Linear Response And path integral I'm following Wen's book on Quantum field theory, and I'm struggling with section 2.2.1 on linear response and response functions.
Specifically I'm unable to reproduce equation 2.2.7 in which the book computes the linear response of the harmonic oscillator for the ground state in which the perturbation is on x , and the observable state is also x
From the response function we have:
$$
D(t,t') = -i\Theta(t-t') <\psi_0| [\hat{O}_1(t), \hat{O}_2(t')]|\psi_0>
$$
And as I mentioned, I'm interested in the computation in which $\hat{O}_1 = \hat{O}_2=\hat{x}$
The following steps are the ones I'm struggling on.
$$
D(t,t') = -i\Theta(t-t') <0| [\hat{x}(t), \hat{x}(t')]|0> = -2\Theta(t-t')<0|\hat{x}^2|0>sin(w(t-t'))
$$
And then, if the perturbation is $\hat{O}_1 = -q\hat{x}\varepsilon(t) = -q\hat{x}\varepsilon e^{-0^+|t|} $
$$
d = \int_{-\infty}^{\infty}D(t-t')e^{-0^+|t'|}(-\varepsilon)dt' = -q^2D_{\omega=0}\varepsilon = \dfrac{2q^2<0|\hat{x}|0>}{\omega_0}\varepsilon
$$
With $D_\omega = \int_{-\infty}^\infty D(t)e^{i\omega t}dt$
I would be very grateful if someone could explain to me the algebra, as I have been unable to reproduce the results.
I have typed the equations as they appear on the book, save for 2 changes:


*

*I replaced the charge $e$ with $q$ to avoid possible confusion with the exponentials

*I wrote all $x$ as operators 
On a final note, the book does not make any mention on the quadratic term of the harmonic oscillator, so I don't know if its written as $\omega$ or $\omega_0$, I can only think that the different angular frecuencies come from a typo.
 A: In my copy of the book, the expression for the harmonic oscillator Hamiltonian is given as $H=\dfrac{1}{2m}(p^2+m^2\omega_0^2x^2)$, so there is no typo, but copies may differ.
I consider Heisenberg operators $x(t)$ and $p(t)$, and define $p_0 \equiv p(0)$ and $x_0 = x(0)$. Standard quantum mechanics textbooks, such as Messiah chapter XII, will give the essential results on the harmonic oscillator, in particular (Messiah XII.43):
$<n|x^2|n> = \dfrac{E_n}{m \omega_0^2}$, so that $<0|x^2|0> = \dfrac{\hbar}{2m \omega_0}$.
Concerning your questions: For the first part, your comment is close to a proper derivation: take $x(t) = x_0 \cos(\omega_0 t) + \dfrac{p_0}{m\omega_0} \sin(\omega_0 t)$ (Messiah XII. 34) but keep in mind that $x_0$ and $p_0$ are operators, and they do not commute: $[x_0,p_0] = i\hbar$. If you take this into account, you find $$[x(t), x(t')] = -i\dfrac{\hbar}{m \omega_0} \sin(\omega_0 (t -  t')) = -2i<0|x^2|0> \sin(\omega_0 (t -  t')).$$ This shows the first relation you were looking for: 
$$D(t-t') = -2<0|x^2|0> \Theta(t-t') \sin(\omega_0 (t -  t'))$$
With this you can handle the second point of your question:
$$d=2e^2 \mathcal{E} <0|x^2|0> \int_{-\infty}^t e^{0^+t'} \sin(\omega_0(t-t'))dt' \\
=2e^2 \mathcal{E} <0|x^2|0> \int_{-\infty}^t \dfrac{e^{i\omega_0(t-t')}-e^{i\omega_0(t-t')}}{2i}e^{0^+t'}=\dfrac{2e^2 \mathcal{E} <0|x^2|0>}{\omega_0}.$$
