# Why is the correlation function for position operator in the harmonic oscillator complex?

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?

• Are you sure that $x(0)x(t)$ is Hermitian? I would say that $[x(0), x(t)] \neq 0$ since $[x(0), H] \neq 0$. – Korf Oct 18 '16 at 7:52

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}