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Expanding the electromagnetic hamiltonian

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xahdoom
#1
May15-12, 10:10 AM
P: 3
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.
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Steely Dan
#2
May15-12, 10:30 AM
P: 317
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).


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