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The Hong Ou Mandel Effect shows photons in the same state turning up in the same detector and not at separate detectors.

https://en.wikipedia.org/wiki/Hong%E2%80%93Ou%E2%80%93Mandel_effect

Does this imply that for any number of identical bosons in exactly the same state, they are either all observed, or none? That there is no way to observe only some? If a detector was set to register exactly one photon - that it would never detect anything, if the state it was detecting had two photons?

For example, if the detector relied on the ejection of an electron from an atom, then the electron would always emerge having absorbed the energy of two photons? That is, all the energy in the photon state?

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No.

For instance, you could build a detector which consists of a beam splitter followed by a perfect detector in one path: This device will only detect the photons which took that path.

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  • $\begingroup$ Cool. But, this is the specific example of the Hong Ou Mandel effect. My question is - can it be proven that Bosons generically in all situations can only be detected all together or not at all. $\endgroup$
    – Ponder Stibbons
    Aug 21, 2018 at 12:37
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    $\begingroup$ @PonderStibbons I just explained in my answer how to detect only part of the photons. There is nothing specific to HOM or any other situation to my construction. So, no, it is not true that "Bosons generically in all situations can only be detected all together or not at all". $\endgroup$
    – Norbert Schuch
    Aug 21, 2018 at 12:42
  • $\begingroup$ Sorry, I misunderstood your comment. But, then your comment is in direct contradiction to the Hong Ou Mandel effect, which says that in precisely that situation both or none of the photons will be detected - or am I still not seeing what you are saying? Or not understanding Hong Ou Mandel? $\endgroup$
    – Ponder Stibbons
    Aug 21, 2018 at 12:47
  • $\begingroup$ @PonderStibbons Wikipedia on HOM: "The effect occurs when two identical single-photon waves enter a 1:1 beam splitter, one in each input port." (Emphasis by me.) In HOM, the two photons are distinguishable, as they enter from different ports. If you only want to measure one of them, simply omit to beam splitter. $\endgroup$
    – Norbert Schuch
    Aug 21, 2018 at 12:54
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The Hong-Ou-Mandel effect for identical photons$^*$ tells you what output state the device will produce. It does not mean that you will always detect all the photons that exist from a particular output port. That is a separate issue related to the quantum efficiency of the detector.

So, if you have a state with $n$ photons directed at a detector that can only at a time detect one photon (assuming such a thing does exist), then with a certain probability, it will detect a photon. The other photons are simply lost.

$^*$ "Indentical photons" means that there are now other degrees of freedom, such as their polarization or spatial modes, that distinguish them. However, for the HOM effect, these idendtical photons need to enter different input ports.

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  • $\begingroup$ Sorry for the delay. For some reason, the activity was not flagged in my email. $\endgroup$
    – Ponder Stibbons
    Sep 2, 2018 at 20:50

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