Simple Feynman diagram

[5614]

Can someone just talk me through this Feynman diagram:

 

1) Firstly can I just check with that virtual positron... is that just how you draw one or is that a virtual positron going backwards in time (so it's an electron)?

 

2) Also that photon (top left) seems to have just split into an electron + virtual positron... why is there on real and one virtual particle? (ie. isn't it 2 real particles? - obviously not, but why?)

 

3) Can someone explain the thing about when you get 2 photons and when you only get 1?

 

Thanks in advance.

[swansont]

Positrons look like electrons travelling backward in time, so that's how they are represented in the diagrams.

 

The positron is virtual because it does not exit the diagram - it doesn't have to be this way, it just is for this particular diagram.

 

Not sure what you mean about 1 vs 2 photons - you can have as many as you want in these diagrams, but IIRC more photons represent a higher-order process, which has a smaller amplitude (i.e. less likely to occur)

[Meir Achuz]

1. FDs are in 4-momentum space. There is no time, so nothing is going back in time.

2. Virtual particles cannot leave the diagram, while real particles do.

A photon cannot decay into two real particles, because that process cnnot conseerve 4-momentum.

3. More photons could be prduced by just adding a photon to any lepton leg.

[swansont]

1. FDs are in 4-momentum space. There is no time' date=' so nothing is going back in time.

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I think you can draw them so that there is a time ordering. In the given diagram the time axis is on the bottom (I've also seen the vertical be the time axis, which adds to the "fun" of figuring out what's going on). In that case there would be a "sister" diagram that had the two vertices in the other order (i.e. the emission of the photon would precede the absorption)

 

You can also draw them so there is no time ordering. Here is more from the SLAC website.

[5614]

Thanks... what confused me with that positron was because I thought it could be a positron travelling backwards in time (which is an electron).

 

What I mean about the number of photons was answered in the Proton/Antiproton thread:

The number of photons will depend on the total spin of the electron/positron pair. They will produce 2 photons if their total angular momentum is 0' date=' and 3 photons if the total angular momentum is 1.

[/quote']

Is that true for all particle annihilations (when you add up the angular momentum)? And then there was the fact that 2 photons are needed in a magnetic field to conserve momentum.

[swansont]

Is that true for all particle annihilations (when you add up the angular momentum)? And then there was the fact that 2 photons are needed in a magnetic field to conserve momentum.

 

It stems from conservation laws. In the center-of-momentum frame, p=0, so having only one photon violates conservation of momentum. If a reaction is impossible in one frame, it is impossible in all frames. So you need two to conserve momentum.

 

The spin-0 state producing 2 photons and the spin-1 state producing 3 photons stems from parity.

[Severian]

Normally I would draw the internal electron/positron line as vertical, so that it would represent both an electron moving backwards in time (i.e. a positron) or an electron moving forwards in time. I just drew it this way because of the original context of the question.

 

Also, on Feynman diagrams, the arrow only denotes the direction of the fermion number flow (in this case, the direction of an electron).

[Severian]

The spin-0 state producing 2 photons and the spin-1 state producing 3 photons stems from parity.

 

Interestingly this will probably be how we show the Higgs boson is spin-0 (or at least not spin-1). If we see it decay into two photons we know that it cannot be spin-1. (Showing it is not spin-2 is a little more difficult.)