# Entanglement Swapping of Werner states

I am provided with two Werner states $$\begin{equation*} \rho=F|\phi^+><\phi^+|+\frac{1-F}{3}(|\phi^-><\phi^-|+|\psi^+><\psi^+|+|\psi^+><\psi^+|) \end{equation*}$$ or equivalently $$\begin{equation*} \rho=x|\phi^+><\phi^+|+\frac{1-x}{4}\mathbb{1} \end{equation*}$$ where $$\mathbb{1}=1/4(|\phi^+><\phi^+|+|\phi^-><\phi^-|+|\psi^+><\psi^+|+|\psi^+><\psi^+|)$$ and $$x=(4F-1)/3$$.

The book I am reading (The Physics of Quantum Information) states that if I connect them with perfect operations I obtain a new Werner state with fidelity $$\begin{equation} F'=\frac{1}{4}\left\{ 1+3\left( \frac{4F-1}{3} \right)^2 \right\} \end{equation}$$ I suppose I have to calculate $$\rho_{1234}=\rho_{12}\otimes\rho_{34}$$ and then find a way to express this product into states of the type $$|\phi^+>_{23}<\phi^+|$$ in such a way that I can then project into the Bell basis. I am trying to decompose the states into the computational basis $$|0>$$ and $$|1>$$ but I cannot regroup the states in a suitable way (I think at some point I get lost in the calculations).

Am I proceeding in a correct way (decomposing into $$|0>$$ and $$|1>$$ and trying to regroup into Bell states of the nodes 2 and 3) or is there a faster way to compute the fidelity? And shouldn't the resulting Werner state be dependent on the measurement outcome?

Edit

I'm not quite sure I have it so I will proceed step by step. Let's consider the contribute $$x|\phi^+><\phi^+|$$. Then on the composite state this should give rise to $$\begin{equation} x^2(|\phi^+>_{14}<\phi^+||\phi^+>_{23}<\phi^+|+|\phi^->_{14}<\phi^-||\phi^->_{23}<\phi^-|+|\psi^->_{14}<\psi^-||\psi^->_{23}<\psi^-|+|\psi^+>_{14}<\psi^+||\psi^+>_{23}<\psi^+|) \end{equation}$$ But when I project onto the Bell basis then I could get a Werner state with fidelity $$x^2$$ with respect to any of the four states, right? I mean, if I project on $$|\phi^->_{23}$$ in my resulting Werner state I will have the contribute $$x^2|\phi^-><\phi^-|$$ and if I project on $$|\phi^+>_{23}$$ I will have $$x^2|\phi^-><\phi^-|$$? So the resulting Werner state will depend on the measurement outcome on the sites 2 and 3.

Decompose the states as in your second equation - a sum of maximally entangled and maximally mixed state. Then carry out a teleportation (entanglement swapping) protocol on sites 2+3. If both states are maximally entangled (with weight $$x^2$$), the protocol succeeds and gives another maximally entangled state. If either of the two states is maximally mixed, you are left with a maximally mixed state (that should be easy to check).
So you should get a new Werner state with $$x'=x^2$$. (If you check the formula for $$x(F)$$ and $$F'(F)$$ you give, you indeed see that this is the case.)
• I edited a bit the best I could. Anyway I studied the teleportation and the swapping protocol. That's why I think that the Werner state we obtain after the measurement on the sites 2 and 3 will depend on the measurement outcome. Am I correct? I mean, depending on the measurement outcome $F'$ will be the resulting fidelity of $F'=<\phi^+|\rho_w|\phi^+>$ or $F'=<\phi^-|\rho_w|\phi^->$... so the Bell state with respect to which we have the computed fidelity $F'$ as in the formulas above depends on the measurement, right? – Luthien May 17 at 14:14
• @Luthien So I understand you understand entanglement swapping with two maximally entangled states $|\phi^+\rangle\langle\phi^+|\otimes |\phi^+\rangle\langle\phi^+|$? Then what you need to consider next is the same protocol, but applied to $|\phi^+\rangle\langle\phi^+|\otimes Id$, and convince yourself that it gives a maximally mixed state. And then to $Id\otimes Id$. Then you add those contributions with weights $x$, $x(1-x)$ and $x^2$, respectively, and you have the full result. – Norbert Schuch May 17 at 15:05