What is the suitable unitary operator for a rotating frame?

[AngeSurTerre]

Hello, I have a Hamiltonian that describes a particle in a rotating cylindrical container at angular frequency ω. In the lab frame the Hamiltonian is time-dependent and takes the form (using cylindrical coordinates)

\mathcal H_o=\frac{\vec P^2}{2m}+V(r,\theta-\omega t,z),

where V(r,\theta,z) is the potential due to the container. In the frame of the rotating container, the Hamiltonian is independent of time and takes the form

\mathcal H=\frac{\vec p^2}{2m}-\vec\omega\cdot\vec{\mathcal L}+V(r,\theta,z),

where \vec{\mathcal L}=\vec r\times\vec P is the angular momentum operator.

I'm looking for a unitary operator \mathcal U such that \mathcal H=\mathcal U^\dagger\mathcal H_o\mathcal U. The operator e^{-i\frac{\vec\omega\cdot\vec{\mathcal L}}{\hbar}t} does the job for the potential,

e^{i\frac{\vec\omega\cdot\vec{\mathcal L}}{\hbar}t}V(r,\theta-\omega t,z)e^{-i\frac{\vec\omega\cdot\vec{\mathcal L}}{\hbar}t}=V(r,\theta,z),

but it leaves the kinetic term unchanged. So what is the unitary operator that adds a shift of \vec\omega\cdot\vec{\mathcal L} to the kinetic term?

 

[George Jones]

I don't know the answer, and I don't think I have time to look into this, but ...

it leaves the kinetic term unchanged.

It seems to me that the kinetic term should not remain unchanged.

Consider two frames in constant relative motion. Classically, both the momentum and kinetic energy of a particle are different in the two frames. A rotating frame is a sequence of comoving frames. How can the kinetic term remain unchanged?

 

[AngeSurTerre]

It seems to me that the kinetic term should not remain unchanged.

There is no doubt that the operator e^{-i\frac{\vec\omega\cdot\vec{\mathcal L}}{\hbar}t} leaves the kinetic term unchanged. This is a straightforward consequence of the well-known relations [\vec P^2,\mathcal L_i]=0, i=x,y,z.

So what I'm telling is that the operator e^{-i\frac{\vec\omega\cdot\vec{\mathcal L}}{\hbar}t} performs a ``static" rotation of the system at time t, that is to say, it expresses all vectors in a coordinate system that has been rotated, but it does not perform a dynamic rotation in the sense that it does not change the total momentum of the system.

So I am looking for an operator \mathcal A that satisfies:

\mathcal A^\dagger\frac{\vec P^2}{2m}\mathcal A=\frac{\vec P^2}{2m}-\vec\omega\cdot\vec{\mathcal L}

 

[George Jones]

There is no doubt that the operator e^{-i\frac{\vec\omega\cdot\vec{\mathcal L}}{\hbar}t} leaves the kinetic term unchanged. This is a straightforward consequence of the well-known relations [\vec P^2,\mathcal L_i]=0, i=x,y,z.

So what I'm telling is that the operator e^{-i\frac{\vec\omega\cdot\vec{\mathcal L}}{\hbar}t} performs a ``static" rotation of the system at time t, that is to say, it expresses all vectors in a coordinate system that has been rotated, but it does not perform a dynamic rotation in the sense that it does not change the total momentum of the system.

So I am looking for an operator \mathcal A that satisfies:

\mathcal A^\dagger\frac{\vec P^2}{2m}\mathcal A=\frac{\vec P^2}{2m}-\vec\omega\cdot\vec{\mathcal L}

Yes, I understand that transforming from one set of fixed axes to a rotated set of fixed does not change the kinetic energy magnitude or the magnitude of the momentum, but, as you have noted, this is not what you need, as you need dynamical rotations.

Now, you have motivated me sufficiently that I have looked up classical Hamiltonians and Lagrangians in rotating frames, and I see the result in Goldstein, but I am clueless, as to what the appropriate quantum operator is, even at the infinitesimal level.

