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In a Mach Zehnder interferometer if we use one at a time photons and tune it so the first several to arrive in channel D1 after BS2 will all next arrive also in D1? Are there reasons that some will not land on D1 f.e. if BS2 is wide and the probability waves do not overlap at BS2 in some cases. If we turn it to Feynman trajectories maybe trajectories in BS2 will not intersect in some cases and photon can land in D2 for that reason? We assume wavelength is fixed by laser and filters!

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  • $\begingroup$ It's all about the path lengths of either arm ... if both are bad then no detector gets light .... if both are good then 50/50 like a bean splitter. Bad path is path length = n(lambda) + 0.5(lambda), good is path length = n(lambda). $\endgroup$ Commented Feb 8 at 23:53
  • $\begingroup$ What a connection has the length of a path? I think only path difference counts. If l1-l2 = lambda we have detector D1, if 0.5£ than D2.????? $\endgroup$
    – Mercury
    Commented Feb 9 at 19:06
  • $\begingroup$ In MZI you can tune both arms to be bad path. No light goes in the MZI … just like in the DSE where no light goes to dark areas/bands because all paths to dark band averages to lambda + 0.5 lambda. $\endgroup$ Commented Feb 9 at 20:03
  • $\begingroup$ The important length is between the input and output beam splitter faces. And also there can be a phase change for the beam that goes thru the glass part of beam splitter. $\endgroup$ Commented Feb 9 at 20:07
  • $\begingroup$ Simple experiment: laser diode to glass plate 1 and then thru glass plate 2 and then ccd. If you adjust path length between the glass to be lambda + 0.5 then no light at detector. $\endgroup$ Commented Feb 9 at 20:10

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In a physical implementation of a Mach-Zehnder interferometer, it is highly unlikely that all the light exits only one of the two output port. The reason is that the laser is a Gaussian beam that propagates and picks up phase curvature. One can compensate for this phase curvature by using lenses, but all lenses have some aberrations. As a result, the two beams that exit the two output port always tend to have some light and dark rings even after one has done one's best to align the system. (Obviously one would want the two beams to overlap in the last beamsplitter to ensure interference.)

The combined effect can be seen as a result of all the different Feynman trajectories that the photons take through the physical system. Provided that the light is fairly coherent we will always see some level of interference at both output port. If for some reason the system causes the light from the two paths to be incoherent when they reach the last beamsplitter we'll just see to incoherent beams leaving the two output port without any interference.

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  • $\begingroup$ It is not to clear for me. I think one must look at the single photon level. No matter what the pulse do the photon interferes only with itself. So each one photon has a definite and same path difference always after the calibration. I think just when the photon can be delayed in one path there will be a phase difference on BS2. But if the medium is vacuum there is no reason for this $\endgroup$
    – Mercury
    Commented Feb 9 at 19:16
  • $\begingroup$ Or do you mean that the two paths do not intersect on BS2 in transverse direction to the paths? $\endgroup$
    – Mercury
    Commented Feb 9 at 19:27
  • $\begingroup$ Single photons are just single excitations of the same field that you get from the classical wave. So it does not make a difference whether you think of a single photon interfering with itself or of a classical wave. Since the wavelength of visible light is so small, it does not take much of a difference in path length to change the interference. The diffraction of the beam during propagation is enough to do that. $\endgroup$ Commented Feb 10 at 3:26
  • $\begingroup$ Do you mean there will be not negative/positive interference to observe behind the PS2? $\endgroup$
    – Mercury
    Commented Feb 10 at 10:06
  • $\begingroup$ Did you mean BS2 instead of PS2? If the light is coherent then you will see interference fringes (constructive and destructive) after the last beam splitter from both ports. $\endgroup$ Commented Feb 11 at 4:01

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