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#omega;) = &epsi#epsilon;(&omega#omega;) ( ,&omega#omega;) where we note the two (forward and backward) fourier transform relations: ( ,t) = 12&pi#pi;&int#int;_-&infin#infty;^&infin#infty; ( ,&omega#omega;)e^-i&omega#omega;t d&omega#omega; ( ,&omega#omega;) = 12&pi#pi;&int#int;_-&infin#infty;^&infin#infty; ( ,t')e^i&omega#omega;t' dt' and of course: ( ,t) = 12&pi#pi;&int#int;_-&infin#infty;^&infin#infty; ( ,&omega#omega;)e^-i&omega#omega;t d&omega#omega; ( ,&omega#omega;) = 12&pi#pi;&int#int;_-&infin#infty;^&infin#infty; ( ,t')e^i&omega#omega;t' dt'

Therefore:
(
,t) & = &12&pi#pi;&int#int;_-&infin#infty;^&infin#infty;
&epsi#epsilon;(&omega#omega;)
(
,&omega#omega;)e^-i&omega#omega;t d&omega#omega;

& = & 12&pi#pi;&int#int;_-&infin#infty;^&infin#infty;
&epsi#epsilon;(&omega#omega;) e^-i&omega#omega;t d&omega#omega;12&pi#pi;&int#int;_-&infin#infty;^&infin#infty;
(
,t')e^i&omega#omega;t' dt'

& = & &epsi#epsilon;_0 {
(
,t) + &int#int;_-&infin#infty;^&infin#infty;
G(&tau#tau;)
(
,t-&tau#tau;)d&tau#tau;}

where we have introduced the *susceptibility kernel*:
G(&tau#tau;) = 12&pi#pi;
&int#int;_-&infin#infty;^&infin#infty;{&epsi#epsilon;(&omega#omega;)&epsi#epsilon;_0 - 1
} e^-i&omega#omega;&tau#tau;d&omega#omega;= 12&pi#pi;
&int#int;_-&infin#infty;^&infin#infty;&chi#chi;_e(&omega#omega;) e^-i&omega#omega;&tau#tau;d&omega#omega;
(noting that
). This
equation is nonlocal in time unless
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: &chi#chi;_e = &epsi#epsilon;&epsi#epsilon;_0 - 1 = &omega#omega;_p^2&omega#omega;_0^2 - &omega#omega;^2 - i&gamma#gamma;_0&omega#omega; so G(&tau#tau;) = &omega#omega;_p^22&pi#pi; &int#int;_-&infin#infty;^&infin#infty;1&omega#omega;_0^2 - &omega#omega;^2 - i&gamma#gamma;_0&omega#omega; e^-i&omega#omega;&tau#tau;d&omega#omega;

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:
&omega#omega;_1,2 = -i&gamma#gamma;±-&gamma#gamma;^2 + 4&omega#omega;_0^22
or
&omega#omega;_1,2 = -i&gamma#gamma;2 ±&omega#omega;_01 -
&gamma#gamma;^24&omega#omega;_0^2 = -i&gamma#gamma;2 ±&nu#nu;_0
where
as long as
(as is
usually the case, remember
and
). Note that these
poles are in the *lower half plane* (LHP) because of the sign of
in the original harmonic oscillator - it was *dissipative*. This is important.

Then G(&tau#tau;) = (2&pi#pi;i)&omega#omega;_p^22&pi#pi; &conint#oint;_C1(&omega#omega;- &omega#omega;_1)(&omega#omega;- &omega#omega;_2) e^-i&omega#omega;&tau#tau;d&omega#omega;

If we close the contour in the upper half plane (UHP), we have to
restrict
(why? because otherwise the integrand will not
vanish on the contour at infinity where
has a positive
imaginary part. Since it encloses no poles,
vanishes, and
we get *no contribution from the future* in the integral above for
. 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
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, and if we do the rest of the (fairly straightforward) algebra we get: G(&tau#tau;) = &omega#omega;_p^2 e^-&gamma#gamma;&tau#tau;2 (&nu#nu;_0)&nu#nu;_0 &Theta#Theta;(&tau#tau;) where the latter is a Heaviside function to enforce the constraint.

Our last little exercise is to use complex variables and Cauchy's
theorem again. We start by noting that
and
and
are all *real*. Then we can integrate by parts and find
things like:
&epsi#epsilon;(&omega#omega;)&epsi#epsilon;_0 - 1 = iG(0)&omega#omega; -
G'(0)&omega#omega;^2 + ...
from which we can conclude that
and the like. Note the even/odd imaginary/real
oscillation in the series.
is therefore analytic in
the UHP and we can write:
&epsi#epsilon;(z)&epsi#epsilon;_0 - 1 = 12&pi#pi;i &conint#oint;_C
&epsi#epsilon;(&omega#omega;')&epsi#epsilon;_0 - 1&omega#omega;' - z d&omega#omega;'

We let where (or deform the integral a bit below the singular point on the Re( ) axis). From the Plemlj Relation: 1&omega#omega;' - &omega#omega;- i&delta#delta; = P1&omega#omega;' - &omega#omega; + i&pi#pi;&delta#delta;(&omega#omega;' - &omega#omega;) (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: &epsi#epsilon;(&omega#omega;)&epsi#epsilon;_0 = 1 + 1i&pi#pi;P &int#int;_-&infin#infty;^&infin#infty;&epsi#epsilon;(&omega#omega;')&epsi#epsilon;_0 - 1&omega#omega;' - &omega#omega; d&omega#omega;'

Although this looks like a single integral, because of the
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:
Re(&epsi#epsilon;(&omega#omega;)&epsi#epsilon;_0) & = & 1 + 1&pi#pi;P
&int#int;_-&infin#infty;^&infin#infty;
Im(&epsi#epsilon;(&omega#omega;')&epsi#epsilon;_0)&omega#omega;' - &omega#omega;
d&omega#omega;'

Im(&epsi#epsilon;(&omega#omega;)&epsi#epsilon;_0) & = & - 1&pi#pi;P
&int#int;_-&infin#infty;^&infin#infty;
Re(&epsi#epsilon;(&omega#omega;')&epsi#epsilon;_0) - 1&omega#omega;' - &omega#omega; d&omega#omega;'

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
is even in
while the imaginary
part is odd. See if you can derive this on your own for the fun of it
all...