14.3: The Response Function and Energy Absorption

Let’s investigate the relationship between the linear response function and the absorption of energy from the external agent. The external agent performs work on the internal variable \(A\), and the energy needed to move \(A\) from equilibrium is ultimately dissipated into other system degrees of freedom. We can calculate the rate of energy absorption, averaged over the nonequilibrium ensemble, from the time-derivative in the expectation value of the Hamiltonian

\[\begin{aligned}
\mathcal{P}_{\text{abs}} &= \frac{\partial \bar H}{\partial t} = - \frac{\partial f}{\partial t}\overline {A(t)} \\
& = \frac{\partial f(t)}{\partial t}\left[ \left\langle A \right\rangle + \int_0^{\infty } d\tau \, R_{AA} ( \tau )f\left( t - \tau \right) \right] \\
\end{aligned} \label{eq15.4.01}\]

For this example, we take the internal variable to be a dipole moment \(\mu \) and the external agent to be an electromagnetic field \(\mathcal{E}\), and we will relate \(\dot E\) to the absorption cross-section \(\sigma\). The Hamiltonian is

\[H = H_0 - f(t) A = H_0 - \mathcal{E}(t)\mu \label{eq15.4.02}\]

We assume that the system is driven by a monochromatic electromagnetic field with a time dependence given by

\[f(t) = \mathcal{E}_0\cos \omega t = \frac{1}{2}\left[ \mathcal{E}_0e^{ - \mathrm{i}\omega t} + \mathcal{E}_0^{*}e^{\mathrm{i}\omega t} \right] \label{eq15.4.03}\]

With such a field, energy will flow in and out of the system as it oscillates, so to obtain a time-averaged rate of energy absorption, we will want to cycle average eq. (\ref{eq15.4.01}) over a period \(T\):

\[\overline {\mathcal{P}} _{\text{abs}} = \frac{1}{T}\int_0^{T} dt \, \frac{\partial \mathcal{E}(t)}{\partial t}\left[ \left\langle \mu \right\rangle + \int_0^{\infty } d\tau \, R_{\mu \mu }(\tau) f\left( t - \tau \right) \right] \label{eq15.4.04}\]

There are two terms in the integral. Over many oscillations of the field, the first term here averages to zero, since \(\langle A \rangle\) is constant. Inserting eq. (\ref{eq15.4.03}) into the second term results in a transformation of the response function into a susceptibility:

\[\frac{1}{2}\int_0^{\infty } d\tau \, R(\tau)\left[ \mathcal{E}_0 e^{ - \mathrm{i}\omega \left( t - \tau \right)} + \mathcal{E}_0^{*} e^{\mathrm{i}\omega \left( t - \tau \right)} \right] = \frac{1}{2}\left[ \mathcal{E}_0 e^{ - \mathrm{i}\omega t}\chi ( \omega ) + \mathcal{E}_0^{*} e^{\mathrm{i}\omega t} \chi (- \omega) \right] \label{eq15.4.05}\]

By differentiating (\ref{eq15.4.03}) and inserting both it and (\ref{eq15.4.05}) into eq. (\ref{eq15.4.04}), we obtain

\[\overline {\mathcal{P}} _{\text{abs}} = - \frac{1}{4T}\int_0^{T} dt \left[ - \mathrm{i}\omega \mathcal{E}_0e^{ - \mathrm{i}\omega t} + \mathrm{i}\omega \mathcal{E}_0^{*}e^{\mathrm{i}\omega t} \right] \left[ \mathcal{E}_0e^{ - \mathrm{i}\omega t}\chi ( \omega ) + \mathcal{E}_0^{*}e^{\mathrm{i}\omega t}\chi (- \omega) \right] \label{eq15.4.06}\]

The result is four terms, of which two oscillate at \(2\omega\) and two at zero frequency. Then we average this expression over one cycle of the field, setting \(T = { 2\pi / \omega }\). The \(2\omega\) terms in the integral vanish, and the zero-frequency terms are retained because \(\tfrac{1}{T}\int_0^{T} dt \, e^{ - \mathrm{i}\omega t} e^{ + \mathrm{i}\omega t} = 1 \) and \(\int_0^{T} dt \, e^{ - \mathrm{i}\omega t} e^{ - \mathrm{i}\omega t} = 0\).

The resulting rate of energy absorption from the field is given by

\[\begin{aligned}
\overline {\mathcal{P}} _{\text{abs}} &= \frac{\mathrm{i}}{4}\omega \left| \mathcal{E}_0 \right|^{2} \left[ \chi (- \omega) - \chi ( \omega ) \right] \\
& = \frac{\omega }{2}\left| \mathcal{E}_0 \right|^{2} \chi '' ( \omega ) \\
\end{aligned} \label{eq15.4.07}\]

Hence the absorption of energy by the system is proportional to the imaginary part of the susceptibility. Now, from the intensity of the incident field, \(I = \tfrac{1}{2}\varepsilon _0c\mathcal{E}_0^{2}\), the absorption cross-section is

\[\sigma ( \omega ) = \frac{\overline {\mathcal{P}} _{\text{abs}}}{I} = \frac{\omega }{\varepsilon _0c}\chi '' ( \omega ) \label{eq15.4.08}\]

This is a general result that is not specific to the current example. An external agent’s interaction with an internal variable induces energy absorption in the system that is proportional to the imaginary part of the susceptibility. Although it is not addressed in the derivation, the energy absorbed must ultimately be dissipated into the system degrees of freedom. Therefore \(\chi'' \) is also associated with describing the dissipation of energy.

Next, let’s relate the imaginary part of the susceptibility to the underlying correlation function \(C_{AA}(t)\). Using the definition of \(\chi'' (\omega)\), eq. (15.1.24), the relation of the complex susceptibility to the response function, eq. (15.1.17), \(\tilde \chi ( \omega ) = \int_0^{\infty } d\tau \, R(\tau)e^{\mathrm{i}\omega \tau }\), and the relationship of \(R(\tau)\) to the imaginary part of \(C_{AA}(t)\), one can show

\[\begin{aligned}
\chi '' ( \omega ) & = \frac{1}{2\hbar }\left[ \int_{ - \infty }^{\infty } d\tau \left( C_{AA} (\tau) - C_{AA} \left( - \tau \right) \right)e^{\mathrm{i}\omega \tau } \right] \\
& = \frac{1}{2\hbar }\left( C_{AA} ( \omega ) - C_{AA} (- \omega) \right) \\
\end{aligned} \label{eq15.4.09}\]

That is, \(\chi'' (\omega)\) is proportional to the difference in Fourier transform power spectra for the correlation function \(C_{AA}(t)\) evaluated at \(\omega\) and \(-\omega\). These terms are related by the detailed balance expression \(C_{AA} (- \omega) = e^{ - \beta \hbar \omega }C_{AA} ( \omega )\), and therefore

\[\chi '' ( \omega ) = \frac{1}{2\hbar }C_{AA} ( \omega )\left( 1 - e^{ - \beta \hbar \omega } \right) \label{eq15.4.10}\]

---