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?
1 Answer
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 $y$.
