Categories:
Complex analysis,
Mathematics,
Optics,
Physics.
Kramers-Kronig relations
Let χ(t) be the response function of a system
to an external impulse f(t), which starts at t=0.
Assuming initial equilibrium, the principle of causality
states that there is no response before the impulse,
so χ(t)=0 for t<0.
To enforce this, we demand that χ(t) satisfies a causality test,
where Θ(t) is the Heaviside step function:
χ(t)=χ(t)Θ(t)
If we take the Fourier transform (FT)
χ(t)→χ~(ω) of this equation,
the right-hand side becomes a convolution in the frequency domain
thanks to the convolution theorem,
where A, B and s are constants determined by
how we choose to define our FT:
χ~(ω)=(χ~∗Θ~)(ω)=B∫−∞∞χ~(ω′)Θ~(ω−ω′)dω′
We look up the full expression for Θ~(ω),
which involves the signum function sgn(t),
the Dirac delta function δ,
and the Cauchy principal value P.
Inserting that, we arrive at:
χ~(ω)=∣s∣ABP∫−∞∞χ~(ω′)(πδ(ω−ω′)+iω−ω′sgn(s))dω′=(22∣s∣πAB)χ~(ω)+isgn(s)(2π2π∣s∣AB)P∫−∞∞ω−ω′χ~(ω′)dω′
From the definition of the FT we know that
2πAB/∣s∣=1, so this reduces to:
χ~(ω)=21χ~(ω)+isgn(s)2π1P∫−∞∞ω−ω′χ~(ω′)dω′
We rearrange this equation a bit to get the final version of the causality test:
χ~(ω)=isgn(s)π1P∫−∞∞ω−ω′χ~(ω′)dω′
Next, we split χ~(ω)
into its real and imaginary parts,
i.e. χ~(ω)=χ~r(ω)+iχ~i(ω):
χ~r(ω)+iχ~i(ω)=isgn(s)π1P∫−∞∞ω−ω′χ~r(ω′)dω′−sgn(s)π1P∫−∞∞ω−ω′χ~i(ω′)dω′
This equation can likewise be split into real and imaginary parts,
leading to the Kramers-Kronig relations,
which enable us to reconstruct χ~r(ω)
from χ~i(ω) and vice versa:
χ~r(ω)χ~i(ω)=−sgn(s)π1P∫−∞∞ω−ω′χ~i(ω′)dω′=sgn(s)π1P∫−∞∞ω−ω′χ~r(ω′)dω′
The sign of these expressions deserves special attention:
it depends on an author’s choice of FT definition via sgn(s),
and, to make matters even more confusing,
many also choose to use the opposite sign in the denominator,
i.e. they write ω′−ω instead of ω−ω′.
In the special case where χ(t) is real,
we can take advantage of the property that
the FT of a real function always satisfies
χ~(−ω)=χ~∗(ω).
Here, this means that χ~r(ω) is even
and χ~i(ω) is odd.
To use this fact, we simultaneously
multiply and divide the integrands by ω+ω′:
χ~r(ω)χ~i(ω)=−sgn(s)π1(P∫−∞∞ω2−ω′2ωχ~i(ω′)dω′+P∫−∞∞ω2−ω′2ω′χ~i(ω′)dω′)=sgn(s)π1(P∫−∞∞ω2−ω′2ωχ~r(ω′)dω′+P∫−∞∞ω2−ω′2ω′χ~r(ω′)dω′)
In χ~r(ω)’s equation, the first integrand is odd,
so the integral’s value is zero.
Similarly, for χ~i(ω), the second integrand is odd, so we drop it too.
We thus arrive at the following common variant of the Kramers-Kronig relations,
only valid for real χ(t):
χ~r(ω)χ~i(ω)=−sgn(s)π2P∫0∞ω2−ω′2ω′χ~i(ω′)dω′=sgn(s)π2P∫0∞ω2−ω′2ωχ~r(ω′)dω′
Note that we have modified the integration limits
using the fact that the integrands are even,
leading to an extra factor of 2.
References
- M. Wubs,
Optical properties of solids: Kramers-Kronig relations, 2013,
unpublished.