 

[George Jones]

As a student many years ago, I worked classically with time dependent rotation operators (matrices) that transform between an inertial frame and a rotating frame. I think that, because of the non-commutativity of the infinitesimal operators, the expression was something like the time-ordering of the exponential of the integral of infinitesimal operators? I completely forget the details.

 

[AngeSurTerre]

Thank you! I will look in this direction.

 

[atyy]

I googled "quantum mechanics in rotating frame" and found http://ptp.oxfordjournals.org/content/85/3/463.full.pdf, whose Eq 3.7 seems similar to what George Jones mentions in post #5.

 

[Haborix]

The Hamiltonian doesn't transform in such a simple way for time-dependent unitary transformations. To discover the proper transformation of the Hamiltonian, let $$|\psi(t)\rangle \rightarrow U(t)|\psi(t)\rangle$$ and substitue this into the Schrodinger equation.
$$i\hbar \frac{\partial}{\partial t}( \hat{U}(t)|\psi(t)\rangle)=\hat{H}\hat{U}(t)|\psi(t)\rangle$$
After some product rule and rearranging you recover the following equation:
$$i\hbar \frac{\partial}{\partial t}|\psi(t)\rangle=(\hat{U}^{\dagger}(t) \hat{H}\hat{U}(t)-i\hbar \hat{U}^{\dagger}(t) \frac{\partial \hat{U}(t)}{\partial t})|\psi(t)\rangle$$
That is, the Hamiltonian transforms like $$\hat{H} \rightarrow \hat{U}^{\dagger}(t)\hat{H}\hat{U}(t)-i\hbar \hat{U}^{\dagger}(t) \frac{\partial \hat{U}(t)}{\partial t}$$

I believe you will now find that your unitary transformation is the correct one and gives the desired result.

 

[AngeSurTerre]

Yes, of course, I forgot that because \mathcal U depends on time, the Hamiltonian that satisfies the Schrödinger equation is not just \mathcal U^\dagger\mathcal H\mathcal U.

Problem solved! Thank you very much!

 

[AngeSurTerre]

Hello, I'm coming back on this problem because I think there is still some confusion. Consider the ground state \big|\psi(t)\rangle in the rotating frame. Then the average of the angular momentum in this frame is given by

\big\langle\psi(t)\big|\mathcal L_z\big|\psi(t)\big\rangle

Now if I want to know this average value measured in the lab frame I can either transform the ground state to the lab frame and sandwich \mathcal L_z in between, or directly measure \mathcal U^\dagger \mathcal L_z\mathcal U in the ground state, which are equivalent. The problem is that [\mathcal L_z,\mathcal U]=0, so I will get exactly the same value in both frames... which cannot be! The two measured values of the angular momentum should differ by a constant proportional to \omega.

What am I missing?

 

[Haborix]

The time evolution in the rotating frame comes from the full Hamiltonian which contains an angle-dependent potential. Therefore, $$[\mathcal L_z,e^{-i \mathcal H t/\hbar}]\neq 0$$ I haven't thought about this aspect of the problem before, so don't trust me too much.

 

[AngeSurTerre]

I tried to think by analogy with the case where the system is in uniform translation instead of rotating. Then everything works well. The unitary operator of interest in this case is \mathcal U=e^{-i\frac{mv}{\hbar}\mathcal X}, and it transforms the momentum as expected, \mathcal U^\dagger\mathcal P\mathcal U=\mathcal P-mv.

This would suggest that the unitary operator for the rotating case would actually be something like \mathcal U=e^{-i\alpha\Theta}.

 

[AngeSurTerre]

I am more and more convinced that the operator \mathcal U=e^{-i\frac{\omega t}{\hbar}\mathcal L_z} is NOT the unitary operator that transforms from the lab frame to the rotating frame. This operator only performs a ``static" transformation to a rotated basis with an angle that depends on time, but this is not a ``dynamical" transformation to a rotating frame.

