2
$\begingroup$

Consider the simple state teleportation gadget below. Measuring the entangled state after the CZ gate in the $X$ eigenbasis (equivalent to first applying $H$ and then measuring in computational basis) teleports the state $|\psi\rangle$ to the second register up to a $X^m H$. teleportation

Now, what if we were to consider a measurement of non-orthogonal operators (for simplicity say a POVM given by $E_1, E_2, E_3$) instead of measuring in the $X$ basis. I am reading an article in which the authors seem to indicate that this will generally leave the state on the second register in a mixed state.

Could someone explain why this is the case?

(I have actually taken this slightly out of context, as this originally concerns a quantum optics measurement, namely a so-called heterodyne measurement, where the measurement operators are given by $E=|\alpha\rangle\langle\alpha|$ but the coherent states $|\alpha\rangle$ are not orthogonal and form an overcomplete basis. However, I believe this should boil down to the same question)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.