[QM] Finding probability current from Hamiltonian and continuity equation

[NanakiXIII]

Homework Statement

Given the Hamiltonian

$$H=\vec{\alpha} \cdot \vec{p} c + mc^2 = -i \hbar c \vec{\alpha} \cdot \nabla + mc^2$$

in which $$\vec{\alpha}$$ is a constant vector. Derive from the Schrödinger equation and the continuity equation what the current is belonging to the density

$$\rho = \psi^{\ast} \psi$$.

Homework Equations

Schrödinger: $$H \psi = i \hbar \frac{\partial \psi}{\partial t}$$

Continuity: $$\frac{\partial \rho}{\partial t} + \nabla \cdot \vec{j} = 0$$

The Attempt at a Solution

Applying Schrödinger and the given Hamiltonian to the density gives

$$-i \hbar c \vec{\alpha} \cdot \nabla \rho + mc^2 \rho = i \hbar \frac{\partial \rho}{\partial t}$$.

The right hand side of the equation can be rewritten, by applying the continuity equation, and you end up with

$$\nabla \cdot \vec{j} = c \vec{\alpha} \cdot \nabla \rho - \frac{m c^2}{i \hbar} \rho$$.

After some tinkering I managed to turn this into

$$\nabla \cdot \vec{j} = \nabla \cdot c \vec{\alpha} \rho - \frac{m c^2}{i \hbar} \rho$$.

This is as far as I and my fellow students managed to get on this problem. Our teacher seemed to think that this was right, but that the second term,

$$\frac{m c^2}{i \hbar} \rho$$,

was supposed to be zero. We've had absolutely no luck, however, in proving any such thing and we don't see how that term should ever evaluate to zero. The hints our teacher gave us didn't get us any further, and I think we may have misinterpreted some of them because they led to very strange results. If anyone can give a hint or point in the right direction, or perhaps spot errors I made in the derivation so far (I hope it is clear enough without me providing the entire derivation), I would greatly appreciate it.

 

[NanakiXIII]

The problem has been solved. If anyone is interested in what I did wrong, I'll post it. If not, I won't go through the trouble of LaTeXing it.

 

[anoreen]

How to quantization photons
the electromagnetic field?