= Solution
If $a_-\psi\ne0$, it has energy $E-\hbar\omega$, distinct from the energy of $\psi$. <Eigenstates> of a <self-adjoint operator> with distinct <eigenvalues> are <orthogonal>. Explicitly,
$$
E\langle\psi,a_-\psi\rangle=\langle H\psi,a_-\psi\rangle=\langle\psi,H(a_-\psi)\rangle=(E-\hbar\omega)\langle\psi,a_-\psi\rangle,
$$
so $\langle\psi,a_-\psi\rangle=0$. If $a_-\psi=0$, the same conclusion is immediate.
Dividing by $\|\psi\|^2$ and using the definition of $a_-$ gives
$$
\mathbb E_\psi(P)-im\omega\mathbb E_\psi(X)=0.
$$
Both <expectation values> are real because $P$ and $X$ are <self-adjoint operators>. Its real and imaginary parts therefore vanish separately:
$$
\boxed{\mathbb E_\psi(X)=0,\qquad \mathbb E_\psi(P)=0.}
$$
An oscillator <energy eigenstate> is centered in both position and momentum, despite having nonzero <quantum uncertainties>.
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