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It ought to be possible with modern technology to replace the two slits of Young's two-slit experiment with a single emitter which is moved laterally (at random coin toss) to a position equivalent to the position of one or other of the slits. When in each position the emitter ejects a single photon toward a detecting screen. The detector is reset after each single photon detection.

Will the interference fringes still appear?

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  • $\begingroup$ You would have to move your source faster than light. $\endgroup$
    – my2cts
    Commented Dec 1, 2020 at 23:09
  • $\begingroup$ If you don't have slits then your sources are free to generate photons that can travel pretty much anywhere. What is meant by the historical meaning of "interference" i.e. a sort of cancelling of energy, only gives you a superficial/outdated understanding. "Interference" is better replaced by acceptable paths or Feynman theory, certain paths resonate based on path length and photon energy. $\endgroup$ Commented Dec 2, 2020 at 0:52
  • $\begingroup$ @PhysicsDave Please post answers as answers, not as comments. $\endgroup$ Commented Dec 2, 2020 at 2:09
  • $\begingroup$ In the original two-slit expt, is it possible that particle trajectories could be correlated in time, not spatially, and still provide the same pattern? If a particle passing through a diaphragm slit left an echo in the vibrations of the diaphragm that gets passed to the next particle in the sequence. But assuming the diaphragm is the culprit could be wrong if it veiled a previously hidden field effect, e.g. particles leaving wakes in a background field, or something not seen before. Links to experimental results that rule out correlation in time would be appreciated. Thx to all. $\endgroup$
    – Neej
    Commented Dec 2, 2020 at 22:49

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This is not the same experiment, as here one tries to study interference between two photons emitted consecutively, rather than the wave nature of a single photon.

Of course, classically photons are electromagnetic waves, whose interference in two-slit geometric does not require quantum approach (just like the interference of waves in water).

Let me also note that the random count toss is not the same as quantum interference. The probability resulting from a random coin toss is $|\psi_1|^2, |\psi_2|^2$, rather than $|\psi_1 + \psi_2|^2$.

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No, the fringes will not appear. Instead you will get the probability distribution you would naively expect from treating the photons as classical billiard balls with uncertain direction.

Your experiment should yield the same results as letting two separate experiments run --- one with just slit 1 and one with just slit 2 --- and then choosing at random which screen you look at each time.

The distribution you will see will then be $$P(\text{photon lands at }x) = P(\text{slit in position 1})P(\text{photon from just slit 1 lands at }x)+P(\text{slit in position 2})P(\text{photon from just slit 2 lands at }x)$$ where $P(\text{photon from just slit }a\text{ lands at }x) = |\psi_{slit\, a}(x)|^2$.

Because all the terms on the right hand side are positive you'd only see dark spots on the screen at locations where both $P(\text{photon from just slit 1 lands at }x)$ and $P(\text{photon from just slit 1 lands at }x)$ are equal $0$. There's no interference that creates new dark spots.

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If you don't have slits then your sources are free to generate photons that can travel pretty much anywhere. What is meant by the historical meaning of "interference" i.e. a sort of cancelling of energy, only gives you a superficial/outdated understanding. "Interference" is better replaced by acceptable paths or Feynman theory, certain paths resonate based on path length and photon energy. –

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