end{aligned}$$

(A1)

$$\begin{aligned} N_{\mathrm{A}({\mathrm{II}})}= & {} \frac{1}{24} \left( (2-8 F) \cos (2 r)\right. \nonumber \\&\left. +\sqrt{-8 (8 F-5) (8 F+1) \cos (2 r)+256 F (2 F-1)+18 \cos (4 r)+86}+8 F-14\right) , \end{aligned}$$

(A2)

$$\begin{aligned} N_{{\mathrm{I}}({{\mathrm{II}}})}= & {} \frac{1}{4} \left( \sqrt{6-2 \cos (4 r)}-2\right) =\frac{1}{2}\left( \sqrt{\sin ^2(2 r)+1}-1\right) . \end{aligned}$$

(A3)

$$\begin{aligned} N_{\mathrm{A} ({\mathrm{I}}\, {{{\mathrm{II}}}})}= & {} -1 + 2 F, \end{aligned}$$

(A4)

$$ \begin{aligned} N_{{\mathrm{I}}(\mathrm{A}\, {{{\mathrm{II}}}})}= & {} \frac{1}{3} \Bigg \{-\text {Root}\Big [2 \#1^3+\#1^2 (2 F \cos (2 r)-6 F+\cos (2 r)-3)\nonumber \\&+\#1 \Big (-28 F^2 \sin ^2(r)-4 F^2 \sin ^2(r) \cos (2 r)\nonumber \\&+32 F \sin ^2(r)+8 F \sin ^2(r) \cos (2 r)-4 \sin ^2(r)-4 \sin ^2(r) \cos (2 r)\Big )\nonumber \\&+16 F^3 \sin ^4(r) \cos ^2(r)-24 F^2 \sin ^4(r) \cos ^2(r)\nonumber \\&+8 \sin ^4(r) \cos ^2(r) \& , 1\Big ]-\text {Root}\Big [2 \#1^3+\#1^2 (2 F \cos (2 r)\nonumber \\&+6 F-2 \cos (2 r)-6)+\#1 \Big (-28 F^2 \cos ^2(r)\nonumber \\&+4 F^2 \cos (2 r) \cos ^2(r)-4 F \cos ^2(r)+4 F \cos (2 r) \cos ^2(r)\nonumber \\&+\cos (2 r) \cos ^2(r)+5 \cos ^2(r)\Big )\nonumber \\&-16 F^3 \sin ^2(r) \cos ^4(r)+12 F \sin ^2(r) \cos ^4(r)+4 \sin ^2(r) \cos ^4(r) \& , 1\Big ]\Bigg \}, \end{aligned}$$

(A5)

$$ \begin{aligned} N_{{{{\mathrm{II}}}}(\mathrm{A}\, {\mathrm{I}})}= & {} \frac{1}{3} \Bigg \{-\text {Root}\Big [2 \#1^3+\#1^2 (-2 F \cos (2 r)+6 F+2 \cos (2 r)-6)\nonumber \\&+\#1 \Big (-28 F^2 \sin ^2(r)-4 F^2 \sin ^2(r) \cos (2 r)\nonumber \\&-4 F \sin ^2(r)-4 F \sin ^2(r) \cos (2 r)+5 \sin ^2(r)-\sin ^2(r) \cos (2 r)\Big )\nonumber \\&-16 F^3 \sin ^4(r) \cos ^2(r)+12 F \sin ^4(r) \cos ^2(r)\nonumber \\&+4 \sin ^4(r) \cos ^2(r) \& , 1\Big ]-\text {Root}\Big [2 \#1^3+\#1^2 (-2 F \cos (2 r)\nonumber \\&-6 F-\cos (2 r)-3)+\#1 \Big (-28 F^2 \cos ^2(r)\nonumber \\&+4 F^2 \cos (2 r) \cos ^2(r)+32 F \cos ^2(r)-8 F \cos (2 r) \cos ^2(r)\nonumber \\&+4 \cos (2 r) \cos ^2(r)-4 \cos ^2(r)\Big )\nonumber \\&+16 F^3 \sin ^2(r) \cos ^4(r)-24 F^2 \sin ^2(r) \cos ^4(r)+8 \sin ^2(r) \cos ^4(r) \& , 1\Big ]\Bigg \}. \end{aligned}$$

(A6)

It should be pointed out that those special symbols # and & that appeared in $N_{{\mathrm{I}}(\mathrm{A}\, {\mathrm{II}})}$ and $N_{{\mathrm{II}}(\mathrm{A}\, {\mathrm{I}})}$ are generated when we solve higher-order polynomial eigenvalue problems, but fortunately they do not affect the final results. On the other hand, we have reverified again why $F\ge 1/2$ based on the result $N_{\mathrm{A} ({\mathrm{I}}\, {\mathrm{II}})}=-1 + 2 F\ge 0$. However, in order to make the $\pi (\rho _{\mathrm{A}({\mathrm{I}}\, {\mathrm{II}})})$ not less than zero, the $F=0.5$ is excluded.