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I am wondering which state leaves the simple interferometer below. The beam splitters are polarizing beam splitters (PBS) which transmit vertical polarization and reflect horizontal polarization. Say an initial photon is prepared in the state $\frac{1}{\sqrt 2}(|H\rangle+|V\rangle)$.

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In my opinion Detector 1 will never see a photon and the state reaching Detector 2 will be the same as the initial state: $\frac{1}{\sqrt 2}(|H\rangle+|V\rangle)$.

But someone told me there would be a problem with unitarity if my prediction is right. Can some one help me?

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  • $\begingroup$ In my experience, the ubiquitous "someone" is almost always wrong. Ask him/her what polarization could reach Detector_1 (and ask what he thinks "unitarity" is). $\endgroup$ Sep 23 '14 at 11:49
  • $\begingroup$ Thanks for the comment. But I would feel a bit impolite if I bother this person again with such a "simple" question (and I also don't want to demonstrate my stupid confusion). So if anyone could just say that my prediction is right or wrong I would be very happy. $\endgroup$
    – thyme
    Sep 23 '14 at 13:26
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I think what you say is correct. By saying it should be unitary, he means it should be time-reversible. It is indeed time-reversible if before reaching the detector 2 it is in the same state as the initial state. It starts off with the superposition, and then only vertical polarization in the top part and horizontal in the lower part of the arms, which finally recombine into the superposition after the beam splitter near detector 2.

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  • $\begingroup$ Thanks. And since the vertical part gets reflected only once while the horizontal part gets reflected three times, the relative phase between $|H\rangle$ and $|V\rangle$ is zero, right? I just ask to justify the $|H\rangle+|V\rangle$ state at the end... $\endgroup$
    – thyme
    Sep 23 '14 at 14:36
  • $\begingroup$ I would think that because the path length is the same and they both hit a mirror at least once the phase shift is equal. The beam splitters could introduce a phase shift for reflected waves, but they are (usually I think) 0 or $\pi$, such that finally it does not matter. $\endgroup$
    – Jasper
    Sep 23 '14 at 14:51

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