Conservation of electric charge in Feynman diagram

Here is a Feynman diagram showing the mutual annihilation of a bound state electron positron pair into two photons:

Is the electric charge conserved at the point A (or B)? What is the "charge" of the virtual fermion "moving" between A and B?

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The figure is not clear enough. Is that an e- on the left? – anna v Jan 5 '12 at 9:07
It is a Feynman diagram showing the mutual annihilation of a bound state electron positron pair into two photons. Electron is on the left side, positron is on the right. – Murod Abdukhakimov Jan 5 '12 at 11:44
Have the virtual particle make a a $1^{\circ}$ angle above the horizontal at point A, rather than be completely horizontal, and then the mystery disappears. Or you could do the same thing at point B. – Jerry Schirmer Jan 5 '12 at 17:17

The Feynman diagram is shown with time going up the y-axis. More commonly time is shown on the x-axis but it is not incorrect to put it on the y-axis so long as people notice.

The diagram shows the tree-level diagram for a negative electron annihilating a positive positron to produce two photons. Total charge is zero before and after so is conserved. The electron line is shown with an arrow that goes forward in time for the electron and backwards for the positron as if a positron is an electron going back in time. This was an interpretation that Feynman liked.

The central line shows the electron/positron line running horizontally as if it moved instantaneously from point A to point B. The diagram just represents a term in a perturbation series and is actually an integral over all possible positions and times for the two events A and B so in some cases it is a virtual electron moving from the point on the left to the point on the right, and sometimes it is a positron moving the other way. In all cases electron charge is conserved on all time slices since there is always one electron and one positron present (with or without photons) or only photons.

You can worry about what happens in the special case where the virtual positron/electron moves instantaneously from one point to another, but however you want to interpret it, this will make a measure zero contribution to the overall integral so it does not really matter.

When dealing with energy and momentum in Feynman diagrams the total is not always conserved at every point and this is explained by the uncertainty principle which allows virtual particles to avoid the conservation laws temporarily. You don't have to take this too literally. It is just an interpretation of an expression for a term with an integral in a perturbation series. What matters is that energy and momentum are conserved in the final result.

Electric charge does not obey an uncertainty relation in the same way as energy and momentum so it is conserved everywhere in the diagram shown.

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Sorry, Philip, but concerning your last sentence, energy and momentum are also preserved at every place of the diagram, despite their uncertainty relations. In the same way, it is not true that the electric charge doesn't obey any uncertainty relation. The phase of a charged field becomes more uncertain when the charge becomes certain, and vice versa. You are attributing incorrect reasons to various properties. – Luboš Motl Jan 5 '12 at 11:12
I'm upvoting because you have the right explanation in your third paragraph, definitely more so than the other answers as I write this, but Lubos does have a point that energy and momentum are explicitly and exactly conserved at all points in a Feynman diagram. – David Z Jan 5 '12 at 11:46
Perhaps its too long since I calculated a FD. Would it be true that the energy/momentum is conserved throughout when you do the calculation in momentum space, but not if you do it in position space? – Philip Gibbs - inactive Jan 5 '12 at 14:50
@Philip: The energy momentum is conserved in position space or momentum space, because you are integrating over all space, so the vertex is translationally invariant. – Ron Maimon Jan 5 '12 at 18:14
@Ron, it is conserved after the integral is done, but the question is whether it is conserved before the integral is done. When you do the calculation in momentum space you have a delta function at each vertex which enforces conservation of the energy momentum four vector. However the diagram above is labelled with space and time so it represents a Feynman diagram in position space. In this form each diagram represents space-time events connected by propogators in position space. There are no delta functions and diagrams where energy and momentum are not conserved contribute to the integral. – Philip Gibbs - inactive Jan 5 '12 at 23:32

The law of conservation of charge has both a field and particle manifestation. In the field manifestation, it states that the charge on any time slice is constant, and this is the answer that people normally give. But this answer interprets the charge in a causal picture, while the Feynman diagrams are acausal particle picture, and in a particle formalism, charge is conserved separately at every vertex.

The situation is exactly analogous to energy-momentum conservation. In a Hamiltonian formalism, energy is not conserved between two finite time slices, because intermediate states in perturbation theory can violate energy conservation by an amount restricted by the time-energy uncertainty principle. The two time-slices break time-translation invariance, and you do not have energy conservation.

But in the particle picture, you are taking an expansion of the S-matrix (or between vacuum insertions with no time boundaries), so that each vertex conserves energy and momentum exactly. This is not a contradiction, because the intermediate lines are off-shell. When you make them be on shell, and you restrict the motion between two time slices, you recover the ordinary Hamiltonian perturbation theory.

The notion of charge conservation at a vertex of the Feynman diagram is that the sum of the charges on the incoming lines must be matched by the charge of the outgoing lines vertex by vertex. The notion of incoming and outgoing here is by the arrows on the diagram, which simply represents the charge carried by the line, so that each incoming arrow must be matched by an outgoing arrow. These arrows do not represent the value of the momentum, which is a separate directional quantity on each line which, when it is on shell, tells you whether the particle is coming from the past or future.

When the particle is in an intermediate line, as in your example, the momentum is integrated over, and you can't covariantly separate the particle from the antiparticle. But the notion of charge entering or leaving a vertex doesn't care about the momentum, only about the arrows at the vertex. The arrow and the momentum can be aligned or antialigned, but whether it's an incoming particle or an outgoing antiparticle, the charge carried away from the vertex is the same. This is the reason that Feynman diagrams require viewing the antiparticle as the particle going back in time--- the two make inseparable contributions to each vertex.

The particle law of conservation of charge says that the number of incoming arrows (times their charge) must equal the number of outgoing arrows.

If you have a charged scalar of charge 3e, it can decay into 3 charged scalars of charge e. The Feynman diagram for this process involves a line with 3 incoming arrows turning into three lines with one outgoing arrow. This process is not paradoxical, it exists in the renormalizable scalar/gauge theory the appropriately charged scalars with a quartic interaction.

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If in your diagram the left is an electron, the diagram is supposed to be about two e- scattering, because if you write e+ going backwards, on the right, that is an e- going forwards in time.

The diagram then cannot be correct because two e- can only to first order interact with a photon and retain their identity.

If the arrow on the positron line goes positively with time, for a correct diagram, then the exchanged particle is an electron going to the right, or a positron going to the left. Conservation of charge at A because the backwards going electron is a positron, and at B because electron meets positron. The reason two photons are necessary for a valid diagram is because of momentum conservation, at CM of the two electrons the momentum is zero, whereas about 1Mev of energy is released as energy of the annihilation products.

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It depends on how the labels are meant to be interpreted. If the label denotes the kind of particle that one would actually see and associate with that line (a fairly common interpretation in my experience), then there's nothing wrong with the diagram. – David Z Jan 5 '12 at 11:48
@DavidZaslavsky I agree that it is a convention, but imo a simple "e" a weak ispin vector, would be sufficient, and the arrow would show the charge. If you name it e+ and it is moving against time then for consistency it is an e- entering the diagram. – anna v Jan 5 '12 at 15:14
@annav thanks for your comment. I actually copied the diagram from wikipedia article on annihilation. – Murod Abdukhakimov Jan 6 '12 at 5:51