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Can photons interact?

  1. May 17, 2010 #1
    Is it possible for photons to interact with each other directly?
     
  2. jcsd
  3. May 17, 2010 #2

    Vanadium 50

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  4. May 17, 2010 #3
    There is only one photon-photon-graviton vertex, but that is exceedingly small. Photon photon scattering is dominated by fermion loops.
     
  5. May 18, 2010 #4

    tom.stoer

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  6. May 18, 2010 #5
    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.
     
  7. May 18, 2010 #6

    Vanadium 50

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    Yes, but the OP said "directly" - I would argue this excludes fermion loops.
     
  8. May 18, 2010 #7

    tom.stoer

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    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.
     
  9. May 18, 2010 #8
    Speaking of fermion loops - what does it mean for us? Can photons attract each other? Is there some change in the usual Coulomb force?
     
  10. May 18, 2010 #9
    in QFT formulation of electrodynamics, photons can fluctuate into particle-antiparticle pairs, thus altering the classical picture of electromagnetism
     
  11. May 18, 2010 #10

    tom.stoer

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    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.
     
  12. May 18, 2010 #11
    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
     
  13. May 18, 2010 #12
    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.
     
  14. May 18, 2010 #13
    http://arxiv.org/abs/physics/0605038" [Broken]

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


    http://arxiv.org/abs/astro-ph/0002442" [Broken]
     
    Last edited by a moderator: May 4, 2017
  15. May 19, 2010 #14
    W bosons couple to photons. Could there be an exchange YW -->YW bosons. if so would that sagest photon interaction.
     
  16. May 20, 2010 #15
    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.
     
  17. May 20, 2010 #16

    tom.stoer

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    This is misleading as it applies to massless gluons as well; but gluons DO interact.
     
  18. May 20, 2010 #17
    If we looked at it as wave function there would be interaction? But as a particle I wouldn’t think it would interact directly.
     
  19. May 20, 2010 #18

    tom.stoer

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    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
     
  20. May 20, 2010 #19
    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.
     
  21. May 20, 2010 #20

    Cthugha

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    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.
     
  22. May 20, 2010 #21
    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.
     
  23. May 20, 2010 #22

    tom.stoer

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    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

    [tex]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[/tex]

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

    Symbolically

    [tex]\psi_\gamma(x) = \langle x|\gamma\rangle[/tex]

    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.
     
    Last edited: May 20, 2010
  24. May 20, 2010 #23
    The bose Einstein model of photon gas Suggested “mysterious non-local interaction” witch is known now as coherent states. im merely saying that I cannot see photons in phase with one another. I wont say they do interact in this context but if two photons cant be in phase I would think there could be an affect preventing the two to be in phase.
     
  25. May 20, 2010 #24

    tom.stoer

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    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).
     
  26. May 20, 2010 #25
    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.
     
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