I calculated correlation function $C(t)=\langle x(t)x(0)\rangle$ for ground state of Simple Harmonic Oscillator (SHO) in two different methods. But the results do not match.

First Attempt:

From Heisenberg equations of motion, $$\mathbf{X}(t)=\mathbf{X}(0)\cos(\omega t)+\frac{\mathbf{P}(0)}{m \omega} \sin(\omega t)$$

So, calculated the required terms,

$$\langle 0|\mathbf{X}^2(0)|0\rangle =\frac{\hbar}{2 m \omega}$$ and

$$\langle 0| \mathbf{P}(0) \mathbf{X}(0) |0\rangle =-~\frac{i \hbar}{2} $$

Using the above two terms and the equation of $\mathbf{X}(t)$ I obtain, $$\langle 0| \mathbf{X}(t)\mathbf{X}(0)|0\rangle =\frac{\hbar}{2 m \omega} \exp(-~i\omega t)$$

This is the required correlation function.

Second Method:

I attempted to solve it using explicit ground state wave function in position basis. In this case, I obtain,

$$\int \psi_0^*(x)\ x^2 \psi_0(x)~ \mathrm{d}x =\frac{\hbar}{2m \omega},$$ which is similar to the first calculation.


$$\int \psi_0^*(x)\ x (-~i \hbar)\frac{\partial}{\partial x} \psi_0 ~\mathrm{d}x=0,$$ which DOES NOT match with the first method.

And hence, using expression for $X(t)$ like the first method I obtained,

$$\langle X(t)X(0)\rangle =\frac{\hbar}{2m \omega} \cos (\omega t)$$

So method 1 and method 2 does not match. But are not they supposed to match up? I can not figure out where I am making mistake. Any help figuring out the mistakes will be very much appreciated.

  • $\begingroup$ Are you sure of your integral for calculating $<PX>$? because it seems to me that you are calculating $<XP>$, remember these do not commute... $\endgroup$ – Saavestro Nov 6 '16 at 1:48
  • $\begingroup$ I just checked again. I think I calculated expectation value of PX. $\endgroup$ – Asaduzz Nov 6 '16 at 1:54
  • $\begingroup$ Oh, you mean second time. yes, I think so. Let me redo that part. $\endgroup$ – Asaduzz Nov 6 '16 at 1:57
  • $\begingroup$ Indeed you should obtain the expected result, I did the integral and found $-i\hbar/2$, so check your calculation. $\endgroup$ – Saavestro Nov 6 '16 at 2:20
  • $\begingroup$ Thank you so much, got the same answer. Now I am wondering what does this imaginary expectation value means for PX operator? P and X are Hermitian, so this is bugging me that expectation value of PX gives imaginary result. $\endgroup$ – Asaduzz Nov 6 '16 at 2:48

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