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It seems that virtual photons also exist in vacuum. So the precise question is:

What is the additional virtual photon density due to the electric field of a unit charge?

Or: How many virtual photons per volume are found around a unit charge?

The answer will depend on distance, but what are the exact numbers?

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It is a strange question. I do not understand what you are going to do with this number. Let us say, there are $10^5$ virtual photons independently from distance. So what? – Vladimir Kalitvianski Feb 26 '11 at 13:45
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here is an excellent descrption of virtual particles -in particular, why we should NOT think of them as particles at all. – FrankH Oct 13 '11 at 13:07

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I do not think that this question has an answer. A photon is a quantum mechanical object.

a) there is no conservation of photons, virtual or not.

b)there is no lower limit to the energy of the photons, so in principle they are infinite ( infrared problem)

c)The energy of the virtual photons will depend on the motion of the charge and or the probe that will be checking for virtual photons.

You can have an energy spectrum for virtual photons, but not a density estimate . See this caluclation for example.

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your a) == there is no conservation of energy ;-) your b) and a photon with wavelength = Radius_of_univ takes how much time to imprint ? ;-) your c) is like: two warped drives at 0.9c : one captain say to the other : you have no energy ;-) – Helder Velez Feb 26 '11 at 21:18
@Helder Velez "no conservation of photons" means that the number is arbitrary, not the energy. See the link I provided. for b) who is imprinting virtual photons? for c)photons are not warped drives. – anna v Feb 27 '11 at 8:04
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It's infinity. This is the soft photon problem, which requires infrared regularization.

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IR regularization in order not to be infinity? Then it is not infinity? – Vladimir Kalitvianski Feb 26 '11 at 15:54
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Apparently you imagine a charge as a point with a Coulomb field around it. Outside the charge there is no charge but there is a field, you say, and it consists of virtual photons. So how many of them are at the distance $r$ from the charge?

I let the others answer this question and here I will give you my vision of that. The charge is not point-like but quantum mechanically smeared. It does not stay at a point but "moves" and creates a quantum mechanical "cloud". This "cloud" is "large and soft" (like a jelly) - you push the charge and the whole system gets deformed inelastically. You break the original state and the new state is a moving now charge and excited quantum oscillators (oscillating cloud). The latter describes photons. So normally the initial state is different from the final state. In experiment they usually add together all particular "pictures" of all scattering events and obtain an "inclusive" picture. It is the inclusive picture that corresponds to the point-like image of the charge.

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Ok. So what is the density of this "cloud"? Or is that not a meaningful question? – user346 Feb 26 '11 at 14:55
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It depends on the boundary conditions. If you have a cavity QED, the cloud size is finite. In a free space the cloud tends formally to infinity but in fact it is determined with the wave packet size (preparation device). It is similar to atomic cloud notion - it can be of any size. – Vladimir Kalitvianski Feb 26 '11 at 15:03
For an atomic cloud (say for Hydrogen) I can formally say that it has a density of "one electron per Bohr volume". The OP's question is about what can we say in a similar manner about the density of virtual photons (and other excitations) around an electron. Your answer doesn't tackle this problem. – user346 Feb 26 '11 at 15:20
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No, and I do not pretend to answer the OP question. Because there is no Coulomb center in space with "virtual" photons "around" it. The Colomb potential $1/r$ is self-sufficient, it should not consist of virtual photons. But the electron is not a point in space, it is omnipresent and not so simple - one needs infinite number of degrees of freedom to describe it completely. It moves as a whole and vibrates "internally", so the picture is quite different. – Vladimir Kalitvianski Feb 26 '11 at 15:28

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