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Kramers-Kronig Relations

We find KK relations by playing looped games with Fourier Transforms. We begin with the relation between the electric field and displacement at some particular frequency $\omega$:

\begin{displaymath}
\mbox{\boldmath$D$}(\mbox{\boldmath$x$},\omega) = \epsilon(\omega)\mbox{\boldmath$E$}(\mbox{\boldmath$x$},\omega)
\end{displaymath} (9.148)

where we note the two (forward and backward) fourier transform relations:
\begin{displaymath}
\mbox{\boldmath$D$}(\mbox{\boldmath$x$},t) = \frac{1}{\sqrt{...
...oldmath$D$}(\mbox{\boldmath$x$},\omega)e^{-i\omega t} d\omega
\end{displaymath} (9.149)


\begin{displaymath}
\mbox{\boldmath$D$}(\mbox{\boldmath$x$},\omega) = \frac{1}{\...
...\mbox{\boldmath$D$}(\mbox{\boldmath$x$},t')e^{i\omega t'} dt'
\end{displaymath} (9.150)

and of course:
\begin{displaymath}
\mbox{\boldmath$E$}(\mbox{\boldmath$x$},t) = \frac{1}{\sqrt{...
...oldmath$E$}(\mbox{\boldmath$x$},\omega)e^{-i\omega t} d\omega
\end{displaymath} (9.151)


\begin{displaymath}
\mbox{\boldmath$E$}(\mbox{\boldmath$x$},\omega) = \frac{1}{\...
...\mbox{\boldmath$E$}(\mbox{\boldmath$x$},t')e^{i\omega t'} dt'
\end{displaymath} (9.152)

Therefore:

$\displaystyle \mbox{\boldmath$D$}(\mbox{\boldmath$x$},t)$ $\textstyle =$ $\displaystyle \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty
\epsilon(\omega)\mbox{\boldmath$E$}(\mbox{\boldmath$x$},\omega)e^{-i\omega t} d\omega$  
  $\textstyle =$ $\displaystyle \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty
\epsilon(\omega) e^{-i...
..._{-\infty}^\infty
\mbox{\boldmath$E$}(\mbox{\boldmath$x$},t')e^{i\omega t'} dt'$  
  $\textstyle =$ $\displaystyle \epsilon_0 \left\{ \mbox{\boldmath$E$}(\mbox{\boldmath$x$},t) + \...
...y}^\infty
G(\tau) \mbox{\boldmath$E$}(\mbox{\boldmath$x$},t-\tau)d\tau \right\}$ (9.153)

where we have introduced the susceptibility kernel:

\begin{displaymath}
G(\tau) = \frac{1}{2\pi}
\int_{-\infty}^{\infty}\left\{\frac...
...\int_{-\infty}^{\infty}\chi_e(\omega) e^{-i\omega\tau}d\omega
\end{displaymath} (9.154)

(noting that $\epsilon(\omega) = \epsilon_0(1 + \chi_e(\omega))$). This equation is nonlocal in time unless $G(\tau)$ is a delta function, which in turn is true only if the dispersion is constant.

To understand this, consider the susceptibility kernel for a simple one resonance model (more resonances are just superposition). In this case, recall that:

\begin{displaymath}
\chi_e = \frac{\epsilon}{\epsilon_0} - 1 = \frac{\omega_p^2}{\omega_0^2
- \omega^2 - i\gamma_0\omega}
\end{displaymath} (9.155)

so
\begin{displaymath}
G(\tau) = \frac{\omega_p^2}{2\pi} \int_{-\infty}^{\infty}\fr...
...ega_0^2
- \omega^2 - i\gamma_0\omega} e^{-i\omega\tau}d\omega
\end{displaymath} (9.156)

This is an integral we can do using contour integration methods. We use the quadratic formula to find the roots of the denominator, then write the factored denominator in terms of the roots:

\begin{displaymath}
\omega_{1,2} = \frac{-i\gamma \pm \sqrt{-\gamma^2 + 4\omega_0^2}}{2}
\end{displaymath} (9.157)

or
\begin{displaymath}
\omega_{1,2} = \frac{-i\gamma}{2} \pm \omega_0\sqrt{1 -
\frac{\gamma^2}{4\omega_0^2}} = \frac{-i\gamma}{2} \pm \nu_0
\end{displaymath} (9.158)

where $\nu_0 \approx \omega_0$ as long as $\omega_0 \gg \gamma/2$ (as is usually the case, remember $\beta$ and $\alpha/2$). Note that these poles are in the lower half plane (LHP) because of the sign of $\gamma$ in the original harmonic oscillator - it was dissipative. This is important.

