Handout on Kramers-Kronig relations
Refer Refer to Chapter Chapter 11 of Kittel. The purpose here is to prove prove the Kramers-Kr Kramers-Kronig onig relations. relations. Further, we show an application of using the K-K relation in an application of analyzing the reflectance spectrum. The result that we would like to get is, e.g., ǫ1 (ω ) − 1 =
2 π
P
0
∞
dω ´
ωǫ ω ´ ǫ 2 (ω ´) , ω ´ 2 − ω2
(1)
and, 2
ǫ2 (ω ) = − P π
∞
0
dω ´
´) ωǫ 1 (ω , ω ´ 2 − ω2
(2)
where P is the principal value. (This application has to do with the fact that ǫ = 1 + 4πχ , that is, χ is actually the response function function of intere interest.) st.) The usefulness usefulness is that if either ǫ1 or ǫ2 is known (experimen (experimentally tally measured), measured), the other can be calculated. calculated. We observe observe from the functional form and learn quickly that only for ´ω near ω would the contribution from ǫ2 be important to construct ǫ1 , and vice versa. versa. In practice, practice, this (limited frequenc frequency y range in integration) is helpful, since it’s impractical to measure the entire spectrum from DC for X-ray. There are several components in the proof. What we will do is to chop the proof in several components, and then use the derived results one by one. First, the linear response theory. Linear response function and susceptibility
An external influence f produces a response y of the system system.. For example example,, an externa externall electric electric field can induce induce the dipole moment moment of a gas atom. We assume here that the response is linear, and y (t) depends only on what happened before y is measured at time t. So, So, we we first assume that (A) the sample system is linear in response and (B) the causality. Thus, we define:
∞
y (t) =
dt′ K (t − t′ )f (t′ ),
(3)
−∞
where the function K is the response function, and f is the force (or called stimulation). Note that instead of a general response function, K (t, t ), we use K (t − t ) to show that the system is invariant under a translation in time. ′
′
Define τ = t − t , we have ′
∞
y (t) =
0
dτ K (τ )f (t − τ ).
(4) 1
Note here that we integrate from zero. This is because by definition, τ is larger than zero. That is, what happens at time t can be affected only by input that occurs earlier. We further assume that: (A) K is real, thus when f is real, y is real; and (B) K = 0 for τ < 0, thus,
∞
y (t) =
dτ K (τ )f (t − τ ).
(5)
−∞
Note that with the definition of K , the integration starts from negative infinity again. The nature of K can be learned by this: letting f (t − τ ) = δ (t − τ ), so
∞
y (t) =
dτ K (τ )f (t − τ ) = K (t).
(6)
−∞
That is, the physical meaning of K is that it is the response of the system under a δ -function excitation. Working in the Fourier-transformed space is often more convenient
The Fourier transform of f is:
∞
f (t) =
dωf (ω )e−iωt ,
(7)
(8)
−∞
and, 1 f (ω ) = 2π
∞
dtf (t)eiωt .
−∞
Now, plug the FT form (of f (t)) into equation (6), we get:
∞
y (t) =
∞
dτ K (τ )
−∞
where χ(ω ) =
dωf (ω )e
−∞
−iω
(t−τ )
∞
=
dωχ (ω )f (ω )e−iωt ,
(9)
−∞
∞
dτ K (τ )eiωτ .
(10)
−∞
So, by observing this functional form, we learn that the function χ(ω ), the susceptibility, is therefore the Fourier Transform of K (t). Symmetry property of the susceptibility χ
2
We learned from equation (9) that y (ω ) = χ(ω )f (ω ).
(11)
Since K (τ ) is real, we have: χ(−(ω )) = χ∗ (ω ).
(12)
So, when considering the real and the complex parts of χ, we have: χ = χ′ + iχ′′ , andχ′ (−ω ) = χ′ (ω ); χ′′(−ω ) = −χ′′ (ω ).
(13)
That is, χ (ω ) is even and χ (ω ) is odd. ′
′′
Complex ω
Normally, ω is real when doing experiments. But, it’s often more convenient to provide additional flexibility by using a complex ω (for mathmatical arguments): χ(ω ) =
∞
dτ K (τ )eiωτ .
(14)
−∞
Let ω ´ =ω ´ r + iω ´ i, then χ(´ ω) =
∞
dτ K (τ )e
´ τ iω
∞
=
−∞
dτ K (τ )eiω´
r
´ i τ τ −ω
(15)
.
−∞
And, since K (τ ) is real, ∗
χ (´ ω) =
∞
dτ K (τ )e
´
´
−iωr τ −ωi τ
−∞
=
∞
dτ K (τ )ei(−ω´
r
´ i τ ) τ +iω
∗
ω ). = χ(−´
(16)
−∞
ω ) = 0 if τ < 0, since K (τ < Because of causality, we observe (from equation (14)) that χ(´ ω ) will diverge (see equation (15)) and 0) = 0. But, if τ > 0, ω ´ i must be > 0, for otherwise χ(´ become unphysical. From this observation, we acknowledge that ω ´ i > 0, i.e., ω ´ i rests on the upper half of the ω complex plane. The nature of the property of χ(ω ) makes ω ´ i positive. So, our conclusion is that τ > 0 and ω ´ i > 0. K-K relations
What’s interesting is that χ and χ are not two independent variables. They are related by the K-K relations. In the following we use results from the above pieces and derive the K-K relations. ′
′′
3
First, if the χ(ω ) = χ (ω ) + iχ (ω ) has no singularities within ”the contour,” then ′
χ(´ ω) dω ´ =0= ω ´ −ω
′′
(1)
+
(2)
+
(3)
+
(4)
(17)
.
