# Maxwell's equations in microscopic and macroscopic forms, and quantization

The macroscopic Maxwell's equations can be put in terms of differential forms as $$\mathrm{d}\mathrm{F}=0,\quad\delta \mathrm{D}=j\implies \delta j=0,\quad \mathrm{D}=\mathrm{F}+\mathrm{P}.$$ $\mathrm{F}$, $\mathrm{D}$, and $\mathrm{P}$ are 2-forms; $j$ is a 1-form. $\mathrm{F}$ is the $electromagnetic\ field$, which has 3-vector components $E_i,B_i$, usually called the $electric\ field$ and the $magnetic\ field$; $\mathrm{D}$ is a generalized electromagnetic field, generalized in the sense that $\mathrm{dD}\not=0$, which has 3-vector components $D_i,H_i$, called, amongst other names, the $electric\ displacement\ field$ and the $magnetizing\ field$; $j$ is a conserved 4-current; and $\mathrm{P}$ is the electromagnetic polarization/magnetization, which has 3-vector components $P_i,M_i$, usually called, up to a factor $\pm 1$, the $polarization$ and the $magnetization$, which in the macroscopic form of Maxwell's equations model the internal degrees of freedom, whereas in the microscopic form of Maxwell's equations we have $\mathrm{P}=0$. The Maxwell's equation Wiki page, together with its many links, does a reasonable job of summarizing all this.

In QED, there are internal degrees of freedom associated with the Dirac field, so should we be working with $\mathrm{D}$ when we quantize? Classically, given a differentiable bivector field such as $\mathrm{D}$, there is inevitably a conserved 4-current $\delta \mathrm{D}$, unless the equations of motion are such that this quantity is identically zero. Given an arbitrary 2-form such as $\mathrm{D}$, there is, also inevitably, a unique Hodge decomposition $$\mathrm{D}=\mathrm{dA}+\delta\mathrm{V}+\mathrm{U},$$ where $\mathrm{A}$ is a 1-form, $\mathrm{V}$ is a 3-form, and $U$ is a harmonic 2-form, satisfying $(\mathrm{d}\delta+\delta\mathrm{d})\mathrm{U}=0$. Looked at this way, the electromagnetic field $\mathrm{F}$ is just the component of the real object of interest, $\mathrm{D}$, that can be expressed as $\mathrm{F}=\mathrm{dA}$, for which, inevitably, $\mathrm{dF}=0$. The other component of $\mathrm{D}$, $\mathrm{D}-\mathrm{F}=\mathrm{P}=\delta\mathrm{V}+\mathrm{U}$, cannot be expressed in terms of a single potential, although $\mathrm{V}$ serves for the non-harmonic part.

If $\mathrm{D}$ is taken to be the real object of interest, with $\delta\mathrm{D}=j$ an independent degree of freedom, then does that mean that we should look for second-order differential equations for $\mathrm{D}$ in the first instance, probably in association with additional degrees of freedom, in view of the additional degrees of freedom that are present in the Standard Model?

I should point out that, for the purposes of this question, I take the Dirac wave function not to be observable in the classical Maxwell-Driac theory, with $\overline{\psi(x)}\psi(x)$, $(\delta\mathrm{D})^\mu=j^\mu=\overline{\psi(x)}\gamma^\mu\psi(x)$, $\overline{\psi(x)}\gamma^{[\mu}\gamma^{\nu]}\psi(x)$, $\overline{\psi(x)}\gamma^{[\mu}\gamma^\nu\gamma^{\rho]}\psi(x)$, and $\overline{\psi(x)}\gamma^5\psi(x)$ being (non-independent) 0-, 1-, 2-, 3-, and 4-form observables, respectively.

This is almost not a question, for which apologies, however I am interested in any critique of details or of whatever minimally elaborated plan there might seem to be to be here, or in any reference that seems pertinent. The ideas lie somewhat in the same territory as an earlier question of mine, Gauge invariance for electromagnetic potential observables in test function form, in that I prefer to disavow unobservable fields unless there are absolutely clear reasons why they are unavoidable.

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 The concept of "D" implies that there are material dipoles present. When you are talking about materials, you need a constitutive equation to relate F to D. In vacuum, you don't have a constitutive equation, and it is best to not introduce D at all, and only use F. So your basic idea that there is a fundamental D is not right. – Ron Maimon Jul 7 '12 at 4:43 @Ron: this question may be naive, but what about vacuum polarization? – Christoph Jul 7 '12 at 7:53 @Ron Thanks. That there apparently is vacuum polarization/magnetization was one of my starting points. Hodge decomposition uniquely defines $\mathrm{F}$ as a component of $\mathrm{D}$, so a solution of a second-order equation for $\mathrm{D}$ determines both $\mathrm{F}$ and the vacuum polarization/magnetization, which determines constitutive relations. One could imagine equations such as $(\mathrm{d}+\delta+m)^2\mathrm{D}=0$, the next to simplest possible classical equation [which admit Dirac algebra presentations because $(\mathrm{d}+\delta)^2\equiv\partial_\mu\partial^\mu$]. – Peter Morgan Jul 7 '12 at 11:36 @PeterMorgan: The thing about vacuum polarization is that it is Lorentz invariant, so that you only get one constant to describe it, called "Z" and it doesn't lead to a constitutive equation, or rather to the stupid constitutive equation F=ZD, and you absorb the Z factors in the renormalization process by defining the A which makes F=dA to produce a physical photon. I guess you are right that you can call Z a constitutive equation, and there is a Z running, so maybe the answer is best formulated in this direction – Ron Maimon Jul 7 '12 at 19:14