Simple Mach-Zehnder Interferometer with Polarizing Beam Splitters

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 $\frac{1}{\sqrt 2}(|H\rangle+|V\rangle)$.

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?

• 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). Sep 23 '14 at 11:49
• 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. Sep 23 '14 at 13:26

• 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... Sep 23 '14 at 14:36
• 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. Sep 23 '14 at 14:51