I'm trying to prove that $$[\hat{L}_i,\hat{H}]=0$$ for $\hat{H}$ the hamiltonian of a central force $$\hat{H}=\frac{\hat{p}^2}{2m}-\frac{\alpha}{r}.$$

I'm getting this:

$$[\hat{L}_i,\hat{H}]=[\hat{L}_i,\frac{\hat{p}^2}{2m}-\frac{\alpha}{r}] = [\hat{L}_i , \frac{\hat{p}^2}{2m}] - [\hat{L}_i , \frac{\alpha}{r}].$$

Already prove that the first one is zero (it's a known result too), but the second one, I dont know what to do with it, there is no way for me. I was doing something like this:

$$[\hat{L}_i , \frac{\alpha}{r}] = [\epsilon_{ijk} r_j p_k , \frac{\alpha}{r}] = \epsilon_{ijk}r_j [p_k ,\frac{\alpha}{r}] + \epsilon_{ijk} [r_j , \frac{\alpha}{r}]p_k.$$

Obviously, the last term is zero, but the other one, I do something that has no sense to be zero. Have any idea for this? I'm I doing wrong? All of this is to prove that Laplace-Runge-Lenz operator commute with hamiltonian of a central force.


Indeed, as you state, $[\hat{L_i},\hat{p^2}]$ and $\epsilon_{ijk}[r_j, \frac{\alpha}{r}]p_k$ are both zero, so we need only consider the commutator:

$$\epsilon_{ijk}\ r_j\ [p_k, \frac{\alpha}{r}]$$

Giving this a test function, we obtain:

$$\epsilon_{ijk}\ r_j\ [p_k, \frac{\alpha}{r}]\ f = \alpha r_j \epsilon_{ijk}\ [\ p_k(\frac{f}{r}) - \frac{1}{r}p_k(f)\ ]$$

The first term on the RHS can be written:

$$p_k(\frac{f}{r}) = -i\hbar[\frac{1}{r}\frac{\partial f}{\partial r_k}-\frac{r_k\ f}{r^3}]$$

Thus we can now write the commutator as (writing out $p_i$ as $-i\hbar\frac{\partial}{\partial r_i}$):

$$\epsilon_{ijk}\ r_j\ [p_k, \frac{\alpha}{r}]\ f = -i\hbar \alpha \epsilon_{ijk} r_j\ [\frac{1}{r}\frac{\partial f}{\partial r_k} - \frac{r_k\ f}{r^3}-\frac{1}{r}\frac{\partial f}{\partial r_k}]$$

Two of the terms cancel, leaving:

$$\epsilon_{ijk}\ r_j\ [p_k, \frac{\alpha}{r}]\ f = i\hbar \alpha \epsilon_{ijk} r_j\ \frac{r_k\ f}{r^3}$$

If we now sum over the repeated indices, and utilising the antisymmetric property of the Levi-Civita symbol, we obtain:

$$\epsilon_{ijk}\ r_j\ [p_k, \frac{\alpha}{r}]\ f = \frac{i\hbar \alpha}{r^3}[r_jr_k-r_kr_j]=0 $$

Thus, the commutator is zero as expected and indeed, $[\hat{L_i},\hat{H}] = 0$ for all of the components of $L$ (as one would expect; the angular momentum vector is conserved under the action of a central force).


Not sure this is what you want, BUT your statement is entirely equivalent to H being invariant under rotations and this is trivial. Why do you need to go through the explicit analysis of commutators? All you need is to remind that $\bf{p}$ is a vector, i.e. $[L_i, p_j] = i \hbar \epsilon_{ijk} p_k$. From this, rotational invariance of H is manifest. Extension to Lenz vector is also made easy exploiting the underlying $SO(4)$ invariance of Kepler ( or 3d isotropic harmonic oscillator ) problem.


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