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I have a slightly out-of line question:

Consider a single electron (or it's current if you please) The STATIC ELECTROMAGNETIC field surrounding it will (no doubt) have a 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 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 $T \simeq F_{\mu\nu}F^{\mu\nu}$. But what can be said about the EM T of a single photon?

Line C: Your choice

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

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related: en.wikipedia.org/wiki/Virtual_particle –  DJBunk Feb 16 '13 at 0:29
    
That article is full of misconceptions. –  Frederic Brünner Feb 16 '13 at 8:27

3 Answers 3

It is an intriguing question if one can formally do this.

Classically the electric field might be considered as a limiting case of the electromagnetic radiation where the wavelength goes to infinity. I have answered a similar question here, where I refer to an analytic demonstration of how classical electromagnetic radiation emerges from a great ensemble of discrete photons.

Virtual photons by construction have all the attributes of real photons except the mass, which can be anything instead of zero. I do not know whether the formalism used to display the connection between classical electromagnetic waves and photon ensembles could be taken over in calculating static electric fields taking a limit of wavelength to infinity and virtual photons. Maybe you could ask the question at this site which is theoretically inclined.

A similar question on the electric field energy was asked at physics forums where the questioner had explicitly calculated the classical field energy of the electron, coming out with infinity because of the 1/r^2.

Quantum electrodynamics and perturbation expansion with renormalization techniques allow us to calculate quantities verified by experiments, ignoring the self energy problem.

Is there any way to express the 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.

But it is not a real photon, its mass can be anything so it is not represented by a consistent four vector as particles have to be. If it is a virtual photon carrying the interaction with another charge/field it will have the energy and momentum balance, but not the mass of zero. Virtual photons exist only within specific interactions.

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?

see above ( I do not know what FF is)

Line C: Your choice

As I said, since the self energy is an open question as far as I know this is also an open question . Possibly for a constant electric field, an operator formalism might be possible on the lines of the link given, i.e. when the energy is not infinite as for the point particle electron.

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Thx for your hints & links. I will come back to this soon. Btw. FF was supposed to be short for $F_{\mu\nu}F^{\mu\nu}$ for the LaTeX-lazy. –  NoEscape Jul 1 at 13:13

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 '13 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 '13 at 12:39
    
HOW is energy "formally assigned" to virtual photons? –  NoEscape Feb 17 '13 at 13:43
    
An internal line in a Feynman diagram corresponds to momentum/an integration over momenta. –  Frederic Brünner Feb 17 '13 at 13:56

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

No. The electromagnetic field is a physical system made of real photons not of virtual photons. The energy of the electromagnetic field is given by

$$H_\mathrm{rad} = \sum_r \hbar w_r a_r^\dagger a_r$$

And virtual photons have nothing to do with real photons

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The link is OK, but you contradict it and yourself... –  NoEscape Jun 23 at 14:40
    
@NoEscape How could it be? My answer states that EM field is made of real photons. And the link explains why the usual argument that real photons are in reality virtual doesn't hold up. –  juanrga Jun 25 at 10:42
    
An 'electromagnetic' field can be static or radiation! You only refer to the latter, which, I agree, is made of real Photons. I was refering to the static part. I made an edit to clarify. –  NoEscape Jun 26 at 14:06
    
@NoEscape The above Hamiltonian describes the whole electromagnetic field was either isolated and static or interacting with matter. I think you are referring to the Coulomb interaction, but this is considered part of the matter-field interaction. –  juanrga Jun 27 at 11:13

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