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[SOLVED] Quantum Angular momentum Question
Show that [tex]\dfrac{{\vec{p}} ^2 }{2m} = \dfrac{{\vec{l} }^2}{2mr^2} - \dfrac{\hbar ^2}{2mr^2}\dfrac{\partial}{\partial r} (r^2 \dfrac{\partial}{\partial r})[/tex] is rotational invariant under the rotation generated by: [tex]\vec{j} = \vec{l} + \vec{s}[/tex] , s is intrinic spin.
[H,J] = 0 and/or [H,J^2] = 0
I think that the second part, the radial operator [tex]\dfrac{\hbar ^2}{2mr^2}\dfrac{\partial}{\partial r} (r^2 \dfrac{\partial}{\partial r})[/tex] commutes with the angular momentas, since it is just a function of radial coordinate, whereas angular momenta depends on the angles (direction) Is that correct?
Homework Statement
Show that [tex]\dfrac{{\vec{p}} ^2 }{2m} = \dfrac{{\vec{l} }^2}{2mr^2} - \dfrac{\hbar ^2}{2mr^2}\dfrac{\partial}{\partial r} (r^2 \dfrac{\partial}{\partial r})[/tex] is rotational invariant under the rotation generated by: [tex]\vec{j} = \vec{l} + \vec{s}[/tex] , s is intrinic spin.
Homework Equations
[H,J] = 0 and/or [H,J^2] = 0
The Attempt at a Solution
I think that the second part, the radial operator [tex]\dfrac{\hbar ^2}{2mr^2}\dfrac{\partial}{\partial r} (r^2 \dfrac{\partial}{\partial r})[/tex] commutes with the angular momentas, since it is just a function of radial coordinate, whereas angular momenta depends on the angles (direction) Is that correct?
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