Exploring Quantum Physics - Week 6 Question 3

Question:

What is $[L \cdot L, \hat{r} \cdot \hat{p} ]$?

Solution:

At this point I still don't know the solution. But I think it might be beneficial at this point to try to typeset whatever I have right now. Maybe that's the key to completion of the complicated piece.

This is just really complicated. Anyway, we need to start somewhere.

$\begin{eqnarray*} & & [L \cdot L, \hat{r} \cdot \hat{p} ] \\ &=& [L_a L_a, x_b p_b ] \\ &=& [L_a L_a, x_b] p_b + x_b[L_a L_a, p_b] \\ \end{eqnarray*}$

To avoid getting too complicated - now we focus on the first term

$\begin{eqnarray*} & & [L_a L_a, x_b] \\ &=& [L_a, x_b] L_a + L_a[L_a, x_b] \\ &=& (\epsilon_{abc}i\hbar x_c) L_a + L_a(\epsilon_{abc}i\hbar x_c) \\ &=& \epsilon_{abc}i\hbar(x_c L_a + L_a x_c) \\ &=& \epsilon_{abc}i\hbar(x_c (\epsilon_{ade} x_d p_e) + (\epsilon_{ade} x_d p_e) x_c) \\ &=& \epsilon_{abc}\epsilon_{ade}i\hbar(x_c x_d p_e + x_d p_e x_c) \end{eqnarray*}$

Now we work on the second term

$\begin{eqnarray*} & & [L_a L_a, p_b] \\ &=& [L_a, p_b] L_a + L_a[L_a, p_b] \\ &=& (\epsilon_{abc}i\hbar p_c) L_a + L_a(\epsilon_{abc}i\hbar p_c) \\ &=& \epsilon_{abc}i\hbar(p_c L_a + L_a p_c) \\ &=& \epsilon_{abc}i\hbar(p_c (\epsilon_{ade} x_d p_e) + (\epsilon_{ade} x_d p_e) p_c) \\ &=& \epsilon_{abc}\epsilon_{ade}i\hbar(p_c x_d p_e + x_d p_e p_c) \end{eqnarray*}$

And then we substitute them back

$\begin{eqnarray*} & & [L \cdot L, \hat{r} \cdot \hat{p} ] \\ &=& [L_a L_a, x_b] p_b + x_b[L_a L_a, p_b] \\ &=& \epsilon_{abc}\epsilon_{ade}i\hbar(x_c x_d p_e + x_d p_e x_c) p_b + x_b\epsilon_{abc}\epsilon_{ade}i\hbar(p_c x_d p_e + x_d p_e p_c) \\ &=& \epsilon_{abc}\epsilon_{ade}i\hbar(x_c x_d p_e p_b + x_d p_e x_c p_b + x_b p_c x_d p_e + x_b x_d p_e p_c) \\ &=& (\delta_{bd}\delta_{ce} - \delta_{be}\delta_{cd})i\hbar(x_c x_d p_e p_b + x_d p_e x_c p_b + x_b p_c x_d p_e + x_b x_d p_e p_c) \\ &=& i\hbar((x_c x_b p_c p_b + x_b p_c x_c p_b + x_b p_c x_b p_c + x_b x_b p_c p_c)-(x_c x_c p_b p_b + x_c p_b x_c p_b + x_b p_c x_c p_b + x_b x_c p_b p_c)) \\ &=& i\hbar((x_b x_c p_b p_c + x_b p_c x_c p_b + x_b p_c x_b p_c + x_b x_b p_c p_c)-(x_c x_c p_b p_b + x_c p_b x_c p_b + x_b p_c x_c p_b + x_b x_c p_b p_c)) \\ &=& i\hbar((x_b p_c x_b p_c + x_b x_b p_c p_c)-(x_c x_c p_b p_b + x_c p_b x_c p_b)) \end{eqnarray*}$

When $b$ and $c$ are equal, the expression above is 0, so we can focus on the case when $b \ne c$, in that case the operator commutes!

$\begin{eqnarray*} & & [L \cdot L, \hat{r} \cdot \hat{p} ] \\ &=& i\hbar((x_b p_c x_b p_c + x_b x_b p_c p_c)-(x_c x_c p_b p_b + x_c p_b x_c p_b)) \\ &=& i\hbar((x_b x_b p_c p_c + x_b x_b p_c p_c)-(x_c x_c p_b p_b + x_c x_c p_b p_b)) \\ &=& i\hbar(2 x_b x_b p_c p_c - 2 x_c x_c p_b p_b) \\ &=& 2i\hbar(x_b x_b p_c p_c - x_c x_c p_b p_b) \end{eqnarray*}$

This is the end - I have no trick to further simplify this. It appears to me no choice matches this.

Now we try something else - discussion forum suggest I can try expanding the operators the other way round.

$\begin{eqnarray*} & & [L \cdot L, \hat{r} \cdot \hat{p} ] \\ &=& [L_a L_a, x_b p_b ] \\ &=& [L_a L_a, x_b p_b ] \\ &=& [L_a, x_b p_b] L_a + L_a [L_a, x_b p_b] \end{eqnarray*}$

Oh - leveraging the previous question - this is just zero?