Then

\begin{displaymath}
G(\tau) = (2\pi i)\frac{\omega_p^2}{2\pi}
\oint_C\frac{1}{(\omega - \omega_1)(\omega - \omega_2) }
e^{-i\omega\tau}d\omega
\end{displaymath} (9.159)

If we close the contour in the upper half plane (UHP), we have to restrict $\tau < 0$ (why? because otherwise the integrand will not vanish on the contour at infinity where $\omega$ has a positive imaginary part. Since it encloses no poles, $G(\tau < 0)$ vanishes, and we get no contribution from the future in the integral above for $\mbox{\boldmath$E$}$. The result appears to be causal, but really we cheated - the ``causality'' results from the damping term, which represents entropy and yeah, gives time an arrow here. But it doesn't really break the symmetry of time in this problem and if our model involved a dynamically pumped medium so that the wave experienced gain moving through it (an imaginary term that was positive) we would have had poles in the UHP and our expression for $\mbox{\boldmath$E$}$ would not be ``causal''. Really it is equally causal in both cases, because the fourier transforms involved sample all times anyway.

If we close the integrand in the LHP, $\tau>0$ and if we do the rest of the (fairly straightforward) algebra we get:

\begin{displaymath}
G(\tau) = \omega_p^2 e^{-\frac{\gamma\tau}{2}} \frac{\sin(\nu_0)}{\nu_0}
\Theta(\tau)
\end{displaymath} (9.160)

where the latter is a Heaviside function to enforce the $\tau>0$ constraint.

Our last little exercise is to use complex variables and Cauchy's theorem again. We start by noting that $\mbox{\boldmath$D$}$ and $\mbox{\boldmath$E$}$ and $G(\tau)$ are all real. Then we can integrate by parts and find things like:

\begin{displaymath}
\frac{\epsilon(\omega)}{\epsilon_0} - 1 = i\frac{G(0)}{\omega} -
\frac{G'(0)}{\omega^2} + ...
\end{displaymath} (9.161)

from which we can conclude that $\epsilon(-\omega) =
\epsilon^*(\omega^*)$ and the like. Note the even/odd imaginary/real oscillation in the series. $\epsilon(\omega)$ is therefore analytic in the UHP and we can write:
\begin{displaymath}
\frac{\epsilon(z)}{\epsilon_0} - 1 = \frac{1}{2\pi i} \oint_...
...rac{\epsilon(\omega')}{\epsilon_0} - 1}{\omega' - z} d\omega'
\end{displaymath} (9.162)

We let $z = \omega + i\delta$ where $\delta \to 0_+$ (or deform the integral a bit below the singular point on the Re($\omega$) axis). From the Plemlj Relation:

\begin{displaymath}
\frac{1}{\omega' - \omega - i\delta} = P\frac{1}{\omega' - \omega} +
i\pi\delta(\omega' - \omega)
\end{displaymath} (9.163)

(see e.g. Wyld, Arfkin). If we substitute this into the integral above along the real axis only, do the delta-function part and subtract it out, cancel a factor of 1/2 that thus appears, we get:
\begin{displaymath}
\frac{\epsilon(\omega)}{\epsilon_0} = 1 + \frac{1}{i\pi}P
\i...
...psilon(\omega')}{\epsilon_0} -
1}{\omega' - \omega } d\omega'
\end{displaymath} (9.164)

Although this looks like a single integral, because of the $i$ in the denominator it is really two. The real part of the integrand becomes the imaginary part of the result and vice versa. That is:

$\displaystyle {\rm Re}\left(\frac{\epsilon(\omega)}{\epsilon_0}\right)$ $\textstyle =$ $\displaystyle 1 + \frac{1}{\pi}P
\int_{-\infty}^\infty \frac{ {\rm
Im}\left(\frac{\epsilon(\omega')}{\epsilon_0}\right)}{\omega' - \omega }
d\omega'$ (9.165)
$\displaystyle {\rm Im}\left(\frac{\epsilon(\omega)}{\epsilon_0}\right)$ $\textstyle =$ $\displaystyle - \frac{1}{\pi}P
\int_{-\infty}^\infty \frac{ {\rm
Re}\left(\frac{\epsilon(\omega')}{\epsilon_0}\right) - 1}{\omega' - \omega } d\omega'$ (9.166)

These are the Kramers-Kronig Relations. They tell us that the dispersive and absorptive properties of the medium are not independent. If we know the entire absorptive spectrum we can compute the dispersive spectrum and vice versa. There is one more form of the KK relations given in Jackson, derived from the discovery above that the real part of $\epsilon(\omega)$ is even in $\omega$ while the imaginary part is odd. See if you can derive this on your own for the fun of it all...


next up previous contents
Next: Plane Waves Assignment Up: Plane Waves Previous: Wave Attenuation in Two   Contents
Robert G. Brown 2014-08-19