Why do different vector potentials in Landau levels problem lead to different quantum mechanical ground state wavefunctions?

Question

Consider a charged particle (electron) moves in xy plane under a magnetic field pointing along the z direction, i.e., $\vec{B}=B\hat{z}$. As a consequence, we can write down three different gauges-

1. symmetric gauge: $\vec{A}=(A_x,A_y)=\frac{B}{2}(-y,x)$,
2. translational invariant gauge along x: $\vec{A}=B(-y,0)$, and
3. translational invariant gauge along y: $\vec{A}=B(0,x)$.

As previously addressed and asked in another post Charged quantum particle in a magnetic field, the answers were the three different ground state wavefunctions resulting from the above three different gauge choices are connected by linear transformations, and giving rise to same energy spectrum. Hence, everything is consistent. Working through the detailed derivation, for instance Hitoshi Murayama's Landau levels note, we see that

1. the symmetric gauge gives rise to ground state wavefuction: $\psi_n (z,\bar{z})=N_n z^n \exp\left(\frac{-eB\bar{z}z}{4\hbar c} \right)$, where $z=x+iy$ and $\bar{z}=x-iy$, while
2. $\vec{A}=B(0,x)$ gives rise to different ground state wavefunction (unnormalized): $\psi_0 (x,y)=\exp\left(ik_y y -\frac{eB}{2\hbar c}\left(x-\frac{\hbar c}{eB}k_y \right)^2 \right)$.

It is obvious that the two ground states are different, and thus $|\psi|^2$. My question is how our different choice of gauge, coming from the same physical magnetic field, results in different $|\psi|^2$, even though it leads to same energy spectrum?

Answer 1

Of course, observables such as $|\psi|^2$ have to be gauge invariant, and the two-wave functions of gives by the OP, in two different gauge, gives obviously two different probability distributions.

The resolution of the paradox comes from the fact that the two wave-functions describe in fact two different ground-states, as it can be seen as follows. $\vec A_1=\frac B2(-y,x)$ and $\vec A_2=B(0,x)$ are related by a gradient of the function $\chi=\frac B2 xy$, $\vec A_2=\vec A_1+\nabla \chi$, which implies that for a given wave-function $\psi_2$ translational invariant gauge along $y$, $\psi_2 =N\exp\left(ik_y y -\frac{eB}{2\hbar c}\left(x-\frac{\hbar c}{eB}k_y \right)^2 \right)$, corresponds a wave-function $\psi_1$ in the other gauge, $\psi_1=e^{i e \chi}\psi_2$. It is easy to show that $$ \psi_1=N' e^{k_y z-\frac{eB}{2}z^2}e^{-\frac{eB}{4}z\bar z} , $$ which can be rewritten as $$ \psi_1=f(z)e^{-\frac{eB}{4}z\bar z} , $$ and is indeed a ground-state wave-function in the symmetric gauge.

One has to remember that the wave-functions given by the OP are just two elements of two basis (corresponding to two gauges) that can describe the massively degenerate states of one particle in a magnetic field.

Edit: To clarify a little bit. The two wave-functions in the OP's question do not describe the same physical state. For a given $n$ or $k_y$, they are all valid ground-state wave-functions, and form a basis to describe the massively degenerated ground-state. But when changing gauge, one will not generally map from one basis state of one gauge onto one basis state of the new gauge, but it will generally be a superposition, as can be seen in the above example.

Answer 2

See old paper of Swenson in American Journal of Physics (magnifying on exactly the above issues) and a new resolution that will appear in a forthcoming paper by G. Konstantinou & K. Moulopoulos..