Submitted by **TheGandPTurtle** t3_111g7s9
in **askscience**

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**ChemicalRain5513**
t1_j8g08qp wrote

Reply to comment by **taphead739** in **Light traveling through a medium that slows it. Does the same photon emerge?** by **TheGandPTurtle**

I don't know if this completely what you mean, but laser works by stimulated emission of photons, since they like to be in the same quantum state.

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**taphead739**
t1_j8g478u wrote

This is unfortunately not what I mean, but thanks anyway. Technically speaking I am talking about the energy contribution in the Hamiltonian integral of a system of multiple identical particles that arises from the requirement that the total wave function must not change its sign upon exchange of particle labels (in the case of bosons). Does this exist for photons?

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**TheoryOfSomething**
t1_j8i3q57 wrote

I believe that the phenomenon known as "photon bunching" could also be described as an exchange interaction (along with the associated anti-bunching effect for fermions). If you consider two points, a and b, in some source that is emitting photons and you set up 2 detectors, A and B, to detect those photons, then for photons (and all bosons) you will see an increase in the probability of simultaneous detection at A&B compared to distinguishable particles and a decrease of simultaneous detection for Fermions: https://en.wikipedia.org/wiki/Hanbury_Brown_and_Twiss_effect

Because photons are non-interacting at the tree-diagram level (that is to say that in the ~~Hamiltonian~~ Lagrangian for QED there is no photon-photon interaction term), this does not lead to the same energy consequences as it does for electrons or alpha particles. Both the Coulomb integral and the exchange integral are proportional to the interaction term in the Hamiltonian, except that for photons there is no interaction term! Metaphorically speaking (because there are technical problems with assigning a wavefunction to single photons), the product basis remains the diagonal basis. As a result, you see interference effects that cause the bunching behavior mentioned earlier, but not the same consequences for energy or spin correlation as with electrons.

A careful reader may object at this point and say, "Ah! But you have neglected the higher-order QED effects. Sure, at tree-level there is no photon-photon interaction, but what about the scattering mediated by virtual electron-positron pains? Surely that gives rise to some interaction which turns out to be either attractive or repulsive." And the careful read is almost correct, almost. There is very weak (starting at 4th order) photon-photon interaction and photon-photon scattering in QED, but it turns out that the effective potential that this interaction gives rise to **has zero range** (one might describe it as "a delta function" although there are technical problems with making this formulation precise in >1 spatial dimension; a complication that comes up repeatedly in my PhD thesis in the context of ultracold atomic physics) and therefore cannot really be described as attractive or repulsive. To expand, to first order in perturbation theory the interaction terms is quartic in the E and B fields, and if you work out what kind of potential it takes to create that you get an operator proportional to a delta function. Of course this is just the effect at first order, the effects at all higher orders will also give rise to interactions that look like they are "zero range", although if it were possible to do the full resummation and go beyond all orders, you might get a spatially-extended potential. I can't think of any reason that that potential should be always attractive or always repulsive, but who knows. Experimentally speaking, attraction/repulsion between beams of photons has not been observed or measured without coupling them to some massive interacting medium.

In the case of photons, there is a purely *classical* explanation for all of this. Essentially, because the classical limit of a multi-particle system of photons is not a system of interacting distinguishable particles but rather an electromagnetic wave (a coherent state superposition of all numbers of photons with a well-defined average number of photons and phase), you can do the math and arrive at correct predictions purely from considering the classical problem of detecting the signal from a spatially extended EM source with 2 nearby detectors.

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**taphead739**
t1_j8i66hg wrote

That was very informative and probably what I was looking for. Thanks a lot!

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**hahahsn**
t1_j8htczx wrote

hey this is a super interesting question! I don't have the answer but in case someone does i'll leave this comment to find it at some point.

Based on my limited understanding I don't see where such an energy contribution would arise. I get it in the case of fermions but boson statistics are unfamiliar. The normalisations are weird.

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**ContaminatedPrime**
t1_j8ggaq1 wrote

Photons are bosons. They have no exchange interaction. Only exists for fermions.

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**taphead739**
t1_j8h8f9d wrote

Exchange interaction exists for bosons and it is attractive: https://en.m.wikipedia.org/wiki/Exchange_interaction

Explains the superfluidity of helium-4, for example

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**eva01beast**
t1_j8gga2w wrote

Photons fall under the category of bosons. They have a spin of 1.

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