# EPR states with permuted qubits

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?

• 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. – Josu Etxezarreta Martinez Mar 7 '19 at 8:41
• I mean permutations betweent qubits 2, 4 and 6, just like in the answer below! – maarkab Mar 7 '19 at 14:50

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).
• 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 $S_{1432} \lvert\Psi\rangle =\frac{1}{2}S_{1432} ( \lvert0000\rangle+\lvert0011\rangle+\lvert1100\rangle+\lvert1111\rangle) = \frac{1}{2} (\lvert0000\rangle+\lvert0110\rangle+\lvert1001\rangle+\lvert1111\rangle)$ ? – maarkab Mar 7 '19 at 15:23