# Hamiltonian of a spin 1/2 particle in a constant mag. field

1. Dec 23, 2005

### quasar987

Why is there no kinetic energy term in said hamiltonian? Suppose I take a magnetic dipole in my hand, and I throw it in the field. Then surely its classical energy is E = p²/2m - $\vec{\mu} \cdot \vec{B}$.

Then why is the p²/2m term absent in the hamiltonian?

2. Dec 23, 2005

### abszero

There is a kinetic energy term if you are dealing with a free particle in a magnetic field, but that problem is slightly more complicated. The Hamiltonian is given by
$$\mathcal{H} = (\hat{\mathbf{p}}/ 2m - e/c \mathbf{A})^2 + \mathbf{\mu}\cdot \mathbf{B}$$
and so you have to solve this problem, which is a little more involved. If you're only interested in spin dynamics (for example if you're interested in dealing with a localized spin on a crystal lattice) then you don't worry about the kinetic energy term. Also, the kinetic energy term is in a different Hilbert space, so you can solve the two problems separately and combine them, if you really wanted to.

3. Dec 24, 2005

### quasar987

You mean you can solve the Schrödinger equation for each term separately and then combine the results? I certainly didn’t know that! What are the rules for combining these results?

4. Dec 24, 2005

### abszero

You end up with a direct product space where one part is a contribution due to the spin-coupling component of the hamiltonian, and the other is due to the kinetic energy term. So you end up with direct product eigenspaces that look like
$$\mathcal{H} |E_{KE} ; E_{\sigma} \rangle = (E_{KE} + E_{\sigma}) |E_{KE} ; E_{\sigma} \rangle$$
So it is in a way like breaking up a classical hamiltonian for multiple particles, where some of the degrees of freedom are not coupled. You can break the system up into parts and solve the individual parts, then put them back together.