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2 photons having sufficient energy can collide and form an electron positron pair (which then annihilate and form a new photon pair - with lower energy?). I assume this means that they can't collide (just pass through each other?) if not energetic enough? As you see I have many small subquestions. My main question, however, is: What is the probability of two photons colliding in a transparent vacuum container? I suspect it would be near zero when talking about sunlight but what about having two high energy laser beams (originating from the same laser) meet head on at a common focal point inside the container?

Jens
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    look at this link https://en.wikipedia.org/wiki/Two-photon_physics – anna v Mar 08 '16 at 07:35
  • As a further subquestion I wonder if the fringe pattern at the common focal point can somehow be explained regarding the photons as particles that interact at the high intensity points but not at the low intensity points? – Jens Mar 08 '16 at 07:38
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    fringes have nothing to do with interacting photons. It is just a superposition of the probability amplitudes of indiviual photons that ends up in contributing macroscopic light behavior. It is not an interaction but a superposition of complex fields – anna v Mar 08 '16 at 07:41
  • Photons are field quanta. "They" don't fly around like little billiard balls and they don't collide like that, either. What experiments that detect photons are measuring are the quantized exchanges between matter and the electromagnetic quantum field. The high intensity regime in which electromagnetism becomes non-linear has not yet been implemented in accelerator experiments. Gamma colliders are on the wish list of high energy physicists, but we aren't quite there, yet. – CuriousOne Mar 08 '16 at 07:49
  • there is no classical counter part to this issue but the framework of quantum field theory allows one to study such processes. – Abhishek Pal Mar 08 '16 at 07:50
  • You just have to use Euler-Heisenberg lagrangian. – Name YYY Mar 08 '16 at 07:51
  • @Jens Possibly of interest http://physics.stackexchange.com/questions/1361/scattering-of-light-by-light-experimental-status ? I know there was a calculation of photon-photon scattering (called "the scattering of light by light" or something like that) dating from the 1950s, which you might be able to dig up with access to a university library. I doubt the paper is available for free anywhere. – PhillS Mar 08 '16 at 11:03

2 Answers2

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Photons are elementary particles. The probability of two photons to interact can be calculated using Feynman diagrams, here is the lowest order diagram for photon photon scattering :

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A Feynman diagram (box diagram) for photon–photon scattering, one photon scatters from the transient vacuum charge fluctuations of the other

Each vertex contributes a factor of (1/137)^1/2, the electromagnetic coupling constant, and that has to be squared so the probability of scattering is very small and negligible for two light beams meeting. (1/137)^2

The crossection goes up with energy and at gamma ray energies there are proposals for gamma gamma colliders.

The interference patterns seen with light does not mean that the photons composing it are interacting. It is just the superposition of the quantum mechanical wave functions that build up the classical wave, and the patterns appear also with one photon at a time.

You state:

One more point: Even if a photon is represented by some probability cloud the "nearest neighbor" probability of two such clouds goes to zero proportional to the square of the distance so the number of photons per unit volume must be extremely high in order for two clouds to be within "touching" distance - assuming this distance is in the micrometer range or less.

Yes, distance at minimum approach is also a factor on when these two photon interactions can have measurable consequences and should be taken into account in designing collider experiments..

anna v
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The amplitudes of such processes can be easily calculated using Field Theory. The Feynman diagram directly gives the M matrix whose amplitude square gives the probability density of the process. Also in the entire process the 4 momentum is conserved.

Abhishek Pal
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  • True enough, but more useful if you gave a sample Feynman diagram and explained how to use it. – Carl Witthoft Mar 08 '16 at 14:17
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    Regardless of whatever labels you might use for photons and their interaction there must be some minimum "simultaneity" and "distance" between them for them to interact? In other words if a photon in Miami can interact with one in New York I'm totally lost. – Jens Mar 08 '16 at 15:00
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    One more point: Even if a photon is represented by some probability cloud the "nearest neighbor" probability of two such clouds goes to zero proportional to the square of the distance so the number of photons per unit volume must be extremely high in order for two clouds to be within "touching" distance - assuming this distance is in the micrometer range or less. I am sorry if this is boring the experts but I find it very difficult to completely abandon some connectedness to classical concepts. – Jens Mar 08 '16 at 15:26