= Solution
The <density operator> is diagonal in the oscillator number basis, so $\langle a^2\rangle=\langle a^{\dagger2}\rangle=0$. With $\bar n=\langle a^\dagger a\rangle$ and $aa^\dagger=a^\dagger a+1$,
$$
\langle x^2\rangle=\frac{\hbar}{2m\omega}(2\bar n+1),\qquad
\langle p^2\rangle=\frac{m\hbar\omega}{2}(2\bar n+1).
$$
Hence
$$
\boxed{\left\langle\frac{p^2}{2m}\right\rangle
=\left\langle\frac{m\omega^2x^2}{2}\right\rangle
=\frac{\hbar\omega}{4}(2\bar n+1)}.
$$
Equality holds even at zero <temperature>, when each contribution is $\hbar\omega/4$. Classical equipartition alone would miss this <zero-point energy>.
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