Mach-Zehnder interferometer reaction to input light with polarization state ±45˚? We have a Mach-Zehnder interferometer with two non-polarizing beam splitters. The polarisation state of the input beam is a superposition of +45˚ and -45˚ polarizations. In the upper arm of the interferometer there is a polarizer oriented at 0˚. In the lower arm there is a polarizer oriented at 90˚. Both output beams are passed through polarizers oriented at 90˚. The upper output beam is detected by detector D1 and the lower output beam by D2. Will the interference fringes, observed at D1 be in phase with the fringes observed at D2 or shifted by π rad and why?

 A: The output will depend on the phase relationship between the two input beams.  If they are mutually coherent and their phase difference is 0 degrees, they will combine to form a beam with vertical polarization.  If they are 180 degrees out of phase, they will combine to form a beam with horizontal polarization.   Various other phase relationships will give right-circular, left-circular, or elliptical polarization.  
If the phase relationship is constant, you will get a stable interference pattern at the output of the interferometer.  Whatever gets past the polarization filters will form the interference pattern.  If, for example, the input beams are in phase so that they amount to a vertically polarized beam with polarization oriented at 0 degrees, nothing will get past the polarizing filters at the output.  Rotate the filters to 0 degrees and you will get a bright interference pattern.  
You did not specify what the light source(s) is(are).  Assuming they aremonochromatic but not mutually coherent, the portion of the beam from each source will form its own interference pattern at the output, and the two interference patterns will overlap.  Their intensities will add, but their amplitudes will not.  That is, the light from the two sources will not interfere unless the sources are mutually coherent.  
