Photon Interaction: Possible or Not?

[DrZoidberg]

Is it possible for photons to interact with each other directly?

 

[Vanadium 50]

No.
.

 

[humanino]

There is only one photon-photon-graviton vertex, but that is exceedingly small. Photon photon scattering is dominated by fermion loops.

 

[tom.stoer]

the graviton is misleading here.

Have a look here: http://arxiv.org/PS_cache/hep-ph/pdf/0512/0512033v1.pdf Fig. 2.2 shows the lowest order of photon-photon coupling due to virtual fermions.

 

[cragar]

If photons can't couple with other photons , And if they obey the principle of superposition , when we fire photons through a double slit , then how are they interacting with each other , I have only begun to study QM so take it easy.

 

[Vanadium 50]

Yes, but the OP said "directly" - I would argue this excludes fermion loops.

 

[tom.stoer]

I agree, "directly" excludes fermion loops or non-linear effects in active materials.

@cragar: "interference" and "coupling" are two different things. Measuring coupling means that you have to prepare (e.g.) a two-photon state ("colliding photons") and check if and how they scatter. Interference in quantum mechanics means that even one single photon = a one photon state can interfer with itself.

In a double slit experiment you do not need different photons to interfere with each other; you will observe an interference pattern even if you send only single photons through the double-sit. You can even do the following: Prepare a huge number of exact copies of one double-slit experiment. Now distribute them all over the Earth in different laboratories. In each lab send exactly one photon through the experiment and register its position on the screen (x and y coordinate). Then collect all (x,y) tupels from all over the Earth and plot them in one diagram. You will find an interference pattern.

 

[haael]

Speaking of fermion loops - what does it mean for us? Can photons attract each other? Is there some change in the usual Coulomb force?

 

[ansgar]

Speaking of fermion loops - what does it mean for us? Can photons attract each other? Is there some change in the usual Coulomb force?

in QFT formulation of electrodynamics, photons can fluctuate into particle-antiparticle pairs, thus altering the classical picture of electromagnetism

 

[tom.stoer]

One can try to express these QFT loop corrections as terms in an effective potential; that would imply quantum corrections to the Coulomb force. But I am not sure if this will always work.

 

[Edgardo]

Two things that I've found:
1. Two photon interference (Hong-Ou-Mandel effect)
Two photons are incident at a beamsplitter and you will observe that they both take the same path (possibility 1 and 4 in the picture).
2. Two-photon physics
Have a look at the external links

 

[Count Iblis]

You can't exclude loops from tree level interactions in a meaningful way. A physical interaction will include all contributions, splitting them up in tree level and higher order loop contributions is unphysical, even though in perturbation theory the contributions appear separately. Well known example: You can have violations of unitarity at tree level while in reality no such violations are possible within quantum mechanics as it is a unitary theory by design.

 

[Count Iblis]

http://arxiv.org/abs/physics/0605038"

Some nice expressions for the effective coefficient of refraction for light propagating through magnetic fields are http://arxiv.org/abs/hep-ph/9806417"


http://arxiv.org/abs/astro-ph/0002442"

 

[threadmark]

W bosons couple to photons. Could there be an exchange YW -->YW bosons. if so would that sagest photon interaction.

 

[Karl Coryat]

DrZoidberg: One way to think of it is, photons cannot interact because each photon occupies a unique point in relativistic spacetime. In that context, photons do not move at all relative to one another and therefore could never achieve local contact.

 

[tom.stoer]

DrZoidberg: One way to think of it is, photons cannot interact because each photon occupies a unique point in relativistic spacetime. In that context, photons do not move at all relative to one another and therefore could never achieve local contact.

This is misleading as it applies to massless gluons as well; but gluons DO interact.

 

[threadmark]

If we looked at it as wave function there would be interaction? But as a particle I wouldn’t think it would interact directly.

 

[tom.stoer]

Photons as wave functions do not interact
1) there is no wave function for a photon (and no Schrödinger equation)
2) wave functions do not interact; they can only interfer, but this is somethign totally different

 

[threadmark]

I do agree with you in this context of a photon interaction being “single”. As a wave function though to say “they interfere” is an interaction. Direct or indirect it can be seen as an interaction.

 

[Cthugha]

I disagree. In order to see interference the photons involved must be indistinguishable. Therefore interference should not be interpreted as the interaction of two or more photons, but as a property of one single state containing more than one excitation.

