muscaria said:
So if you end up with a real wavefunction, the gauge potential coupled to the charge must then be identical to the mechanical momenta along the streamlines (integral curves/thinking quantum Hamilton Jacobi formalism) traced out by every initial infinitesimal volume of the wave, right? If one is to describe any real non-stationary wavefunction that is. So that the gauge potential effectively serves as a velocity field for the wave?
Looks that way for Klein-Gordon-Maxwell, not sure about Dirac-Maxwell, probably not.
[QUOTE="muscaria]It's just I'm finding it a bit hard because usually the vector potential couples to the current and simply curves the path taken by the current without changing its magnitude, but you need current in the first place for this to take effect.[/QUOTE]
Let me note that current arises in the Maxwell equations as well, not just in the Klein-Gordon equation.
[QUOTE="muscaria]Or is the idea to find a condition for a dynamic scalar potential field which somehow constrains the phase to 0 everywhere, and the vector potential then serves as a mechanical momentum field for the wave?[/QUOTE]
I am not sure I quite understand this phrase. Scalar potential does not seem to define the phase uniquely: for example, you can change the phase by a constant without changing the scalar potential. Even the expression "scalar potential" is a bit strange in this context - I guess you mean a component of the 4-vector potential.
As for the idea... It is the same as in Dirac-1951 and Schroedinger -1952: the Maxwell equations in a certain gauge can describe both the electromagnetic field and the matter field.
[QUOTE="muscaria]Thinking again in H-J/de Broglie-Bohm terms, a real wavefunction would give the following scenario for some free space H in the non-relativistic limit:
Taking $$i\partial_t\Psi = \left[\frac{(\textbf{p}-\textbf{eA})^2}{2m} -e\phi\right]\Psi$$
and $$\Psi=Re^{i\theta}$$
where I'll take the phase to be constant over space and time after our calculation. Otherwise we end up simply with a continuity equation and no quantum H-J equation and we lose potential insight into a possible relation existing between the gauge potential and the quantum potential which would yield an effectively real wavefunction. So doing that the usual pair of equations of continuity and QHJ: $$\nabla\cdot(R^2\textbf{v})+\frac{\partial R^2}{\partial t}$$ $$\frac{\partial\theta}{\partial t}+\frac{(\nabla\theta -e\textbf{A})^2}{2m} -e\phi + Q =0$$ reduce to
$$\nabla\cdot\left(R^2\frac{\textbf{A}}{m}\right)=\frac{\partial R^2}{\partial t}$$ $$\frac{(e\textbf{A})^2}{2m} -e\phi + Q =0.$$
Is the last equation the gauge condition which gives rise to a real wavefunction? I'm not sure if what I've done is wrong, but if it isn't then it would seem like the quantum potential ##Q\propto\frac{\nabla^2R}{R}##, which depends on the form of the wave, and the electromagnetic potentials are constrained, giving rise to a situation where the form of the wave constrains the E-M field potentials which in turn dictates the current of the wave from the role taken on by the vector potential as a velocity field. Hmm...seems like a non-linear gauge condition in any case, with some effective feedback between wave and EM field?
I'm guessing carrying this out relativistically may lead to a more symmetric gauge condition, which must be what Schrodinger did for the K-G field? Any thoughts? Also, does this have any significant implications for the divergence of A? Is this complete rubbish? :p[/QUOTE]
I did not read this very carefully, but I don't like the idea of making the wave function real in a non-relativistic equation: on the one hand, it gives a gauge choice that is different from the choice required to make real the wave function in a relativistic wave equation, on the other hand, the relativistic equation is more realistic.