# Displacement current quantum mechanical interpretation?

while there are quite many classical explanations of displacement current to make Maxwell's equations work, see e.g. here: Displacement current - how to think of it , it sounds just a little bit like an accounting trick. Therefore, I was thinking about a more plausible explanation and came up with the following idea:

(1) The vacuum is quantum mechanically a seething sea of virtual particles and has defintely a electric polarizability. Examples for this are positron-electron generation by high electric fields by the Schwinger effect or by photon-photon collisions (Breit-Wheeler-effects) and as well as the Uehling effect (that has an impact on the Lamb-shift of the electron).

(2) A changing electric field in time will start to polarize the virtual (charged) particles, even in vacuum, - very similar to polarization effects in dielectrics. This "motion" of virtual charges in turn consitutes an effective "virtual" current that should have the same value as the displacement current from Maxwell's equations. And also, as soon as the electric field stops to change with time, the virtual polarization will stop changing and hence no effective virtual current will flow.

Is my interpretation at least somewhat correct? And if not where did I go wrong? Thanks a lot!

Or in other words (in case my interpretation is too wrong or too confusing):

How does the standard model describe the displacement current of Maxwell's equations?

The concept of displacement current in vacuum solved a specific problem with the EM law

$$\nabla \times \mathbf B = \mu_0 \mathbf j.$$ The problem was, this equation cannot describe situations where electric current is such that concentration/removal of electric charges from fixed volume is occuring. In such situations, $$\nabla \cdot \mathbf j$$ is not zero, but the above equation requires it to be zero.

By adding displacement current, we have the equation

$$\nabla \times \mathbf B = \mu_0 \mathbf j + \mu_0\epsilon_0 \frac{\partial \mathbf E}{\partial t}$$

where that deficiency is removed.

Any reinterpretation of displacement current as an actual electric current due to electric charge motion has to also explain how the original problem is to be resolved anew, or why it is not a problem.

In other words, what do we do with the first equation when in vacuum where there are no usual charges, only these hypothetical charges, and adding or removal of those charges from some fixed volume is occuring, so that

$$\nabla \cdot (\epsilon_0 \frac{\partial \mathbf E}{\partial t})$$

is non-zero?

• Unfortunately, I don't understand what you try to say with your last paragraph, since I think that the net charge of polarized virtual particles would still be 0. I'm also pretty certain - while no expert on the standard model - that Maxwell's equations do not contain the elements of perturbation theory needed for description of the interactions I sketched in my question. But while I think that my interpretation sounds plausible, I am not sure if perturbation calculations using standard model would give the right values for the B-field induced by time dependent E-field – Quit007 Oct 14 '18 at 20:10
• > "net charge of polarized virtual particles would still be 0" How do you know that density of such charge is zero everywhere? In general, in polarizable medium $\nabla\cdot \mathbf P$ does not vanish (usually, this happens on the surface of the dielectric). If you want to model vacuum as polarizable medium, why exclude this common situation? – Ján Lalinský Oct 14 '18 at 20:24
• I am not excluding it. The net charge of an ideal dielectric medium inside a capacitor is still 0, while the dipole moments of its atoms add up. In my interpretation the alignment process of microscopic dipoles (or the generation of dipoles and then their alignment) parallel to the E-field would be the same as the displacement current. – Quit007 Oct 14 '18 at 20:54
• Ok, I'll say it different way. If you pronounce displacement current density in vacuum as actually due to motion of polarizable medium containing new, hitherto ignored charges (and this is not how things work even in dielectrics), you are also getting back to square one in solving the above problem with the Ampere law. Now net current density is sum of current density of normal charges plus polarization current density of the new charges. So what happens when this net current has non-zero divergence? How do you fix the first equation? – Ján Lalinský Oct 14 '18 at 21:38
• I'm sorry, maybe I'm dense today, but why should ∇⋅(ϵ0∂E∂t) or the divergence of the net current be nonzero? More precisely: why should the divergence of the displacement current be nonzero? Because div(rot(B)) must be zero and if there are no real charges then the "real charge current" is zero and, consequently, the divergence of the displacement current, that is ∇⋅(ϵ0∂E∂t) must be zero, too. Do you think that using my interpretation the divergence would be non-zero? If so, could you please tell me how you arrive at that reasoning? Maybe this is what I'm missing... – Quit007 Oct 15 '18 at 8:29