Correlation function for the ground state of simple harmonic oscillator

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

Most of the computational mistakes for the second method have been pointed out in the comments, therefore I only focus on the matter of time dependent correlator. What you essentially want to compute are multi-time correlators, which are not so easy to formulate in the Schrödinger picture, since you only have time dependent wavefunctions.

One of the strategies to follow is to reformulate it as

$ <0|x(t) x | 0 > = < 0 | x(t) |\Psi> = < 0 (t) | x | \Psi(t) > \qquad \text{with} \quad |\Psi> = x | 0 > $

Note that $|\Psi>$ in this case is not normalised. We can then write with ladder operators and $x_0 = \sqrt{\frac{\hbar}{2 m \omega}}$

$ |\Psi> = \hat{x} | 0 > = x_0 (a^\dagger + a ) | 0 > = x_0 | 1 > \\ |\Psi(t)> = x_0 U(t,0) | 1 > = x_0 e^{-i \omega t 3/2} | 1 > \\ \hat{x} | \Psi(t) > = x_0^2 (a^\dagger + a) e^{-i \omega t 3/2} |1 > = x_0^2 e^{-i \omega t 3/2} ( | 0 > + \sqrt{2} |2> ) \\ <0| U(0,t) = <0| e^{+i \omega t / 2} $

Therefore in total we get the same result as in the Heisenberg picture $ <0|x(t) x | 0 > = x_0^2 e^{-i \omega t} $

For generic operator strings with time dependencies one should think about the Schrödinger picture as a chain of time-evolutions and application of the operator.