 

[IcedEcliptic]

Define interaction? When I hear that in such a context I usually think of an exchange of energy/momentum, or "information" in the general sense. This does not occur with photons directly.

 

[tom.stoer]

Interaction means that there is a term in the time evolution operator (coming from the Hamilton operator) that changes an initial state. So if there should be an interaction in a two-photon state that means that at least the two photons
1) exchange momentum,
2) change into two different particles, e.g. an electron-positron pair
...

Symbolically

$$U|\gamma_{p_1}, \gamma_{p_2}\rangle = \alpha_1 |\gamma_{p^\prime_1}, \gamma_{p^\prime_2}\rangle + \alpha_2 |e^-_{p^\prime_1}, e^+_{p^\prime_2}\rangle + \ldots$$

Interference means that there is an interference pattern when observing e.g. a single-photon state or its wave function

Symbolically

$$\psi_\gamma(x) = \langle x|\gamma\rangle$$

That can happen w/o a second photon and even in a free theory where nothing "happens" with a two-photon state. So interference is an "intrinsic property" of the state.

In practice, observation of interference patterns of course requires interaction with a screen; w/o this screen the interference pattern does not become visible.

 

[threadmark]

The bose Einstein model of photon gas Suggested “mysterious non-local interaction” witch is known now as coherent states. I am merely saying that I cannot see photons in phase with one another. I won't say they do interact in this context but if two photons can't be in phase I would think there could be an affect preventing the two to be in phase.

 

[tom.stoer]

If you look at a LASER all photons are in phase (coherent state), but they do not interact with each other. They interact with the LASER medium which "forces" them to be in phase. Later on they can interfere with each other (because they are coherent), but as I said this interference is no interaction (basically because interference is possible already for one single photon)

But if you look at the maths you see the difference as well. Interaction is due to a change of the state whereas interference is due to observation (or projection).

 

[threadmark]

Coherence in relation to a laser is a rough description of its properties. Being its beam is in phase achievable with optics. But the idea of two photons being in phase is not valid in quantum theory.

 

[tom.stoer]

But the idea of two photons being in phase is not valid in quantum theory.

Why?

 

[threadmark]

Because it is defined in coherent states in quantum optics.

 

[tom.stoer]

I don't understand, but that doesn't really matter in this context.

The question was if photons can interact. The answer is: not directly, only via fermion loops. And interference is not to be confused with interaction; it is a different phenomenon that applies already to single particle states.

 

[Cthugha]

I am merely saying that I cannot see photons in phase with one another. I won't say they do interact in this context but if two photons can't be in phase I would think there could be an affect preventing the two to be in phase.
[...]
Coherence in relation to a laser is a rough description of its properties. Being its beam is in phase achievable with optics. But the idea of two photons being in phase is not valid in quantum theory.

Rough description? Glauber got the Nobel prize for rough descriptions?

However, the reason why two photons are not "in phase" is that phase is not really a property of a particle, but of the underlying fields. If you have two coherent emitters A and B (emitting fields with a fixed phase relation) and detect a photon at some position inside the coherence volume, the consequence is that you cannot in principle know whether the photon you detected originated from A, B or of both fields. You can apply the concept of phase to the em fields or to probability amplitudes, but usually not to intensities or particles. Accordingly there is no "effect preventing photons from being in phase".

 

[IcedEcliptic]

Interaction means that there is a term in the time evolution operator (coming from the Hamilton operator) that changes an initial state. So if there should be an interaction in a two-photon state that means that at least the two photons
1) exchange momentum,
2) change into two different particles, e.g. an electron-positron pair
...

Symbolically

$$U|\gamma_{p_1}, \gamma_{p_2}\rangle = \alpha_1 |\gamma_{p^\prime_1}, \gamma_{p^\prime_2}\rangle + \alpha_2 |e^-_{p^\prime_1}, e^+_{p^\prime_2}\rangle + \ldots$$

Interference means that there is an interference pattern when observing e.g. a single-photon state or its wave function

Symbolically

$$\psi_\gamma(x) = \langle x|\gamma\rangle$$

That can happen w/o a second photon and even in a free theory where nothing "happens" with a two-photon state. So interference is an "intrinsic property" of the state.

In practice, observation of interference patterns of course requires interaction with a screen; w/o this screen the interference pattern does not become visible.

That was the gist of what I was asking. I'm confused by the OP's confusion, which seems to be baseless.