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.

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    $\begingroup$ what is the meaning of $\langle x^3\rangle$ behaves like a classical harmonic oscillator?” $\endgroup$ Commented Aug 16, 2019 at 2:06

1 Answer 1


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.

  • $\begingroup$ This is extremely helpful, thank you! $\endgroup$
    – user502382
    Commented Aug 18, 2019 at 11:48

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