Wavefunction in rotating frame

[Coffee_]

Hello, I have the following problem:

A system in the lab frame is described by a time dependent rotating potential $V(\vec{r},t)$.

So $H_{lab}=\frac{\boldsymbol{p}^{2}}{2m} + V(\vec{r},t)$.

My book says that the Hamiltonian in the rotating frame is given by

$H_{rot}=\frac{\boldsymbol{p'}^{2}}{2m} + V(\vec{r}) - w\boldsymbol{L'_z}$

Where $\boldsymbol{p'}$ and $L'_z$ are operators in the rotating frame. So to clarify, in this frame the potential is stationary.

(Let's take this as an ansatz! Assume this is the correct rotating Hamiltonian for all reasoning to follow.)

So in the rotating frame a $\Psi'$ will solve the Schrödinger equation with the adjusted Hamiltonian.

How to conver from $\Psi'$ to the wavefunction in the lab frame?Edit:

I see some papers proposing the following: (in hbar=1 units)

$\Psi'(\vec{r'},t)=e^{i\boldsymbol{L'_z}wt}\Psi_{lab}(\vec{r'},t)$

(1)

Let me show why I don't see how this is correct:

Writing out the time evolution for $\Psi'$ in the rotating frame Schrödinger equation gives:

$\Psi'(\vec{r'},t)=e^{-i\boldsymbol{H''}t + i\boldsymbol{L'_z}wt} \Psi'(\vec{r},0)$

Where $\boldsymbol{H''}=\frac{\boldsymbol{p'}^{2}}{2m} + V(\vec{r'})$

Now if $\boldsymbol{L'z}$ and $\boldsymbol{H''}$ commute which I'm pretty sure they do then:

$\Psi'(\vec{r'},t)=e^{i\boldsymbol{L'_z}wt} e^{-i\boldsymbol{H''}t} \Psi'(\vec{r},0)$

Since $\Psi'(\vec{r},0) = \Psi_{lab}(\vec{r},0)$ we have this final equation :

$\Psi'(\vec{r'},t)=e^{i\boldsymbol{L'_z}wt} e^{-i\boldsymbol{H''}t} \Psi(\vec{r},0)$

Eq(2)

Now now! Compare eq(1) and eq(2), it does look like:

$\Psi_{lab}(\vec{r},t)=e^{-i\boldsymbol{H''}t} \Psi(\vec{r},0)$

Which is a result I can not believe. If you take a look at $\boldsymbol{H''}$ it has a time INDEPENDENT potential in there. This would mean that to find the evolution in the Lab frame all we have to do is ignore the time dependence of the potential and just let it evolve as if the potential was stationary in the lab frame all along.

What is wrong?

 

[DrDu]

Your potential V(r') is not isotropic, hence the commutator with the operator L=L' can't vanish as L is the generator of rotations. So L' and H'' certainly don't commute.

 

[Coffee_]

Great. This was exactly what I thought a few hours after posting it (the non-commuting part) but thanks for confirming. So the unitary transformation I wrote down is correct for rotating frames in general? This means both the lab frame and the rotating frame agree upon the values of the WF in a certain point in real space. By this I mean, for example if the person in the lab frame points a laser pointer at a certain location and yells the value of te WF for me is ''...'' and the person in the rotating frame looks at this same spot that is lit up, he agrees with this value?

Sounds correct but I'd like to be sure.

 

[Coffee_]

Your potential V(r') is not isotropic, hence the commutator with the operator L=L' can't vanish as L is the generator of rotations. So L' and H'' certainly don't commute.

Hey DrDu. I don't mean to persist, but just a headsup to ask if you have seen my response? If you don't have time right now it's totally okay just making sure.

 

[DrDu]

Yes, this sounds correct to me. It is not too long that I understood this transformation to a rotating frame myself when I tried to give an answer in another thread after several wrong attempts:
https://www.physicsforums.com/threa...unitary-operator-for-a-rotating-frame.780800/