- #1
AngeSurTerre
- 25
- 4
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)
[itex]\mathcal H_o=\frac{\vec P^2}{2m}+V(r,\theta-\omega t,z)[/itex],
where [itex]V(r,\theta,z)[/itex] is the potential due to the container. In the frame of the rotating container, the Hamiltonian is independent of time and takes the form
[itex]\mathcal H=\frac{\vec p^2}{2m}-\vec\omega\cdot\vec{\mathcal L}+V(r,\theta,z)[/itex],
where [itex]\vec{\mathcal L}=\vec r\times\vec P[/itex] is the angular momentum operator.
I'm looking for a unitary operator [itex]\mathcal U[/itex] such that [itex]\mathcal H=\mathcal U^\dagger\mathcal H_o\mathcal U[/itex]. The operator [itex]e^{-i\frac{\vec\omega\cdot\vec{\mathcal L}}{\hbar}t}[/itex] does the job for the potential,
[itex]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)[/itex],
but it leaves the kinetic term unchanged. So what is the unitary operator that adds a shift of [itex]\vec\omega\cdot\vec{\mathcal L}[/itex] to the kinetic term?
[itex]\mathcal H_o=\frac{\vec P^2}{2m}+V(r,\theta-\omega t,z)[/itex],
where [itex]V(r,\theta,z)[/itex] is the potential due to the container. In the frame of the rotating container, the Hamiltonian is independent of time and takes the form
[itex]\mathcal H=\frac{\vec p^2}{2m}-\vec\omega\cdot\vec{\mathcal L}+V(r,\theta,z)[/itex],
where [itex]\vec{\mathcal L}=\vec r\times\vec P[/itex] is the angular momentum operator.
I'm looking for a unitary operator [itex]\mathcal U[/itex] such that [itex]\mathcal H=\mathcal U^\dagger\mathcal H_o\mathcal U[/itex]. The operator [itex]e^{-i\frac{\vec\omega\cdot\vec{\mathcal L}}{\hbar}t}[/itex] does the job for the potential,
[itex]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)[/itex],
but it leaves the kinetic term unchanged. So what is the unitary operator that adds a shift of [itex]\vec\omega\cdot\vec{\mathcal L}[/itex] to the kinetic term?