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)?
Consideration #1: The 3 fictitious forces should somehow be present already in the Schrodinger equation (not under the direct form of "forces" of course). I suppose that 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 system subject to apparent forces in QM is the rotating oscillator, see e.g. this arxiv 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).
