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Commutator maths

  1. Jan 28, 2007 #1

    quasar987

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    Is it true that [tex]ABC = CBA[/tex] implies [A,B]=[A,C]=[B,C]=0 ??

    The converse is of course true, and I cannot find a counter-exemple (ex: no 2 of the above commutation relation above are sufficient), but how is this proven?? :confused:
     
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  3. Jan 28, 2007 #2

    Dick

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    A trivial example is B=0. This can't imply [A,C]=0. Obviously.
     
  4. Jan 28, 2007 #3

    quasar987

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    Ok, I forgot to add the hypothesis that A,B,C are not null operators.
     
  5. Jan 28, 2007 #4

    Dick

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    Ok, A=C=[[0,1],[0,0]], B=[[0,0],[1,0]]. ABC=CBA is obvious but [A,B]!=0.
     
  6. Jan 28, 2007 #5

    Dick

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    Actually no need for explicit matrices. Just take any A and B that don't commute and set C=A.
     
  7. Jan 28, 2007 #6

    quasar987

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    Sorry for wasting your time Dick but twice you've demonstrated that I am trying to show something stronger than what I really need for this physics problem.

    I need to find an iff condition on A, B, C three non equal and non null hermitian operators (they're Sx,Sy and Sz, the spin operators) that makes their product ABC hermitian.

    I came up with (ABC)+=C+B+A+=CBA. And now I want to find the iff condition on A,B,C that will make this equal to ABC.

    (+ denotes hermitian conjugation)
     
  8. Jan 28, 2007 #7

    Dick

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    Yeah, 'false' is sort of by definition 'too strong'.

    Well, Sx.Sy.Sz is non-hermitian, isn't it? Are A,B,C supposed to be linear combinations of the S's? Ok, so to sum up for the product AB to be hermitian we need that A and B commute. For ABC to be hermitian we need ABC=CBA. I guess I'm fuzzing out on what the actual problem is here... Can you be more specific?
     
  9. Jan 28, 2007 #8

    quasar987

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    Yes, Sx.Sy.Sz is non-hermitian. Precisely, I need to find for which integers l,m,n is the operator [tex]S_x^lS_y^mS_z^n[/tex] an observable. I should have said that in the first place, huh. :p

    My hypothesis is that it is hermitian when at most 1 of the exponent is odd.
     
  10. Jan 28, 2007 #9

    Dick

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    You are clearly right. Since (S_i)^2=1 for i=x,y,z. There are relatively few cases to consider.
     
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