18
$\begingroup$

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.

Improve this question
$\endgroup$
1

4 Answers 4

Reset to default
10
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Corrections due to light-by-light scattering can happen and are in fact necessary to obtain refined theoretical values for the anomalous magnetic moment of the muon in the Standard Model. Actually, it is also necessary to compute hadronic light-by-light corrections to get this refined value, which are even more difficult to deal with and have taken up a huge amount of theoretical effort. $\endgroup$
    – Tom
    Jul 7, 2021 at 18:05
7
$\begingroup$

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

Improve this answer
$\endgroup$
0
6
$\begingroup$

By time symmetry, two photons with sufficient mutual energy necessarily are capable of annihilating each other to produce an electron-positron pair, since one of the decay modes for positron-electron annihilation is the production of two gamma photons. It's just harder to arrange experimentally, since unlike the electron and positron the uncharged photons have no attraction for each other.

Here's an experimental exploration of two-photon positron production: D.L Burke et al, Positron Production in Multiphoton Light-by-Light Scattering, SLAC, June 1997.

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

Improve this answer
$\endgroup$
5
  • 3
    $\begingroup$ 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. $\endgroup$
    – anna v
    Sep 18, 2013 at 9:59
  • $\begingroup$ 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! $\endgroup$
    – Terry Bollinger
    Sep 30, 2013 at 1:38
  • $\begingroup$ I shortened it greatly but did not (yet) delete it. Hopefully this version better conveys the only point I was really trying to make, which is that the very common perception that chargeless photons "don't interact" is not accurate physics. $\endgroup$
    – Terry Bollinger
    Sep 20, 2015 at 21:33
  • $\begingroup$ Is annihilating the correct word choice here? $\endgroup$
    – Magic Octopus Urn
    Nov 28, 2018 at 19:07
  • $\begingroup$ Sure. The two photons disappear and are replaced (most commonly) by a positron and an electron. Since the photons cease to exist, "annihilation" accurately describes the outcome of the interaction. $\endgroup$
    – Terry Bollinger
    Nov 28, 2018 at 19:16
3
$\begingroup$

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.

Improve this answer
$\endgroup$
7
  • 8
    $\begingroup$ 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. $\endgroup$
    – QuantumDot
    Feb 18, 2013 at 18:17
  • $\begingroup$ 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. $\endgroup$
    – Frederic Brünner
    Feb 18, 2013 at 18:20
  • $\begingroup$ You mean that they are bosons? $\endgroup$
    – Mozibur Ullah
    Feb 18, 2013 at 18:38
  • $\begingroup$ Yes, that they are bosons and carry zero charge. $\endgroup$
    – Frederic Brünner
    Feb 18, 2013 at 18:39
  • 2
    $\begingroup$ 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. $\endgroup$
    – QuantumDot
    Feb 18, 2013 at 18:40

Not the answer you're looking for? Browse other questions tagged or ask your own question.