# Are photons the antiparticle of itself?

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1. Feb 22, 2014

### PhotonicBoom

Hey guys, I was thinking about this for a while now and I seem to be on a dead end. So here it is, this is my speculation. I would love some feedback, tell me which parts are correct/false and if they are false, guide me towards the right path! :)

1. Photons can technically be their own antiparticle.

2. They don't annihilate because they are bosons and can occupy the same quantum state.

3. In many Feynman diagrams we see a particle & antiparticle annihilation producing another particle and antiparticle pair. I was thinking thats why we get 2 photons in some other annihilation interactions (one photon, one antiphoton)

I must be missing something. Is my logic flawed?

2. Feb 22, 2014

### D H

Staff Emeritus
Sure they are, and they can be viewed as being their own antiparticle.

3. Feb 22, 2014

### WannabeNewton

You cannot have a QED Feynman diagram with two photon lines at a vertex. The fundamental vertex of the QED interaction consists of two electron (charged particle) lines and a photon line. You can certainly have the scattering $\gamma + \gamma \rightarrow \gamma + \gamma$, which is called Delbruck scattering, but this is a four vertex diagram at the lowest order and consists of 4 external photon lines and 4 internal electron lines since higher order QED diagrams can only be built up from the fundamental vertex.

None of this is contradictory to the (correct) statement that photons are their own antiparticles.

4. Feb 22, 2014

### Bill_K

Photons ARE their own particle. Literally, not just technically.

Two photons don't annihilate because there is no direct interaction between two photons. Indirectly, through their interaction with other particles such as electrons, two photons can scatter, or turn into a neutrino-antineutrino pair.

It's conservation of momentum, angular momentum and parity that requires two photons to be produced, or sometimes three of them.

5. Feb 22, 2014

### PhotonicBoom

Thanks a lot to everyone for the responses!

Can two boson pairs (boson - antiboson) annihilate in general? Since they can occupy the same quantum state I thought they wouldn't.

6. Feb 22, 2014

### ChrisVer

7. Feb 22, 2014

### Bill_K

Sure, no problem. The simplest example comes from e+ + e- → γ + γ. Just run it backwards: γ + γ → e+ + e-.

Most examples are unobserved, but that's just because it's more difficult experimentally to collide two bosons, since they tend to decay! But in principle, π+ + π-, or W+ + W- are annihilation examples.

Having two bosons just means that their two-particle wavefunction is symmetric, so e.g. they could orbit each other in an S state.

8. Feb 22, 2014

### ChrisVer

just a remark:
Wouldn't it be better to avoid the pions? fundamentally the annihilation still occurs between two fermions.
The rest OK

9. Feb 22, 2014

### Bill_K

Are you saying that pions are not bosons??

10. Feb 22, 2014

### ChrisVer

No I am not saying that.. they are not fundamental (consisting of (anti)quarks which are fermions), and the annihilation happens between the quarks (fermions).

Also
I am not sure whether they are bosons or not...since they have inner structure, they will tend to behave like bosons, but only effectively...there are also bosonic nuclei but they also behave like that effectively because they have fermionic inner structure (nucleons)
maybe I'm wrong or missing something

11. Feb 22, 2014

### PhotonicBoom

Pions obey Bose-Einstein statistics - > Bosons :)

12. Feb 23, 2014

### unusualname

Yeah, but only at energies so their constitute fermions aren't relevant - a composite boson can not break the exclusion principle for its constituents!