Energy of virtual photons

1. Sep 4, 2010

exponent137

I suppose that momentum of virtual photons is known precisely and energy of them is uncertain. But how uncertain? Maybe let us look the simplest example: scattering on heavy nuclei. Or, there are some more simple examples as scalar photons.

2. Sep 4, 2010

xepma

In any process you will encounter virtual photons that have all allowed momentum and energies. The reason is that in higher order perturbation theory one sums over all "possible" intermediate states. This includes so-called loop diagrams, in which photons or electrons or whatever essentially are created and annihilated with themself. The "weight" that is given to a certain intermediate state is given by the exponential of the action evaluated wrt that specific intermediate state: exp(iS). By summing over all possible intermediate states you get the total cross section, $$\int$$exp(iS).

The fact that these intermediate states sum over all allowed momentum is also the reason why perturbation theory beyond first order is divergent -- the virtual photons with very high energies give divergent contributions. One The trick is to get rid of these contributions in a systematic way, which is what renormalization means.

Now, one point that should be made clear, again and again: virtual photons do not exist. For one, they cannot be measured. But more specifically is that they resemble mathematical objects -- higher order terms in a perturbative approach. If we would have a non-perturbative way of determing cross-sections (which exist in some specific theories) we would never encounter these intermediate states and the whole idea of a virtual photon becomes obsolete.

3. Sep 5, 2010

exponent137

But scattering on heavy nuclei: here divergence does not exist?

"virtual photons do not exist": But calculations give their presence. They cannot be measured directly, but their consequences are measured.

"If we would have a non-perturbative way of determing cross-sections (which exist in some specific theories) we would never encounter these intermediate states and the whole idea of a virtual photon becomes obsolete."
But for now we do not have non-perturbative way and this perturbative way gives good results?!?! I think that working mathematics is blueprint of physical state.

4. Sep 5, 2010

tom.stoer

Virtual photons do not even exist mathematically as single particles with fixed energy and momentum; they exist only in an integral which integrates over dE d³p.

At every vertex energy and momentum is conserved.

In the loops virtual photons violate E²-p²=m², they are off-shell!

@xepma: I agree with you that virtual photons are mathematical artefacts. But one could turn things round and say that ONLY virtual photons can be measured (they are absorbed by the detector, so their lines of the Feynman diagrams end). Then real photons do NOT exist, as they are the asymptotic states only which are NEVER detected :-)

5. Sep 5, 2010

exponent137

So renormalization means that mass of electron also becomes infinite, so impact of self photons is finite.

But, what is average energy and uncertainty of it at scattering on heavy nuclei? Average energy is not zero, otherwise there would not be problems at quantization of gravity.