I'm dealing with the harmonic oscillator. I'm calculating the correlation function $$c(t) = \langle 0 | x(t) x(0) |0 \rangle$$ where $|0\rangle$ is the ground state.

I'm dealing in the Heisenberg picture, where $$x(t) = \exp(i H t / \hbar) x(0) \exp(-i H t / \hbar) \, .$$

Now, I have calculated the above using the ladder operator approach, and by explicit integration in the position representation, and I keep on getting the answer:

$$c(t) = \frac{\hbar}{2 m \omega} \exp(i\omega t)$$

How is it possible that my answer has a complex part? Surely the answer should be real since $x(t) x(0)$ is Hermitian. What is going on?

  • 1
    $\begingroup$ Are you sure that $x(0)x(t)$ is Hermitian? I would say that $[x(0), x(t)] \neq 0$ since $[x(0), H] \neq 0$. $\endgroup$
    – Korf
    Oct 18, 2016 at 7:52

1 Answer 1


Consider two Hermitian matrices \begin{align} A = A^{\dagger} \qquad B = B^{\dagger} \end{align} The product of two Hermitian matrices is not always Hermitian since $[A, B] \neq 0 $ \begin{align} (AB)^{\dagger} = B^{\dagger} A^{\dagger} = BA \neq AB \end{align}


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