
#1
Apr1211, 10:04 AM

P: 80

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 



#2
Apr1211, 10:33 AM

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Hi Chronos000!
(try using the X^{2} and X_{2} icons just above the Reply box ) Maybe the second one should be ih(bar) X_{k} e_{ljk} X_{l} ? 



#3
Apr1211, 03:26 PM

P: 80

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? 



#4
Apr1211, 04:04 PM

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quantum, commutation, suffix
That is equal to 0 because X_{o}X_{m} is symmetric while e_{jmo} is antisymmetric.




#5
Apr1211, 04:19 PM

P: 80

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




#6
Apr1211, 04:25 PM

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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}.




#7
Apr1211, 04:56 PM

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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}) 



#8
Apr1211, 05:09 PM

P: 80

thanks for your replies. So my expression is just 0 + 0. I don't do anything like you have done in your final comment tinytim?
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. 



#9
Apr1211, 05:15 PM

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I only did that to prove that it works, but you can just say "it's obvious!" 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 antisymmetric indices, in which case you can if you multiply by 1). 



#10
Apr1211, 05:22 PM

P: 80

thanks I'm pretty happy with this now



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