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A massless spin $1$ particle in 4D has 2 degrees of freedom. However, we usually describe it using four-vectors, which have four components. Hence, somehow we must get rid of the superfluous degrees of freedom. This job is done by the Maxwell equations. To quote from Gilmore's "Lie Groups, Physics, and Geometry"

“In some sense, Maxwell's equations were a historical accident. Had the discovery of quantum mechanics preceded the unification of electricity and magnetism, Maxwell's equations might not have loomed so large in the history physics. … Since the quantum description has only two independent components associated with each four momentum, there are four dimensions worth of linear combinations of the classical field components that do not describe physically allowed states, for each four momentum. Some mechanism must be derived for annihilating these superpositions. This mechanism is the set of equations discovered by Maxwell. In this sense, Maxwell's equations are an expression of our ignorance.”

Is there an analogous interpretation for the Dirac equation and possibly other equations of motion?

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  • $\begingroup$ Firstly, the author seems to be describing the role of Maxwell's equation for light--or "photons", in this context. The classical analogue of photons are E&M waves which are solutions to the vacuum Maxwell equations, but the sourced equations describe also massive charges which surely have more than 2 degrees of freedom. Secondly, the Dirac equation describes particles of ("fundamental") spin 1/2 which is a distinctly quantum phenomenon--unless your notion of "classical fields" extendeds to ones which cannot be represented by complex numbers (en.wikipedia.org/wiki/Grassmann_number) $\endgroup$ – user143410 Sep 27 '17 at 15:18
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  1. There are several things wrong/imprecise in the initial claims of the question about electromagnetism: In the Lagrangian formulation where the four-potential is the fundamental dynamical variable, the Maxwell equations are not equations of motion - only half of them are. The other half is just a consequence of $F=\mathrm{d}A$ and $\mathrm{d}^2 = 0$, and holds off-shell. Of all of Maxwell's equations, only Gauß' law represents a constraint related to gauge degrees of freedom. The other degree of freedom (the one also having to be eliminated for a massive vector field with three degrees of freedom) is eliminated by a constraint setting the timelike Hamiltonian canonical momentum to zero, which has nothing to do with Maxwell's equations (it's simply the almost trivial statement that $F^{00} =0 $).

  2. To answer the actual question: Dirac's equation is not an equation for a gauge theory, there are no gauge degrees of freedom to be eliminated. In general, it is not the equations of motion that eliminate gauge degrees of freedom. In the Lagrangian formalism, one has to introduce explicit auxiliary gauge fixing terms to eliminate them, which would admittedly lead to "equations of motion" that kill off the superfluous degrees of freedom. In the Hamiltonian formalism, the gauge degrees of freedom are eliminated by constraints incurred from the non-invertibility of the Legendre transform for gauge theory, cf. e.g. this answer of mine.

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The four polarisation degrees of freedom of the EM field correspond to the four DoFs of the charge-current vector. One is required for charge, two are for photon polarisation and the third is eliminated by the Lorenz relation, which corresponds to charge conservation.

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The answer to the question is: Yes!

Electromagnetic interactions are mediated by photons. Since a photon is massless it only has 2 physical degrees of freedom (there is no longitudinal polarization). Historically, electromagnetic interactions were described in terms of electric $\vec E$ and magnetic fields $\vec B$ or alternatively, in terms of the field strength tensor $F_{\mu \nu}$. Thus, in total 6 variables are used to describe only 2 physical degrees of freedom. Maxwell's equations are simply statements about the redundant components, similarly to how $x^2+y^2=r$ is a statement for the redundant variables $x$ and $y$ if we describe a two-dimensional rotationally symmetric system.

A nice discussion can be found at page 444ff in Spin in Particle Physics by Elliot Leader

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  • $\begingroup$ It would be much better if this answer actually explained why the answer is supposed to be yes instead of referring to a book. $\endgroup$ – ACuriousMind Oct 21 '17 at 13:53

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