# Deriving an Angular Momentum Commutator Relation using $ϵ_{ijk}$ Identities

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 $$\epsilon_{ijk}$$?

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