How to Prove the Commutator Relationship for Angular Momentum Operators?

[WendysRules]

Homework Statement

Show that $[\hat{L} \cdot \vec{a}, \hat{L} \cdot \vec{b}] = i \hbar \hat{L} \cdot (\vec{a} \times \vec{b})$

Homework Equations

$[\hat{L}_i, \hat{L}_j]= i \hbar \epsilon_{ijk} \hat{L}_k$

The Attempt at a Solution

[/B]
Maybe a naive attempt, but it has been a while. I have two ways I think this can work, but both I'm not sure. they "work", but without a more trained eye, I'm not sure if they're valid.

Start with $[\hat{L} \cdot \vec{a}, \hat{L} \cdot \vec{b}] \rightarrow [\hat{L}_ia_i, \hat{L}_jb_j] =i \hbar \epsilon_{ijk} \hat{L}_k a_ib_i \rightarrow i \hbar\hat{L}_k \epsilon_{ijk} a_i b_i = i \hbar \hat{L} \cdot (\vec{a} \times \vec{b})$

The other one starts similarly, but once we put it in the $[\hat{L}_ia_i, \hat{L}_jb_j]$ form, we can take out the $a_i$ and $b_j$ giving us $a_ib_j[\hat{L}_i,\hat{L}_j]$ which from here the rest follows as above, but not sure if either approach is "valid".

 

[Orodruin]

How are these different ways? They look identical to me.

 

[WendysRules]

How are these different ways? They look identical to me.

Yes, i believe they are. Is it true that I am allowed to take out the $a_i$ from the brackets? I think that is what i conclude from above.

 

[Orodruin]

Is it true that I am allowed to take out the aiaia_i from the brackets?

Yes, the commutator is linear in both arguments.