# Seeing gauge-covariance of the wavefunction for a uniform magnetic field

Consider the Hamiltonian $$H=\frac1{2m}(\mathbf p-q\mathbf A)^2+q\phi$$ and let $$\psi$$ be a solution to the Schrödinger equation $$i\hbar\frac{\partial\psi}{\partial t}=H\psi$$ Then if we gauge transform $$\phi\rightarrow\phi'=\phi-\frac{\partial\phi}{\partial t} \quad\quad \mathbf A\rightarrow\mathbf A'=\mathbf A+\nabla\Lambda$$ for any scalar field $$\Lambda$$, the corresponding Hamiltonian has a solution $$\psi'=e^{iq\Lambda/\hbar}\psi$$ This is the gauge-covariance of the Schrödinger equation for the minimally-coupled Hamiltonian, which I buy. I am trying to see this in practice for the case of a uniform magnetic field $$\mathbf B=B\hat{\mathbf z}$$. If I choose a gauge $$\mathbf A=-By\hat{\mathbf x}$$, I get the solutions $$\psi_n(\mathbf x) = e^{i(k_xx+k_zz)}e^{-\rho^2/2}H_n(\rho)$$ where $$\rho=\sqrt{\frac{qB}\hbar}\left(y+\frac{\hbar k_x}{qB}\right)$$ and the $$H_n$$ are Hermite polynomials. Now, this is manifestly gauge-dependent, which we can see in the asymmetry between the $$x$$ and $$y$$ directions. Moreover, if I had chosen instead the gauge $$\mathbf A=Bx\hat{\mathbf y}$$, I would have gotten the solutions $$\psi_n'(\mathbf x) = e^{i(k_yy+k_zz)}e^{-\rho'^2/2}H_n(\rho')$$ where now $$\rho'=\sqrt{\frac{qB}\hbar}\left(x-\frac{\hbar k_y}{qB}\right)$$ The gauge transformation taking me to this second case is $$\Lambda=Bxy$$, but clearly, $$\psi'_n(\mathbf x)\neq e^{iqBxy/\hbar}\psi_n(\mathbf x)$$ Perhaps more to the point, we learn as we solve that in the first gauge we have a harmonic oscillator in the $$y$$-direction, and in the second gauge it is in the $$x$$-direction. How can these be equivalent up to a gauge transformation? What am I missing?

Take the gauge $$\mathbf{A}=Bx\mathbf{y}$$ first. In this case, eigenstates are labeled not just by $$n$$, but also by $$k_x$$. In other words, one should really write $$\psi_{n, k_x}(\mathbf{x})$$. For each $$n$$, there is actually an infinite number (if the system is infinitely large) of degenerate eigenstates, all with energy $$\hbar \omega_c(n+1/2)$$.
Similarly, in the other gauge $$\mathbf{A}=-By\mathbf{x}$$, the eigenstates should be labeled as $$\psi'_{n, k_y}(\mathbf{x})$$. Again, fixing $$n$$ there is an infinite number of them.
The statement is then the gauge transformed wavefunction $$e^{iqBxy/\hbar}\psi_{n,k_x}(\mathbf{x})$$ is an energy eigenstate of the Hamiltonian in the other gauge, but it doesn't have to be exactly one of those $$\psi'_{n,k_y}$$. In fact, it must be an superposition of different $$k_y$$'s to create something localized in $$y$$.