# 12.7: Relation to Spectra

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

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

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

\begin{align*} \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} \\[4pt] &=\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} \end{align*}

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

Define a frequency-dependent susceptibility by

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

then

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

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

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

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

Recall that

\begin{align*} \Phi_{AB}(t) &= {i\over\hbar}\langle [A(t),B(0)]\rangle _0 \\[4pt] &= {i \over \hbar}\langle[e^{iH_0t/\hbar} Ae^{-iH_0t\hbar},B]\rangle _0 \end{align*}

Under time reversal, we have

\begin{align*} \Phi_{AB}(-t) &= {i \over \hbar} \langle \left[e^{-iH_0t/\hbar}Ae^{iH_0t/\hbar},B\right]\rangle _0 \\[4pt] &= { {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 } \\[4pt] &= {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 \\[4pt] &= {i \over \hbar} \langle \left(AB(t)-B(t)A\right)\rangle _0 \\[4pt] &= -{i \over \hbar} \langle \left[B(t),A\right]\rangle \\[4pt] &= -\Phi_{BA}(t) \end{align*}

Thus,

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

and if $$A = B$$, then

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

Therefore

\begin{align*} \chi_{BB}(\omega) &= \lim_{\epsilon\rightarrow 0^+}\int_0^{\infty} dt\;e^{-i(\omega-i\epsilon t)}\Phi_{BB}(t) \\[4pt] &=\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] \\[4pt] &={\rm Re}(\chi_{BB}(\omega)) - i{\rm Im}(\chi_{BB}(\omega)) \end{align*}

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

\begin{align*} {\rm Re}(\chi_{BB}(\omega) &= {\rm Re}(\chi_{BB}(-\omega) \\[4pt] {\rm Im}(\chi_{BB}(\omega) &= -{\rm Im}(\chi_{BB}(-\omega) \end{align*}

so that $${\rm Im}(\chi_{BB}(\omega))$$ is positive for $${ \omega > 0 }$$ and negative for $${ \omega < 0 }$$. 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

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

This page titled 12.7: Relation to Spectra is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Mark Tuckerman.