Recent content by 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...

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...

This sounds like a proof "by authority".

But how can the angular momentum be the same in both frames? Even classicaly, this is not correct!

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.

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.

This is precisely the part that I don't understand! We have ##\quad\mathcal T=\frac{-\hbar^2}{2m}\Big(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}-\frac{\mathcal L_z^2}{\hbar^2 r^2}+\frac{\partial^2}{\partial z^2}\big)##, and ##\quad\Big[\frac{\partial^2}{\partial...

I looked at the BCH formula you mentioned, and I still don't see how you can show that \quad\mathcal U^\dagger\frac{\mathcal P^2}{2m}\mathcal U=\frac{\mathcal P^2}{2m}-\omega\mathcal L_z, with ##\mathcal U=e^{-i\frac{\omega}{\hbar}(t\mathcal L_z+mr^2\theta)}##.

Then I guess I need to extend my background about covariance. Thank you!

Yes, but then the problem is that it does not transform the Hamiltonian as expected. There are some residual terms, unless I made a mistake in my calculation. Or unless we can show that the exact residual terms just lead to a redefinition of the zero of the energy.

Which \mathcal U are we talking about? I was talking about \mathcal U=e^{-i\frac{\omega t}{\hbar}\mathcal L_z}. It commutes with \mathcal L_z.

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...

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...

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...

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...