# Correlation function for the ground state of simple harmonic oscillator

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.

Again,

$$\int \psi_0^*(x)\ x (-~i \hbar)\frac{\partial}{\partial x} \psi_0 ~\mathrm{d}x=0,$$ 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.

• Are you sure of your integral for calculating $<PX>$? because it seems to me that you are calculating $<XP>$, remember these do not commute... Nov 6, 2016 at 1:48
• I just checked again. I think I calculated expectation value of PX. Nov 6, 2016 at 1:54
• Oh, you mean second time. yes, I think so. Let me redo that part. Nov 6, 2016 at 1:57
• Indeed you should obtain the expected result, I did the integral and found $-i\hbar/2$, so check your calculation. Nov 6, 2016 at 2:20
• 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 $\langle XP \rangle =i \frac{\hbar}{2}$. Nov 6, 2016 at 4:32

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

$$\left<\mathbf{X}(t)\mathbf{X}(0)\right>=\left$$ 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 $$\psi\left(x,\ t\right)=\psi\left(x,\ 0\right) \mathrm{e}^{-i\omega t}$$