- #1

Nikitin

- 735

- 27

## Homework Statement

Hi.

As you all know, ##<F> = <F>^*##, where ##F## is an observable linked with the operator ##\hat{F}##. This means ## <F> = \int \Psi^* \hat{F} \Psi d\tau = <F>^* = [\int \Psi^* (\hat{F} \Psi) d \tau]^* = \int \Psi (\hat{F} \Psi)^* d \tau \Rightarrow \int \Psi (\hat{F} \Psi)^* d \tau = \int \Psi^* \hat{F} \Psi d\tau## (btw, what theorem allows you to ignore the integral sign when using complex conjugation on a function under an integral?).

OK. But my question is, why does that mean

##\int (\hat{F} \Psi_1)^* d \Psi_2\tau = \int \Psi_1^* \hat{F} \Psi_2 d\tau##, if the the wavefunctions ##\Psi_1## and ##\Psi_2## are quadratically integrable? This is what my lecture notes say, and it confuses me.

## The Attempt at a Solution

Have no idea.