# Charged quantum particle in a magnetic field - choosing a different gauge leads to different wavefunctions

Consider a charged quantum particle confined to the $xy$ plane, subject to a magnetic field $\mathbf{B}=B\hat{z}$.
The Hamiltonian is:

$$H = \frac{1}{2m} \left( \mathbf{p} - \frac{e \mathbf{A}}{c}\right)^2~,$$

where $\mathbf{A}$ is the vector potential, for which we have a freedom of gauge in choosing. One possible choice is $\mathbf{A} = B x \hat{y}$, which leads to the Hamiltonian:

$$H = \frac{1}{2m} \left[ p_x^2 + \left(p_y- m \omega_c x\right)^2\right],$$

where $\omega_c = eB/mc$ is the cyclotron frequency. Following the usual derivation of Landau quantization, we get the wavefunctions:

$$\psi(x,y) = f_n ( x- k_y / m \omega_c ) e^{i k_y y},$$

where $f_n$ are the eigenfunctions of the simple harmonic oscillator ($n=0,1,2...$).

However, we can also choose $\mathbf{A} = -B y \hat{x}$, which following the same lines would lead to the wavefunctions:

$$\psi(x,y) = f_n ( y+ k_x / m \omega_c ) e^{i k_x x}.$$

The two outcomes seems different by more then just a global phase, yet all we did was use our freedom in picking a gauge.
How does this work out?

No, you did another thing: choosing the complete set of commuting observables. In the first case you choose $\left\{H,p_y\right\}$ and the second $\left\{H,p_x\right\}$. The difference in operators ($p_x$ vs $p_y$) leads to difference in eigenvectors.