Causality and the Kramers-Kronig Relations

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3/20/14 Causality and the Kramers-Kronig relations lamp.tu-graz.ac.at/~hadley/ss2/linearresponse/causality.php?print 1/7 Causality and the Kramers-Kronig relations Causality describes the temporal relationship between cause and effect. A bell rings after you strike it, not before you strike it. This means that the function that describes the response of a bell to being struck must be zero until the time that the bell is struck. Consider a particle of mass moving in a viscous fluid . The differential equation that describes this system is, Here is the damping constant, is the velocity, and is a driving force . A special case for the driving force is a δ-function force which strikes the system at . The solution to the differential equation for a δ-function drive force is called the impulse response function . The symbol is used because the impulse response function is sometimes called the Green's function. The solution to this equation is, Where is the decay time . The utility of the impulse response function is that any driving force can be thought of as being built up of many δ-function forces. -3 -2 -1 0 1 2 3 0.0 0.2 0.4 0.6 0.8 1.0 1.2

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Transcript of Causality and the Kramers-Kronig Relations

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3/20/14 Causality and the Kramers-Kronig relations

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Causality and the Kramers-Kronig relations

Causality describes the temporal relationship between cause and effect. A bell rings after you strike

it, not before you strike it. This means that the function that describes the response of a bell to being

struck must be zero until the time that the bell is struck. Consider a particle of mass moving in a

viscous fluid. The differential equation that describes this system is,

Here is the damping constant, is the velocity, and is a driving force. A special case for the

driving force is a δ-function force which strikes the system at . The solution to the differential

equation for a δ-function drive force is called the impulse response function . The symbol

is used because the impulse response function is sometimes called the Green's function.

The solution to this equation is,

Where is the decay time .

The utility of the impulse response function is that any driving force can be thought of as being built

up of many δ-function forces.

m

m + bv = F (t).dvdt

b v F(t)t = 0

g(t) g

m + bg = δ(t).dg

dt

g(t) = exp(−t/τ),1m

τ τ = m/b

g

m

-3 -2 -1 0 1 2 30.0

0.2

0.4

0.6

0.8

1.0

1.2

t/τ

F (t) = δ(t − )F( )d∞

′ ′ ′

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By superposition, the response to a driving force is a sum of the impulse response functions.

A special driving force is a harmonic driving force, . The response will occur at the

same frequency as the driving force, . To show this, insert a harmonic force into the

equation above.

Since the integral is over , a factor of can be put inside the integral.

Make a change of variables: , , and reverse the limits of integration.

The only time dependence of is the factor of because the variable gets integrated out.Thus a harmonic driving force produces a harmonic response where,

The generalized susceptibility is the ratio of response to driving force.

The generalized susceptibility is the Fourier transform of the impulse response function. For the caseof a particle moving in a viscous fluid,

F (t) = δ(t − )F( )d∫−∞

∞t′ t′ t′

F(t)

v(t) = g(t − )F( )d∫−∞

∞t′ t′ t′

F(t) = F(ω)eiωt

v(t) = v(ω)eiωt

v(t) = g(t − )F (ω) d∫−∞

∞t′ eiωt′

t′

t′ e−iωt

v(t) = g(t − )F (ω) deiωt ∫−∞

∞t′ e−iω(t− )t′

t′

= t −t′′ t′ d = −dt′′ t′

v(t) = F (ω) g( ) deiωt ∫−∞

∞t′′ e−iωt′′

t′′

v(t) eiωt t′′

F(ω)eiωt v(ω)eiωt

v(ω) = F (ω) g(t) dt∫−∞

∞e−iωt

χ

χ(ω) = = g(t) dtv(ω)

F(ω)∫

−∞

∞e−iωt

χ(ω) = .τm

(1−iωτ)

1+ω2τ 2

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Another way to calculate the generalized susceptibility is to assume that the driving force and the

response both have a harmonic time dependence , . Substitutingthis form into the differential equation yields,

This can be solved for the generalized susceptibility.

ωτ

There is a subtle issue with minus signs here. It is equally valid to assume that the harmonicdependencies of the drive and the response have the form , .

Notice the minus sign that has appeared in the exponent. With this choice, the imaginary part of thesusceptibility changes sign:

Either descriptions of the harmonic dependence or are equally valid and there is no

consistent choice made in the literature. Here we continue with assuming a harmonic dependence of . Be aware that the sign of the imaginary part of the susceptibility might be different from

formulas found in other sources.

The causal nature of the impulse response function (it has to be zero for ) has consequencesfor the form of the susceptibility. Any function can be written in terms of an even component

and an odd component .

v(t) = v(ω)eiωt F(t) = F(ω)eiωt

iωmv(ω) + bv(ω) = F (ω).

χ(ω) = = = .v(ω)

F(ω)1

iωm+bτm

1−iωτ

1+ω2τ 2

χ(ω)m

τ

-4 -2 0 2 4-1.0

-0.5

0.0

0.5

1.0

1.5Re

Im

v(t) = v(ω)e−iωt F(t) = F(ω)e−iωt

χ(ω) = =1−iωm+b

τm

1+iωτ

1+ω2τ 2

eiωt e−iωt

eiωt

t < 0E(t)

O(t)

g(t) = E(t) + O(t).

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Since the impulse response function must be zero for , the even and the odd components must

add to zero for .

Note that if we know the either the even component or the odd component we construct the other.