Let me explain what I mean with the example of a Galilean tranformation, for which there is no problem. Consider the one-dimensional Hamiltonian of a single particle in a potential, \mathcal H_o=\frac{\mathcal P^2}{2m}+\mathcal V(\mathcal X). And consider the operator \mathcal A=e^{i\frac{vt}{\hbar}\mathcal P}. It is straightforward to check that this operator performs a time-dependent translation of the potential:

\mathcal A^\dagger\mathcal V(\mathcal X)\mathcal A=\mathcal V(\mathcal X-vt).

However, it is clear that it does not perform a Galilean transformation, as it commutes with the momentum operator and leaves it unchanged:

\mathcal A^\dagger\mathcal P\mathcal A=\mathcal P.

On the other hand, the operator \mathcal B=e^{-i\frac{mv}{\hbar}\mathcal X} leaves the potential unchanged and performs a translation of the momentum:

\mathcal B^\dagger\mathcal V(\mathcal X)\mathcal B=\mathcal V(\mathcal X),
\mathcal B^\dagger\mathcal P\mathcal B=\mathcal P-mv.

Therefore the unitary operator that performs a transformation from the lab frame to a frame moving a constant velocity, that is to say a Galilean transformation, is given by:

\mathcal G=e^{i\frac{v}{\hbar}(t\mathcal P-m\mathcal X)}.

So you see that the \mathcal A operator only translates the coordinates as a function of time, but it does not change to the moving frame. And \mathcal A is the direct analog of \mathcal U=e^{-i\frac{\omega t}{\hbar}\mathcal L_z}. The latter rotates the coordinates, but it does not change the angular momentum, so it does not change to the rotating frame.

Actually, if we are interested at time t=0, it is sufficient to consider the \mathcal B operator, which is exactly the operator used to perform Galilean transformations in the article Superfluid density in continuous and discrete spaces: Avoiding misconceptions (http://journals.aps.org/prb/abstract/10.1103/PhysRevB.90.134503).

I just don't know how to generalize this to a rotating frame.

 

[Haborix]

This is an interesting problem, and I'd like to help you figure it out. I'm a little busy right now, but I will try and think through it when I get time. I still don't believe that commutator is zero since the Hamiltonian has angular dependence in the potential. With respect to your last post, have you tried letting $$m \rightarrow mr^2, \, \, v \rightarrow \omega, \, \, \hat{p} \rightarrow \hat{L}_z, \, \, \hat{x} \rightarrow \hat{\theta}$$ Good luck!

 

[AngeSurTerre]

Yes I tried. In that case, the unitary operator becomes \mathcal U=e^{-i\frac{\omega}{\hbar}(t\mathcal L_z+mr^2\theta)}. It does the expected transformation on the potential,

\mathcal U^\dagger V(r,\theta-\omega t,z)\mathcal U=V(r,\theta,z),

that is to say it removes the time dependence of the potential (since the axes are rotating with the container). For the kinetic term, let us consider the Laplacian in cylindrical coordinates:

\Delta=\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}+\frac{\partial^2}{\partial z^2}.

Then the kinetic term transforms as

\mathcal U^\dagger \frac{-\hbar^2}{2m}\Delta\mathcal U=\frac{-\hbar^2}{2m}\Delta-\omega\mathcal L_z+4i\hbar\omega\Big(1+r\frac{\partial}{\partial r}\Big)\theta,

for an infinitesimal \omega. I guess we have to integrate over something for the above equation to be valid for a finite \omega. But the point is that the last term in the above equation is "residual", since I expect the Hamiltonian in the rotating frame to be:

\mathcal H=-\frac{\hbar^2}{2m}\Delta-\omega\mathcal L_z+V(r,\theta,z).

I will try to derive the exact expression for a finite \omega.

