- #1
MisterX
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Homework Statement
[itex]\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} [/itex]
Derive
[itex]\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 [/itex]
Homework Equations
[itex]\left(\mathbf{X} \times \mathbf{Y} \right)_i = \epsilon_{ijk}X_jY_k[/itex]
[itex]\left[L_i, p_j \right] = \epsilon_{ijk}p_k[/itex]
[itex]\sum_i \epsilon_{ijk}\epsilon_{i\ell m} = \delta_{j\ell } \delta_{km} -\delta_{jm} \delta_{k\ell }[/itex]
The Attempt at a Solution
I found that
[itex]-\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 [/itex]
So another way to write A
[itex]\mathbf{A} =-\hbar\mathbf{p} \times \mathbf{L} +\hbar \mathbf{p} + Ze^2m \frac{\mathbf{r}}{r} [/itex]
My attempt ( [itex]\left\{\;,\;\right\}[/itex] indicates anti-commuator with dot product)
[itex]\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)[/itex]
[tex]=\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[/tex]
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
[itex]\left[L_k, p_j\right] = \epsilon_{kj\ell}p_\ell =\epsilon_{j\ell k}p_\ell [/itex]
\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 [itex] \mathbf{p}^2 \mathbf{L}^2 [/itex] term without too much effort. And I see the second term is going to cancel with part of the anti-commutator [itex]\left\{\mathbf{p} \times \mathbf{L}, \mathbf{p}\right\}[/itex]. 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.
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