A R

(6.114)

we then get the relation between the correlation function and the imaginary part of

the response function

ω

˜ (j)

K±β (x, x , ω) = coth mK±β (x, x , ω) . (6.115)

2kT

We have established the relationship between the imaginary part of the linear re-

sponse function, governing according to Eq. (6.88) the dissipation in the system, and

the equilibrium ¬‚uctuations, the ¬‚uctuation“dissipation theorem.16

According to the ¬‚uctuation“dissipation theorem we can express the change in

energy of a system in an external ¬eld of frequency ω, Eq. (6.88), in terms of the

current ¬‚uctuations

T

dEω 1 — ˜j

= dx dx E± (x, ω) K±β (x, x , ω)Eβ (x , ω) . (6.116)

ω

dt 2 ω coth 2kT ±β

For the current ¬‚uctuations we have (recall Eq. (6.50))

1p ˜ (j)

p

{j± (x, ω), jβ (x , ’ω)} = K±β (x, x , ω)

2

15 If we introduced

i i ˜ (j)

p

K p

K±β (x, t; x , t ) = {δj± (x, t), δjβ (x , t )} = 2 K±β (x, t; x , t )

0

we would be in accordance with the standard notation of the book.

16 Formally the ¬‚uctuation“dissipation theorem expresses the relationship between a commu-

tator and anti-commutator canonical equilibrium average. The ¬‚uctuation“dissipation relation,

Eq. (6.115), is also readily established by comparing the spectral representation of the imaginary

part of the retarded current response function, Eq. (6.37), with that of K (j) . The ¬‚uctuation“

˜

dissipation relationship expresses that the system is in equilibrium and described by the canonical

ensemble.

172 6. Linear response theory

ω

= ω coth e σ±β (x, x , ω) (6.117)

2kT

and the equal-time current ¬‚uctuations are speci¬ed by

∞ ∞

dω ˜ (j) dω ω

˜ (j)

K±β (x, t; x , t) = K (x, x , ω) = coth mK±β (x, x , ω)

2π ±β 2π 2kT

’∞ ’∞

(6.118)

and Eq. (6.79) guarantees the positivity of the equal-time and space current density

¬‚uctuations.

In a macroscopic description we have a local relationship between ¬eld and current

density, Ohm™s law,

j± (x, ω) = σ±β (x, ω) Eβ (x, ω) (6.119)

or equivalently

σ±β (x, x , ω) = σ±β (x, ω) δ(x ’ x ) . (6.120)

The equilibrium current density ¬‚uctuations at point x are then speci¬ed by

1

≡ d(x ’ x ) {j± (x, ω), j± (x , ’ω)}

2 p p

j± xω 0

2V

1 ˜ (j) 1 ω

= K±± (x, ω) = ω coth e σ±± (x, ω) . (6.121)

V V 2kT

We note that the factor

ω ω 1

coth = ω n(ω) + (6.122)

2 2kT 2

is the average energy of a harmonic oscillator, with frequency ω, in the thermal state.

The average energy consists of a thermal contribution described by the Bose function,

and a zero-point quantum ¬‚uctuation contribution.

In the high-temperature limit where relevant frequencies are small compared to

the temperature, ω kT , we get for the current ¬‚uctuations in a homogeneous

conductor with conductivity σ, Johnson noise,

2kT σ

2

j± = (6.123)

ω

V

independent of the speci¬c nature of the conductor.

In the linear response treatment we have assumed the ¬eld ¬xed, and studied the

¬‚uctuations in the current density. However, ¬‚uctuations in the current (or charge)

density gives rise to ¬‚uctuations in the electromagnetic ¬eld as well. As an example

of using the ¬‚uctuation“dissipation theorem we therefore turn the point of view

around, and using Ohm™s law obtain that the (longitudinal) electric ¬eld ¬‚uctuations

are given by

1

2 2

E± xω = j± xω . (6.124)

|σ(x, ω)|2

6.6. Time-reversal symmetry 173

According to Eq. (6.121) we then obtain for the (longitudinal) electric ¬eld ¬‚uctua-

tions

1 ω eσ±± (x, ω)

2

E± xω = ω coth . (6.125)

2kT |σ±± (x, ω)|2

V

In the high-temperature limit, ω kT , we have for the (longitudinal) electric

¬eld ¬‚uctuations, Nyquist noise,

2kT

2

E± = . (6.126)

ω

σV

6.6 Time-reversal symmetry

Hermitian operators will by suitable phase choice have a de¬nite sign under time

reversal: position and electric ¬eld have positive sign, and velocity and magnetic

¬eld have negative sign.17 The following considerations can be performed for any

pair of operators (see Exercise 6.2), but we shall for de¬niteness consider the current

operator, and show that Eq. (6.114) is a consequence of time-reversal invariance.

