What is the suitable unitary operator for a rotating frame?

[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 $mv^2/2$.