I can show that $$ [\hat L_i,\hat L_j] = i\hbar\epsilon_{ijk} \hat L_k $$ where $\hat L$ is the angular momentum operator. But I'm struggling to show that $$[\vec a \cdot \hat L , \vec b \cdot \hat L] = i(\vec a \times \vec b) \cdot \hat L$$ where two vectors $\vec a$ and $\vec b$ commute with each other and with $\hat L$, that is, $[\vec a, \vec b] = [\vec a, \hat L] = [\vec b, \hat L] = 0$.

I can do it in three dimensions by writing each component, but how can I show the mentioned relation using $\epsilon_{ijk}$?


1 Answer 1


Starting with $$[a\cdot L, b\cdot L] = a_iL_ib_jL_j - b_jL_ja_iL_i$$

Note that since $a$ and $b$ commute with one another and $L$, we can factor this as $$= a_ib_j(L_iL_j - L_jL_i)$$

The second term is simply just $[L_i,L_j]$, so we have $$[a\cdot L, b\cdot L] = a_ib_j[L_i,L_j]$$

And then using $[L_i,L_j] = i\hbar \epsilon_{ijk}L_k$, $$ = i\hbar a_ib_j \epsilon_{ijk}L_k$$

Note that $a_ib_j\epsilon_{ijk} = (a\times b)_k$, so that $$ = i\hbar (a\times b)_k L_k$$

Which is just $i\hbar(a\times b)\cdot L$.

Thus $$[a\cdot L, b\cdot L] = i\hbar(a\times b)\cdot L$$

  • $\begingroup$ You're answer was super-helpful. Thanks! $\endgroup$ Oct 21, 2020 at 21:21

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