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Is a projection operator hermitian?

  1. Dec 23, 2009 #1
    I was reading Lie Algebras in Physics by Georgi.......................second edition...

    Theorem 1.2: He proves that every finite group is completely reducible.

    He takes

    [tex]PD(g)P=D(g)P[/tex]


    ..takes adjoint...and gets..

    [tex]P{D(g)}{\dagger} P=P {D(g)}{\dagger} [/tex]

    So..does this mean that the projection operator P is hermitian?
     
    Last edited: Dec 23, 2009
  2. jcsd
  3. Dec 23, 2009 #2
    I think he is assuming P is Hermitian (as it must be; if you want to think of this in terms of QM, it leaves all states unchanged, so the eigenvalue associated with the operator is '1', a real number - any operator which outputs a real eigenvalue is Hermitian)
     
  4. Dec 23, 2009 #3

    Fredrik

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    All projection operators are hermitian. You might want to review the properties of projection operators here.
     
  5. Dec 24, 2009 #4
    Thanks a lot!!!:smile:
     
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