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Using Furry's theorem, we know that the time ordered expectation value of odd number of currents in QED is zero. I was wondering taking into account Furry's theorem, will there be light by light scattering for the following process: $\gamma\gamma\gamma \rightarrow\gamma\gamma$?

Can someone please explain the above scenario?

Qmechanic
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SSS
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  • Do you mean $\gamma\gamma\rightarrow\gamma\gamma$? That has a four vertex loop and a non-zero amplitude, whike the $5\gamma$ one in the question is identically zero. – PhillS May 26 '18 at 15:19
  • No I mean what I wrote above i.e 3 photons going to 2 photons. Does it mean there is no such scattering involving 3 photons going to 2 photons? – SSS May 26 '18 at 15:24
  • AI believe there is no such process: the amplitude had to be zero for a five vertex loop and I can't think of a way to construct a Feynmann diagram that doesn't involve such a loop. – PhillS May 26 '18 at 15:27
  • I know there is a diagram with 3 photons vertex and a loop but that one also goes to zero. Are there other possibilities? Does that imply that any scattering with odd numbers of photons eventually go to zero? – SSS May 26 '18 at 15:30
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    Looks like you answered your own question. This doesn’t happen, by Furry’s theorem. – knzhou May 26 '18 at 18:25
  • Okay that's what I wanted to confirm. So aren't there any other diagrams which give non zero amplitude to the process? – SSS May 27 '18 at 06:23

2 Answers2

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Furry's theorem say's that there are no effective vertex that can poder your photon number by an odd number. Because of this, if you have a dyagram that goes from an even number to an odd number of photons, somewhere you will have an effective vertex that does that dirty thing, and you know it gives you a zero factor, so your dyagram amplitude is zero.

Iliado Odiseo
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  • this answer is relevant https://physics.stackexchange.com/questions/79215/proof-of-furrys-theorem – anna v May 27 '18 at 03:49
  • So, are there any other diagrams other than the one which makes the amplitude go to zero? I mean are there other contributions to the amplitude of the scattering or there is only one diagram and that diagram itself gives zero? – SSS May 27 '18 at 06:21
  • @SSS No: for a given diagram with the virtual electron circulating in a certain direction, there is the diagram with the electron circulating in the opposite direction. Both diagrams are identical up to a sign, and they cancel off each other. – AccidentalFourierTransform May 27 '18 at 13:25
  • yes, so both diagrams cancel each other. Other than these 2 diagrams are there any other possibilities? If no, so then this cross-section becomes zero since the amplitude goes to zero – SSS May 27 '18 at 14:43
  • For every dyagram you could need, it will involve at least a fermión loop with an evento number of atached photons, and you will have another dyagram identical to that, but with that fermion loop un the reverse direction, so it will have the opposite amplitude, and you'll need yo sum both of them, so they will give you zero. Because of that, the amplitude forma the process you are interested is a sum of zeros, so it's zero. There is no way out, it's a forbiden process. What do you not understand about the answer? – Iliado Odiseo May 28 '18 at 21:30
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What you propose is not scattering as one photon is annihilated. Scattering can not do that. Also such reactions are indeed not possible. For example the reaction $\gamma\rightarrow \gamma \gamma $ is disallowed.

Light by light scattering is known as Delbrück scattering. It involves an even number of currents.

my2cts
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