# Feynman diagram of electron and positron annihilation

1. Jul 16, 2012

### ianhoolihan

Hi all,

I am learning Feynman diagrams, and I have a quick question: in diagrams such as the one for electron-positron annihilation (see http://en.wikipedia.org/wiki/Feynman_diagram#Electron-positron_annihilation_example) why is it that the line "in the middle" is that of an electron/positron? (Either or, as, in the link above, it is horizontal.) This seems to imply to me that there is an electron/positron acting as a mediator between the two, which doesn't seem to make sense. To me, it would make more sense if there was no middle line, and it looked like an X.

Could someone explain if this is just a rule for how they're drawn, or if I'm missing something fundamental?

Cheers.

2. Jul 16, 2012

### booster

Hi

An internal line in a Feynman diagram represents the exchange of a virtual particle. In this diagram, there is a virtual electron / positron involved in the process. A virtual particle is a short lived excitation which is allows you to realize a sub-process which would normally violate energy and momentum conservation, so long as at the end of the process, the total energy and momentum have been conserved.

The reason this process must involve a virtual electron is as follows.

An electron and an anti-electron cannot annihilate to produce just a single photon, because this would violate energy / momentum conservation (try it, bearing in mind that the mass of the photon is zero). Hence the annihilation must produce two photons - in this case energy/momentum conservation can be satisfied (again I recommend trying it explicitly to see how).

However, in quantum electrodynamics, there are only 3-particle vertices. Hence the diagram drawn here is the simplest way to represent a process where an e+e- pair turns into two photons. You will note, if you try to balance energy and momentum at each vertex assuming the electron in the middle has the usual electron mass that you can't get it to work, just like e+e- to 1 photon didn't work. This is because the internal electron is virtual and not on its mass shell, that is, has energy and momenta that do not balance to give E^2-p^2=m^2. This is only allowed for virtual, or off-shell, particles.

3. Jul 16, 2012

### ianhoolihan

I still don't understand why it has to be a virtual electron/positron. I understand why one cannot have the annihilation of positron and electron to give one photon, but the reason you've given why I cannot simply draw an X shaped diagram (no virtual particle) is that QED deals only with 3-particle vertices. I haven't read into the actual calculations, but if they only work for 3-particle vertices, then I guess inserting a virtual particle is necessary for the calculations, though not physically. However, is there any empirical evidence to suggest that there is actually a virtual particle exchanged, and that it is not just a necessary tool to make the theory work?

Furthermore, your final paragraph does not require it to be an electron. Given energy conservation can be (temporarily) violated, why not make it some other particle? (It appears conservation of charge is not the issue, as the particle is both positive and negative, in a sense.) If it can be called another particle, then why not represent it by a (purely symbolic) line alone?

Actually, there must be something wrong with my argument, as I could make the same argument for any virtual particle --- and then what would distinguish the force mediating bosons from eachother?

Cheers

4. Jul 16, 2012

### booster

Ah, I think I see what your question is now.

QED is a gauge theory, and in a gauge theory, some guiding symmetry principal determines all the types of vertices allowed in its Feynman diagrams. It turns out that the symmetry principal which leads to QED, U(1) symmetry, produces only 3-particle vertices.

QED represents a subset of the known interactions in nature. Other forces, like the strong force, or quantum chromodynamics, do have 4-particle vertices. However, photons are gauge bosons of the electromagnetic force, not the strong force, so the the process e+e- -> photons involves the QED interactions and hence has only allowed QED vertices in its Feynman diagrams.

5. Jul 17, 2012

### Bill_K

This is (one of) your mistakes. Energy conservation cannot be violated, and is not, even temporarily. In a Feynman diagram, energy and momentum are exactly conserved at each vertex. What booster said was, the virtual electron represented by the internal line is not on the mass shell, and therefore violates the usual relation between energy and momentum, E2 = p2 + m2.

The reason the internal line must be an electron and not something else is that each vertex must represent a possible interaction, and there is no interaction that turns an electron into a photon plus "something else." There is an interaction that turns an electron into a photon plus another electron.

