# Feynman diagram of annihilation-pair production question

• throneoo
In summary, virtual photons are allowed to exist in interactions because they do not have to follow the standard energy-momentum relation and do not have a classical motion in spacetime. They are simply a mathematical tool for calculating energy and momentum transfer in interactions, and should not be viewed as actual particles.

#### throneoo

Consider the lowest order interaction e.g. e- e+ -> virtual photon-> muon anti muon.

I appreciate that the electron-positron pair cannot annihilate into a real photon due to conservation of 4-momentum, but why is the pair permitted to produce a virtual photon? I know that virtual particles are "off the mass shell" insofar as energy,momentum and charge are conserved at the vertices and thus virtual photons are allowed to have non-zero mass, but how can I proceed from here to convince myself?

I could argue that since this virtual photon has non-zero mass, it can stay at rest in the COM frame unlike the massless real photons, but why do physicists represent this intermediate stage with a photon in the first place? I could also argue that since it only exist for a short duration of time Δt and can only be localized with uncertainty Δx=cΔt the HUP tells me there would be a minimum uncertainty in momentum Δp such that the virtual photon can appear to have zero momentum, but it doesn't sound very convincing either and that I might have misused HUP here.

throneoo said:
but why is the pair permitted to produce a virtual photon?
Why not? The 4-vector of the virtual photon can have arbitrary entries, here they are simply the sum of the entries of the incoming electron and positron.

The other quantum numbers are typical for a photon, so it has to be a photon (or Z, with additional interference between the two).

mfb said:
Why not? The 4-vector of the virtual photon can have arbitrary entries, here they are simply the sum of the entries of the incoming electron and positron.

The other quantum numbers are typical for a photon, so it has to be a photon (or Z, with additional interference between the two).
I was worried that the virtual photon need to move at the speed of light in all inertial frames, hence it would cause the same issue as real photons do. But it looks like all the properties of this virtual photon are dependent on the reactants

It is a virtual particle, it doesn't have a classical motion.

vanhees71
What do you mean and why?

In the case of electron collision, I understand that the virtual photons is just a means for the electrons to exchange energy and momentum, therefore this virtual particle carries whatever energy and momentum that is transferred, so the standard energy momentum relation need not apply.

On the other hand, the diagram of the annihilation-production process seem to imply there exists an intermediate stage where there's only one virtual photon hanging around until it decays into a pair of particles, which is different from the case above.

It is the same case, and you can even rotate the Feynman diagram to transform one into the other. Virtual particles don't have to follow the standard energy-momentum-relation, and they don't have a motion in spacetime in the classical way either.

vanhees71
The thing to remember is that virtual particles really are not particles. They are a convenient mathematical trick for computations, nothing more. Trying to view them as actual particles will generally lead to misconceptions and faulty interpretations.

vanhees71 and mfb