 

[DrDu]

Use the BCH (Baker, Campbell, Hausdorff) formula

 

[AngeSurTerre]

Here is another very simple argument that supports the fact that \mathcal U(\vec\omega)=e^{i\frac{\vec\omega t}{\hbar}\cdot\vec{\mathcal L}} does not transform the system from the lab frame to the rotating frame. Consider a general Galilean transformation with velocity \vec v. We can easily check that the unitary operator \mathcal G(\vec v)=e^{i\frac{\vec v}{\hbar}\cdot(t\vec{\mathcal P}-m\vec{\mathcal R})} does the job:
\quad\mathcal G^\dagger(\vec v)\vec{\mathcal R}\mathcal G(\vec v)=\vec{\mathcal R}-\vec v t
\quad\mathcal G^\dagger(\vec v)\vec{\mathcal P}\mathcal G(\vec v)=\vec{\mathcal P}-m\vec v
Now, it is straighforward to check that \mathcal G(\vec\omega\times\vec{\mathcal R})=\mathcal U(\vec\omega). Thus, the operator \mathcal U(\vec\omega) appears as a particular case of a Galilean transformation, hence it cannot describe a transformation from the inertial lab frame to the non-inertial rotating frame.

 

[DrDu]

Nobody doubts that the operator you derived is correct. For the kinetic energy it implements the substitution $L_z \to L_z+mr^2\omega$.

 

[AngeSurTerre]

Nobody doubts that the operator you derived is correct. For the kinetic energy it implements the substitution $L_z \to L_z+mr^2\omega$.

But the problem is that the operator commutes with \mathcal L_z, so I measure the same angular momentum in the lab frame and in the rotating frame:
\quad\big\langle\mathcal L_z\big\rangle_{\textrm{lab}}=\big\langle\psi_o\big|\mathcal L_z\big|\psi_o\big\rangle
\quad\big\langle\mathcal L_z\big\rangle_{\textrm{rot}}=\big\langle\psi_o\big|\mathcal U^\dagger\mathcal L_z\mathcal U\big|\psi_o\big\rangle=\big\langle\psi_o\big|\mathcal L_z\big|\psi_o\big\rangle
How is that possible?

 

[DrDu]

But the problem is that the operator commutes with \mathcal L_z, so I measure the same angular momentum in the lab frame and in the rotating frame:
\quad\big\langle\mathcal L_z\big\rangle_{\textrm{lab}}=\big\langle\psi_o\big|\mathcal L_z\big|\psi_o\big\rangle
\quad\big\langle\mathcal L_z\big\rangle_{\textrm{rot}}=\big\langle\psi_o\big|\mathcal U^\dagger\mathcal L_z\mathcal U\big|\psi_o\big\rangle=\big\langle\psi_o\big|\mathcal L_z\big|\psi_o\big\rangle
How is that possible?

I have problems to follow your argument. U does not commute with $L_z$ so that
\big\langle\psi_o\big|\mathcal U^\dagger\mathcal L_z\mathcal U\big|\psi_o\big\rangle=\big\langle\psi_o\big|\mathcal L_z+mr^2\omega\big|\psi_o\big\rangle

 

[AngeSurTerre]

I have problems to follow your argument. U does not commute with $L_z$ so that
\big\langle\psi_o\big|\mathcal U^\dagger\mathcal L_z\mathcal U\big|\psi_o\big\rangle=\big\langle\psi_o\big|\mathcal L_z+mr^2\omega\big|\psi_o\big\rangle

Which \mathcal U are we talking about? I was talking about \mathcal U=e^{-i\frac{\omega t}{\hbar}\mathcal L_z}. It commutes with \mathcal L_z.

 

[DrDu]

I was talking about the operator you proposed in #15.

 

[AngeSurTerre]

I was talking about the operator you proposed in #15.

