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Could someone explain to me what the collision of two photons would look like? Will they behave like,

  1. Electromagnetic waves: they will interfere with each other and keep their wave nature
  2. Particles: they will bounce like classical balls

I assume that energy of that system is too small to make creation of pairs possible.

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see also physics.stackexchange.com/q/1361 . They used terawatt pulses of laser light to experimentally demonstrate light by light scattering. –  Andre Holzner Feb 18 '13 at 19:42

4 Answers 4

up vote 9 down vote accepted

Your assumption that pair production is ruled out, rules out* that two photons interact through higher-order processes. Quantum electrodynamics tells us that two photons cannot couple directly. That leaves us with classical electromagnetism, which tells us that electromagnetic waves pass through each other without any interference.

*Edit. The photons can interact through higher-order processes. As pointed out in the comments (and I hope I'm getting this right), there is a (quite small) probability amplitude for two photons to get absorbed in, and two photons be emitted by, e.g. an fermion-antifermion virtual pair (which is the leading contributor to the combined amplitude of all such processes). Whether (and this is my cop-out) the emitted photons can be considered the same photons as the absorbed photons, I leave to the, certainly more knowledgeable, commentators.

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Though this answer is good enough for essentially all practical purposes, it is worth noting that insufficient energy for the production of on-shell pairs does not mean that loop diagrams do not contribute. –  dmckee Feb 18 '13 at 19:02
@dmckee I'm considering an edit, but just to sure: Are you alluding to absorption and emission by a quantum fluctuation already there? –  Glen The Udderboat Feb 19 '13 at 1:38
To get the first loop diagram take the Feynman diagram in Andre Holzner's answer and hook the fermion lines (they needn't be electrons: all the quarks and heavier leptons contribute too) into those from a mirror image of the same diagram. You get a diagram with two incoming photons, a loop of fermions and two outgoing photons. It's contribution to the cross-section is suppressed by the two extra couplings (a factor of more than 18000) and the off-shell nature of the fermions in the loop, but it is still the leading diagram that contributes. –  dmckee Feb 19 '13 at 1:48
@dmckee I got that far. But wouldn't that take more energetic photons than the question allows for and wouldn't you therefore need (to borrow from) the vacuum energy to create a virtual pair and have the photons bump into that pair? Or am I just making things worse? –  Glen The Udderboat Feb 19 '13 at 2:05
Well, I wouldn't phrase it that way. You don't borrow from some existing pool, you just overdraw your account for bit, or another way to look at it is the virtual particles aren't on-shell which means they don't have the right mass and it don't take extra energy. Any way, I just a dumb experimenter and I don't claim to understand the ontology or epistemology of it, I just draw the diagrams, figure the amplitudes and add them up. Or actually, mostly let other people draw the diagrams, figure the ... –  dmckee Feb 19 '13 at 2:18

A lowest order QED Feynman diagram for the process photon + photon $\rightarrow$ electron + positron looks like shown below (the time axis is the horizontal axis).

From the point of view of energy conservation, this process is only possible if sum of the energy of the photons is above twice the electron mass. In the center of mass frame of the di-photon system, the photons need to have at least 511 keV.

gamma gamma scattering

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Well, yes, but the poster explicitly disavowed interest in pair production processes. –  dmckee Feb 18 '13 at 20:10

Photons do not have the feature of self interaction, meaning that two photons can neither attract nor repell each other. Therefore two photons can not collide.

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Actually, there is a small amplitude $\mathcal{O}(\alpha)^2$ for photons to interact with each other because of a tiny polarization induced by virtual $e^+$ $e^-$ pairs. I think the OP wants to know what such an event would look like. –  QuantumDot Feb 18 '13 at 18:17
While this is true, I do not think that this would have been the appropriate answer in this case. It seems that the question was asked due to lack of more fundamental knowledge. –  Frederic Brünner Feb 18 '13 at 18:20
You mean that they are bosons? –  Mozibur Ullah Feb 18 '13 at 18:38
Yes, that they are bosons and carry zero charge. –  Frederic Brünner Feb 18 '13 at 18:39
The OP's final sentence "I assume that energy of that system is too small to make creation of pairs possible" suggests that he/she is aware to some extent of this effect. –  QuantumDot Feb 18 '13 at 18:40

Just to make sure this is clearly understood: Two photons absolutely can collide head on, annihilating themselves to produce an electron-positron pair in their place. In principle, sufficiently high-energy colliding gammas could produce pretty much any particle pairs, but of course it's tough enough just to get two photons whose net energy is twice the mass of an electron.

Photon annihilation has to be true because of time symmetry. That is, since one of the decay modes for positron-electron annihilation is the production of two gamma photons, time symmetry requires that the inverse process also be possible. It's just harder to arrange, particularly since uncharged photons have no guiding attraction for each other.

This idea is not hypothetical. Photon annihilation was demonstrated in experiments done several years ago by shining a green laser directly into a beam of gamma rays. A few of the green and gamma photons mutually annihilated to produce fast-moving electron-positron pairs with leftover momentum in the gamma propagation direction. Looking for the reference now...

I think this was it, but it seems earlier than the one I was thinking of:

D.L Burke et al, Positron Production in Multiphoton Light-by-Light Scattering, SLAC, June 1997.

If you allow photon-photon interactions (scattering) via virtual particle pairs you get two-photon physics, which looks at the probabilities of photon-photon production of particle pairs much heavier than electrons.

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Just noticed this: "Two photons absolutely can collide head on". It is not head on, it is through the exchange of a virtual particle. Head on would mean a vertex of photon on photon. For example e+e- make a vertex with Z, that is head on collision. –  anna v Sep 18 '13 at 9:59
Wow... you are correct and I stand corrected. I was being way too sloppy in terminology, trying to emphasize that photons can interact without matter, and thereby said it poorly. As best I can see, @AndreHolzner's figure above nicely captures the interaction you just described. Thanks! –  Terry Bollinger Sep 30 '13 at 1:38

protected by Qmechanic Mar 8 at 20:43

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