2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ You already found the answer yourself! $\endgroup$ – Vibert Mar 2 '14 at 3:58
  • 1
    $\begingroup$ What does QHM stand for? Do you mean QHO (quantum harmonic oscillator)? $\endgroup$ – JeffDror Mar 2 '14 at 16:30
1
$\begingroup$

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, \begin{equation} \left\langle 0 \right| \hat{x} ( t ) \hat{x} ( t - \tau ) \left| 0 \right\rangle \end{equation} 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.

Every position measurement collapses the wavefunction into delta function like state and then it spreads out depending on how long afterwords you wait before the next measurement (since the ground state wavefunction of a harmonic oscillator is not a position eigenstate).

The Fourier transform of the amplitude is given by, \begin{equation} \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) \end{equation} 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''.

$\endgroup$
  • $\begingroup$ The correlation function should actually have a $t+\tau$ not a $t-\tau$. Would this explain the difference? $\endgroup$ – yankeefan11 Mar 3 '14 at 0:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.