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Simple question concerning Hermitian operators

  1. Dec 8, 2014 #1
    Hi. This might sound like a stupid question, but is it, in general, true that ##(\hat{H} \psi)^* \psi'= \psi^* \hat{H}^*\psi'##? Here ##\hat{H}## is a hermitian operator and ## \psi## a wave function.

    I.e. do they switch places even when not inside an inner product? I am aware of the fact that you can move hermitian operators around inside an inner product, i.e. <H X| Y> = <X | H Y>, but I am unsure as to why you can do that. Can somebody link me to some proofs that explain all this?

    thanks.

    edit: fixed mistake in my question..
     
    Last edited: Dec 8, 2014
  2. jcsd
  3. Dec 8, 2014 #2

    Orodruin

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    This is the definition of a self-adjoint operator so naturally a self-adjoint operator is going to fulfil it.

    With regards to your original question, if you see the wave function simply as a function here you may be in trouble. Consider instead ##\langle \psi| H## as a linear map from the state space to complex numbers.

    Writing down the inner product in position representation, you will get that the linear maps defined by multiplication with ##(H\psi)^*## and acting with ##\psi^* H^*## on any wave function before integration will give the same result (by definition of the adjoint, that H is self-adjoint gives ##H^* = H##.
     
  4. Dec 8, 2014 #3
    arghh I forgot to put a function to the right of the expression I was asking about, sorry about that. It's fixed now. Anyway:

    what is position representation? I'm not sure if I follow your explanation..
     
    Last edited: Dec 8, 2014
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