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Suppose I prepare following state consisting of (for example) three EPR pairs:

$$\lvert\Psi\rangle = \frac{\lvert00\rangle+\lvert11\rangle}{\sqrt{2}}\otimes\frac{\lvert00\rangle+\lvert11\rangle}{\sqrt{2}}\otimes\frac{\lvert00\rangle+\lvert11\rangle}{\sqrt{2}}$$

Then I shuffle the second qubits of all pairs before giving you the full (shuffled) state. You would not know which pairs of qubits form EPR pairs. Does this mean you would have a mixed stated over all possible permutations? How can I write such state?

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  • $\begingroup$ What do you mean by shuffling the the second qubits of all pairs? Using SWAP gates between them? and if so, which are the qubits you are acting on with such gates. $\endgroup$
    – Josu Etxezarreta Martinez
    Mar 7, 2019 at 8:41
  • $\begingroup$ I mean permutations betweent qubits 2, 4 and 6, just like in the answer below! $\endgroup$
    – maarkab
    Mar 7, 2019 at 14:50

1 Answer 1

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Let $S_x$ be a permutation between qubits 2,4 and 6, and acts as identity on the other qubits. Use $x$ to index all possible permutations of that form, of which there are 6. Then you do indeed have a mixed state overall, and it's of the form $$ \rho=\frac{1}{6}\sum_{x=1}^6S|\Psi\rangle\langle\Psi|S^T $$ (Note that the $S$ matrix will be real so I can use the transpose in the place of the Hermitian conjugate).

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  • $\begingroup$ Thank your for your answer! Then for the two EPR states case, if I choose to apply permutation that switches the 2nd and 4th qubits, will the resulting state be $S_{1432} \lvert\Psi\rangle =\frac{1}{2}S_{1432} ( \lvert0000\rangle+\lvert0011\rangle+\lvert1100\rangle+\lvert1111\rangle) = \frac{1}{2} (\lvert0000\rangle+\lvert0110\rangle+\lvert1001\rangle+\lvert1111\rangle)$ ? $\endgroup$
    – maarkab
    Mar 7, 2019 at 15:23
  • $\begingroup$ @maarkab yes, that looks right. If you're wanting to look at different numbers of EPR pairs, then it might be handy to have a more efficient way of constructing the density matrices. I suspect you want to find out about how to construct a projector onto the symmetric subspace. $\endgroup$
    – DaftWullie
    Mar 7, 2019 at 16:09
  • $\begingroup$ I think these states should not be symmetric since the permutations are only acting on the even qubits, or am I missing something here? $\endgroup$
    – maarkab
    Mar 8, 2019 at 19:29
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    $\begingroup$ @maarkab yes, so I was meaning the projector onto the symmetric subspace of the even qubits. $\endgroup$
    – DaftWullie
    Mar 8, 2019 at 20:49

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