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

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.

• \begin{aligned}x\mapsto x\cos \left( \Omega \,t\right) -y\sin \left( \Omega \, t\right) \\ y\mapsto x\,\sin\left( \Omega \, t\right) +y\,\cos\left( \Omega t\right) \end{aligned} – Eli Jun 23 '19 at 13:13
• Thank you for your comment. I don't think this transformation is enough... because at $t=0$ the two wavefunctions, $\psi$ and $\psi^\prime$ would coincide. I agree that the square moduli of $\psi$ and $\psi^\prime$ should coincide at $t=0$, but there should be something different in their phases. Namely, their phases should be different because the currents measured in the two frames should be different. – AndreaPaco Jun 24 '19 at 10:28

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 $$\Omega$$.

• If you take for example the harmonic oszillazor hamiltonian as an example, what is sigma z in that case? Also what would H rot be then? – lalala Jun 23 '19 at 7:00
• Thank you for your answer! Anyway, I fear you haven’t answered my question. I’m interested in how the wave function modifies, not in the Hamiltonian. Moreover, at t=0, the transformation you’ve proposed reduces to the identity, which -I think- is wrong. – AndreaPaco Jun 23 '19 at 9:39
• @AndreaPaco As rotation is quantized in quantum mechanics. Wavefunction in rotation frame will be $\phi_{rot}(t) = e^{-i\Omega t L_z}\phi(t)$. $L_z$ can be matrix if you have finite dimensions or $L_z=xp_y-yp_x$ if you have infinite dimensional Hilbert space. – Jitendra Jun 23 '19 at 11:21
• @lalala for quantum harmonic oscillator $U=e^{i\Omega t L_z}$ where $L_z= xp_y-y_px$. You can do the rest of the maths. – Jitendra Jun 23 '19 at 11:23
• I thank you again for your interest in my question. I'm sorry, but your explanation does not fully convince me. For me it's very strange that, at $t=0$ the wavefunctions in the two frames coincide. As I said (and you agreed with me) the angular momentum measured in the two frames should be different, also at $t=0$. – AndreaPaco Jun 24 '19 at 10:34