Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    ..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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook