Quantum, commutation, suffix

  1. 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. jcsd
  3. tiny-tim

    tiny-tim 26,043
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    Hi Chronos000! :smile:

    (try using the X2 and X2 icons just above the Reply box :wink:)
    That's just 2ih(bar) Xk ekjl Xl. :redface:

    Maybe the second one should be ih(bar) Xk eljk Xl ? :confused:
     
  4. I done it again

    got ih Xo ejmo Xm + ih ejmo Xm Xo

    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?
     
  5. vela

    vela 12,726
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    That is equal to 0 because XoXm is symmetric while ejmo is antisymmetric.
     
  6. 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
     
  7. vela

    vela 12,726
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    XoXm is symmetric because you can swap the subscripts without changing anything, i.e. XoXm=XmXo. The tensor ejmo on the other hand is antisymmetric because it changes sign when you exchange a pair of subscripts, i.e. ejmo=-ejom.
     
  8. tiny-tim

    tiny-tim 26,043
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    Hi Chronos000! :smile:

    sorry, I wasn't thinking straight earlier :redface:

    vela is right, ejmo is antisymmetric, so if you multiply it by something symmetric (in 2 of the 3 indices in ejmo), you get 0

    (when you do all the adding, for example you add ej12X1X2 and ej21X2X1 … the latter is -ej12X2X1 = -ej12X1X2)
     
  9. 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.
     
  10. tiny-tim

    tiny-tim 26,043
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    That's right …

    I only did that to prove that it works, but you can just say "it's obvious!" :wink:
    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).
     
  11. thanks I'm pretty happy with this now
     
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