Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Say you have a magnetic field $\vec{B}=(0,0,B_0)$. Then the Schrodinger Equation Hamiltonian for a spin-2 particle of charge $e$ moving in this field is:

$$H = \frac{1}{2m}[\vec{p}-e\vec{A}]^2-\vec{\mu}\cdot\vec{B},$$

where $\vec{A}=(-\tfrac{1}{2}y,\tfrac{1}{2}x,0)$ is the magnetic vector potential.

You can find the speed by finding this commutator: $\frac{d\vec{r}}{dt}=i[H,\vec{r}]$.

When I did out this calculation, I ended up with $\frac{\vec{p}-e\vec{A}}{m}$. However, I'm told that I should have gotten $\frac{\vec{p}}{m}+\vec{\omega}_c\times\vec{r}$, where $\vec{\omega}_c$ is the cyclotron frequency. This has the value $\vec{\omega}_c=-\frac{e\vec{B}}{m}$, but if you plug that in, you get $\vec{\omega}_c\times\vec{r}=-\frac{2e\vec{A}}{m}$. Why is this off by a factor of 2?

share|cite|improve this question
Did you include the gyromagnetic ratio (which is not = 1 in quantum)? – honeste_vivere Jan 30 '15 at 0:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.