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Permutation operator

  1. Nov 16, 2007 #1

    cks

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    I can't really imagine how this was approached.

    Let [tex] P_{\alpha0} [/tex] fixed

    [tex] P_{a0}A=\frac{1}{N!}\sum_{\alpha}\epsilon_{\alpha}P_{\alpha0}P_{\alpha}=\frac{1}{N!}\epsilon_{\alpha0}\sum_{\alpha}\epsilon_{\beta}P_{\beta}=\epsilon_{\alpha0}A

    [/tex]


    I can understand that [tex] P_{\alpha0}P_{\alpha} = P_{\beta} [/tex] is a new permutation operator.

    [tex] P_{a0}A=\frac{1}{N!}\sum_{\alpha}\epsilon_{\alpha}P_{\alpha0}P_{\alpha}=\frac{1}{N!}\sum_{\alpha}\epsilon_{\alpha}P_{\beta} [/tex]
     
  2. jcsd
  3. Nov 18, 2007 #2
    can you tell me where you got this from or what is this about?

    what i think happened was that

    Ea = EB x Ea0 (sorry i dont know how to put in the greek words and the subscripts here)

    so since Ea0 is a constant it was brought out of the equation, then the summation divided by the N! was equal to A.... and hence we get the answer
     
  4. Nov 20, 2007 #3

    cks

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    Actually, P is the permutation operator that we frequently come across from a chapter of identical particles of any quantum textbook.
     
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