Why is there an arrow mediating a process in a Feynman diagram?

[Phys12]

In the following Møller scattering process, two electrons enter, exchange a photon and then leave (and if I understand this correctly, we say that both of the electrons emitted a photon).

However, in this case:

We have an electron scattering off a photon, but the interaction happens by an exchange of electron, is that correct? Why is it not the case that photon mediates the interaction between the photon and the electron since photons are the mediators of the EM force?

 

[Orodruin]

That would violate charge and Lorentz invariance in both vertices.

 

[Orodruin]

Also;

we say that both of the electrons emitted a photon).

No. The electrons exchanged a (virtual) photon. There is no emission of photons.

 

[Phys12]

That would violate charge and Lorentz invariance in both vertices.

Can you please elaborate on how exactly that will violate charge and Lorentz invariance?

 

[protonsarecool]

Can you please elaborate on how exactly that will violate charge and Lorentz invariance?

Take a photon-photon-electron vertex. You could have the time-ordering such that:

• electron spontaneously is destroyed/annihilated and 2 photons "come out"
• electron and photon interact and 1 photon comes out

Even as a virtual process (ie. at least one internal line coming out of the vertex), charge is destroyed, which is non-sensical.

The more technical explanation is of course, that there is simply no term in the QED Lagrangian which would generate such diagrams.

 

[PeterDonis]

Take a photon-photon-electron vertex.

There is no such vertex in the diagrams shown in the OP. The vertices in those diagrams all have two electron lines and one photon line meeting at them, so they are electron-electron-photon vertices, not photon-photon-electron vertices.

there is simply no term in the QED Lagrangian which would generate such diagrams.

The single electron-electron-photon vertex in itself is fine; it is directly generated by the ##i e \bar{\psi} \gamma^\mu A_\mu \psi## term in the QED Lagrangian. The issue is that this vertex can only occur if the photon line that comes from it is an internal line, not an external one, because a diagram that had three external lines, two electron and one photon, would violate charge conservation and Lorentz invariance.

 

[protonsarecool]

There is no such vertex in the diagrams shown in the OP. The vertices in those diagrams all have two electron lines and one photon line meeting at them, so they are electron-electron-photon vertices, not photon-photon-electron vertices.

Right, i have no idea how i could see any such diagrams in the OP. Guess i was tired. OP, please forget that i said anything.

 

[Orodruin]

There is no such vertex in the diagrams shown in the OP.

But the OP explicitly asks why those diagrams look the way they do instead of being mediated by a photon propagator. A photon propagator would imply a photon-photon-electron vertex instead of the normal QED vertex. The ”new” vertex would violate charge (net charge one) and Lorentz invariance (half-integer spin). I believe post #5 is completely appropriate for the OP.

 

[Phys12]

But the OP explicitly asks why those diagrams look the way they do instead of being mediated by a photon propagator. A photon propagator would imply a photon-photon-electron vertex instead of the normal QED vertex. The ”new” vertex would violate charge (net charge one) and Lorentz invariance (half-integer spin). I believe post #5 is completely appropriate for the OP.

Oh yeah. At first when the photon-photon-electron vertex was brought up, it threw me off too. Now that I see it as something shown to prove why it wouldn't make sense, I can see why you'd have a virtual electron mediating the process.