Hamiltonian/Angular Momentum Commuter

1. Apr 1, 2006

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!

Last edited: Apr 1, 2006
2. Apr 1, 2006

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 $$r$$.

3. Apr 1, 2006

Thank you!

4. Apr 1, 2006

nrqed

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

5. Apr 3, 2006

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.