Quantum Hamiltonian commuting with the Pauli-Runge vector. I have to prove that $[A_j, H] = 0$, with;
$$\vec{A} = \frac{1}{2Ze^{2}m}(\vec{L} \times \vec{P} - \vec{p} \times \vec{L}) + \frac{\vec{r}}{r}$$
$$H = \frac{p^2}{2m} - \frac{Ze^2}{r}$$
And, $Z, e, m$ are all constants.
Now, after expanding out the commutator bracket, I'm left with six terms, which I need to prove are all equal to zero.
Now, I can do this for $[\vec{L} \times \vec{p}, p^2]$, $[\vec{p} \times \vec{L}, p^2]$, and $[\frac{\vec{r}}{r}, \frac{1}{r}]$, but I'm having trouble with $[\vec{L} \times \vec{p}, \frac{1}{r}]$, $[\vec{p} \times \vec{L}, \frac{1}{r}]$, and $[\frac{\vec{r}}{r}, p^2]$.
I'm sure once I've $[\vec{L} \times \vec{p}, \frac{1}{r}]$, I can prove $[\vec{p} \times \vec{L}, \frac{1}{r}]$, but, I'm not really sure what to do.
With $[\vec{L} \times \vec{p}, \frac{1}{r}]$, I've done the following;
$$[\vec{L} \times \vec{p}, \frac{1}{r}] = \varepsilon_{ijk}[L_{j}p_{k}, \frac{1}{r}]$$
$$ = \varepsilon_{ijk}(L_j[p_k, \frac{1}{r}] + [L_j, \frac{1}{r}]p_k)$$
$$ = \varepsilon_{ijk}L_{j}[p_k, \frac{1}{r}]$$
$$ = \frac{i \varepsilon_{ijk} L_j q_k}{r^3}$$
But from here, I'm not sure where to go. I was hoping to use the cyclic nature of the Levi-Civita symbol, but I can't really see how that helps here. Any input would be fantastic!!
 A: Actually, I was just being dumb, and I totally worked it out.
In the expansion of $[A_j, H]$, we can cancel down to $$[A_j, H] = [\frac{\vec{r}}{r}, p^2] - [(\vec{L} \times \vec{p})_j, \frac{1}{r}] + [(\vec{p} \times \vec{L})_j, \frac{1}{r}]$$
Now, using the fact that $$[p_k, r^{-1}] = \frac{i q_k}{r^3}$$
Along with a few other commutator identities, obtain the following;
$$[\frac{\vec{r}}{r}, p^2] = \frac{i (\vec{L} \times \vec{r})_j}{r^3} - \frac{i (\vec{r} \times \vec{L})_j}{r^3}$$
Then, noting that $q_i$ is representative of position, using my working previously, we can obtain the following;
$$[(\vec{L} \times \vec{p})_j, r^{-1}] = \frac{i \varepsilon_{jkl} L_k q_l}{r^3} = \frac{i (\vec{L} \times \vec{r})_j}{r^3}$$
$$[(\vec{p} \times \vec{L})_j, r^{-1}] = \frac{i \varepsilon_{jkl} q_k L_l}{r^3} = \frac{i (\vec{r} \times \vec{L})_j}{r^3}$$
Putting it all together, we get;
$$[\frac{\vec{r}}{r}, p^2] - [(\vec{L} \times \vec{p})_j, \frac{1}{r}] + [(\vec{p} \times \vec{L})_j, \frac{1}{r}] = \frac{i (\vec{L} \times \vec{r})_j}{r^3} - \frac{i (\vec{r} \times \vec{L})_j}{r^3} - \frac{i (\vec{L} \times \vec{r})_j}{r^3} + \frac{i (\vec{r} \times \vec{L})_j}{r^3} = 0$$
