# Correlation Function of ground state; Physical Meaning

I was asked to find the correlation function of the ground state of the QHM: $$\langle0|\hat x(t)\hat x(t-\tau)|0\rangle$$ I found that this evaluated to $\frac{\hbar}{2m\omega}e^{i\omega \tau}$.

I was then asked to take the fourier transform of this (w.r.t $\tau$), which evaluates to $\delta(x-\omega)$ (times constants). I am curious as to what the interpretation of this would be? The correlation function shows that the number states of the harmonic oscillator oscillate, even though expectation of x and p vanish. But the fourier spectrum of x would suggest that it only oscillates at a given frequency?

• You already found the answer yourself! – Vibert Mar 2 '14 at 3:58
• What does QHM stand for? Do you mean QHO (quantum harmonic oscillator)? – JeffDror Mar 2 '14 at 16:30

First I should mention that I got a slightly different result for the correlation function, namely, $\frac{ \hbar }{ 2 m \omega } e ^{ \frac{ 3 }{ 2} \hbar \omega \tau }$ (though I may have made a mistake).
The correlation function, $$\left\langle 0 \right| \hat{x} ( t ) \hat{x} ( t - \tau ) \left| 0 \right\rangle$$ has the following interpretation: If the position of the ground state is measured once, then again a time $\tau$ then the correlation function gives the amplitude of measuring the ground state again.
The Fourier transform of the amplitude is given by, $$\int \frac{ d \omega' }{ 2\pi } e ^{ -i \omega ' \tau } \frac{ \hbar }{ 2 m \omega' } e ^{ \frac{ 3 }{ 2} \hbar \omega \tau } = \frac{ \hbar }{ 2 m \omega } \delta \left( \omega ' - \omega \right)$$ I purposely use $\omega'$ to denote the conjugate variable to $\tau$ since frequency (or equivalently, energy) is the conjugate (not position) to time.
What you found is that the amplitude for the correlation function oscillates and the oscillation frequency is a single frequency, $\omega$. I don't think I would interpret this as that the number of states of the harmonic oscillator oscillate''.
• The correlation function should actually have a $t+\tau$ not a $t-\tau$. Would this explain the difference? – yankeefan11 Mar 3 '14 at 0:48