Question

For a motion of a particle of mass μ in a spherically symmetric potential show that
L2 and Lz commute with the Hamiltonian.

Answer 1

Question #85012—Physics — Quantum Mechanics

For a motion of a particle of mass $\mu$ in a spherically symmetric potential show that $L^{\wedge}2$ and $L_{-}z$ commute with the Hamiltonian.

Solution

Hamiltonian of the particle in spherically symmetric potential $V(r)$ have form in spherical coordinates

$$H = - \frac {\hbar^ {2}}{2 \mu} \Delta + V (r) = - \frac {\hbar^ {2}}{2 \mu} \frac {\partial^ {2}}{\partial r ^ {2}} + \frac {1}{2 \mu r ^ {2}} \hat {L} ^ {2} + V (r),$$

where $\hat{L}^2 = -\hbar^2\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin \theta \frac{\partial}{\partial\theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\right]$ .

and $L_{z}$ in spherical coordinates $L_{z} = -i\hbar \frac{\partial}{\partial\phi}$.

Functions $L_{z}$, $\hat{L}^2$ commute with any function of $r$

$$[ \hat {L} ^ {2}, f (r) ] = [ L _ {z}, f (r) ] = 0,$$

So $[H,\hat{L}^2 ] = [\frac{1}{2\mu r^2}\hat{L}^2,\hat{L}^2 ] = 0,$

$$[ H, L _ {z} ] = [ \frac {1}{2 \mu r ^ {2}} \hat {L} ^ {2}, L _ {z} ] = 0,$$

because $\hat{L}^2\neq \hat{L}^2 (\phi)$

Answer: $[H,L_z] = [H,\hat{L}^2 ] = 0$

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