ω 2 ’ ω2

π 0

∞

2 ωχ (ω )

χ (ω) = dω . (18.16)

ω2 ’ ω 2

π 0

Since the integrands diverge when ω approaches ω, the integrals are not

well-de¬ned. They must be interpreted as the principal value: exclude an

interval ω ’ µ to ω + µ from the integration and then take the limit µ ’ 0.

The relations arise from the fact that both χ and χ follow from the same

delta-response function through equations (18.11) and (18.12). The proof

of (18.16) is given below. The reader is challenged to prove (18.15).

Proof Start from (18.9). This is a one-sided Fourier (or Fourier“Laplace)

1 These relations were ¬rst formulated by Kronig (1926) and Kramers (1927) and can be found

in many textbooks, e.g., McQuarrie (1976).

510 Linear response theory

transform, but taking into account that ¦(„ ) = 0 for „ < 0, the integral can

be taken from ’∞ instead of zero. Thus we obtain

∞

¦(„ )e’iω„ d„,

χ(ω) = χ ’ iχ =

’∞

with inverse transform

∞

1

χ(ω)eiω„ dω.

¦(„ ) =

2π ’∞

Using the symmetry properties of χ and χ (see (18.11) and (18.12)):

χ (’ω) = χ (ω) and χ (’ω) = ’χ (ω),

we can write

∞ ∞

1 1

¦(„ ) = χ (ω) cos ω„ dω + χ (ω) sin ω„ dω.

π π

0 0

Now, using the fact that ¦(„ ) = 0 for „ < 0, we see that both integrals on

the r.h.s. must be equal (for negative „ the last integral changes sign and

the total must vanish). Therefore

∞ ∞

π

χ (ω) cos ω„ dω = χ (ω) sin ω„ dω = ¦(„ ).

2

0 0

The equality is a direct result of the causality principle. Now insert the ¬rst

expression for ¦(„ ) into (18.12) and obtain

∞ ∞

2

χ (ω) = d„ sin ω„ dω χ (ω ) cos ω „

π 0 0

∞ ∞

2

= dω χ (ω ) d„ sin ω„ cos ω „.

π 0 0

Note that we have changed the integration variable to ω to avoid confusion

with the independent variable ω. After rewriting the product of sin and cos

as a sum of sines, the last integral can be evaluated to

1 1 1 ω

+ =

ω’ω ω2 ’ ω 2

2 ω+ω

Here the primitive function has been assumed to vanish at the limit „ ’ ∞

because of the in¬nitely rapid oscillation of the cosines (if this does not sat-

isfy you, multiply the integrand by a damping function exp(’µ„ ), evaluate

the integral and then take the limit µ ’ 0). Equation 18.16 results.

18.3 Relation to time correlation functions 511

18.3 Relation to time correlation functions

In this section we consider our black box as a gray box containing a sys-

tem of mutually interacting particles. We assume that the system obeys

the classical Hamilton equations and is “ in the absence of a disturbance “

in an equilibrium state. In the linear regime the perturbation is so small

that the system deviates only slightly from equilibrium. When after a delta-

disturbance some observable Y deviates slightly from its equilibrium value, it

will relax back to the equilibrium value through the same intra-system inter-

actions that cause spontaneous thermal ¬‚uctuations of Y to relax. Therefore

we expect that the time course of the relaxation of Y after a small perturba-

tion (i.e., the time course of ¦(t)) is directly related to the time correlation

function of Y .

Let us be more precise. Consider the time correlation function of Y :

Y (0)Y (t) . The triangular brackets stand for an ensemble average. For a

system in equilibrium, presumed to be ergodic, the ensemble average is also

a time average over the initial time, here taken as the origin of the time

scale. If the probability of Y in the equilibrium ensemble is indicated by

Peq (Y ), we can write

Y (0)Y (t) = Y0 Y (t)|Y (0) = Y0 Peq (Y0 ) dY0 , (18.17)

where Y (t)|Y (0) = Y0 is the conditional ensemble-averaged value of Y at

time t, given the occurrence of Y0 at time 0. But that is exactly the response

function after an initial disturbance of Y to Y0 :

¦(t)

Y (t)|Y (0) = Y0 = Y0 . (18.18)

¦(0)

Here ¦(0) is introduced to normalize ¦. Inserting this into (18.17), we arrive

at the equality:

¦(t) Y (0)Y (t)

= . (18.19)

Y2

¦(0)

This relation is often called the ¬rst ¬‚uctuation“dissipation theorem (Kubo,

1966); see for a further discussion page 258.

— —

The response Y (0) after a delta-disturbance X(t) = X0 δ(t) equals X0 ¦(0)

(see (18.1)). Therefore:

—

¦(0) = Y (0)/X0 . (18.20)

This ratio can normally be computed without knowledge of the details of

the intra-system interactions, as the latter have no time to develop during

a delta-disturbance. The value of Y 2 follows from statistical mechanical

512 Linear response theory

considerations and appears to be related to the delta-response. This relation

between Y (0) and Y 2 , which we shall now develop, forms the basis of the

Green-Kubo formula that relates the integral of time correlation functions

to transport coe¬cients.

—

Following a delta-disturbance X0 δ(t), the point in phase space z (we use

symplectic notation, see Section 17.8 on page 492) will shift to z + ”z.

—

The shift ”z is proportional to X0 . For example, when the perturbation

is an electric ¬eld E — δ(t), the ith particle with (partial) charge qi will be

0

subjected to a force qi E — δ(t), leading to a shift in momentum ”pi = qi E — .

0 0

The phase-point shift leads to a ¬rst-order shift in the response function Y ,

which is simply a property of the system determined by the point in phase

space:

2n

‚Y (z)

”Y = ”zi . (18.21)

‚zi

i=1

—

The delta-response Y (0) = X0 ¦(0) is the ensemble average of ”Y :

2n

‚Y (z)

’βH(z)