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I was wondering how to derive the 3 typical "fictitious" forces that arise in a rotating frame. Their derivation in classical mechanics is clear to me, but not in quantum mechanics.

Classical point particle: we have an inertial frame centered in $O$ and a rotating one in $O'$, 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 $$ \dot{{\bf x}} = \dot{\bf x}' + {\bf w} \times {\bf x}' $$ $$ \ddot{{\bf x}} = \ddot{\bf x}' + 2 {\bf w} \times \dot{\bf x}'+ {\bf w} \times ( {\bf w} \times {\bf x}')+\dot{ {\bf w} }\times{\bf x}' $$ Question: How does this work in quantum mechanics (assuming, for simplicity, a spin-$0$ wave function)?

I suppose that the 3 apparent forces should somehow manifest themselves in the Schrodinger equation (not under the direct form of "forces" of course). I also suppose we should find something that resembles the last two equations 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).

Edit: I may be wrong, but I have the feeling that the answer of Qmechanic below is not the whole story. 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, I have the feeling that $U_t$ can be expressed in terms of a T-ordered exponential, otherwise a simple

$$ U_t = e^{\frac{-i}{\hbar} L_z \int_0^t \Omega(t') dt'} $$

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 new term $U_t \partial_t U_t^*$. Please correct me if my line of reasoning is over-complicated and there is no real need to consider any unitary transformation.

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    $\begingroup$ Can you be more specific as to what you want to know? Quantum mechanics is largely based on the classical Hamiltonian formalism, not on notions of "forces" or "frames". If there is a Hamiltonian description of your rotating systems, you can quantize that. What is the question? $\endgroup$ – ACuriousMind Apr 21 at 23:17
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    $\begingroup$ Quantum mechanics doesn't know anything about "forces" or "frames", so I don't really know what you're looking for. If you want to "go to a rotating reference frame", you should probably just transform your observables by some time-dependent rotation operator $R(t)$. Whether or not that yields something similar to the classical pseudo-forces in Ehrenfest's theorem should be a straightforward calculation. $\endgroup$ – ACuriousMind Apr 22 at 10:47
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    $\begingroup$ @ACuriousMind I found this article: doi.org/10.1006/aphy.1997.5720 "Quantum Mechanics in Nonintertial Reference Frames", I will try to read and see if it helps, but it seems quite technical. $\endgroup$ – Quillo Apr 22 at 11:53
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    $\begingroup$ @Quillo Yes, that is the reference. It is merely glossed over in standard QM because standard QM is specifically designed for inertial reference frames, whereas what you ask about is an extension of QM of little use to QFT (for example) or many-body micro/mesoscopic physics. $\endgroup$ – DanielC Apr 22 at 17:19
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    $\begingroup$ I find this statement -- Quantum mechanics doesn't know anything about "forces" or "frames" -- is rather controversial. The wavefunction do response to the translation and rotation of the coordinate frame, and potentail generally comes from a conservative force. $\endgroup$ – ytlu Apr 27 at 19:54
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Here is one approach:

  1. 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.

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

  3. Next write down the TDSE.

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$^1$ 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.

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  • $\begingroup$ Thank you, this is a useful sketch, but I am not able to figure out where the Euler force appears (I see no time derivatives of $\Omega$). Should we conclude that there is no Euler "force" in QM? $\endgroup$ – Quillo Apr 22 at 12:37
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    $\begingroup$ The Euler force is present here. It shows up when you write down the Euler-Lagrange equations/Hamilton's equations. (I suggest working through this exercise if you are not convinced of this.) $\endgroup$ – d_b Apr 22 at 17:31
  • $\begingroup$ @d_b I do not have to write the Euler-Lagrange, that is for the classical equation of motion. According to Qmechanic I have to write the TDSE. If I just write $i \hbar \dot \psi = H \psi$ then there is no Euler "term". $\endgroup$ – Quillo Apr 22 at 23:12
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    $\begingroup$ @Quillo You have missed d_b's point. If you write down the Euler-Lagrange equations and observe a term you recognize as the "Euler force," then that immediately tells you this Lagrangian, and hence Hamiltonian, correctly includes the effect. There are many sources online explaining how to obtain the Lagrangian here, in addition to the references in the linked Phys.SE post Qmech included in the answer. Doing quantum mechanics does not mean you should forget everything about classical mechanics. $\endgroup$ – Richard Myers Apr 25 at 6:52
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    $\begingroup$ If you define a unitary operator to implement the rotation and define the Hamiltonian in the rotating frame as $H' = UHU^\dagger +i\hbar \dot{U}U^\dagger$, then the dynamics is described by the TDSE. I am willing to bet that working through the computations gives back the same Hamiltonian as derived classically above. Then the Ehrenfest theorem should give the right expression for $\partial_t^2 \langle x \rangle$ (with the 3 inertial terms) $\endgroup$ – SolubleFish Apr 27 at 15:30
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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.

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  • $\begingroup$ Thank younfor the references. Is this $H$ equivalent to the one proposed by @Qmechanic in their answer? It seems to me that by expanding the square in Qmechanic's kinetic part of the Hamiltonian one also gets a "potential" for the centrifugal force that is missing here. $\endgroup$ – Quillo Apr 28 at 5:34
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    $\begingroup$ In the rotating frame: $\textbf{p}=m\textbf{v}+m\boldsymbol{\omega}\times\textbf{r}$ and $\textbf{L}=\textbf{r}\times\textbf{p}$ so the $\textbf{L}\cdot\boldsymbol{\omega}$ contains terms in $\omega^2$. See Landau/Lifshits on Mechanics, Eq.(39.13) $\endgroup$ – ZeroTheHero Apr 28 at 12:36
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    $\begingroup$ see also Anandan, J. and Suzuki, J., 2004. Quantum mechanics in a rotating frame. In Relativity in rotating frames (pp. 361-370). Springer, Dordrech, with preprint available on arxiv. $\endgroup$ – ZeroTheHero Apr 28 at 12:40

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