Why is the expectation value of Hermitian conjugate operators $RR^\dagger$ always real and non-negative?

Question

I've been reading through a derivation of the wavefunctions and energy levels for the quantum harmonic oscillator. It defines

$$\hat R^\pm=\frac{1}{\sqrt{2}}[\hat p \pm \mathrm{i}\omega \hat q]$$ in mass-weighted coordinates $q=x\sqrt{\mu},$ such that $\hat p= -\mathrm{i}\hbar(\mathrm d/\mathrm dq)$ and $\hat q$ is the position operator.

It then asserts that that $\langle \psi|\hat R^+\hat R^-|\psi\rangle$ must be real and non-negative, since $\hat R^+$ and $\hat R^-$ are hermitian conjuagtes. I know that $(\hat A\hat B)^\dagger = \hat B^\dagger \hat A^\dagger$ and so $\hat R^+\hat R^-$ is a hermitian operator meaning that the bra-ket must be real, but I can't see why it should be non-negative. Could anyone shed some light on this?

Answer 1

Let $\hat{R}^-|\psi\rangle = |\psi'\rangle$ (a new ket; we don't care what it is). If you take the adjoint of this equation you get

$$\langle\psi'| = \langle\psi|(\hat{R}^-)^\dagger = \langle\psi|\hat{R}^+$$

and hence

$$\langle\psi|\hat{R}^+\hat{R}^-|\psi\rangle = \langle\psi'|\psi'\rangle$$

which by definition of the inner product must be real and non-negative.