# Relation to spectra

Suppose that $$F_e (t)$$ is a monochromatic field

$F_e(t) = F_{\omega}e^{i\omega t}e^{\epsilon t}$

where the parameter $$\epsilon$$ insures that field goes to 0 at $$\underline { t = - \infty }$$. We will take $$\underline {\epsilon\rightarrow 0^+ }$$ at the end of the calculation. The expectation value of  $$B$$ then becomes

 $$\langle B(t)\rangle$$ $$\langle B\rangle _0 + \int_{-\infty}^t\;ds\;\Phi_{BB}(t-s)F_{\omega}e^{i\omega s}e^{\epsilon s}$$ $$\langle B\rangle _0 + F_{\omega}e^{(i\omega + \epsilon)t} \int_0^{\infty}d\tau\Phi_{BB}(\tau)e^{-i(\omega-i\epsilon)\tau}$$

where the change of integration variables $$\underline {\tau=t-s }$$ has been made.

Define a frequency-dependent susceptibility by

$\chi_{BB}(\omega-i\epsilon) = \int_0^{\infty}d\tau \Phi_{BB}(\tau) e^{-i(\omega-i\epsilon)\tau}$

then

$\langle B(t)\rangle = \langle B\rangle _0 + F_{\omega}e^{i\omega t}e^{\epsilon t}\chi_{BB}(\omega-i\epsilon)$

If we let $$z=\omega-i\epsilon$$, then we see immediately that

$\chi_{BB}(z) = \int_0^{\infty}d\tau\;\Phi_{BB}(\tau) e^{-iz\tau}$

i.e., the susceptibility is just the Laplace transform of the after effect function or the time correlation function.

Recall that

$\Phi_{AB}(t) = {i\over\hbar}\langle [A(t),B(0)]\rangle _0 = {i \over \hbar}\langle[e^{iH_0t/\hbar} Ae^{-iH_0t\hbar},B]\rangle _0$

Under time reversal, we have

 $$\Phi_{AB}(-t)$$ $$\underline { {i \over \hbar} \langle \left[e^{-iH_0t/\hbar}Ae^{iH_0t/\hbar},B\right]\rangle _0}$$ $$\underline { {i \over \hbar} \langle \left(e^{-iH_0t/\hbar}Ae^{iH_0t/\hbar}B -Be^{-iH_0t/\hbar}Ae^{iH_0t/\hbar}\right)\rangle _0 }$$ $$\underline { {i \over \hbar} \langle \left(Ae^{iH_0t/\hbar}Be^{-iH_0t/\hbar} -e^{iH_0t/\hbar}Be^{-iH_0t/\hbar}A\right)\rangle _0 }$$ $$\underline { {i \over \hbar} \langle \left(AB(t)-B(t)A\right)\rangle _0 }$$ $$\underline {-{i \over \hbar} \langle \left[B(t),A\right]\rangle = -\Phi_{BA}(t) }$$

Thus,

$\Phi_{AB}(-t) = -\Phi_{BA}(t)$

and if  $$A = B$$, then

$\Phi_{BB}(-t) = -\Phi_{BB}(t)$

Therefore

 $$\chi_{BB}(\omega)$$ $$\lim_{\epsilon\rightarrow 0^+}\int_0^{\infty} dt\;e^{-i(\omega-i\epsilon t)}\Phi_{BB}(t)$$ $$\lim_{\epsilon\rightarrow 0^+}\int_0^{\infty}dt\;e^{-\epsilon t}\left[\Phi_{BB}(t)\cos\omega t - i\Phi_{BB}(t)\sin\omega t\right]$$ $${\rm Re}(\chi_{BB}(\omega)) - i{\rm Im}(\chi_{BB}(\omega))$$

From the properties of $$\Phi_{BB}(t)$$ it follows that

 $${\rm Re}(\chi_{BB}(\omega)$$ $${\rm Re}(\chi_{BB}(-\omega)$$ $${\rm Im}(\chi_{BB}(\omega)$$ $$-{\rm Im}(\chi_{BB}(-\omega)$$

so that $${\rm Im}(\chi_{BB}(\omega))$$ is positive for $$\underline { \omega > 0 }$$ and negative for $$\underline { \omega < 0 }$$. It is a straightforward matter, now, to show that the energy difference $$Q (\omega )$$ derived in the lecture from the Fermi golden rule is related to the susceptibility by

$Q(\omega) = 2\omega\vert F_{\omega}\vert^2{\rm Im}(\chi_{BB}(\omega))$