Change of reference frame for a wavefunction: same modulus but different currents?

Question

Suppose that, at a certain $t=0$, one has a wavefunction $$ \psi=\psi(x,y) $$ defined on a plane and well normalized to $1$. Coordinates (x,y) refer to the frame $xOy$.

How does the wavefunction change if, at time $t=0$, one jumps on a rotating frame $x\prime O\prime y\prime $ such that

1. The origin coincides with that of the initial frame (i.e, $O\,\equiv \, O^\prime$);
2. At time $t=0$, $x \, \equiv x^\prime$ and $y \, \equiv \, y^\prime$;
3. The second reference frame has angular velocity $\Omega$ with respect to the first one.

N.B. 1: I'm not interested in the wavefunction at different times, i.e. for $t>0$, but just in $$ \psi'=\psi'(x',y') $$ at $t=0$.

N.B. 2: I expect (please correct me if I am wromg) $\psi$ and $\psi^\prime$ to be such that $|\psi|^2=|\psi^\prime|^2$ but their phases should be different because the currents, e.g. the probability current $$ \vec{j}= \frac{\hbar}{2mi}(\psi^*\nabla\psi-\psi\nabla\psi^*) $$ should be different in the two frames.

Answer 1

Change of coordinate system in quantum mechanics can be easily done in Heisenberg approach of quantum mechanics. You can very easily transform(as I will demonstrate below) the Hamiltonian from one frame of reference to the other and then plug this new Hamiltonian in Schrodinger equation to find the transformed wavefunctions.

Hamiltonian in the rest/ground frame- $H$

Hamiltonian in rotation frame(rotating with angular velocity $\Omega$) - $H_{rot}$

$H_{rot} = UHU^\dagger -iU\frac{dU^\dagger}{dt}$

where $U$ is the transformation operator. $U = e^{-i\Omega tJ_z }$

else, $\phi_{rot}(t) = e^{-i\Omega t J_z}\phi(t)$

where $J_z$ is the $z$-component of total angular momentum for rotation about the $z$-axis with angular velocity $\Omega$.