Yes, but then the problem is that it does not transform the Hamiltonian as expected. There are some residual terms, unless I made a mistake in my calculation. Or unless we can show that the exact residual terms just lead to a redefinition of the zero of the energy.

 

[DrDu]

In #16 I tried to give you a hint how to proceed. There are no residual terms besides the covariant replacement $\partial /\partial \theta \to \partial/\partial \theta +imr^2 \omega/\hbar$.

 

[AngeSurTerre]

In #16 I tried to give you a hint how to proceed. There are no residual terms besides the covariant replacement $\partial /\partial \theta \to \partial/\partial \theta +imr^2 \omega/\hbar$.

Then I guess I need to extend my background about covariance. Thank you!

 

[DrDu]

See here:
http://en.wikipedia.org/wiki/Gauge_covariant_derivative
The rotation acts in many respect like a gauge transformation due to a magnetic field. E.g. the flux tubes in a rotating superfluid are similar to the magnetic fluxt tubes in a type II superconductor.

 

[AngeSurTerre]

In #16 I tried to give you a hint how to proceed. There are no residual terms besides the covariant replacement $\partial /\partial \theta \to \partial/\partial \theta +imr^2 \omega/\hbar$.

I looked at the BCH formula you mentioned, and I still don't see how you can show that
\quad\mathcal U^\dagger\frac{\mathcal P^2}{2m}\mathcal U=\frac{\mathcal P^2}{2m}-\omega\mathcal L_z,
with $\mathcal U=e^{-i\frac{\omega}{\hbar}(t\mathcal L_z+mr^2\theta)}$.

 

[DrDu]

Use the BCH formula to decompose your transformation U into a product of two transformations V and W where V depends on Lz and W on $mr^2 \theta$. The transformation V commutes with the kinetic energy operator, so you are left with the transformation V. The only part of T which does not commute with W is the operator $L_z^2$. Use $WL_z^2W^\dagger=(WL_zW^\dagger)^2$ to calculate it. See e.g. post #7 in
https://www.physicsforums.com/threads/simplified-hausdorf-campbell-formula.445564/

 

[AngeSurTerre]

The only part of T which does not commute with W is the operator $L_z^2$

This is precisely the part that I don't understand! We have
$\quad\mathcal T=\frac{-\hbar^2}{2m}\Big(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}-\frac{\mathcal L_z^2}{\hbar^2 r^2}+\frac{\partial^2}{\partial z^2}\big)$,
and
$\quad\Big[\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r},mr^2\theta\Big]\neq 0$.
You are telling me that the above commutator is actually zero !?

 

[DrDu]

That's a good point indeed! :-)
Anyhow, at a first sight it looks as if only two commutators [T,W] and [[T,W],W] are nonvanishing.

 

[DrDu]

This article by Anandan looks interesting:
http://www.google.de/url?sa=t&rct=j...NvwHWRvPXsgZgwN7w&sig2=OUyguAjVkE7AQq2RXsnjaA

 

[DrDu]

I think that in contrast to what I said before, the correct unitary transformation is $U=\exp(i L_z \omega t/\hbar)$ and the correct transformation rule has been given already by harborix in post #6: While U commutes with the kinetic energy T, it does not commute with the time derivative in the time dependent Schroedinger equation, whose transformation yields the additional factor $L \omega$. Nevertheless I would find it rewarding to derive the transformation from that for Galilean boosts.

 

[DrDu]

Ok, I think I just figured out what is wrong with the Galilean argument: The Galilean boost contains a term $m \vec{v}\cdot \vec{r}=m (\vec{r}\times \vec{\omega})\cdot \vec{r}=0$ which vanishes for rotations as v is always perpendicular to r. So there won't be any term $mr^2 \omega$ and certainly not a term $mr^2\theta$.

 

[AngeSurTerre]

According to Phys. Rev. D 54, 4 (1996), there are misconceptions about this subject, and the correct unitary transformation involves a continuous product of infinitesimal rotations followed Lorentz boosts.

