Can a static EM field consist of photons?

[fxdung]

We can analysis a static EM field into Fourier serie. Then we can consider a static EM field as a superposition of many running EM wave. So why we could not consider static EM field as a superposition of many photons(maybe virtue photons)?

 

[BvU]

What would be the fun of that ?

$\$

 

[fxdung]

Can it be applyed in QED?

 

[hutchphd]

This is more than I know:
https://en.wikipedia.org/wiki/Stati...le_exchange#The_Coulomb_potential_in_a_vacuum

 

[HomogenousCow]

You can certainly add a source term $J^{\mu} A_{\mu}$ to the EM lagrangian with a static source and see what becomes of the groundstate.

 

[tech99]

I think you will need a charge to be present for a static field to be set up.

 

[fxdung]

Of course we need a charge to be present, but I think static EM field itself consist of many photons.

 

[Demystifier]

We can analysis a static EM field into Fourier serie. Then we can consider a static EM field as a superposition of many running EM wave. So why we could not consider static EM field as a superposition of many photons(maybe virtue photons)?

The Fock basis of photon states is complete, so formally any quantum state of EM field can be expanded in the photon basis. And no, those photons are not virtual. But quantum states of EM field are not classical states of EM field. Quantum states obey uncertainty relations similar to position and momentum in nonrelativistic QM. This means that very certain electric field implies very uncertain magnetic field, and vice versa. The EM state corresponding to a static macroscopic charge is in a coherent state of photons in which both electric and magnetic field are uncertain, but their average values are equal to the classical values.

 

[PeterDonis]

We can analysis a static EM field into Fourier serie. Then we can consider a static EM field as a superposition of many running EM wave.

The fact that you can do this mathematically does not mean it has any physical meaning. There are an infinite number of ways to express any particular field in terms of a basis of functions (or states in QM).

So why we could not consider static EM field as a superposition of many photons(maybe virtue photons)?

Mathematically, you could; @Demystifier described one way to do it in QFT. But as above, the fact that you can do it mathematically does not mean it has any physical meaning.

 

[Demystifier]

But as above, the fact that you can do it mathematically does not mean it has any physical meaning.

Here is one possible physical meaning. In path integral formulation of QFT, one first derives the general path integral formula by inserting unit operator expressed in terms of eigenstates of field and its canonical momentum. Then one derives the Feynman rules, i.e. expands everything in the Fock basis. That's quite related (though not exactly the same) to expanding field eiegenstates into photon number eigenstates.

 

[vanhees71]

A static electric field is a semi-classical approximation for a very heavy charge with a light charge via the electromagnetic interaction at not too high energies (i.e., scattering energies much less than the mass of this particle). When calculating, e.g., the elastic scattering you realize that you have to resum an entire set of "ladder" diagrams with photon lines as rungs. The net result is that you describe the scattering process as the scattering of the light particle in the Coulomb field of the heavy charge.

More details about this and related topics about "soft photons" can be found in Weinberg, QT of Fields, Vol. 1.

 

[PeterDonis]

Here is one possible physical meaning.

A basis doesn't necessarily have any physical meaning; there are an infinite number of possible basis sets for any Hilbert space.

In the case of, e.g., scattering experiments, the physical arrangement of the experiment picks out a basis, but that basis, for, say, photon experiments, is not necessarily the Fock basis.

 

[HomogenousCow]

I messed around with this problem for a scalar field just now since it's an interesting question.
Consider a Klein-Gordon system with a perturbed Hamiltonian $$H = H_{0} + \int d^{3}x ~J(\mathbf{x}) \phi(\mathbf{x}),$$ where the source term is constant in time. Treating this perturbatively, the first order change in the groundstate will look something like $$|0\rangle ^{(1)} = \frac{1}{(2\pi)^3}\int d^3\mathbf{k} \frac{1}{2 E_{\mathbf{k}}} \frac{\langle \mathbf{k}| \int d^{3}x ~J(\mathbf{x}) \phi(\mathbf{x}) |0\rangle}{-E_{\mathbf{k}}} |\mathbf{k}\rangle,$$ note the first factor in the integrand which comes from the relativistic completeness-relation. The amplitude for this state to be in a one-particle state $|\mathbf{p}\rangle$ is then $$\langle \mathbf{p} |0\rangle ^{(1)} = \frac{1}{(2\pi)^3}\int d^3\mathbf{k} \frac{1}{2 E_{\mathbf{k}}} \frac{\langle \mathbf{k}| \int d^{3}x ~J(\mathbf{x}) \phi(\mathbf{x}) |0\rangle}{-E_{\mathbf{k}}} 2 E_{\mathbf{p}} (2\pi)^3 \delta^3 (\mathbf{p} - \mathbf{k}),$$ evaluating the delta function gives us $$\langle \mathbf{p} |0\rangle ^{(1)} = - \frac{\langle \mathbf{p}| \int d^{3}x ~J(\mathbf{x}) \phi(\mathbf{x}) |0\rangle}{E_{\mathbf{p}}}.$$ Using $\langle \mathbf{p}|\phi(\mathbf{x})|0\rangle = e^{-i \mathbf{p} \cdot \mathbf{x}}$ the above simplifies to $$\langle \mathbf{p} |0\rangle ^{(1)} = - \frac{\tilde J(\mathbf{p})}{E_{\mathbf{p}}}.$$ For a point source at the origin $\tilde J(\mathbf{p}) = 1$.
I'm sure you could do something analogous for a photon field.

 

[cmb]

We can analysis a static EM field into Fourier serie. Then we can consider a static EM field as a superposition of many running EM wave. So why we could not consider static EM field as a superposition of many photons(maybe virtue photons)?

My explanations are more from an engineer's rough and ready back pocket of explanations rather than refined maths, but maybe the way I look at it will help; EM radiation (photons) only manifest when there is an acceleration/state change of a charge.

i.e. No rate of change of position/velocity/energy of a charge = no EM radiation.

So taking that as a guide, a static field has nothing changing so no radiation (photons). The photons are the mediating particle for EM fields, but if there is nothing to mediate (there is no change in the EM field that needs to be mediated) then no photons.

 

[HomogenousCow]

My explanations are more from an engineer's rough and ready back pocket of explanations rather than refined maths, but maybe the way I look at it will help; EM radiation (photons) only manifest when there is an acceleration/state change of a charge.

i.e. No rate of change of position/velocity/energy of a charge = no EM radiation.

So taking that as a guide, a static field has nothing changing so no radiation (photons). The photons are the mediating particle for EM fields, but if there is nothing to mediate (there is no change in the EM field that needs to be mediated) then no photons.

This is not a take on things that aligns with how QFT actually works. Photons are just a basis for expressing the state of a quantized EM field, they make sense in the absence of any coupling to matter. A static EM field can be expressed in terms of a series of photon states the same way the hydrogen groundstate can be expressed in a momentum basis. Whether or not this means that the static field state “consists of photons” is a matter of language. I think one way we could justify this view is by looking at light by light scattering via an external field, one could interpret the scattered photon as having interacted with the photons making up the external field.