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I am reading David Deutsch's The Beginning of Infinity. In chapter 11, he explains the basics of quantum theory and gives evidence for the many worlds (Everett) interpretation.

The argument uses a Mach-Zehnder interferometer:

enter image description here

It is explained that if a photon is introduced between the beam splitter and either of the mirrors, it will be detected at one of the two detectors at random. However, if it is introduced before the beam splitter, it will always appear at the same detector. From this follows that the (unobservable) split histories of the photon influence where it will be detected, so an explanation like splitting and interference of universes is necessary.

The question Deutsch doesn't answer is: Why does the photon always appear at the same detector?

(I have a strong mathematical background, but am not familiar at all with the notation used in quantum physics.)

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"Why does the photon always appear at the same detector?"

I had the same question when I read about it in Penrose's The Road To Reality, but I found in https://arxiv.org/pdf/quant-ph/9610033.pdf that the apparatus must be carefully tuned to get the destructive interference.

Apparently, Mach-Zehnder interferometers are used in other applications where this is not the case.

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    $\begingroup$ That's correct. It's not at all unusual for a set of fringes to appear at BOTH output ports of an M-Z interferometer. However, the fringes at one output port are always 180 degrees out of phase with the fringes at the other port. If the alignment is such as to show a "bull's eye" pattern, when the center of the bull's eye at one port is bright, it will be dark at the other port. $\endgroup$ – S. McGrew May 17 '18 at 2:15
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Well, the quick and dirty answer is down to phase changes. In the upper path (re your diagram)to upper detector, the photon has bounced three times at a phase change of π each time. In the lower path the photon has bounced only once, hence a 2π difference and destructive interference. At the right hand detector, the upper photon bounces only twice as does the lower photon hence constructive interference. Whether or not this constitutes the full quantum explanation is still, maybe, open to debate.

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  • $\begingroup$ 2π is effectively zero. π is what causes destructive interference. $\endgroup$ – S. McGrew May 17 '18 at 3:45
  • $\begingroup$ Yep, because of phase changes at mirrors, but no phase change at reflection of the BS. e.g. this video demonstrates the phase changes at different points, resulting in destructive interference for 1 detector, but constructive at the other - youtube.com/watch?v=M6y_igUpyCg $\endgroup$ – Brett Apr 30 at 10:55
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The key, I think, is to understand that a beam-splitter represents a unitary process. If one writes down the unitary matrix that would represent the operation of the beam-splitter, one finds that the unitarity implies that there would always have to be a relative minus sign between the components of one output port compared to the other. As a result, if the paths give constructive interference for one output port, they will give destructive interference for the other. Hence, the photon only appears in one of the output ports.

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The probability of a photon coming out of each port depends on the phase difference between the different versions of the photon at the final beamsplitter. For a mathematical description see Quantum Computation and Quantum Information by Nielsen and Chuang, Section 7.4 or this paper:

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.620.682&rep=rep1&type=pdf

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