# 11.3: The Response Function and Energy Absorption

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Let’s investigate the relationship between the linear response function and the absorption of energy from the external agent—in this case an electromagnetic field. We will relate this to the absorption coefficient $$\alpha = \dot {E} / I$$ which we have described previously. For this case,

$H = H _ {0} - f (t) A = H _ {0} - \mu \cdot E (t) \label{10.64}$

This expression gives the energy of the system, so the rate of energy absorption averaged over the nonequilibrium ensemble is described by:

$\dot {E} = \dfrac {\partial \overline {H}} {\partial t} = - \dfrac {\partial f} {\partial t} \overline {A (t)} \label{10.65}$

We will want to cycle-average this over the oscillating field, so the time-averaged rate of energy absorption is

\begin{align} \dot {E} & = \dfrac {1} {T} \int _ {0}^{T} d t \left[ - \dfrac {\partial f} {\partial t} \overline {A (t)} \right] \\[4pt] & = \dfrac {1} {T} \int _ {0}^{T} d t \dfrac {\partial f (t)} {\partial t} \left[ \langle A \rangle + \int _ {0}^{\infty} d \tau R ( \tau ) f ( t - \tau ) \right] \label{10.66} \end{align}

Here the response function is

$R ( \tau ) = - i \langle [ \mu ( \tau ) , \mu ( 0 ) ] \rangle / \hbar.$

For a monochromatic electromagnetic field, we can write (and expand)

\begin{align} f (t) &= E _ {0} \cos \omega t \\[4pt] &= \dfrac {1} {2} \left[ E _ {0} e^{- i \omega t} + E _ {0}^{*} e^{i \omega t} \right] \label{10.67} \end{align}

which leads to the following for the second term in Equation \ref{10.66}:

$\dfrac {1} {2} \int _ {0}^{\infty} d \tau R ( \tau ) \left[ E _ {0} e^{- i \omega ( t - \tau )} + E _ {0}^{*} e^{i \omega ( t - \tau )} \right] = \dfrac {1} {2} \left[ E _ {0} e^{- i \omega t} \chi ( \omega ) + E _ {0}^{*} e^{i \omega t} \chi ( - \omega ) \right] \label{10.68}$

By differentiating Equation \ref{10.67}, and using it with Equation \ref{10.68} in Equation \ref{10.66}, we have

$\dot {E} = - \dfrac {1} {T} \langle A \rangle [ f ( T ) - f ( 0 ) ] - \dfrac {1} {4 T} \int _ {0}^{T} d t \left[ - i \omega E _ {0} e^{- i \omega t} + i \omega E _ {0}^{*} e^{i \omega t} \right] \left[ E _ {0} e^{- i \omega t} \chi ( \omega ) + E _ {0}^{*} e^{i \omega t} \chi ( - \omega ) \right] \label{10.69}$

We will now cycle-average this expression, setting $$T = 2 \pi / \omega$$. The first term vanishes and the cross terms in second integral vanish, because

$\dfrac {1} {T} \int _ {0}^{T} d t e^{- i \omega t} e^{+ i \omega t} = 1$

and

$\int _ {0}^{T} d t e^{- i \omega t} e^{- i \omega t} = 0.$

The rate of energy absorption from the field is

\left.\begin{aligned} \dot {E} & = \dfrac {i} {4} \omega \left| E _ {0} \right|^{2} [ \chi ( - \omega ) - \chi ( \omega ) ] \\[4pt] & = \dfrac {\omega} {2} \left| E _ {0} \right|^{2} \chi^{\prime \prime} ( \omega ) \end{aligned} \right. \label{10.70}

So, the absorption of energy by the system is related to the imaginary part of the susceptibility. Now, from the intensity of the incident field,

$I = \dfrac{c \left| E _ {0} \right|^{2}}{8 \pi}$

the absorption coefficient is

$\alpha ( \omega ) = \dfrac {\dot {E}} {I} = \dfrac {4 \pi \omega} {c} \chi^{\prime \prime} ( \omega ) \label{10.71}$