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Photon its own anti-particle?

  1. Jun 7, 2015 #1
    I see that a photon is said to be its own anti-particle. How is that possible? For example, how is it consistent with the idea that a particle and its anti-particle annihilate each other leaving only energy?

    PS. The links to other questions about this don't seem to answer the question, at least not in any way that I understand.
     
  2. jcsd
  3. Jun 7, 2015 #2

    Nugatory

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    There's quite a bit more to the definition of an anti-particle than whether it annihilates with its counterpart particle, and an annihilation reaction is not a required part of the definition - the wikipedia article is worth reading.
    The particles that make up the normal matter around us (electrons, neutrons, protons) do annihilate with their anti-particle counterparts, so all normal matter will annihilate with antimatter - but that's not quite the same thing as saying that every particle must annihilate with its antiparticle.
     
  4. Jun 7, 2015 #3

    A.T.

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    It's better to say that the distinction between matter and anti-matter only applies to massive particles, while photons are neither matter nor anti-matter.

    This applies only to massive particles which are converted to photons during annihilation. Photons on the other hand can combine to generate massive particles.
     
  5. Jun 7, 2015 #4
    In a sense, two photons do cancel, leaving only two photons--they pass through each other unchanged.

    In terms of conserved quantities, when a particle and its antiparticle annihilate, what remains is energy, momentum, and spin.

    A particle and its antiparticle have opposite sign in things like lepton number, baryon number, and electric charge. These quantities can all cancel leaving only light. Light has zero baryon number, lepton number and charge.

    I believe that in a more accurate way, light doesn't interact with other light, because it has no conserved quantities to exchange. Someone might correct me on this.
     
  6. Jun 7, 2015 #5
    Photons don't interact directly with other photon, this is, there is no Feynmann diagram at tree level with a photon-photon vertex, however there is an effective interaction of four photons via a loop of charged particles, that is more or less the idea behind Euler-heisenberg lagrangian.
     
  7. Jun 7, 2015 #6
    what does "leaving only energy" means?, perhaps you can benefit from reading this insight post

    https://www.physicsforums.com/insights/what-is-energy/
     
  8. Jun 8, 2015 #7
  9. Jun 9, 2015 #8
    It's just semantics, but I'd say a photon is its own antiparticle in the same way that 0 is its own additive inverse. 3 and -3 "annihilate" each other, leaving 0.
     
  10. Jun 9, 2015 #9
    Consider an annihilation reaction of some massive particle-antiparticle pair.
    There is a good probability that this will produce a pair of photons.
    The inverse reaction will destroy the photons and produce a massive particle antiparticle pair.
    This proves that a photon with spin +h is the antiparticle of a photon with spin -h and the same frequency.
    The total energy of the photons needs to exceed at least the rest masses of an electron positron pair, otherwise the photon pair cannot be annihilated.
     
  11. Jun 12, 2015 #10
    Two-photon_physics
     
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