Exploring Quantum Physics - More commutators

In this post we will talk about the angular momentum operator and their commutators. The angular momentum operator is $\hat{L} = \hat{r} \times \hat{p}$, so we can write $\hat{L_c} = \epsilon_{abc}\hat{x_a}\hat{p_b}$, and we wanted to compute $[\hat{L_a}, \hat{x_d}]$ and $[\hat{L_a}, \hat{p_b}]$.

Given the principle we learnt in the last post. I will try to work in abstract as much as I can. It works great this way.

$\begin{eqnarray*} [\hat{L_c}, \hat{x_d}] &=& [\epsilon_{abc}\hat{x_a}\hat{p_b}, \hat{x_d}] \\ &=& \epsilon_{abc}[\hat{x_a}\hat{p_b}, \hat{x_d}] \\ &=& \epsilon_{abc}([\hat{x_a}, \hat{x_d}]\hat{p_b} + \hat{x_a}[\hat{p_b}, \hat{x_d}]) \\ &=& \epsilon_{abc}\hat{x_a}[\hat{p_b}, \hat{x_d}] \\ &=& -\epsilon_{abc}\hat{x_a}[\hat{x_b}, \hat{p_d}] \\ &=& -\epsilon_{abc}\hat{x_a}(\delta_{bd}i\hbar) \\ &=& -i\hbar\epsilon_{adc}\hat{x_a} \\ &=& i\hbar\epsilon_{cda}\hat{x_a} \\ \end{eqnarray*}$

Let's also solve the momentum problem:

$\begin{eqnarray*} [\hat{L_c}, \hat{p_d}] &=& [\epsilon_{abc}\hat{x_a}\hat{p_b}, \hat{p_d}] \\ &=& \epsilon_{abc}[\hat{x_a}\hat{p_b}, \hat{p_d}] \\ &=& \epsilon_{abc}([\hat{x_a}, \hat{p_d}]\hat{p_b} + \hat{x_a}[\hat{p_b}, \hat{p_d}]) \\ &=& \epsilon_{abc}[\hat{x_a}, \hat{p_d}]\hat{p_b} \\ &=& \epsilon_{abc}(\delta_{ad}i\hbar)\hat{p_b} \\ &=& i\hbar\epsilon_{dbc}\hat{p_b} \\ &=& i\hbar\epsilon_{cdb}\hat{p_b} \end{eqnarray*}$

These match perfectly with the standard result. Imagine how complicated and convoluted it would look like without all these simplifying identity and symbols.