Hamiltonian/Angular Momentum Commuter

[eep]

I'm having trouble proving that the Hamiltonian commutes with each component of angular momentum as long as the potential only depends on r.

I have gotten to the following step:

$$[H,L_x] = [\frac{p^2}{2m} + V(r), L_x] = [V(r), L_x]$$

$$[V(r), L_x] = [V(r), yp_z - zp_y]$$

$$= V(r)yp_z - V(r)zp_y - yp_zV(r) + zp_yV(r)$$

$$= y[V(r), p_z] - z[V(r), p_y]$$I'm not sure where to go from here... the problem states that V depends only on r but I'm not sure if I should interperet that as V being linear in terms or r or if there can be higher powers. Help, please!

 

[Physics Monkey]

$$V(r)$$ is meant to be a general function of $$r = \sqrt{x^2 + y^2 + x^2}$$. The simplest way to evaluate those commutators is to use the position representation where $$\vec{p} = - i \hbar \vec{\nabla}$$. You will find that the key feature is that $$V$$ depends on $$x$$, $$y$$, $$z$$ only through $$r$$.

 

[eep]

Thank you!

 

[nrqed]

Thank you!

Physics Monkey already gave you the answer..Let me just add that it will be useful to use

$${\partial V \over \partial z} = {\partial V \over \partial r} {\partial r \over \partial z}$$

and so on.

Patrick

 

[dextercioby]

The fact that $\hat{H}$ is rotationally invariant follows simply from the fact that $V$ depends only on $|\vec{r}|$. Since the rotation group is $SO(3)$ and its generators are the angular momentum operators, it follows by definition that

$$[\hat{L}_{i},\hat{H}]_{-} =\hat{0} , \forall \ i=1,2,3.$$.


Daniel.