 

[DrDu]

I can't find this citation and I don't see why in a Galilean context, a Lorentz transformation should be necessary.

 

[AngeSurTerre]

I can't find this citation and I don't see why in a Galilean context, a Lorentz transformation should be necessary.

Sorry, I mixed up the information that I read. This is actually a citation of a message that I received from Pr. Jun Suzuki , one of the authors of arXiv:quant-ph/0305081.

Regarding your question: Yes, the unitary operator U=exp(-i\omega t.L) does not transform an inertial frame to a rotating frame. As we stated in our paper, the correct one cannot be written as a single unitary transformation: It should be sequences of the Lorentz transformation and rotation.

 

[DrDu]

Yes, that's more or less what they also have written in their paper. However, I am not convinced. In contrast to Lorentz boosts, Galilean boosts commute. But Anandan is quite renowned, so probably he has some sound arguments.

 

[DrDu]

I did some reading and some calculations and I am quite convinced now, that the operator $U=\exp(i \omega t L_z/\hbar)$ is the correct one to describe the transition to a uniformly rotating coordinate system. The transformed hamiltonian is
$H'=H_r+\frac{1}{2mr^2} (L_z-m\omega r^2)^2-\frac{m\omega^2 r^2}{2}$, where $H_r$ is the part of the Hamiltonian not depending on $\theta$ or it's derivatives.
I also find $L'_z=L_z$.
Note that the kinetic energy appearing in H' can also be written as
$\frac{mr^2}{2} v'^2_\theta$ as $v'_\theta=\dot{\theta}'=v_\theta-\omega$.
Hence both H' and L'_z coincide with the expressions given in Landau Lifshetz, Vol. 1 in classical mechanics.
Hence it is quite strange that Anandan and Suzuki take the equality $L'_z=L_z$ as an argument against the correctness of the transformation:
"

 

[AngeSurTerre]

I also find $L'_z=L_z$.

But how can the angular momentum be the same in both frames? Even classicaly, this is not correct!

 

[DrDu]

Landau lifshetz states it is correct

 

[AngeSurTerre]

Landau lifshetz states it is correct

This sounds like a proof "by authority".

 

[DrDu]

This sounds like a proof "by authority".

Admittedly, however, it is easy to see that it is correct. The angular momentum operator is defined to be the generator of an infinitesimal rotation: $f(\theta'+d\phi)=(1+i d\phi L'_z/\hbar)f(\theta')=f(\theta')+df/d\theta 'd\phi$, so $L'_z=\hbar/i d/d\theta'=\hbar/i d/d\theta$.

 

[AngeSurTerre]

Admittedly, however, it is easy to see that it is correct. The angular momentum operator is defined to be the generator of an infinitesimal rotation: $f(\theta'+d\phi)=(1+i d\phi L'_z/\hbar)f(\theta')=f(\theta')+df/d\theta 'd\phi$, so $L'_z=\hbar/i d/d\theta'=\hbar/i d/d\theta$.

Then I could use the same argument for the unitary operator $\mathcal G=e^{i\frac{v}{\hbar}(t\mathcal P-m\mathcal R)}$ that performs a Galilean transformation and conclude that $\mathcal G^\dagger\mathcal P\mathcal G=\mathcal P$, which is obviously wrong. It is clear that we actually have $\mathcal G^\dagger\mathcal P\mathcal G=\mathcal P-mv$.

In other words, the position operstor $\mathcal R$ is the generator of an infinitesimal translation in momentum space. I still believe that the angular momentum just performs a "static" rotation of the axes, but that it doesn't reflect the dynamical effect of the accelerated frame. In the example of the $\mathcal G$ operator, $\mathcal P$ performs the static translation of the axes, resulting in $x^\prime=x-vt$, and $R$ incorporates the dynamical effect resulting in $p^\prime=p-mv$.

