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:

<br /> [H,L_x] = [\frac{p^2}{2m} + V(r), L_x] = [V(r), L_x]<br />

<br /> [V(r), L_x] = [V(r), yp_z - zp_y] <br />

<br /> = V(r)yp_z - V(r)zp_y - yp_zV(r) + zp_yV(r)<br />

<br /> = y[V(r), p_z] - z[V(r), p_y]<br />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.