- #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.