Categories:
Complex analysis,
Mathematics,
Optics,
Physics.
Kramers-Kronig relations
Let χ(t) be a complex function describing
the response of a system to an impulse f(t) starting at t=0.
The Kramers-Kronig relations connect the real and imaginary parts of χ(t),
such that one can be reconstructed from the other.
Suppose we can only measure χr(t) or χi(t):
χ(t)=χr(t)+iχi(t)
Assuming that the system was at rest until t=0,
the response χ(t) cannot depend on anything from t<0,
since the known impulse f(t) had not started yet,
This principle is called causality, and to enforce it,
we use the Heaviside step function
Θ(t) to create a causality test for χ(t):
χ(t)=χ(t)Θ(t)
If we Fourier transform this equation,
then it will become a convolution in the frequency domain
thanks to the convolution theorem,
where A, B and s are constants from the FT definition:
χ~(ω)=(χ~∗Θ~)(ω)=B∫−∞∞χ~(ω′)Θ~(ω−ω′)dω′
We look up the FT of the step function Θ~(ω),
which involves the signum function sgn(t),
the Dirac delta function δ,
and the Cauchy principal value P.
We arrive at:
χ~(ω)=∣s∣ABP∫−∞∞χ~(ω′)(πδ(ω−ω′)+isgnω−ω′1)dω′=(21∣s∣2πAB)χ~(ω)+i(2πsgn(s)∣s∣2πAB)P∫−∞∞ω−ω′χ~(ω′)dω′
From the definition of the Fourier transform we know that
2πAB/∣s∣=1:
χ~(ω)=21χ~(ω)+sgn(s)2πiP∫−∞∞ω−ω′χ~(ω′)dω′
We isolate this equation for χ~(ω)
to get the final version of the causality test:
χ~(ω)=−sgn(s)πiP∫−∞∞ω−ω′χ~(ω′)dω′
By inserting χ~(ω)=χ~r(ω)+iχ~i(ω)
and splitting the equation into real and imaginary parts,
we get the Kramers-Kronig relations:
χ~r(ω)χ~i(ω)=sgn(s)π1P∫−∞∞ω′−ωχ~i(ω′)dω′=−sgn(s)π1P∫−∞∞ω′−ωχ~r(ω′)dω′
If the time-domain response function χ(t) is real
(so far we have assumed it to be complex),
then we can take advantage of the fact that
the FT of a real function satisfies
χ~(−ω)=χ~∗(ω), i.e. χ~r(ω)
is even and χ~i(ω) is odd. We multiply the fractions by
(ω′+ω) above and below:
χ~r(ω)χ~i(ω)=sgn(s)(π1P∫−∞∞ω′2−ω2ω′χ~i(ω′)dω′+πωP∫−∞∞ω′2−ω2χ~i(ω′)dω′)=−sgn(s)(π1P∫−∞∞ω′2−ω2ω′χ~r(ω′)dω′+πωP∫−∞∞ω′2−ω2χ~r(ω′)dω′)
For χ~r(ω), the second integrand is odd, so we can drop it.
Similarly, for χ~i(ω), the first integrand is odd.
We therefore find the following variant of the Kramers-Kronig relations:
χ~r(ω)χ~i(ω)=sgn(s)π2P∫0∞ω′2−ω2ω′χ~i(ω′)dω′=−sgn(s)π2ωP∫0∞ω′2−ω2χ~r(ω′)dω′
To reiterate: this version is only valid if χ(t) is real in the time domain.
References
- M. Wubs,
Optical properties of solids: Kramers-Kronig relations, 2013,
unpublished.