Expanding the electromagnetic hamiltonian

xahdoom

This isn't a homework problem - I can't understand a particular statement in my professor's notes. As such, I hope it's in the correct forum.

1. The problem statement, all variables and given/known data

The Hamiltonian for a charged particle in a potential field A is

[itex]\hat{H}[/itex] = (1/2m) ( -i [itex]\hbar[/itex] [itex]\nabla[/itex] - q A)[itex]^{2}[/itex]

The square bracket can be expanded.

2. Relevant equations

In my professor's notes, this expands to [itex]\hat{H}[/itex] = (1/2m) ( -[itex]\hbar[/itex][itex]^{2}[/itex][itex]\nabla[/itex][itex]^{2}[/itex] + q[itex]^{2}[/itex]A[itex]^{2}[/itex] + 2 q i [itex]\hbar[/itex] A[itex]\bullet \nabla[/itex] + q i [itex]\hbar[/itex] ( [itex]\nabla \bullet [/itex] A )

3. The attempt at a solution

When I attempt the expansion myself, I don't get the factor of 2 present in the 3rd term of the expansion. I know that it must be there - subsequent proofs using the Landau gauge don't work without it - but I don't understand where it came from.

Any help in understanding the reasoning behind this would be greatly appreciated.

Steely Dan

Always be careful when you have an operator like the gradient operator. It is not meaningful to just expand out the binomial product and write [itex]A\cdot \nabla[/itex], because that must be acting on something. So the way to expand the Hamiltonian is to write it acting on a wavefunction, [itex]\hat{H}f = \frac{1}{2m} (-i \hbar \nabla - q\vec{A})^2 f[/itex]; if you do that, you'll see where the factor of 2 came from (if you correctly apply the product rule).