# Paramagnetic term of the hamiltonian

1. Dec 27, 2009

### dd331

The Hamiltonian for particle in an EM field is

H = 1/2m (p - qA)^2 + q phi

If we take the cross-terms, which corresponds to the paramagnetic term, we have
$$H para = -q/2m * (p.A + A.p ) = iqh/2m * (\nabla .A + A.\nabla)$$

What I do not understand is how this simplifies into $$iqh/m * A.\nabla$$?

assuming that $$\nabla .A = 0$$ (i.e. Coulomb gauge). Why does the factor of 1/2 disappears? I'm only a first year undergraduate and I'm learning this on my own. I will appreciate it if you give a fuller answer. Thank you.

2. Dec 31, 2009

### clem

In QM, H is assumed to act on a wave function \psi.
This means that del.A really means del.(A \psi)=(del.A)\psi + (A.del)\psi, so the (A.del) comes in twice.