# Expanding the electromagnetic hamiltonian

1. May 15, 2012

### 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

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

The square bracket can be expanded.

2. Relevant equations

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

2. May 15, 2012

### 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 $A\cdot \nabla$, because that must be acting on something. So the way to expand the Hamiltonian is to write it acting on a wavefunction, $\hat{H}f = \frac{1}{2m} (-i \hbar \nabla - q\vec{A})^2 f$; if you do that, you'll see where the factor of 2 came from (if you correctly apply the product rule).