Exploring Quantum Physics - More angular momentum commutators

The goal of this post is to compute $[L_a, L_d]$.

We'll prove something more basic first, let's start with $[x_a p_b, x_c]$. This one is basic, we have

$[x_a p_b, x_c] = x_a p_b x_c - x_c x_a p_b = x_a p_b x_c - x_a x_c p_b = x_a ( p_b x_c - x_c p_b ) = x_a [p_b, x_c] = -i\hbar \delta_{bc} x_a$.

Similarly we can also do $[x_a p_b, p_c]$, using essentially the same trick

$[x_a p_b, p_c] =  x_a p_b p_c - p_c x_a p_b = x_a p_c p_b - p_c x_a p_b = (x_a p_c - p_c x_a) p_b = [x_a p_c] p_b = i\hbar \delta_{ac} p_b$.

I thought I will use these relations to prove the $[L_a, L_d]$ relation, but I was wrong. We should always use closest established result, which is $[L_a, x_b]$ and $[L_a. p_b]$.

$\begin{eqnarray*} [L_a, L_d] &=& [L_a, \epsilon_{bcd}x_b p_c] \\ &=& \epsilon_{bcd}[L_a, x_b p_c] \\ &=& \epsilon_{bcd}([L_a, x_b] p_c + x_b [L_a, p_c]) \\ &=& \epsilon_{bcd}(i\hbar\epsilon_{abe} x_e p_c + i\hbar\epsilon_{acf} x_b p_f ) \\ \end{eqnarray*}$

The expression seems complicated but it could be simplified. Let focus on one term at a time

$\begin{eqnarray*} & & i\hbar\epsilon_{bcd}\epsilon_{abe} x_e p_c \\ &=& i\hbar\epsilon_{bcd}\epsilon_{bea} x_e p_c \\ &=& i\hbar(\delta_{ad}\delta_{ce} - \delta_{ac}\delta_{de}) x_e p_c \end{eqnarray*}$

Similarly, we can get the right hand side as

$\begin{eqnarray*} & & i\hbar\epsilon_{bcd}\epsilon_{acf} x_b p_f \\ & & i\hbar\epsilon_{cdb}\epsilon_{cfa} x_b p_f \\ &=& i\hbar(\delta_{ab}\delta_{df} - \delta_{ad}\delta_{bf}) x_b p_f \end{eqnarray*}$

The key idea to simplification is that $b$ and $f$ are just running index in the second term, we can rename it as we wish. Apparently, we want $b \to e$ and $f \to c$, so we can write the second term as

$\begin{eqnarray*} & & i\hbar(\delta_{ab}\delta_{df} - \delta_{ad}\delta_{bf}) x_b p_f \\ &=& i\hbar(\delta_{ae}\delta_{cd} - \delta_{ad}\delta_{ce}) x_e p_c \end{eqnarray*}$

So if we put them back together, we get:

$\begin{eqnarray*} & & [L_a, L_d] \\ &=& i\hbar(\delta_{ad}\delta_{ce} - \delta_{ac}\delta_{de}) x_e p_c + (\delta_{ae}\delta_{cd} - \delta_{ad}\delta_{ce}) x_e p_c \\ &=& i\hbar(\delta_{ae}\delta_{cd} - \delta_{ac}\delta_{de}) x_e p_c \\ &=& i\hbar(x_a p_d - x_d p_a) \\ \end{eqnarray*}$

To complete the circle, let's prove the answer is $i\hbar\epsilon_{adb}L_b$

$\begin{eqnarray*} & & i\hbar\epsilon_{adb}L_b \\ &=& i\hbar\epsilon_{adb}\epsilon_{ceb}x_c p_e \\ &=& i\hbar\epsilon_{bad}\epsilon_{bce}x_c p_e \\ &=& i\hbar(\delta_{ac}\delta_{de} - \delta_{ae}\delta_{cd})x_c p_e \\ &=& i\hbar(x_a p_d - x_d p_a) \end{eqnarray*}$

No wonder professor said this is complicated.