Landau levels in symmetric gauge

Question

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

$$H_0=\frac{(\bf{p}+e\mathbf{A})^2}{2m}$$

$$e\mathbf{A}=-\frac{\hbar}{2l^2}\hat{z}\times \mathbf{r}$$ 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

Answer 1

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

$$\left(\boldsymbol{p}-\dfrac{\hbar}{2l^{2}}\hat{z}\times\boldsymbol{r}\right)^{2}=\left(p_{x}+\dfrac{\hbar}{2l^{2}}y\right)^{2}+\left(p_{y}-\dfrac{\hbar}{2l^{2}}x\right)^{2}=$$

$$=\dfrac{\hbar^{2}}{l^{4}}\left[\left(\dfrac{1}{2}x-\dfrac{l^{2}}{\hbar}p_{y}\right)^{2}+\left(\dfrac{1}{2}y+\dfrac{l^{2}}{\hbar}p_{x}\right)^{2}\right]=\dfrac{\hbar^{2}}{l^{4}}\left(\dfrac{1}{2}\boldsymbol{r}+\dfrac{l^{2}}{\hbar}\hat{z}\times\boldsymbol{p}\right)^{2}=\dfrac{\hbar^{2}\eta^{2}}{l^{4}}$$