# What is the suitable unitary operator for a rotating frame?

1. Nov 8, 2014

### 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?

2. Nov 10, 2014

### George Jones

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

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?

3. Nov 10, 2014

### AngeSurTerre

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:

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

4. Nov 10, 2014

### George Jones

Staff Emeritus
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.

5. Nov 10, 2014

### George Jones

Staff Emeritus
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.

6. Nov 10, 2014

### AngeSurTerre

Thank you! I will look in this direction.

7. Nov 10, 2014

### atyy

8. Nov 10, 2014

### 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.

9. Nov 11, 2014

### 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!

10. Nov 14, 2014

### 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?

11. Nov 14, 2014

### 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.

12. Nov 14, 2014

### AngeSurTerre

I tried to think by analogy with the case where the sytem 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 $\mathcal U=e^{-i\alpha\Theta}$.

13. Nov 17, 2014

### 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.

Last edited: Nov 17, 2014
14. Nov 17, 2014

### 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!

15. Nov 17, 2014

### 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 $\omega$.

Last edited: Nov 17, 2014
16. Nov 18, 2014

### DrDu

Use the BCH (Baker, Campbell, Hausdorff) formula

Last edited: Nov 18, 2014
17. Nov 18, 2014

### 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.

18. Nov 19, 2014

### DrDu

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

19. Nov 19, 2014

### AngeSurTerre

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?

20. Nov 19, 2014

### DrDu

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

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