Quantum correlation swapping

Abstract

Quantum correlations (QCs), including quantum entanglement and those different, are important quantum resources and have attracted much attention recently. Quantum entanglement swapping as a kernel technique has already been applied to quantum repeaters for successfully generating long-distance shared maximally entangled qubit states. Long-distance shared QCs containing shared entanglements are useful and important for some quantum information processing in future quantum networks. In this paper, the concept of quantum entanglement repeater is extended to that of QC repeater by generalizing quantum entanglement swapping to QC swapping. Specifically, the swapping of QCs in a pair of Werner states through a local bipartite von Neumann measurement is treated. Four different QC measures, i.e., entanglement of formation (William in Phys Rev Lett 80:2245, 1998), quantum discord (Ollivier and Zurek in Phys Rev Lett 88:017901, 2001), measurement-induced disturbance (MID) (Luo in Phys Rev A 77:022301, 2008) and ameliorated MID (Girolami et al. in J Phys A 44:352002, 2011), are employed to characterize and quantify QCs. Properties and thresholds of all QCs which occur in the swapping process are revealed, and two different phenomena are exposed and explained. It is found that a long-distance shared QC can be generated from two short-distance ones via QC swapping indeed; however, its amount cannot exceed the minimum one among the QCs in the two initial states and in the measuring state as far as the four quantifiers are concerned.

Appendices

Appendix 1

Partial derivative in Sect. 3.2 for OQD measure. Let $\varDelta _0 =q_0^2 -4\mathcal{G}(f_2,f_0,\alpha ) \mathcal{G}(f_3,f_1,\alpha ) +\zeta \cos ^2\alpha \sin ^2\alpha$ and $\varDelta _1 =q_1^2 -4\mathcal{G}( f_0,f_2,\alpha ) \mathcal{G}(f_1,f_3,\alpha ) +\zeta \cos ^2\alpha \sin ^2\alpha$. Easily, one can arrive at

$$\begin{aligned} \frac{\hbox {d}\varDelta _0}{\hbox {d} \alpha }&= \frac{1}{4}\sin \alpha \cos \alpha \left\{ [ 2(f_0f_3+f_1 f_2- 2f_2 f_3 ) +f_{23}(f_{23}-f_{01})]\sin ^2 \alpha \right. \nonumber \\&\left. \quad +\,[2(2f_0 f_1 -f_1f_2-f_0f_3)+ f_{01}(f_{23}-f_{01} ) ]\cos ^2 \alpha \right\} \nonumber \\&\quad +\,4\kappa (1-\kappa )x^{2}y^{2} \sin 2\alpha \cos 2\alpha , \end{aligned}$$

(64)

$$\begin{aligned} \frac{\hbox {d}\varDelta _1}{\hbox {d} \alpha }&= -\frac{1}{4}\sin \alpha \cos \alpha \left\{ [2(2f_0 f_1 -f_1 f_2- f_0f_3 )+( f_{23}-f_{01} )f_{01} \sin ^2 \alpha \right. \nonumber \\&\left. \quad +\,\ [2( f_0 f_3 + f_1 f_3-2f_2 f_3)+( f_{23}-f_{01} )f_{23} ] \cos ^2 \alpha \right\} \nonumber \\&\quad +\,\ 4\kappa (1-\kappa )x^{2}y^{2} \sin 2\alpha \cos 2\alpha . \end{aligned}$$

(65)

