Meaning of Harmonic oscilator correlation function In Quantum Mechanics What is the physical interpretation of the quantity $$\xi(t)=\langle 0|x(t)x(0)|0\rangle=\frac{\hbar}{2m\omega}e^{-i\omega t}$$ where $|0\rangle$ is the ground state of the harmonic oscillator? I know it is some kind of correlation function but no being able to interpret it physically. If I could interpret the states $x(0)|0⟩$ and $x(t)|0⟩$ the rest should follow.
 A: This question is rather interesting and has never been clarified in textbooks. It could be used for didactical aims to explain the Dyson-Schwinger set of equations in an introductory course of quantum mechanics. 
That expectation value $\langle x(t)x(0)\rangle$ represents the Green function of the corresponding classical equation for the harmonic oscillator. This can be understood using the technique devised by Carl M. Bender, Kimball A. Milton and Van M. Savage here. So, let us consider the quantum theory of a harmonic oscillator having action
$$
 S = \int dt\left[\frac{1}{2}m(\dot x(t))^2-\frac{1}{2}m\omega^2(x(t))^2+j(t)x(t)\right]
$$
where we have introduced a source $j(t)$ that is a c-number. Varying the action, one gets the equation of motion
$$
   m\ddot x(t)+m\omega^2x(t)=j(t).
$$
Now, using the technique of Bender at al., we take the expectation value of this equation on the ground state obtaining
$$
    m\frac{d^2}{dt^2}\langle x(t)\rangle+m\omega^2\langle x(t)\rangle = j(t).
$$
We divide this by the partition function $Z[j]=\langle 0|0\rangle$ obtaining
$$
   m\frac{d^2}{dt^2}G_1^j(t)+m\omega^2 G_1^j(t)=Z^{-1}[j]j(t)
$$
that is the one-point function of the harmonic oscillator. For $j(t)=0$ we recover the known classical equation of motion for $G_1^0(t)$. We have used the definition of correlation functions as
$$
    G_n^j=\frac{\delta^n}{\delta j(t_1)\delta j(t_2)\ldots\delta j(t_n)}\ln Z[j]
$$
and we set $j=0$ at the end of the computation. Now, we take the functional derivative with respect to $j(t)$ obtaining the equation for the two-point function. It is easy to see that, deriving the equation for $G_1^j(t)$ yields immediately
$$
   m\frac{d^2}{dt^2}G_2^j(t,t')+m\omega^2G_2^j(t,t')=-Z^{-1}[j]G_1^j(t)j(t)+Z^{-1}[j]\delta(t-t')
$$
and so, for $j=0$,
$$
   m\frac{d^2}{dt^2}G_2^0(t,t')+m\omega^2G_2^0(t,t')=\delta(t-t')
$$
where we have reabsorbed the factor $Z^{-1}[0]$ multiplying the delta with a redefinition of $G_2^0(t,t')$. We see that the case $\langle x(t)x(0)\rangle$ represents the solution for $t>0$ of this equation when $t'=0$. So, it propagates the motion of a particle under an elastic force in time when an external source is present. In quantum mechanics, it arises from the deviation of the average path that is that of a classical particle. We note that higher order correlation functions are just trivial for this case as it should be expected. It would be more interesting for a cubic potential as in the paper of Bender et al.
A: In a comment you ask:

How do you interpret the states $x(0)|0⟩$ and $x(t)|0⟩$?

Answer: You don't. You should not read the correlation function as the overlap of these two states. Instead, you should view it as the expectation value of the (Heisenberg) operator $x(t)x(0)$ with respect to the state $\lvert 0 \rangle$. Then it becomes clearly a measure of statistical correlation in the usual sense: Both $x(0)$ and $x(t)$ may be viewed as random variables, and the expectation values of both individually are zero, so $\langle x(t)x(0)\rangle$ is the covariance of these two variables. If you divide by the standard deviations you obtain Pearson's correlation coefficient.
