I am given the following problem about fluctuation dissipation theorem:
Consider an external force $f(t)= \frac{f_0}{2}(e^{i\omega_0 t}+e^{-i\omega_0 t})$ acting on a particle with momentum $p=mv$ in some environment. The total Hamiltonian reads:
$$H = \frac{p^2}{2m}+U(\{q,p\}_{env})-\frac{p f(t)}{m},$$
where $env$ stands for the environment positions and momenta.
I am asked to find the average dissipation rate. And I am given that $\langle v(\omega)\rangle=\chi_v(\omega)f(\omega)$, where $\chi_v(\omega)$ is the velocity response function. The answer should be as follows:
$$\Big\langle \frac{dE}{dt}\Big\rangle=\frac{1}{2}f_0^2 \chi_v^{\prime\prime}(\omega_0)\omega_0$$ where $\chi^{\prime\prime}_v(\omega_0)$ denotes the imaginary part of $\chi_v(\omega_0)$.
My trial:
We can find $\langle v(t)\rangle$ by using the inverse Fourier transform:
$$\langle v(t)\rangle=\frac{1}{2\pi}\int^{\infty}_{-\infty}\chi_v(\omega)f(\omega) e^{-i\omega t} d\omega.$$
We note that the Fourier transform of the force $f(t)$ is given by:
$$f(\omega) = f_0 \pi[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)].$$
We therefore find that:
$$\langle v(t) \rangle = \frac{f_0}{2}\int^{\infty}_{-\infty}\chi_v(\omega)[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)] e^{-i \omega t}d\omega.$$
From this I find that $\langle v(t) \rangle$ is given by:
$$\langle v(t) \rangle = \frac{f_0}{2} (\chi_v(\omega_0)e^{-i\omega_0 t}+\chi_v(-\omega_0)e^{i\omega_0 t}).$$
Now I used the following to find the dissipation of the energy of the particle:
$$\frac{dE}{dt} = f(t)\langle v(t) \rangle = \frac{f^2_0}{4}(e^{i\omega_0 t}+e^{-i\omega_0 t})(\chi_v(\omega_0)e^{-i\omega_0 t}+\chi_v(-\omega_0)e^{i\omega_0 t}).$$
We can average this over some time period $T$ as follows:
$$\Big\langle \frac{dE}{dt} \Big\rangle = \frac{1}{T}\int^T_0 \frac{dE}{dt} dt = \frac{f_0^2}{4T} \int^T_0 (\chi_v (\omega_0 )+\chi_v(-\omega_0)+e^{2i \omega_0 t}\chi_v(-\omega_0)+e^{-2i \omega_0 t}\chi_v (\omega_0))dt.$$
Since we take $T = 2\pi/\omega_0$ the last two terms give a vanishing contribution to the integral. We therefore find that:
$$\Big\langle \frac{dE}{dt} \Big\rangle = \frac{f^2_0}{4}(\chi_v(\omega_0)+\chi_v(-\omega_0)).$$
Then we see that because of $\chi_v(\omega_0) = \chi^\prime_v(\omega_0)+i\chi^{\prime \prime}_v (\omega_0)$ we can write:
$$\chi_v(\omega_0)+\chi_v(-\omega_0) = 2 \chi^\prime_v(\omega_0),$$ where $\chi^\prime$ denotes the real part of $\chi$, we can do this because the real part is even and the imaginary part is odd in $\omega_0$. But then I find the following:
$$\Big\langle \frac{dE}{dt} \Big \rangle = \frac{f^2_0}{2}\chi_v^\prime(\omega_0),$$
but this is different from what I have to proof.
Question: can somebody spot the difference or help me with this problem?