Time Derivatives of Expectation Value of X^2 in a Harmonic Oscillator

[isochore]

Homework Statement

Prove that $\frac{d^2}{dt^2} \langle \psi (t) \vert X^2 \vert \psi (t) \rangle = 4 \omega^2 \langle \psi (t) \vert X^2 \vert \psi (t) \rangle + \frac{1}{m}\langle \psi (t) \vert H \vert \psi (t) \rangle$ for the simple harmonic oscillator.

Relevant Equations

$\frac{i}{\hbar}[H, X] = \frac{1}{m} P$
$\frac{i}{\hbar}[H, P] = -m\omega^2X$
$\frac{d}{dt} \langle \psi (t) \vert A \vert \psi (t) \rangle = \frac{i}{\hbar} \langle \psi (t) \vert [H,A] \vert \psi (t) \rangle$ if A is time-independent.

I can show that $\frac{d}{dt} \langle \psi (t) \vert X^2 \vert \psi (t) \rangle = \frac{1}{m} \langle \psi (t) \vert PX+XP \vert \psi (t) \rangle$.

Taking another derivative with respect to time of this, I get $\frac{d^2}{dt^2} \langle \psi (t) \vert X^2 \vert \psi (t) \rangle = \frac{i}{m \hbar} \langle \psi (t) \vert [H, PX+XP] \vert \psi (t) \rangle$.

However, I'm not sure how to manipulate this into something that matches the right side of the equation.
I can expand the commutator in the middle to get
$[H, PX + XP] = [H, PX] + [H, XP] = [H, P]X + P[H, X] + [H, X]P + X[H,P]$

I can then substitute in the commutator values known from above (ignoring the factor of $\frac{i}{\hbar}$ momentarily):
$[H, PX + XP] = (-m\omega^2 X)X + P(\frac{1}{m}P) + (\frac{1}{m}P)P + X(-m\omega^2 X) = -2m\omega^2 X^2 + \frac{2}{m}P^2$

I see a way to get from $P^2$ to $H$ using the time-independent Schrödinger equation, but I took derivatives with respect to time to arrive at this equation. Is there another path besides using the time-independent Schrödinger equation, or is it okay to use the time-independent form here?

 

[Antarres]

You don't use time independent Schrodinger equation to pass from $P^2$ to $H$, but you use the definition of your Hamiltonian(I assume this is possibly what you meant, but saying it just in case). Hamiltonian for harmonic oscillator is, as you know:
$$H = \frac{P^2}{2m} + \frac{m\omega^2X^2}{2}$$

Edit: Also, you seem to have done this correctly so far, but I don't see the final formula appearing in that form, although it would be similar. So I'm not sure whether there's a mistake in the text of the exercise, as your formulas look all good(or maybe I missed a mistake of yours somehow).

 

[isochore]

Thanks!
I was trying something complicated with the Schrödinger equation because I thought I would get something similar but not quite the final formula if I just used the straight Hamiltonian.
Turns out all you need is a few messy algebraic substitutions to get the two sides equivalent :)