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.
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.
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