# Quantum Cyclotron Frequency - Why is it off by a factor of 2?

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?

If you calculate $\vec B$ from your (corrected) vector potential $\vec A$ =(−By/2,Bx/2,0) you get the correct value for the cyclotron frequency ω and velocity. I suspect you have made a calculation error with the factors 1/2 in the expression for $\vec A$.