# Field energy of/from virtual Photons

I have a slightly out-of line question:

Consider a single electron (or it's current if you please) The EM field surrounding it will (no doubt) have an EM field energy (T) to go with.

The standard description of interaction is by exchanging a virtual photon. (For simplicity suppose, that only one virtual photon (of arbitrary momentum) can be exchanged)

Question: Is there any way to express the EM field energy "in terms of" virtual photons???

Line A: Every virtual photon has energy-momentum. The EM field energy-momentum is some weighted sum over all virtual photons.

Line B: A photons energy-momentum is very different from the EM field energy, because T ~ FF. But what can be said about the EM T of a single photon?

NOTE: Please note that I own a PhD in physics, so you can answer on any level you like.

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–  DJBunk Feb 16 at 0:29
That article is full of misconceptions. –  Frederic Brünner Feb 16 at 8:27

The assumption that one can express the energy of an electric field in terms of energy carried by separate virtual photons is fundamentally misguided. A virtual photon is a mathematical object that plays a role in calculating various physical quantities by making use of a perturbation series. One formally assigns energy and momentum to such objects, but what matters on a physical level is only the result of the calculation. Therefore it doesn't make much sense to talk about virtual photons in this way.

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Quite the expected reaction. Please think again. It is virtually IMPOSSIBLE that there is NO connection. One virtual photon (or the assembly of all possible virtual photons) for sure has energy-momentum that is not 0 and not nonsense. –  NoEscape Feb 16 at 12:19
As I said, energy is formally assigned to virtual photons, but the connection to physical quantities is obscure. Please prove your statement. –  Frederic Brünner Feb 16 at 12:39
HOW is energy "formally assigned" to virtual photons? –  NoEscape Feb 17 at 13:43
An internal line in a Feynman diagram corresponds to momentum/an integration over momenta. –  Frederic Brünner Feb 17 at 13:56
$$H_\mathrm{rad} = \sum_r \hbar w_r a_r^\dagger a_r$$