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I What is photon phase?

  1. Mar 5, 2017 #1
    It occurs to me that I've probably been confusing "phase" in a number of contexts. In particular:

    1) If we write a photon's polarization as [itex]\psi = |x\rangle + e^{i\theta}|y\rangle[/itex], then we can call [itex]\theta[/itex] a "phase."

    2) When a photon bounces off a mirror, it picks up a relative phase of [itex]i[/itex]. If I understand correctly, this is the same "phase" as above. It is also the "phase" referred to in the two-slit experiment.

    3) Inside a Mach-Zehnder interferometer, after a beam splitter, we represent the state as something like [itex]\psi = |0\rangle + i|1\rangle[/itex]. If I understand, the phase shift happens because of (2). But it's not clear to me in what sense "which-path" information has a phase.

    And I should add:

    4) We can perform the two-slit experiment with electrons and get a similar result. How can we explain it with "phase" (like we did in 2) if there's no polarization basis in which "phase" is meaningful?

    Can anyone help me clarify?
  2. jcsd
  3. Mar 5, 2017 #2


    Staff: Mentor

    Last edited: Mar 5, 2017
  4. Mar 6, 2017 #3
    Thanks bhobba. I think most of that stuff is over my head though :(

    From more reading, it seems that the photon version is a result of EM phase and the electron result is due to de Broglie wave interference.

    Still hoping someone can enlighten me in very simple terms.
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