The ”contour” for the Cauchy principal value integral and the definition of the four segments are definedx in Kittel (see drawing in Fig. 2, chapter 11, page 311, 7th edition) and repeated here:
(1)
(2)
(3)
= =
(18)
,
−R
is the half circle above ´ω = ω, integrated clockwise,
=
and
ω−ε
C
R
ω +ε
(4)
=
(19)
,
C ′
is the half circle above ´ω = ω , integrated counter-clockwise.
ω)| goes to zero faster than | ω1´ | as ω We further take a physical assumption that |χ(´ ´ goes to ∞. Given this assumption, we have for the 4th term:
lim
R→∞,ε→0 C ′
dω ´
χ(´ ω) = 0. ω ´ −ω
(20)
The integration of half circle C can be treated by assuming that ω ´ = ω + εeiφ ,
(21)
and, thus, dω ´ = iεeiφ dφ, ω ´ − ω = εeiφ .
(22)
The 2nd term is therefore obtained to be: lim
R→∞,ε→0
χ(´ ω) dω ´ = lim ω ´ − ω R→∞,ε→0 C
0
dφ
π
χ(ω + εeiφ ) iφ iεe = −iπχ (ω ). εeiφ
(23)
The principal value is the sum of the 1st and the 3rd term: Term1 + term3 = 0 − term2 − term4 = iπχ =
χ(´ ω) ω d´ = lim P ω ´ − ω R→∞,ε→0 −∞ ∞
ω −ε
R
−
+
R
ω +ε
′
′′
= iπχ (ω ) = iπ (χ(ω ) + iχ(ω )). 4
(24)
Now, separate equation (24) into the real and imaginary parts, we reach the important conclusion: ′
χ (ω ) =
1 π
P
∞
ω d´
−∞
and
1
χ′′ (´ ω) , ω ´ −ω
(25)
χ′ (´ ω) χ (ω ) = − P ω . d´ π ω ´−ω −∞ ′′
∞
(26)
We would like to integrate over only the positve ´ω by using the facts that χ (ω ) is even and χ (ω ) is odd. So, by multiplying the expressions in (25) and (26) by ′
′′
ω ´ +ω , ω ´ +ω
(27)
and use the property that
∞
(oddfunction) = 0,
(28)
−∞
we get (from equation (25)):
1
χ′′ (´ ω)(´ ω + ω) 2 χ (ω ) = P ω d´ = P π ω π ´ 2 − ω2 −∞ ′
∞
∞
0
χ′′ (´ ω)´ ω ω 2 , d´ ω ´ − ω2
(in the last equation the integrand has this symmetry
1
χ′ (´ ω)(´ ω + ω) 2 χ (ω ) = P ω d´ = − P π ω π ´ 2 − ω2 −∞ ′′
∞
(the last integrand has symmetry
even even
∞
0
odd×odd even
χ′ (´ ω )ω ω 2 . d´ ω ´ − ω2
(29) =
even even
= even) and (30)
= even.)
Equation (29) and (30) are “one form of the K-K relations,” in the sense that the K-K relation has been widely applied. We can “identify” a physical quantity Q = Qreal + iQimaginary (a response function of a system) with χ. For example: [ǫ (ω ) − 1] → χ ǫ (ω ) → χ n(ω ) − 1 → χ κ(ω ) → χ . ′
′′
′
′′
(31)
′
′′
Application in reflectance spectrum.
5
The reflection coefficient can be represented by its amplitude and phase: r (ω ) =
E reflection n + iκ − 1 = ρ(ω )eiθ(ω) = . E incident n + iκ + 1
(32)
n is called the index of refraction, and κ is the extinction coefficient. The reflectance R = |r |2 = ρ2 . Now,
1 ln R(ω ) + iθ (ω ). 2
ln r (ω ) =
(33)
We then identify: 1 ln R(ω ) 2
→χ
′
(34)
θ(ω ) → χ′′ .
So, applying the K-K relations, θ (ω ) = −
2ω π
P
∞
0
ω d´
1 ω) lnR(´ 2 ω ´ 2 − ω2
1 =− 2π
∞
0
ω ln | d´
ω ω )) ´ + ω d(ln R(´ ω, d´ | ω ω ´ −ω d´
(35)
as shown in Kittel. So, when R is experimentally measured, θ can be calculated. Once R and θ are known, the reflection coefficient r is calculated, and serveral the other macroscopic optical constants can be subsequently calculated: n(ω ) =
1−r2 , 1+r2 −2rcos(θ) 2rsin(θ) , 1+r2 −2rcos(θ) 2 2
κ(ω ) = ǫ1 (ω ) = n − κ , ǫ2 (ω ) = 2nκ.
(36)
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