Using these two derivatives, one can further get

$$\begin{aligned}&\frac{\hbox {d} \{S[\rho '_{d}(x,y,\kappa )]- S[\rho '_{bd}(x,y,\kappa )|\{\Pi _{b}^{(j)}\}]\}}{\hbox {d} \alpha }\nonumber \\&\quad = \frac{1}{2} (-f_{01} + f_{23} )\sin \alpha \cos \alpha [ (- \lambda _0 \log _2 \lambda _0 - \lambda _1 \log _2 \lambda _1 +\eta _0 \log _2 \eta _0 +\eta _1 \log _2 \eta _1)\nonumber \\&\qquad -\,\frac{\sqrt{\varDelta _0}}{2q_0} (- \log _2 \lambda _0+ \log _2 \lambda _1)+ \frac{\sqrt{\varDelta _1}}{ 2q_1}(- \log _2 \eta _0+ \log _2 \eta _1)]\nonumber \\&\qquad +\, \frac{\varDelta _0^{-1/2}}{4q_0}[- \log _2 \lambda _0+ \log _2 \lambda _1]\frac{d\varDelta _0}{d \alpha } + \frac{\varDelta _1^{-1/2}}{4q_1}[- \log _2 \eta _0+ \log _2 \eta _1]\frac{d\varDelta _1}{d \alpha }. \end{aligned}$$

(66)

Obviously, if $\alpha = 0$ or $\alpha = \frac{\pi }{4}$, then $\frac{\hbox {d} \{S[\rho '_{d}(x,y,\kappa )]- S[\rho '_{bd}(x,y,\kappa )|\{\Pi _{b}^{(j)}\}]\}}{\hbox {d} \alpha }=0$.

Appendix 2

Partial derivatives in Sect. 3.4 for AMID measure.

$$\begin{aligned} \frac{\partial \mathcal{C} [\rho '_{bd}(x,y,\kappa )]}{\partial \alpha _1}&= - \frac{1}{2} \sin {\alpha _1}\cos {\alpha _1}( f_{23}-f_{01})\left( \log _2 p_b^{(0)}-\log _2 p_b^{(1)}\right) \nonumber \\&\quad +\,\left\{ \frac{1}{2}\left[ ( f_2 -f_0) \cos ^2{\alpha _2}+( f_3 -f_1)\sin ^2{\alpha _2}\right] \sin \alpha _1 \cos \alpha _1\nonumber \right. \\&\quad \left. +\,\sqrt{\kappa (1-\kappa )}xy \cos 2\alpha _1\sin {2\alpha _2}\cos \omega \right\} \left( \log _2 p_{bd}^{(00)}-\log _2 p_{bd}^{(10)} \right) \nonumber \\&\quad +\,\left[ \frac{1}{2} \left[ ( f_2 -f_0) \sin ^2{\alpha _2}+( f_3 -f_1)\cos ^2{\alpha _2}\right] \sin \alpha _1 \cos \alpha _1\nonumber \right. \\&\quad \left. -\,\sqrt{\kappa (1-\kappa )}xy \cos 2\alpha _1\sin {2\alpha _2}\cos \omega \right] \left( \log _2 p_{bd}^{(01)} -\log _2 p_{bd}^{(11)}\right) ,\end{aligned}$$

(67)

$$\begin{aligned} \frac{\partial \mathcal{C} [\rho '_{bd}(x,y,\kappa )]}{\partial \alpha _2 }&= - \frac{1}{2} \sin {\alpha _2}\cos {\alpha _2} ( f_{13}-f_{02})\left( \log _2 p_d^{(0)}-\log _2 p_d^{(1)}\right) \nonumber \\&\quad +\,\left\{ 2 [\mathcal{G}(f_3,f_1,\alpha _1)-\mathcal{G}(f_2,f_0,\alpha _1)]\sin {\alpha _2}\cos {\alpha _2}\nonumber \right. \\&\quad \left. +\,\sqrt{\kappa (1-\kappa )}xy\sin {2\alpha _1}\cos {2\alpha _2}\cos \omega \right\} \left( \log _2 p_{bd}^{(00)}-\log _2 p_{bd}^{(01)} \right) \nonumber \\&\quad +\,\left\{ 2 [\mathcal{G}(f_1,f_3,\alpha _1)-\mathcal{G}(f_0,f_2,\alpha _1)]\sin {\alpha _2}\cos {\alpha _2}\nonumber \right. \\&\quad \left. -\,\sqrt{\kappa (1-\kappa )}xy\sin {2\alpha _1}\cos {2\alpha _2}\cos \omega \right\} \left( \log _2 p_{bd}^{(10)} -\log _2 p_{bd}^{(11)}\right) ,\end{aligned}$$

