1. The problem statement, all variables and given/known data The question asks for [Xi^2, Lj] I can get to the line: ih(bar) Xk ekjl Xl + ih(bar) ekjl Xk Xl this line must be zero but I don't see how it is. It looks like an expansion of a commutation between Xk and Xl but not quite right
Hi Chronos000! (try using the X^{2} and X_{2} icons just above the Reply box ) That's just 2ih(bar) X_{k} e_{kjl} X_{l}. Maybe the second one should be ih(bar) X_{k} e_{ljk} X_{l} ?
I done it again got ih X_{o} e_{jmo} X_{m} + ih e_{jmo} X_{m} X_{o} can I switch the order of the last X operators and put a negative in front? which would give me a commutation why did u say the previous solution was 2ih(bar) Xk ekjl Xl when the order was different in the last expression?
I'm sure I've seen something like this before but I really don't have a clue how you can see which is symmetric or not
X_{o}X_{m} is symmetric because you can swap the subscripts without changing anything, i.e. X_{o}X_{m}=X_{m}X_{o}. The tensor e_{jmo} on the other hand is antisymmetric because it changes sign when you exchange a pair of subscripts, i.e. e_{jmo}=-e_{jom}.
Hi Chronos000! sorry, I wasn't thinking straight earlier … vela is right, e_{jmo} is antisymmetric, so if you multiply it by something symmetric (in 2 of the 3 indices in e_{jmo}), you get 0 (when you do all the adding, for example you add e_{j12}X_{1}X_{2} and e_{j21}X_{2}X_{1} … the latter is -e_{j12}X_{2}X_{1} = -e_{j12}X_{1}X_{2})
thanks for your replies. So my expression is just 0 + 0. I don't do anything like you have done in your final comment tiny-tim? Are you allowed to change the position of the epsilon yes? as it does really "act" on anything but just tells you what the cross product is.
That's right … I only did that to prove that it works, but you can just say "it's obvious!" You can change the position of any whole thing … what you can't do is change the position of indices (unless they're symmetric indices, in which case you can, or unless they're anti-symmetric indices, in which case you can if you multiply by -1).