Repeating what was stated above, the susceptibility is the Fourier transform of the impulse response

function,

The integral of an odd function over an even interval ( ) is zero so the real part of thesusceptibility is the Fourier transform of the even component,

and the imaginary part is the Fourier transform of the odd component,

t < 0t < 0

g

m

-3 -2 -1 0 1 2 3-0.50

-0.25

0.00

0.25

0.50

0.75

1.00

1.25g

E

O

t/τ

E(t) = sgn(t)O(t) = (g(−t) + g(t))12

O(t) = sgn(t)E(t) = (−g(−t) + g(t))12

χ(ω) = g(t) dt = (E(t) + O(t))(cos(−ωt) + i sin(−ωt))dt.∫−∞

∞e−iωt ∫

−∞

−∞, ∞

Re[χ] = = E(t) cos(ωt)dt,χ′ ∫−∞

Im[χ] = = − O(t) sin(ωt)dt.χ′′ ∫−∞

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Moreover is an even function while is an odd function .

The Kramers-Kronig relationsThe Kramers-Kronig relations describe how the real and imaginary parts of the susceptibility are

related to each other. If either the real part or the imaginary part of the susceptibility is known for

positive frequencies , the entire susceptibility can be calculated at all frequencies. Suppose we

know for . Then for all frequencies can be constructed because . Theeven component of the impulse response function can be found by inverse Fourier transforming .

The odd component of the impulse response function is related to the even component by

. The imaginary part of the susceptibility can then be constructed since it isthe Fourier transform of the odd component. Collectively these formulas are known as the

Kramers-Kroning relations.

Many observable quantities obey the Kramers-Kroning relations. For instance the electric

susceptibility describes the electric polarization of a material responds to an applied electric field.

This response must be causal so the real and imaginary parts of the electric susceptibility must berelated by the Kramers-Kronig relations. This is also true for the magnetic susceptibility, the

electrical conductivity, the thermal conductivity, and the dielectric constant.

A plane wave moving in the positive -direction has the form . If the frequency is negative,the wave moves in the negative -direction. Typically in an experiment, only the positive frequencies

are measured where the waves move from a source to the detector. This presents no difficulty since

all of the information is contained in the positive frequencies.

Sometimes it is experimentally easier to measure the real part or the imaginary part of the

susceptibility. The Kramer-Kronig relations can then be used to calculate the part that is difficult to

measure. If both real and imaginary parts can be measured, it is possible to check for experimental

errors using the Kramers-Kronig relations. If a susceptibility is calculated theoretically, it is a goodidea to check and see if it satisfies the Kramers-Kronig relations. It is considered a serious error to

present a result that violates causality.

It is traditional to rewrite the Kramers-Kronig relations in a more compact form which unfortunatelyintroduces a singularity in the formula. The singularity in the integral makes the form that is given

below less suitable for a numerical evaluation of the Kramers-Kronig relation. Nevertheless, it

χ′ (ω) = (−ω)χ′ χ′ χ′′ (ω) = − (−ω)χ′′ χ′′

ω > 0χ′ ω > 0 χ′ (ω) = (−ω)χ′ χ′

χ′

O(t) = sgn(t)E(t) χ′′

(ω) = E(t) cos(ωt)dtχ′ ∫−∞

∞E(t) = (ω) cos(ωt)dω1

2π∫

−∞

∞χ′

O(t) = sgn(t)E(t) E(t) = sgn(t)O(t)

(ω) = − O(t) sin(ωt)dtχ′′ ∫−∞

∞O(t) = (ω) sin(ωt)dω−1

2π∫

−∞

∞χ′′

x eikx−ωt

x

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commonly appears in the literature and is given for completeness. Consider the relationship between

the even component and the odd component of the impulse response function.

Use the convolution theorem to take the Fourier transform of both sides of this equation.

Here the before the integral indicates that one should use the Cauchy principle value of theintegral. This is necessary because of the singularity that the integral contains.

The convolution theorem can also be used to transform to find a similar

expression for in terms of . The Kramers-Kronig relations can then be written as,

This form has the advantage that one sees immediately that the real part of the susceptibility can be

determined from the imaginary part and vice versa.

The Kramers-Kronig relations are often put in another form where the integrals only involve positive

frequencies. The integral for is split into two parts.

In the first term make a change of variables , use the fact that is an odd function:

, and reverse the limits of integration.

The integrals can be combined.

E(t) = sgn(t)O(t)

= ∗ i = P d .χ′ −iπω χ′′ −1

π ∫−∞

∞ ( )χ′′ ω′

−ωω′ ω′

P

O(t) = sgn(t)E(t)χ′′ χ′

= P d ,χ′ −1π ∫

−∞

∞ ( )χ′′ ω′

−ωω′ ω′

= P d .χ′′ 1π ∫

−∞

∞ ( )χ′ ω′

−ωω′ ω′

χ′

= − P d − P d ,χ′ 1π ∫

−∞

0 ( )χ′′ ω′

−ωω′ ω′ 1π ∫

0

∞ ( )χ′′ ω′

−ωω′ ω′

→ −ω′ ω′′ χ′′

(−ω) = − (ω)χ′′ χ′′

= − P d − P d ,χ′ 1π ∫

0

∞ ( )χ′′ ω′′

+ωω′′ ω′′ 1π ∫

0

∞ ( )χ′′ ω′

−ωω′ ω′

= − P ( + ) ( )d′∞

′′ ′ ′

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Rewriting the factor,

the Kramers-Kronig relations can also be written,

Note that the singularity is stronger in this form making it less suitable for a numerical evaluation.

= − P ( + ) ( )dχ′ 1π ∫

0

∞1+ωω′

1−ωω′ χ′′ ω′ ω′

( + ) = ,1+ωω′

1−ωω′

2ω′

−( )ω′2

ω2

= P d ,χ′ −2π ∫

0

∞ ( )ω′χ′′ ω′

−( )ω′2

ω2ω′

= P d .χ′′ 2ωπ ∫

0

∞ ( )χ′ ω′

−( )ω′2

ω2ω′