How is the CSCO in an harmonic oscillator?

[Dario SLC]

Hi everyone, I have a great doubt in this problem:
Let a mass m with spin 1/2, subject to the following central potencial V(r):
V(r)=1/2mω²r²
Find the constants of motion and the CSCO to solve the Hamiltonian?

This is my doubt, I can't find the CSCO in this potencial. Is a problem in general quantum physics, not use the Dirac's notation brakets and ket.

Thanks a lot!
Dar

 

[vanhees71]

What the heck is CSCO? Define your acronyms!

 

[Dario SLC]

What the heck is CSCO? Define your acronyms!

Yes I am sorry, CSCO= Complete Set of Conmuting Observables. This I need for resolve the hamiltonian.

 

[hilbert2]

It's the same set as with a hydrogen atom, as this is rotation symmetric. Isn't that term just a "centrifugal" force that actually belongs to the kinetic energy but can also be thought of as an effective noninertial part of the potential field?

 

[vanhees71]

For the symmetric 3D harmonic oscillator you have of course several possibilities. On one hand you can use Cartesian coordinates and use the three phonon-number operators $\hat{N}_j=\hat{a}_j^{\dagger} \hat{a}_j$ as the complete set. The relation to the Hamiltonian is
$$\hat{H}=\hbar \omega \left (\sum_{j=1}^3 \hat{N}_j+\frac{3}{2} \right ).$$
On the other hand, as any central potential you can as well use $\hat{H}$, $\hat{\vec{L}}^2$, and $\hat{L}_z$ as the complete set. The corresponding quantum numbers are then as for any central potential $E$, $\ell$ and $m$.

 

[Dario SLC]

Ok thanks a lot!

Then in addition, the $\hat{L}_z$ and $\hat{L}^2$ they are motion constants because the hamiltonian is invariant to rotations and translations, and the energy $E$ only depends of the quantum number $n$just because the hamiltonian conmute with the operator rotator $\hat{R}$ and translator $\hat{T}$. It is true no?

 

[Dario SLC]

For the symmetric 3D harmonic oscillator you have of course several possibilities. On one hand you can use Cartesian coordinates and use the three phonon-number operators $\hat{N}_j=\hat{a}_j^{\dagger} \hat{a}_j$ as the complete set. The relation to the Hamiltonian is
$$\hat{H}=\hbar \omega \left (\sum_{j=1}^3 \hat{N}_j+\frac{3}{2} \right ).$$
On the other hand, as any central potential you can as well use $\hat{H}$, $\hat{\vec{L}}^2$, and $\hat{L}_z$ as the complete set. The corresponding quantum numbers are then as for any central potential $E$, $\ell$ and $m$.

Hi, a question about the spin, this add a new constant of movement, ie., in addition to $L^2$, $L_z$ and $H$, $S_z$ also is a constant of movement? My doubt is because this problem have in reality four constant of movement if we take into account that spin.