Is every density moment of a quantum harmonic oscillator a classical harmonic oscillator? Suppose we have the usual harmonic oscillator:
$$ \hat{H}=\frac{\hat{p}^2}{2m}+\frac{1}{2}m\omega^2\hat{x}^2 $$
with an arbitrary initial state. It is well known that the first density moment $\langle\hat{x}\rangle$ behaves like a classical harmonic oscillator as a function of time, and this can be shown with Ehrenfest's Theorem.
I suspect the same applies for higher-order density moments - that is, that $\langle \hat{x}^n\rangle$ for $n\in\mathbb{N}_0$ behaves like a classical harmonic oscillator. Is this true? How can you derive it? I tried using the Heisenberg equation of motion, but there didn't seem to be a nice way to simplify the Heisenberg-picture commutators.
Essentially, I want to show that:
$$\frac{d^2\langle\hat{x}^n\rangle}{dt^2}\propto-\langle\hat{x}^n\rangle$$
which is definitely true for $n=1$, and intuition leads me to suspect is true for higher moments.
 A: I will show that the higher moments do not behave like classical harmonic oscillators by giving an explicit counterexample.
The counterexample is about a coherent state in the quantum harmonic oscillator.
The coherent state $|\alpha\rangle$ is defined as the eigenstate
$$ \hat a |\alpha\rangle = \alpha |\alpha\rangle $$
of the annihilation operator for some complex $\alpha$.
It is well known that the time evolution keeps a coherent state coherent; in fact,
$$ |\psi(t)\rangle = |\alpha_t\rangle , \quad \alpha_t = \alpha_0\, \mathrm e^{\mathrm i \omega t} $$
is a solution of the Schrödinger equation.
For simplicity, I will assume that $\alpha_0$ is real.
The first moment is
$$ \langle \hat q \rangle = \sqrt{\frac{\hbar}{2m\omega}} \langle \alpha_t | (\hat a + \hat a^\dagger) | \alpha_t \rangle = \sqrt{\frac{2\hbar}{m\omega}} \alpha_0 \cos(\omega t) . $$
This is obviously a harmonic motion, as demanded by the Ehrenfest theorem.
However, already the second moment
\begin{align}
  \langle \hat q^2 \rangle &= \frac{\hbar}{2m\omega} \langle \alpha_t | (\hat a^2 + (\hat a^\dagger)^2 + 2\hat a^\dagger \hat a + 1) | \alpha_t \rangle \\
  &= \frac{\hbar}{2m\omega} \left( 2\alpha_0^2 \cos(2\omega t) + 2\alpha_0^2 + 1 \right)
\end{align}
is not harmonic.