(68)

$$\begin{aligned} \frac{\partial \mathcal{C} [\rho '_{bd}(x,y,\kappa )]}{\partial \omega }&= \frac{1}{2}\sqrt{\kappa (1-\kappa )}xy\sin {2\alpha _1}\sin {2\alpha _2}\sin \omega \nonumber \\&\left[ \log _2 p_{bd}^{(01)}+ \log _2 p_{bd}^{(10)} -\log _2 p_{bd}^{(00)}-\log _2 p_{bd}^{(11)}\right] . \end{aligned}$$

(69)

If $(\alpha _1, \alpha _2, \omega )=(0,0,0)$ or $(\alpha _1, \alpha _2, \omega )=(\frac{\pi }{4},\frac{\pi }{4},0)$, then $\frac{\partial \mathcal{C} [\rho '_{bd}(x,y,\kappa )]}{\partial \alpha _1 }$ $=\frac{\partial \mathcal{C} [\rho '_{bd}(x,y,\kappa )]}{\partial \alpha _2 } =\frac{\partial \mathcal{C} [\rho '_{bd}(x,y,\kappa )]}{\partial \omega }=0$.

Appendix 3

Prove $\frac{ \partial \theta _{bd}(x,y,\kappa )}{\partial x} > 0$, where $\theta _{bd}= 2 xy \sqrt{\kappa (1-\kappa )}-\frac{1}{2} \sqrt{(1-x^2)(1-y^2)+4\kappa (1-\kappa )(x-y)^2}$. Let $\varDelta =\sqrt{(1-x^2)(1-y^2)+4\kappa (1-\kappa )(x-y)^2}$. Easily, one can get

$$\begin{aligned} \frac{\partial \theta _{bd}(x,y,\kappa )}{\partial x} =\frac{8\varDelta y \sqrt{\kappa (1-\kappa )}+2x(1-y^2)- 8\kappa (1-\kappa )(x-y)}{4\varDelta }. \end{aligned}$$

(70)

Since $\varDelta >2\sqrt{\kappa (1-\kappa )}|x-y|>0$, one is readily to find

$$\begin{aligned} \frac{ \partial \theta _{bd}(x,y,\kappa )}{\partial x}>\frac{N}{4\varDelta }, \end{aligned}$$

(71)

where

$$\begin{aligned} N =16 y \kappa (1-\kappa )|x-y|+2x(1-y^2)-8\kappa (1-\kappa )(x-y). \end{aligned}$$

(72)

Easily, one can conclude that,

• (1) If $x\le y$, then $N\ge 0$;

• (2) If $x>y$, then $N$ can be rewritten as $N \equiv 8(2y-1) \kappa (1-\kappa )(x-y)$ $+2x(1-y^2)$. If $y\ge \frac{1}{2}$, then $N\ge 0$. Otherwise, $y<\frac{1}{2}$, then $N > 2[(2y-1)(x-y)+x(1-y^2)]>2[\frac{2}{3} -2y^2+y-xy^2]$, where $3xy>1$ is used. Because $x\in (\frac{1}{3},1]$, it is obvious that $F(y)=\frac{2}{3} -2y^2+y-xy^2 >0$ provided that $y\in \left( \frac{1-\sqrt{1+\frac{8}{3}(2+x)}}{2(2+x)}, \frac{1+\sqrt{1+\frac{8}{3}(2+x)}}{2(2+x)}\right)$. Furthermore, because $y\in (\frac{1}{3},1]$, one can easily verify that $y\in \left( \frac{1}{3}, \frac{1}{2}\right) \subset \left( \frac{1-\sqrt{1+\frac{8}{3}(2+x)}}{2(2+x)}, \frac{1+\sqrt{1+\frac{8}{3}(2+x)}}{2(2+x)}\right)$ for any $x\in (\frac{1}{3},1]$. This means that in this case $N>0$, too.

