Question

If we arrange the Maxwell's equations in a fasion that all first-order time deriviatives are on the LHS, we obtain a time-evolution equation for E and B. Does this correspond to the Heisenburg's equation of motion in quantum physics for E and B? If yes, what is the form of the Hamiltonian?

Answer 1

i think it doesn't

Answer 2

In QM we deal with electronmagnatism by modifying the momentum operator and energy (eigen value of a hamiltonian) as follows:

\[\mathbf{p} \to \mathbf{p} - (e/c)\mathbf{A}\]
\[E \to E - e \phi\]

where $$e,c$$ are electron charge (negative) and speed of light, and $$\mathbf{A},\phi$$ are vector and scalar potentials from electronmagnatism. Then our hamiltonian is just,

\[\mathbf{H}=\frac{\mathbf{p}^2}{2m} +V_0 \to \mathbf{H} = \frac{(\mathbf{p}-(e/c)\mathbf{A})^{2}}{2m}+e \phi+V_0\]

For more info see Gauge invariance,

Answer 3

I guess to answer your question, no this does not correspond to the Heisenburg representation. Recall there are generally three forms of representation:
Schroedinger where the kets evolve in time, Heisenburg where the operators evolve in time, and interaction representation where they both evolve in time and the evolution is usually due to a perturbing interaction that is explicitly dependant on time.

Answer 4

It should, yes, if you assume no sources. I believe you should be able to deduce the Hamiltonian because the RHS of the Heisenberg equations should be something like the commutator of the operator on the LHS with the Hamiltonian.