Feynman Diagram-Momentum conservation in primitive vertex

[Phys12]

In the first Feynman diagram, an electron comes in, emits a photon and then leaves. Is this an allowed process?

Because if you rotate the diagram by 90^o, the diagram should be just as valid, but it doesn't seem to be since it would violate the law of conservation of momentum. So is the primitive vertex by itself a valid phenomenon?

 

[PeterDonis]

In the first Feynman diagram, an electron comes in, emits a photon and then leaves. Is this an allowed process?

Do you mean, does this diagram, all by itself, describe something you can actually observe happening? No, it doesn't. No single Feynman diagram, by itself, ever describes something you can actually observe happening. Anything you can actually observe happening will be described by adding amplitudes derived from multiple Feynman diagrams (in principle an infinite number of them, but in many cases all but a manageably small finite number give contributions to the amplitude that are too small to make a difference).

 

[protonsarecool]

I don't think it is a good habit to think of Feynman diagrams in terms of actually "happening" dynamical processes. They are just useful shortcuts in a perturbative expansion of time-ordered correlation functions. One should first learn enough foundations of QFT (either in the functional integral or canonical formalism) to learn what that means, before learning Feynman diagram methods. Doing it the other way around is (in my opinion) recommendable.

Edit: In relation to your question: Why do you think that the first proposed process conserves momentum while the second does not?

 

[Phys12]

Edit: In relation to your question: Why do you think that the first proposed process conserves momentum while the second does not?

I know that the second process does not conserve momentum since the Feynman diagrams do not describe the physical motion of the particles and only give information about which particles went in, interacted and which ones came out. So, If I consider a particle-antiparticle pair approaching each other head on, then their momentum will be 0, but the photon emitted will not have a zero momentum, hence momentum will not be conserved.

I thought that the first process conserved momentum since moving charges produce photons and the diagram seemed to describe just that-- a moving electron emitting a photon.

 

[protonsarecool]

I thought that the first process conserved momentum since moving charges produce photons and the diagram seemed to describe just that-- a moving electron emitting a photon.

Classically, moving charges produce magnetic fields, sure. But notice that even then, a classical moving electron does not change momentum by "producing" a magnetic field (which is just the Lorentz transformed electric field anyway). The quantum version of this is slightly more complicated, in particular it is not described by a single diagram, but by an infinite family of diagrams.

In any case, if you think of the problem being in the rest frame of the electron in (1). you will see that momentum conservation is violated in both diagrams the same way.

 

[Phys12]

In any case, if you think of the problem being in the rest frame of the electron in (1). you will see that momentum conservation is violated in both diagrams the same way.

So, just to make sure I am imagining this correctly, I am an electron and I see a positron coming towards me, we annihilate to produce one photon, which would be a violation since there really should be two. Is that correct?

 

[Nugatory]

So, just to make sure I am imagining this correctly, I am an electron and I see a positron coming towards me, we annihilate to produce one photon, which would be a violation...

You can also consider the time reversal of that process: pair production from a single photon, which is only possible if some other heavy particle (typically an atomic nucleus) that can soak up the excess momentum is part of the interaction.

 

[Vanadium 50]

So, just to make sure I am imagining this correctly, I am an electron and I see a positron coming towards me

And now we know what the problem is.

Feynman diagrams are not cartoon representations of particle motion. (And even if they were, they would be in momentum space, not position space) They are mnemonic devices for calculating matrix elements.

 

[Phys12]

And now we know what the problem is.

Feynman diagrams are not cartoon representations of particle motion. (And even if they were, they would be in momentum space, not position space) They are mnemonic devices for calculating matrix elements.

Matrix of what? Is there a way to picture the momentum space?

 

[Vanadium 50]

Matrix of what?

Matrix elements in quantum mechanics. If you haven't seen that yet, you don't have the background to understand Feynman diagrams, which are a mathematical tool to calculate them.

 

[Phys12]

Matrix elements in quantum mechanics. If you haven't seen that yet, you don't have the background to understand Feynman diagrams, which are a mathematical tool to calculate them.

So will it be a good idea to take QM first and then read Griffith's Introduction to Elementary Particles?

 

[PeterDonis]

I thought that the first process conserved momentum since moving charges produce photons and the diagram seemed to describe just that-- a moving electron emitting a photon.

"Moving charges produce photons" is not the same as "we can directly observe a single moving electron emitting a single photon". We can't, since it's impossible for such a process to conserve energy and momentum. This is easy to see by writing down the equations for energy and momentum conservation in the original rest frame of the electron. The same goes for the reverse process of a single electron absorbing a single photon.

 

[Vanadium 50]

So will it be a good idea to take QM first

Yes. And Griffiths agrees with me. See his preface.