 Solution

If $a_{-}\psi \ne 0$, it has energy $E-\hslash \omega$, distinct from the energy of $\psi$. Eigenstates of a self-adjoint operator with distinct eigenvalues are orthogonal. Explicitly,so $⟨\psi ,a_{-}\psi ⟩=0$. If $a_{-}\psi =0$, the same conclusion is immediate.

Dividing by $\Vert \psi {\Vert }^2$ and using the definition of $a_{-}$ givesBoth expectation values are real because $P$ and $X$ are self-adjoint operators. Its real and imaginary parts therefore vanish separately:An oscillator energy eigenstate is centered in both position and momentum, despite having nonzero quantum uncertainties.