Determining the commutation relation of operators - Einstein summation notation

[jmcelve]

Determining the commutation relation of operators -- Einstein summation notation

Homework Statement

Determine the commutator [L_i, C_j].

Homework Equations

L_i = \epsilon_{ijk}r_j p_k
C_i = \epsilon_{ijk}A_j B_k
[L_i, A_j] = i \hbar \epsilon_{ijk} A_k
[L_i, B_j] = i \hbar \epsilon_{ijk} B_k

The Attempt at a Solution

To be clear, the goal of this procedure is to become familiar with Einstein summation notation. That said, I've broken open C_j = \epsilon_{jmn}A_m B_n and expanded the commutator accordingly. My problem is opening up L_i in a meaningful way that gives me a nice identity with four deltas. Will I have to expand two epsilons with 6 different indices into deltas, or is there any way to get the epsilons to share an index that will give me the result (which I know to be [L_i, C_j] = i \hbar \epsilon_{ijk}C_k)?

 

[Hypersphere]

Well, since the two free indices i and j are expressedly independent you can't set i=j (it's more interesting to calculate the commutator for two specific terms than the sum anyway). And the other four indices are already contracted, so you can't contract them with either i or j. You will just have to use all six indices.

 

[jmcelve]

Well, since the two free indices i and j are expressedly independent you can't set i=j (it's more interesting to calculate the commutator for two specific terms than the sum anyway). And the other four indices are already contracted, so you can't contract them with either i or j. You will just have to use all six indices.

That's super ugly. We really only talked about expanding epsilons in terms of deltas when there were four differing indices, not six. Is there a method to do this problem that requires using only four indices?

 

[Hypersphere]

That's super ugly. We really only talked about expanding epsilons in terms of deltas when there were four differing indices, not six. Is there a method to do this problem that requires using only four indices?

It's not pretty, I agree, but then again the Levi-Civita symbol is what you use to hide the ugly things. It's possible to group the terms somewhat, but you can't remove degrees of freedom unless you now that one or more of the vector components happen to be zero... Just be happy that you don't work in four dimensions.

 

[jmcelve]

Just be happy that you don't work in four dimensions.

Noted. Thanks for the help.