Response functions for the quantum harmonic oscillator I'm going through problems in Quantum Field Theory for the Gifted Amateur, and have been trying to understand a problem on the forced quantum oscillator [$L = \frac{1}{2}\dot{x}(t)^2-\frac{1}{2}m\omega^2x(t)^2+f(t)x(t)$ ] and response functions. 
The response function is
$$
\langle\psi(t)|\hat{x}(t)|\psi(t)\rangle = \int_{-\infty}^\infty \mathrm{d}t'\chi(t-t')f(t')
$$
I want to show, using the interaction representation, that to first order in the force function $f_I(t)$ 
$$
|\psi_I(t)\rangle = |0\rangle + i\int_{-\infty}^t \mathrm{d}t'f_I(t')\hat{x}_I(t')|0\rangle
$$
Here is what I've done so far:
I started by taylor expanding the interacting ket:
$$ |\psi_I(t)\rangle = e^{i \hat{H_0}(t)t}|\psi(t)\rangle = |\psi(t)\rangle + i \hat{H_0}(t)t|\psi(t)\rangle+O(H_0^2)
$$
but I am confused about how to relate the wave function to the ground state, and how to use the information I have about the response function. When you have an expression for $|psi\rangle$ there is a procedure for finding the expectation value. I don't know how to go the other way and around and extract a ket from the response function.
I also note that I can convert the response function to the interaction picture and it will have the same value, and that I can change f(t) to the interaction picture $f_I(t) = e^{i H_0 t}f(t)e^{-i H_0 t} =f(t) + O(H_0^2)$ since $H_0$ and f(t) commute.
Related: linear response for a simple harmonic oscillator
 A: I notice the problem was raised 5 years ago, however here is my solution.  This question is actually kubo formula, here is a reference if you are interested: https://www.damtp.cam.ac.uk/user/tong/kintheory/four.pdf.
$$|\psi_I (t)\rangle =\hat{U}_I(t,-\infty)|0\rangle =\hat{T}[e^{-\int_{-\infty}^{t}dt' \hat{H}_{I}(t')}]|0\rangle$$
Note the interaction Hamiltonian is given by: $$\hat{H}_{I}(t)=-e^{-iH_0 t}fxe^{iH_{0}t}=-f\hat{x}_I(t)$$ note that $$f(t)$$ is a real function but not an operator so $$f(t)=f_{I}(t)$$ as time evolution operator has no effect on it.
Substitute the expression into the dyson series, expand to first order gives: $$|\psi_{I}(t) \rangle=\hat{T}[|0\rangle+i\int_{-\infty}^{t}f_{I}(t)\hat{x}_{I}(t)|0\rangle]$$
Here is the trick to link up expectation value and response function:$$\langle \psi (t)|\hat{x}(t)|\psi(t) \rangle=\langle \psi_I (t)|\hat{x}_{I}(t)|\psi_I (t) \rangle $$ now substitute the approximation of previous step and taking only first order consideration you get $$\langle 0| \hat{x}_{I}(t)|0\rangle +i\int_{-\infty}
^{t}f(t) \langle 0| [\hat{x}_I(t),\hat{x}_{I}(t')]|0\rangle dt'$$ .
The first term vanish as the expectation value of displacement in ground state is zero, massage the expression by adding a heaviside step function gives $$i\theta(t-t')\int_{-\infty}^{\infty}f(t) \langle 0| [\hat{x}_I(t),\hat{x}_{I}(t')]|0\rangle dt'$$comparing with the expression Blundell provided you obtain the expression he provided.
A: The insight I was missing was that the $e^{i \hat{H_0}(t)} t|\psi(t)\rangle$ can be thought of more abstractly as $U(t)|\psi(0)\rangle $, where $U(t) = e^{-i \hat{H}(t)t}$. When the ket starts out in the ground state, doing the Taylor expansion and substitution of the interaction picture for $f$ and $x$ works.
