Simple question concerning Hermitian operators

1. Dec 8, 2014

Nikitin

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. Dec 8, 2014

Orodruin

Staff Emeritus
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 $H^* = H$.

3. Dec 8, 2014

Nikitin

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