6. Jul 17, 2012

### ianhoolihan

Right, so the virtual particle is only necessary as a calculation tool for the Feynman diagrams to work? (I'm being deliberately inflammatory here, but I don't see why a virtual electron is necessary physically. In fact is there any evidence for virtual particles, given they blip in and out of existing? Are they just a heuristic way of describing force at a distance? Sorry, new thread probably.)

I understand that. This kind of gets to the virtual particle thing I just went on about...if a virtual particle is only presumed because there is a difference between an initial and a final state, then does it matter if it was an electron or some other virtual particle? They both exist only fleetingly, so we'd be none the wiser?

So, back to the point: why an electron as the virtual particle in electron positron annihilation? Charge isn't conserved at each vertex in the original diagram, so why not just have a virtual photon?

7. Jul 17, 2012

### ianhoolihan

Sorry, I was referring to the energy-time uncertainty relation. I thought this implied that something could have greater energy than it should (violating energy conservation) as long as the uncertainty relation was obeyed. Am I wrong?

I see where you are going with your point. I guess my counterpoint is what I was trying to say before --- given the virtual particle is never observed, why can't it be anything? While there may no be no such "interaction that turns an electron into a photon plus "something else" " is not the reaction $e^{+} + e^{-} = 2\gamma$, so whatever the something else is does not matter. (Again, I'm being a little naive intentionally, as I can't clearly give an answer to my own questions.)

8. Jul 17, 2012

### Bill_K

Yes.
The laws of physics still apply at each step, even though the intermediate products are not observed. All of the conservation laws (including energy conservation) must be obeyed at each vertex.

9. Jul 17, 2012

Because there is no photon-photon-electron vertex in QED. Also where is charge not conserved?

10. Jul 17, 2012

### booster

Just to clarify two things for ianhoulian:

Whether a virtual particle represents a brief violation of energy / momentum conservation, or the creation of a particle which no longer obeys the mass shell equation E^2-p^2=m^2 is, as far as I am concerned, just a question of semantics - but I chose the latter since I saw that people in this forum agreed to prefer that description and I didn't want to cause confusion.

Also, a guiding rule of quantum mechanics is, anything that can happen does, with some amplitude. There are also other processes, with a different set of particles exchanged, which also contribute to e+e- -> photons. However, because of the rules of the game, they have smaller amplitudes and are less likely. The total amplitude for the process can be written as a sum of many Feynman diagrams with different virtual stuff in the middle (there are in fact an infinite number of possible diagrams), and this one you are looking at is just the leading contribution.

11. Jul 17, 2012

### Bill_K

Ok, then you and I disagree on that point. Not just semantics, the "temporary violation of energy conservation" is a part of pop physics mythology.

Particularly when you come to general relativity, gravitation demands a locally conserved source, the energy momentum tensor, just as electromagnetism requires a locally conserved source, the current vector. It is absolutely inconsistent to imagine that in quantum mechanics, energy (or momentum) can be borrowed "if you're quick about it"!

How far do you think ianhoulian will get, calculating a Feynman diagram, if he does not accept that energy is conserved at the vertices?

12. Jul 17, 2012

### Staff: Mentor

Take the Higgs decay channel H->ZZ*->4 leptons, for example, assuming the 126GeV-boson is the Higgs. One Z has to be virtual, as the Higgs is not heavy enough to produce two real Z. This decay has been observed, and two leptons can be added to the correct Z mass, while the other two can be added to something less.
It is hard to interpret this as "force", similar to most processes of the weak interaction.

It does. If multiple different virtual particles (or multiple feynman diagrams) are possible, you have to consider them all. The total process is then sum of the individual diagrams, each with its own magnitude and phase.

