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... $\endgroup$
    – rsaavedra
    Nov 6, 2016 at 1:48
  • $\begingroup$ I just checked again. I think I calculated expectation value of PX. $\endgroup$
    – Asaduzz
    Nov 6, 2016 at 1:54
  • $\begingroup$ Oh, you mean second time. yes, I think so. Let me redo that part. $\endgroup$
    – Asaduzz
    Nov 6, 2016 at 1:57
  • 1
    $\begingroup$ Indeed you should obtain the expected result, I did the integral and found $-i\hbar/2$, so check your calculation. $\endgroup$
    – rsaavedra
    Nov 6, 2016 at 2:20
  • 1
    $\begingroup$ Just noticed, initially when I posted the question: in the second method, I not only calculated $\langle XP \rangle$ instead of $\langle PX \rangle$, but also calculated that quantity wrong. That should be $\langle XP \rangle =i \frac{\hbar}{2}$. $\endgroup$
    – Asaduzz
    Nov 6, 2016 at 4:32

2 Answers 2


Because you should calculate $\int \Psi^*_0(x)(-i\hbar)\frac{\partial}{\partial x}x\Psi_0(x)$

  • $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Oct 20, 2023 at 19:05

You have to consider that in the second case, the wave functions need to carry the time dependence as the operators in Schrödinger-Picture are assumed to be time independent. Thus the equation should look like:

$$ \left<\mathbf{X}(t)\mathbf{X}(0)\right>=\left<x,\ t\right| \mathbf{X}^2\left|x,\ 0 \right> $$ This however would in position representation with wave-functions reduce to

$$ \int\psi^*\left(x,\ t \right) x^2\psi\left(x,\ 0\right) \mathrm{d} x $$ and thus the time-dependence from $$ \psi\left(x,\ t\right)=\psi\left(x,\ 0\right) \mathrm{e}^{-i\omega t} $$

should remain, giving the same result for both cases.


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