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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?

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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.

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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.)

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  • $\begingroup$ Thank you for your answer Norbert, you gave me a hint on how to proceed. I'm still a bit confused, so I edited my answer, if you want to take a look at it (it was too long for a comment) $\endgroup$ – Luthien May 17 at 11:44
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    $\begingroup$ @Luthien This is very hard to read, maybe you can TeX it up more neatly? In any case, you should try to go through the "standard" entanglement swapping protocol, where you have two maximally entangled states and measure in a Bell basis. Or even better, start by studying teleportation. Once you understand teleportation, you understand entanglement swapping. $\endgroup$ – Norbert Schuch May 17 at 13:24
  • $\begingroup$ 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? $\endgroup$ – Luthien May 17 at 14:14
  • $\begingroup$ @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. $\endgroup$ – Norbert Schuch May 17 at 15:05
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    $\begingroup$ @Luthien Maybe an added comment about probabilities: The probabilities will be independent of the outcome, since the reduced states of 2 and 3 are both maximally mixed for Werner states, and the state on 2+3 is all what determines the measurement outcomes. $\endgroup$ – Norbert Schuch May 18 at 12:24

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