How Do You Derive the Squared Quantum Runge Lenz Vector?

[MisterX]

Homework Statement

\mathbf{A} = \frac{1}{2}\left(\hbar\mathbf{L} \times \mathbf{p} \right) - \frac{1}{2}\left(\mathbf{p} \times \hbar\mathbf{L} \right) + Ze^2m \frac{\mathbf{r}}{r}

Derive
\mathbf{A}^2 = \left(\mathbf{p}^2 - 2m\frac{Ze^2}{r} \right)\left(\hbar^2 \mathbf{L^2} + \hbar^2 \right) + m^2Z^2e^4

Homework Equations

\left(\mathbf{X} \times \mathbf{Y} \right)_i = \epsilon_{ijk}X_jY_k
\left[L_i, p_j \right] = \epsilon_{ijk}p_k
\sum_i \epsilon_{ijk}\epsilon_{i\ell m} = \delta_{j\ell } \delta_{km} -\delta_{jm} \delta_{k\ell }

The Attempt at a Solution

I found that
-\left(\mathbf{p} \times \mathbf{L} \right)_i = \left(\mathbf{L} \times \mathbf{p} \right)_i^\dagger = \left(\mathbf{L} \times \mathbf{p} \right)_i - 2p_i

So another way to write A

\mathbf{A} =-\hbar\mathbf{p} \times \mathbf{L} +\hbar \mathbf{p} + Ze^2m \frac{\mathbf{r}}{r}

My attempt ( \left\{\;,\;\right\} indicates anti-commuator with dot product)

\mathbf{A} ^2 = \left(-\hbar\mathbf{p} \times \mathbf{L} + \hbar\mathbf{p} + Ze^2m \frac{\mathbf{r}}{r}\right)\left(-\hbar\mathbf{p} \times \mathbf{L} + \hbar\mathbf{p} + Ze^2m \frac{\mathbf{r}}{r}\right)
=\hbar^2\left(\mathbf{p} \times \mathbf{L}\right)^2 + \hbar^2\mathbf{p}^2 -\hbar^2\left\{\mathbf{p} \times \mathbf{L}, \mathbf{p}\right\} - \hbar Ze^2m\left\{\mathbf{p} \times \mathbf{L}, \frac{\mathbf{r}}{r}\right\} + \hbar Ze^2m\left\{ \mathbf{p}, \frac{\mathbf{r}}{r}\right\} + m^2Z^2e^4
Now I get that
\begin{align*}\left(\mathbf{p} \times \mathbf{L}\right)^2 &= \epsilon_{ijk}\epsilon_{i\ell m}p_jL_kp_\ell L_m = \left(\delta_{j\ell} \delta_{km} -\delta_{jm} \delta_{k\ell} \right)p_jL_kp_\ell L_m \\
&= p_jL_kp_j L_k - p_jL_kp_k L_j\end{align*}
using
\left[L_k, p_j\right] = \epsilon_{kj\ell}p_\ell =\epsilon_{j\ell k}p_\ell

\begin{align*}\left(\mathbf{p} \times \mathbf{L}\right)^2 &= p_j(p_jL_k + \epsilon_{j\ell k}p_\ell)L_k - p_jp_kL_k L_j\\ &= \mathbf{p}^2 \mathbf{L}^2 + \mathbf{p} \cdot \left(\mathbf{p} \times \mathbf{L} \right) - p_jp_kL_k L_j\end{align*}

So if I do it this way, at least I get the \mathbf{p}^2 \mathbf{L}^2 term without too much effort. And I see the second term is going to cancel with part of the anti-commutator \left\{\mathbf{p} \times \mathbf{L}, \mathbf{p}\right\}. But I am not sure what to do with the third term, or how the other half of that anti-commutator would go away. I am not sure if this is the right approach and perhaps I have made a mistake. I'd appreciate any help with this problem.

 

[Oxvillian]

Hi MisterX!

You missed an i in one of your commutation relations:

\left[L_i, p_j \right] = i\epsilon_{ijk}p_k

Then things should work out ok. Also note that

\mathbf{p} \cdot \left(\mathbf{p} \times \mathbf{L} \right) = 0

and

- p_jp_kL_k L_j = 0.

 

[MisterX]

Thanks, however I already found out those things (and solved this problem after a significant amount of work).

Perhaps somewhere we should have a catalog of vector calculus identities for vectors of quantum mechanical operators.\mathbf{p} \cdot \left(\mathbf{p} \times \mathbf{L} \right) = 0
and
\left(\mathbf{p} \times \mathbf{L} \right) \cdot \mathbf{p} =i2\mathbf{p}^2