The 3 fictitious forces of the rotating frame in Quantum Mechanics I am wondering how the 3 typical fictitious forces (centrifugal, Coriolis, Euler) typical of a rotating frame manifest themselves in Quantum Mechanics.

Background (classical point particle): we have an inertial frame centered in $O$ and a rotating one centered at $O'$, such that the axis are mutually oriented as $\hat{{\bf e}}_i = R \, \hat{{\bf
 e}}'_i$, where $R$ is a rotation matrix and $i=1,2,3$. Given a point  $\bf x$ as seen by $O$, we have ${\bf{x}} = {\bf{r}}+R \, {\bf{x}}'$,  where $\bf r$ is the position of $O'$ measured by $O$.  Now, we can introduce the "angular velocity matrix" $W=R^{-1}\dot{R}$, namely  $\dot{R} = R \,W$ and  $\ddot{R} = R(W^2+\dot{W})$. In this way, for a given vector  $\bf u$, we have that $W{\bf u} ={\bf w} \times {\bf u}$, where ${\bf w}$ is the usual "angular velocity vector" associated to that fact that $R$ may have a temporal dependence ($W$ and $\bf{w}$ are related by Hodge duality).  Now, we just have to take temporal derivatives ($\bf{r}$ is constant):  $$ \dot{{\bf x}} = R (\dot{\bf x}' + W{\bf x}')  $$ $$ \ddot{{\bf x}} = R (\ddot{\bf x}' + 2 W\dot{\bf x}'+W^2{\bf x}'+\dot{W}{\bf x}')  $$ where the last term $\dot{W}{\bf x}'$ is the so-called "Euler force" (it is less famous than Coriolis and the centrifugal because you need a non-constant angular velocity of the rotating frame). Setting $R=1$ at the given time, the above equations read $$ \dot{{\bf x}} =  \dot{\bf x}' + {\bf w} \times {\bf x}'   $$ $$ \ddot{{\bf x}} =  \ddot{\bf x}'+ 2 {\bf w} \times \dot{\bf x}'+ {\bf w} \times ( {\bf w} \times {\bf x}')+\dot{ {\bf w}  }\times{\bf x}'  $$ namely, $$ \ddot{{\bf x}} = \ddot{\bf x}' +``Coriolis"+ ``centrifugal"+ ``Euler"  $$

Question: How do ''Coriolis'', ''centrifugal'' and ''Euler'' manifest themselves in Quantum mechanics? Assuming, for simplicity, a spin-$0$ wave function, I expect the final answer to be consistent with the classical-field-theory result for a complex scalar field described in Section V here.
Consideration #1: The combined effect of the 3 fictitious forces should somehow be already present in the Schrodinger equation (not under the direct form of "forces"). We should find something resembling the classical equations above when the Ehrenfest theorem is applied, or when working in the Heisenberg picture (in particular, I am thinking about the time derivative of the momentum operator: in this case, some "fictitious force" operator should appear). A concrete example of a system subject to apparent forces in QM is the rotating oscillator, see this paper.
Consideration #2: the change of frame (to an inertial or a non-inertial one) should preserve the probability, and so it should be implemented by means of a unitary transformation $U_t$, which is basically a rotation parametrized by time. If the rotation axis is not constant, $U_t$ should be expressed in terms of a T-ordered exponential, otherwise a simpler
$$
U_t = e^{\frac{-i}{\hbar} L_z \int_0^t \Omega(t') dt'}
$$
could do the job (assuming that the non-inertial frame is rotating along the $z$-axis). Now we can start from the Schrodinger equation in the inertial frame,
$$
i \hbar \partial_t \psi( {\bf x} ,t) = H \psi( {\bf x} ,t)
$$
and obtain
$$
i \hbar (\partial_t \psi' + \psi' U_t \partial_t U_t^*)= H'  \psi'
$$
where $ \psi ' = U_t    \psi $ and $H' = U_t H U_t^* $. So, there is an extra term related to $ \partial_t U_t $, that we usually don't have when we perform a time-independent rotation (some sign may be wrong, this is just to give the idea). The Euler effect is probably encoded (also) into the term $U_t \partial_t U_t^* \propto L_z \Omega(t)$. Please correct me if my line of reasoning is wrong (I see the analogy with this answer).
Related: clearly, if $U$ is not a time-dependent rotation but a boost, then we're just moving from one inertial frame to another (see this, this and this or these notes).
 A: Here is one approach:

*

*The Lagrangian for a point particle in an accelerated reference frame is$^1$
$$ L ~=~\frac{1}{2}m\vec{v}^2+m\vec{v}\cdot (\vec{\Omega} \times \vec{r})-V(\vec{r}),$$
where
$$ V(\vec{r})~=~m\vec{A}\cdot \vec{r} -\frac{m}{2} (\vec{\Omega} \times \vec{r})^2,$$
cf. my Phys.SE answer here.


*So the Hamiltonian becomes$^1$
$$H~=~ \frac{1}{2m}(\vec{p}- m\vec{\Omega} \times \vec{r})^2 + V(\vec{r}). $$


*Next write down the TDSE.
--
$^1$ If the angular velocity $\vec{\Omega}$ of the reference frame depends explicitly on time, then the Lagrangian $L$ and the Hamiltonian $H$ depend explicitly on time, and there will be an Euler force.
A: In a rotating frame, the Hamiltonian picks up an additional term proportional to $\vec\omega\cdot \vec L$.  Thus typically
\begin{align}
\hat H=\frac{p^2}{2m}+V(r)-\vec\omega(t) \cdot \vec L\, ,
\end{align}
and the extra term completely based on classical mechanics.
This type of additional non-inertial term comes up quite a bit in the study of nuclear and molecular rotational bands.
Indeed, quoting from [1]:

nuclei provide a unique laboratory to study
rapidly rotating quantum systems under strong Coriolis and
centrifugal fields.

The Coriolis term is often held to be the source of "backbending", a change in the coupling scheme with noticable effect in the rotational spectra.  [2] contains a simple and short discussion of this.
Solutions are largely numerical.

[1] Nakatsukasa T, Matsuyanagi K, Matsuzaki M, Shimizu YR. Quantal rotation and its coupling to intrinsic motion in nuclei. Physica Scripta. 2016 Jun 27;91(7):073008.
[2] Verhaar BJ, Schulte AM, de Kam J. The connection of the nuclear “Coriolis” force with classical mechanics. Zeitschrift für Physik A Atoms and Nuclei. 1976 Sep;277(3):261-4.
