In part of a problem I'm asked to obtain an expression for the norm of $\hat{A}\psi$ in terms of the norm of $\psi$, i.e. I would like to simplify
$\int_{-\infty}^\infty (\hat{A}\psi)^*(\hat{A}\psi)dx$
where $*$ denotes complex conjugation.
The operator $\hat{A}$ is defined as $\hat{A} = \hat{p} - im\omega \hat{x}$.
($\hat{p}$ and $\hat{x}$ are the usual momentum and position operators)
Now I get
$(\hat{A}\psi)^*(\hat{A}\psi) = (\hat{p}\psi - im\omega \hat{x}\psi)^*(\hat{A}\psi)$
= $(\hat{p}\psi^*+im\omega \hat{x}\psi^*)(\hat{p}\psi - im\omega \hat{x}\psi)$
(since $\hat{x}^* = \hat{x}$ and $\hat{p}^* = \hat{p}$ - they are Hermitian)
= $\hat{p}\psi^*\hat{p}\psi - im\omega\hat{p}\psi^*\hat{x}\psi + im\omega \hat{x}\psi^*\hat{p}\psi + m^2\omega\hat{x}\psi^*\hat{x}\psi$
I suspect I have made a mistake somewhere because I can't see how to obtain the norm of $\psi$ from this expression.
What am I doing wrong?
You misunderstood Hermiticity. $\hat A$ is not hermitian, so $\hat A^\dagger = \hat A^*\neq \hat A$. Hence $$\int_{-\infty}^\infty (\hat{A}\psi)^*(\hat{A}\psi)dx=\int_{-\infty}^\infty \psi^*(\hat{A}^* ~\hat{A}\psi)~dx\\. =\int_{-\infty}^\infty \psi^*((\hat{p}^2+m^2\omega^2 \hat x^2-m\hbar \omega ~ )\psi)~dx , $$ from where you should be able to pick it up if the functions ψ are eigenfunctions of the oscillator hamiltonian...