Concerning "temporary violation of energy conservation":
"Temporary violation of the energy-momentum-relation"

13. Jul 17, 2012

### ianhoolihan

What I was originally was thinking was that if you take the vertices as 1) electron in virtual particle and photon out, and 2) positron in and virtual particle and photon out, then charge will only be conserved at one vertex e.g. if virtual particle is electron, won't be conserved at positron vertex. I guess the solution is then to say that virtual electron has a direction (as in original diagram), and is going in to the positron vertex, so net charge on both sides is zero. But, then it does depend which way you draw your arrow pointing, as to whether it is a virtual electron or virtual positron. Are these two different diagrams?

Ah, I feel a little silly now. Yes, the violation of conservation is what I hear in popular culture, but also from my lecturers as well! I knew the energy-time uncertainty relation was different, but I need to go and review it. A question though: when virtual particle pop in and out of the "vacuum" does this not violate conservation of energy? OR does "vacuum energy" have something to do with this?

Thanks for the example, though it is beyond me. Clearly I'm missing something about the distinction between "virtual" and "real" particle: is it that real particle obey the energy-momentum-relation, whereas virtual ones do not (are "off shell" I think the term is)?

Secondly, you say "if multiple different virtual particles ... are possible", does this mean there are different virtual particles possible? Or just the electron/positron (or are those distinct, as above?).

Finally, thanks for the clarification on energy conservation.

As for why the virtual particle must be an electron/positron, I guess my objection is more philosophical. If I have two apples which become two oranges, then it doesn't matter if I say they were pears or plums in between, as I can't measure that.

14. Jul 17, 2012

### Bill_K

Yes, that's the idea. Wikipedia has a picture of it. You place an arrow on each fermion (electron/positron) line segment; and at each vertex, charge conservation requires that one arrow must point in while the other points out.
Right again. These are considered two different ways of drawing the same diagram, and so it is just counted once.
If the initial state is vacuum the energy is zero, and if several particles are then temporarily created, the sum of their energies must still be zero. One or more of the virtual particles will have negative energy, but that's fine. As mfb said, virtual particles are off the mass shell and don't have to obey the usual relationship between energy, momentum and rest mass.
Again per mfb, if pears or plums were possible then you'd have to add all the possibilities together, and the overall result would be observably different.

15. Jul 17, 2012

### ianhoolihan

Oh, that was what I was getting at --- it is only counted once. So, in this case, there is nothing to distinguish between the virtual electron or positron?

This page helped http://en.wikipedia.org/wiki/Propagator#Propagators_in_Feynman_diagrams, and I'm guessing that the propagator takes care of what I just mentioned. A question though: does the propagator depend on what type of virtual particle the internal line represents? If it doesn't, then the internal lines are just "tools" and there type doesn't matter. (So I guess the propagator must depend on the type?)

Right: "off shell" can give negative energy too.

OK, we're thinking of this in different ways. I'm saying that since I only know the before and after, and can say whatever I want happened in the middle. Maybe theory constrains it, but how do we know if we can't observe it? I understand your point though --- if it were possible, there'd be a Feynman diagram, which you'd have to add.

16. Jul 18, 2012

### Staff: Mentor

Right.

Right. You can view it as an electron going one way or a positron going the other way - it is the same process.

It does. And the vertex depends on the type of virtual particle, too.

Take N electron/positron collisions in a collider. Theory predicts that M muon+antimuon-pairs are produced via a virtual photon. But now you observe more than this number! Either your theory is wrong, or you forgot the channel with a Z-boson. Add this in the calculation (and some other diagrams of higher order), and you get the correct result.

17. Jul 18, 2012

### ianhoolihan

This is just an example of needing to include all possible processes, correct? I think I finally see the point, however. If there was some other intermediary virtual particle that acted in the same way as the virtual photon and Z-boson, then some reactions would have to occur through this channel, and you would expect to see even more. Given you don't, there can't be this unknown process happening.

So, to reconcile my prior point of view, while we may never observe the virtual particle itself, the current theory (of which virtual particles are used etc) agrees so well with so many experiments that one can assume it likely enough to be correct.

18. Jul 19, 2012

### Staff: Mentor

At least with the current experimental precisions, yes.