# Cross-product within commutator

I've been trying to prove some commutator identities of angular momentum, and I don't want to go brute force and prove for each coordinate seperately. So I tried using the Levi-Civita formalism for the cross product-

$$[\mathbf{a}\times \mathbf{b}]_i=\epsilon _{ijk}a_jb_k$$

My question is, how do I treat $\epsilon_{ijk}$ within a commutator.

I tried reading some proofs on this site, and follow the apparent rules they used. But I still don't understand exactly how it is done, as I got stuck here:

$$[P_i, L_j]=[P_i, \epsilon_{ijk} X_k P_i]=\epsilon_{ijk}([P_i,X_k]P_i+X_k[P_i,P_i])=\epsilon_{ijk}[P_i,X_k]P_i=\epsilon_{ijk}(-i\hbar\delta_{ik})P_i$$

It seems to me like this equals zero, since the $\delta$ forces the $\epsilon_{ijk}$ to be $\epsilon_{kjk}=0$.

I know this is wrong, but what is right?

• Your expression has $i$ as an "external" index, if you also use it as a mute index with summation implied it gets confusing and you are bound to make mistakes. – fqq Jun 7 '16 at 9:43
• @fqq are you talking about $i$ the imaginary unit? I think that seeing it next to $\hbar$ prevents the confusion. But thanks, your point is taken. I still don't understand what happens here with the $\delta$ – Whyka Jun 7 '16 at 9:45
• No, I'm talking about the index. – fqq Jun 7 '16 at 9:47
• Are they not the same?.. Where should I note a difference? – Whyka Jun 7 '16 at 9:48

$$[P_i,L_j]=\varepsilon_{jkl} ([P_i, X_k]P_l+X_k[P_i,P_l])=-\mathrm{i}\hbar \varepsilon_{jkl}\delta_{ki}P_l = -\mathrm{i} \varepsilon_{jil} P_l = \mathrm{i}\varepsilon_{ijl} P_l$$
Which shows that as expected $P_k$ is a vector.
The lesson here is that you should use not use the same letter twice as an external index and a dummy index. Here $i, j$ are external indices, so when writing the cross product with implied sum you should use something like $L_j=\sum_{kl}\varepsilon_{jkl} X_kP_l=\varepsilon_{jkl} X_kP_l$.