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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!!

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  • $\begingroup$ Have you tried writing $L_i=\epsilon_{ibc}q_b p_c$, and play with the summation of $\epsilon_{ijk}\epsilon_{jab}$ and the commutation relation of $q$ and $p$ ? $\endgroup$
    – Adam
    Commented Oct 11, 2013 at 4:59
  • $\begingroup$ I have, yes. Ignoring that $r^3$ bit on the bottom, we can write $L_j = \varepsilon_{jab} q_a p_b$. Then, the top line becomes $$i \varepsilon_{ijk} \varepsilon_{jab} q_a p_b q_k = -i \varepsilon_{jik} \varepsilon_{jab} q_a p_b q_k = -i (\delta_{ia}\delta_{kb} - \delta_{ib}\delta_{ka}) q_a p_b q_k = i \delta_{ib} \delta_{ka} q_a p_b q_k - i \delta_{ia}\delta_{kb} q_a p_b q_k = i q_a p_b q_a - i q_a p_b q_b$$ But I'm not sure how that helps... $\endgroup$
    – Jack
    Commented Oct 11, 2013 at 5:15
  • $\begingroup$ Another commutator question by OP: physics.stackexchange.com/q/79599/2451 $\endgroup$
    – Qmechanic ♦
    Commented Oct 11, 2013 at 5:30
  • $\begingroup$ Yeah, I should really delete that. I've moved on from it up to this point. $\endgroup$
    – Jack
    Commented Oct 11, 2013 at 5:59

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