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If you do an electron double slit experiment (with one electron at a time), you get an interference pattern as usual. I am wondering what would happen if you scattered a photon (with a given energy) off each electron after it had passed through the slits. Would the interference pattern change?

What is confusing me is the fact that the electron does not have a definite momentum (direction) after passing through the slits, so I am not sure what effect the photon has. Does it cause the wavefunction to collapse to a definite momentum, in which case the interference pattern disappears? Or does it do something in between, reducing the visibility but not entirely? And how does the photon energy affect the result?

Also, in terms of the principle of complementarity, because the photon is scattered after the electron has already passed through both the slits, it doesn't seem to reveal anything about which path the electron took, so why should the photon make any difference?

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  • $\begingroup$ No matter how you design your experiment, if you know which slit the electron went through, you get a particle result. Every variation of the double slit experiment to date (and there have been MANY) has consistently shown that result. $\endgroup$
    – David White
    Commented Apr 1, 2021 at 21:07
  • $\begingroup$ I know, but I'm saying that in this experiment you don't know which slit the electron went through. $\endgroup$
    – Ghorbalchov
    Commented Apr 2, 2021 at 8:22

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The interference pattern disappears if you mess with the electron any time after the slits. You can think of it this way ...

  1. the electron has chosen a path in advance (before it has even left the emitter), the excited electron in the emitter is already generating changes/fluctations in the EM field (called virtual forces or virtual photons, virtual particles cause force but not energy transmission).
  2. there are similarly electrical virtual forces being generated in the screen also in the EM field, there are many possible landing targets for the electron ...
  3. the electron has wave properties, these properties allow propagation along certain probable paths ( especially when there are slits or other obstacles present), the final chosen path appearing random to us.
  4. the probable paths are generated based on the geometry or setup of the experiment i.e. big slits, small slits, wide slits, screen distance etc.

The electron will travel to the target and generate the "interference" pattern if it it is not disturbed when there are slit(s) .... a typical wave DSE function.

Any electron or any photon can have its path disturbed along the way ... this will cause it to "recalculate" a new path or wave function. For example a photon from a star can hit a mirror on earth and be sent back into space.

The concept of a wave function can be used in different ways: 1) a wave function could be developed to describe all the possible paths and is a probability graph, this is the typical DSE pattern, it shows the average of many impacts; 2) a wave function could be developed for a single photon based on a single known or theorized path.

There is an experiment where the scientist claimed that when light was shone on the slits and the camera was unplugged the interference pattern appeared and when the camera was recording the pattern disappeared .... kind of amazing!

So its not just the interaction with photons but the interaction with the electron with the photon and with the sensor in the camera (not sure of this experiment was ever verified or if it was just a statement based on extrapolation of other experiments.) But in any case if it is true it may just mean that for the electron to "recalculate" its path requires that the the photon must be absorbed at a known position (i.e the photon path or wave function collapses or becomes known).

Also you say in your question the electrons momentum is not known after the slits, that is not true if we consider the electrons one at a time ... the electron has chosen one of the probable paths based on probability and QM .... and it will help show the eventual DSE pattern that emerges.

The energy of a photon is never zero .... the photon energy would affect the probability of interacting with the electron .... a bunch of gamma rays ( high energy ) may never interact ... there would be an interference pattern. At the other end of the spectrum ... say radio waves ... may also have low probability. It may be that 1um light gives the highest probability and thus the pattern is reduced.

The intensity (number of photons) is also important, weak light would have fewer interactions (pattern visible) but strong intensity would reduce the pattern.

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  • $\begingroup$ I think you are explaining in terms of field theory, but I was thinking more in terms of usual wavefunctions. Even then, this doesn't answer much of what I asked, for example how the photon energy would affect it. Clearly as the photon energy goes to zero, the interference pattern would reappear. $\endgroup$
    – Ghorbalchov
    Commented Apr 2, 2021 at 8:25
  • $\begingroup$ @AlexGhorbal I added some more to the answer. See above. $\endgroup$
    – PhysicsDave
    Commented Apr 2, 2021 at 14:37

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