= Solution
Twice applying the <creation operator> or twice applying the <annihilation operator> changes the <energy eigenvalue> by $2\hbar\omega$, unless they give zero. The same <orthogonality> argument as in part (i) therefore gives
$$
\mathbb E_\psi(a_+^2)=0=\mathbb E_\psi(a_-^2).
$$
Also the factorization in part (a) gives
$$
\mathbb E_\psi(a_+a_-)=2m(E-\hbar\omega/2),\qquad \mathbb E_\psi(a_-a_+)=2m(E+\hbar\omega/2).
$$
Now $P=(a_++a_-)/2$ and $X=(a_+-a_-)/(2im\omega)$. Squaring in the correct operator order yields
$$
\begin{aligned}
\mathbb E_\psi(P^2)&=\frac14\mathbb E_\psi(a_+^2+a_+a_-+a_-a_++a_-^2)=mE,\\
\mathbb E_\psi(X^2)&=-\frac1{4m^2\omega^2}\mathbb E_\psi(a_+^2-a_+a_--a_-a_++a_-^2)=\frac{E}{m\omega^2}.
\end{aligned}
$$
By part (i), the means are zero, so these are the <variances>. Taking positive square roots gives the requested <standard deviations>:
$$
\boxed{\Delta_\psi(X)=\sqrt{\frac{E}{m\omega^2}},\qquad \Delta_\psi(P)=\sqrt{mE}.}
$$
These <harmonic-oscillator number-state second moments> also show equal mean <kinetic energy> and <potential energy>, each $E/2$. Their uncertainty product is $E/\omega=(n+\tfrac12)\hbar$, so the <ground state> saturates the <Heisenberg uncertainty relation>.
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