# Using the commutation relation [AB,C]=A[B,C]+[A,C]B canonical H

1. Mar 24, 2014

### AlexCdeP

Under the effect of an electric and magnetic field the momentum in the Hamiltonian becomes the canonical momentum, p-qA where p is the linear momentum and A is the vector potential so H=(1/2m)(p-qA)^2 + qV where V is the scalar potential. I am trying to find [H,(p-qA)].

My main question arises because as I expand the commutator out in different ways I seem to be getting different answers. For example if I begin with [H,(p-qA)]=[H,p]-q[H,A] I get a different answer to when I use go [H, (p-qA)] = (1/2m)[(p-qA)^2,(p-qA)] + [qV,(p-qA)] and use the relation mentioned in the title to get (p-qA)[(p-qA),(p-qA)] + [(p-qA),(p-qA)](p-qA) which i'm pretty sure is zero.

So what's going on? I can only assume that I have to expand out all the brackets for the operators first to get them individually and then use the relation in the title. If I'm right, then why?

2. Mar 24, 2014

### TSny

Yes, (1/2m)[(p-qA)^2,(p-qA)] is zero. If you don't use the relation in the title and expand out the terms individually, you should be able to show that all the terms cancel. But it's a tedious process.