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Let's say I have two arbitrary quantum states $|\psi\rangle$ and $|\phi\rangle$. I apply the SWAP test circuit (something like what is given here), and measure the first qubit to get either $0$ or $1$. Is there any way to get back the states $|\psi\rangle$ and $|\phi\rangle$, in an unentangled form, after this measurement? From what I understand, measuring the output entangles the two states.

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  • $\begingroup$ It's not so much the measurement of the control bit that entangles $\vert \psi\rangle$ and $\vert\phi\rangle$; merely performing the CSWAP operation entangles them. $\endgroup$
    – Mark S
    Sep 13 '20 at 15:29
  • $\begingroup$ Is there a way to get back the unentangled form? $\endgroup$
    – BlackHat18
    Sep 13 '20 at 15:35
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The output of the qubits is either $$ (I\pm\text{SWAP})|\phi\rangle|\psi\rangle $$ (up to normalisation). As a general rule, information is lost from this compared to $|\phi\rangle|\psi\rangle$ so you cannot reuse the states (unless the states happened to be eigenstates of SWAP, but if you know that, you wouldn't have bothered doing the SWAP test).

To demonstrate this, imagine that $|\phi\rangle|\psi\rangle$ is either $|01\rangle$ or $|10\rangle$. If you get the $I+\text{SWAP}$ answer, the output state is the same for both, $|01\rangle+|10\rangle$. You cannot tell which of the two inputs it was, so you've lost information.

It could be that whatever further processing you want to do can be done on that output. For example, you might be using the SWAP test to deliberately create a symmetric state that you need for something else.

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  • $\begingroup$ Why is it a general rule that information is lost from the output, compared to the initial state? $\endgroup$
    – BlackHat18
    Sep 14 '20 at 8:02
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    $\begingroup$ does my edit answer your question? $\endgroup$
    – DaftWullie
    Sep 14 '20 at 9:14

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