Integrating all the possible cases above, one is readily to get $\frac{ \partial \theta _{bd}(x,y,\kappa )}{\partial x} > 0$. $\square$

Appendix 4

Prove $\frac{\partial \varDelta }{\partial \kappa } > 0$, $\frac{\partial \varDelta }{\partial x} > 0$ and $\frac{\partial \varDelta }{\partial y} > 0$. From the definition $\xi \equiv f_3-f_0+\varDelta$ and some definitions below Eq. (25), one can get

$$\begin{aligned} \varDelta ^2+ 4\varDelta (1-2\kappa )(x+y) - 64\kappa (1-\kappa )x^{2}y^{2}=0. \end{aligned}$$

(73)

Considering $\varDelta > 0$, by solving Eq. (73), one can get

$$\begin{aligned} \varDelta = -2(1-2\kappa )(x+y) + 2\sqrt{(1-2\kappa )^2 (x+y)^2+ 16\kappa (1-\kappa )x^{2}y^{2} }. \end{aligned}$$

(74)

The partial differential of $\varDelta$ about $\kappa$ is

$$\begin{aligned} \frac{\partial \varDelta }{\partial \kappa }&= 4(x+y) + \frac{-4(1-2\kappa )(x+y)^2 + 16(1-2\kappa )x^{2}y^{2} }{\sqrt{(1-2\kappa )^2 (x+y)^2+ 16\kappa (1-\kappa )x^{2}y^{2} }}\nonumber \\&= \frac{ 4(x+y)\sqrt{(1-2\kappa )^2 (x+y)^2+ 16\kappa (1-\kappa )x^{2}y^{2}} -4(1-2\kappa )(x+y)^2 + 16(1-2\kappa )x^{2}y^{2} }{\sqrt{(1-2\kappa )^2 (x+y)^2+ 16\kappa (1-\kappa )x^{2}y^{2} }}\nonumber \\&> \frac{ 16(1-2\kappa )x^{2}y^{2} }{\sqrt{(1-2\kappa )^2 (x+y)^2+ 16\kappa (1-\kappa )x^{2}y^{2} }}. \end{aligned}$$

(75)

Considering $\kappa \in (0,\frac{1}{2}]$, $x,y \in (0,1]$, one can get $\frac{\partial \varDelta }{\partial \kappa } > 0$.

The partial differential of $\varDelta$ about $x$ is

$$\begin{aligned} \frac{\partial \varDelta }{\partial x}&= -2(1-2\kappa ) +\frac{2(1-2\kappa )^2(x+y) + 32\kappa (1-\kappa )x y^{2} }{\sqrt{(1-2\kappa )^2 (x+y)^2+ 16\kappa (1-\kappa )x^{2}y^{2} }}\nonumber \\&= \frac{2(1-2\kappa )^2(x+y) + 32\kappa (1-\kappa )x y^{2} -2(1-2\kappa )\sqrt{(1-2\kappa )^2 (x+y)^2+ 16\kappa (1-\kappa )x^{2}y^{2} } }{\sqrt{(1-2\kappa )^2 (x+y)^2+ 16\kappa (1-\kappa )x^{2}y^{2} }}.\nonumber \\ \end{aligned}$$

(76)

Notice that there are two factors in the numerator in Eq. (76). Let the former’s square subtracts the latter’s, one can get

$$\begin{aligned}&\left[ 2(1-2\kappa )^2(x+y) + 32\kappa (1-\kappa )x y^{2}\right] ^2 -4(1-2\kappa )^2\nonumber \\&\quad \left[ (1-2\kappa )^2 (x+y)^2+ 16\kappa (1-\kappa )x^{2}y^{2}\right] \nonumber \\&\quad = 64 \left[ 16\kappa ^2(1-\kappa )^2 x^2 y^{4} + (2 x y^{2}-x^{2}y^{2})\kappa (1-\kappa )(1-2\kappa )^2\right] . \end{aligned}$$

(77)

Considering $\kappa \in (0,\frac{1}{2}]$, $x,y \in (0,1]$, one can conclude that the function in Eq. (77) is positive. Finally, one can get $\frac{\partial \varDelta }{\partial x} > 0$. By using the similar method, one can obtain $\frac{\partial \varDelta }{\partial y} > 0$. $\square$