On Shankar’s Quantum Many body page 394 it says for one electron in a magnetic field, ignoring spin,


$$e\mathbf{A}=-\frac{\hbar}{2l^2}\hat{z}\times \mathbf{r}$$ where $l=\sqrt{\hbar/eB}$ is the magnetic length.

He then says that

$$H_0=\frac{\hbar^2 \mathbf{\eta}^2}{2ml^4}$$ where $\mathbf{\eta}=\frac{1}{2}\mathbf{r}+\frac{l^2}{\hbar}\hat{z}\times \mathbf{p}$ where $\eta$ is the cyclotron coordinate.

It seems like something was left out here. I’m not seeing how squaring this $\eta$ would give me back the original Hamiltonian. Am I missing something?

Any help greatly appreciated


1 Answer 1


The problem is two-dimensional. You thus don't have dynamical $z$ and $p_{z}$ components. Therefore, direct calculation proves




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