 

[DrDu]

That's an interesting discussion and you bring up good points.
I think that, basically, we can't distinguish p from p-mv as mv is a constant and we can only measure changes of momentum (at least, in classical mechanics). What we can measure, is (angular)velocity, but it transforms correctly under the transformation U.
I am speculating when I say that the different transformation properties of momentum and angular momentum have to do with the fact that we can define uniquely whether a system is not rotating but can't define an absolute rest frame with respect to translations.

 

[AngeSurTerre]

Here is another trivial argument showing that $\mathcal U(t)=e^{-i\frac{\omega t}{\hbar}\mathcal L_z}$ does not transform the system from the inertial lab frame to the rotating frame. Simply consider the time $t=0$. Then we have $\mathcal U(0)=\mathcal I$, which means that the operator transforms from an inertial frame to... the same intertial frame!

On the other hand, if we consider again the operator that performs a Galilean transformation, $\mathcal G(t,\vec v)=e^{i\frac{\vec v}{\hbar}\cdot(t\vec{\mathcal P}-m\vec{\mathcal R})}$, then $\mathcal G(0,\vec v)=e^{-i\frac{m}{\hbar}\vec v\cdot\vec{\mathcal R}}$, and we have

$\quad \mathcal G^\dagger(0,\vec v)\vec{\mathcal R}\mathcal G(0,\vec v)=\vec{\mathcal R}$,
$\quad \mathcal G^\dagger(0,\vec v)\vec{\mathcal P}\mathcal G(0,\vec v)=\vec{\mathcal P}-m\vec v$,

which shows that, while the transformed frame coincides in position with the lab frame, it is moving at velocity $\vec v$. In the case of the above $\mathcal U$ operator, the transformed frame is not moving, since we clearly have $\mathcal U^\dagger(0)\vec{\mathcal P}\mathcal U(0)=\vec{\mathcal P}$.

 

[DrDu]

That's the same argument as before in other disguise. Obviously you get a change of velocity, but not of angular momentum.

 

[DrDu]

I think I now understand this:
When changing the coordinates in a Galilei transformation $x'=x+vt$ and $t'=t$ then the partial derivatives change as $\partial_x=\partial_{x'}$ and $\partial_t=\partial_{t'}-v\partial_{x'}$.
Hence the Schroedinger equation $i\hbar \partial_t \psi=-\frac{\hbar^2}{2m}{\partial_x}^2 \psi$ becomes
$i\hbar \partial_{t'}\psi=(-\frac{\hbar^2}{2m}(\partial_{x'}+imv/\hbar)^2+mv^2/2) \psi$.
However, due to Galilean relativity, we want the Schroedinger equation to have the same form in the frame moving with speed v than in the original system. This can be accomplished by including the gauge transformation
$\psi'=\exp(-imvx'/\hbar +i\hbar m v^2t'/2)\psi$ into the definition of the boosts, from which we also get $p'=p-mv$. I.e. if we demand p to vanish in one inertial frame if the velocity of the particle vanishes, we have to demand this also in a boosted system.

However, if we transform the Schroedinger equation into a rotating frame, we have no reason to expect it to be of the same form in the non-intertial system nor and in deed it is impossible to find a pure gauge transformation which would allow us to get rid everywhere of the additional terms in the rotating hamiltonian. Hence if we don't apply any gauge transformation at all, we get the rotating frame hamiltonian with it's strange centrifugal and corriolis terms, on the other hand, the angular momentum will also remain untransformed, i.e. $L_z=L'_z$.

This does not preclude the possibility to transform the hamiltonian locally. E.g. usually we disregard in our laboratories the fact that Earth is rotating and rotating around the sun. This can be formalized using an Inönü Wigner contraction, assuming that $r=R+r_0$ where R is the radius of the earth.
Formally, we can take $R\to \infty$ with $v=\omega R=const$. In the space of our small laboratory we can then apply a gauge transformations to get rid of spurious terms like $mv$ or the centrifugal potential $mv^2/2$.