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I know that for a Fock State you can calculate the interference pattern, and that the distribution does not change with the intensity, in this case with n.

In this case, the intensity pattern has the same shape and only increase the intensity with n, thus for a fock state $|n\rangle$ the intensity follows the following distribution.

$I (r, t ) = n | f (r)|^2 [1 + cos \theta] $

Actually, the shape is the same for coherent state $|\alpha\rangle$

$I (r, t) = |α|^2 | f (r)|^2 [1 + cos\theta] $

Thus, the distribution of the interference doest not change,

What I wonder is, if you send a state $|n\rangle$ trough a double slit, could you see n simultaneous detections in the screen?

For example, for one photon $|1\rangle$, the way to measure this pattern is repeat the experiment several times, and as you record the detections the pattern appears. But you only detect a photon each time. In this case, since we have n photons ($|n\rangle$). I wonder if it is possible to detect in each iteration of the experiment n photons, or if the n photons hit the same point in the screen.

We can consider we have a screen that can detect 1 photon in each point of the screen.

Tarek Peña
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  • Not all photons will make it through the slit - otherwise, there would no need to have a slit in the first place. So you will detect less than n photons. What is a meaningful question eg is what happens if you send a Fock state to the slit, and then conditioned on the fact that e.g. all n photons go through the slit, what will happen. – Norbert Schuch Sep 21 '23 at 14:25
  • Sure, I mean what would happen in the scenario all photons go through. On the other hand, since it's a Fock state I think that what should happen is that either all photons go through the double slit or none of the does it. Since this n photons are part of the same state. – Tarek Peña Sep 21 '23 at 15:23
  • Certainly not; why should this be the case? (With the same reasoning, you could argue that a Fock state always goes to one arm of a beam splitter. See https://physics.stackexchange.com/questions/128365/output-of-a-beamsplitter-with-photon-number-fock-state-inputs) – Norbert Schuch Sep 21 '23 at 17:58
  • I would assume a reasonable way to model the scenario is a beamsplitter with three outputs, two corresponding to the two slits, and one to the absorbed photons. This should effectively amount to a mixture of different Fock states impinging on a normal beam splitter (or the double slit). Then, the interference would be given by mixing those beams again at another beamsplitter (i.e., an interferometer), where the relative phase of the two paths depends on the position on the screen. – Norbert Schuch Sep 21 '23 at 18:38
  • Oh I see what you mean. I updated my question to make it more clear.

    I like the idea of model it as a beam splitter. However, since we have a whole screen of detection, I think the model should contain at least m detectors in the screen, didn't it?

     – Tarek Peña Sep 21 '23 at 18:55

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