In case the Hamiltonian is time-reversal invariant,

†

T [j± (x, t), jβ (x , t )] T † [T jβ (x , t ) T † , T j± (x, t) T † ]

p p

p p

=

p

’ [j± (x, ’t), jβ (x , ’t )]

p

= (6.127)

and

T ψ|T [j± (x, t), jβ (x , t )]† T † |T ψ ,

p p

p p

ψ|[j± (x, t), jβ (x , t )]|ψ = (6.128)

where |T ψ is the time-reversed state of |ψ . Consequently,

p p

Tr(ρ(H)[j± (x, ’t), jβ (x , ’t )]) = ’Tr(ρ(H)[j± (x, t), jβ (x , t )])

p p

(6.129)

and we therefore ¬nd that time-reversal invariance implies

—

R A

K±β (x, x ; ω) = K±β (x, x ; ω) (6.130)

i.e. we have established Eq. (6.114).

Exercise 6.2. Consider two physical quantities represented by the operators A1 (x, t)

and A2 (x, t), which transform under time reversal according to

T Ai (x, t) T † = si Ai (x, ’t) si = ±1, i = 1, 2.

, (6.131)

Show that when the Hamiltonian is invariant under time reversal, the response func-

tion

Aij (x, x , t ’ t ) ≡ Tr(ρ(H)[Ai (x, t), Aj (x , t )]) (6.132)

17 For a discussion of time-reversal symmetry we refer the reader to chapter 2 of reference [1].

174 6. Linear response theory

satis¬es the relations

Aij (x, x , t ’ t ) = ’si sj Aij (x, x , t ’ t) = si sj Aij (x , x, t ’ t ) (6.133)

and thereby18

Aij (x, x , ω) = ’si sj Aij (x, x , ’ω) = si sj Aij (x , x, ω) . (6.134)

6.7 Scattering and correlation functions

In measurements on macroscopic bodies only very crude information of the micro-

scopic state is revealed. For example, in a measurement of the current only the

conductance is revealed and not any of the complicated spatial structure of the con-

ductivity. To reveal the whole structure of a correlation function takes a more indi-

vidualized source than that provided by a battery. It takes a particle source such as

the one used in a scattering experiment, using for example neutrons from a spallation

source.

In this section we shall consider transport of particles (neutrons, photons, etc.)

through matter. To be speci¬c we consider the scattering of slow neutrons by a

piece of matter. A neutron interacts with the nuclei of the substance (all assumed

identical). The interaction potential is short ranged, and we take for the interaction

with the nucleus at position RN 19

V (rn ’ RN ) = a δ(rn ’ RN ) . (6.135)

We have thus neglected the spin of the nuclei (or consider the case of spin-less

bosons).20 For the interaction of a neutron with the nuclei of the substance we

then have

V (rn ’ RN ) = a δ(rn ’ RN ) .

V (rn ) = (6.136)

N N

The interaction is weak, and the scattering can be treated in the Born approximation.

For the transition rate between initial and ¬nal states we then have

2π ˆ

| f| V (ˆn ’ RN )|i |2 δ(Ef ’ Ei ) .

“¬ = (6.137)

r

N

For simplicity we assume that the states of the substance can be labeled solely by

their energy

(i) (i) (f) (f)

|i = |p , ES = |p |ES |f = |p, ES = |p |ES

, , (6.138)

18 Ifthe Hamiltonian contains a term coupling to a magnetic ¬eld, the symmetry of the correlation

function is Aij (x, x , ω, B) = ’si sj Aij (x, x , ’ω, ’B) = si sj Aji (x , x, ω, ’B).

19 We thus exclude the possibility of any nuclear reaction taking place.

20 However, it is precisely the magnetic moment of the neutron that makes it an ideal tool to

investigate the magnetic properties of matter. The subject of neutron scattering is thus vast, and

for a general reference we refer the reader to reference [21].

6.7. Scattering and correlation functions 175

where the initial and ¬nal energies are

p2 p2

(f) (i)

Ef = ES + , Ei = ES + (6.139)

2mn 2mn

and mn is the mass of the neutron. We introduce the energy transfer from the neutron

to the material

p2 p2 (f) (i)

’ = ES ’ ES

ω= (6.140)

2mn 2mn

and we have for the transition probability per unit time

2π (f) (i) (f) (i)

ˆ

| p, ES |a δ(ˆn ’ RN )|p , ES |2 δ(ES ’ ES ’ ω) .

“¬ = (6.141)

r

N

Since the interaction is inelastic, the di¬erential cross section of interest, d2 σ/dˆ d ,

p

is the fraction of incident neutrons with momentum p being scattered into a unit

solid angle dˆ with energy in the range between and + d . Noting that

p

”p = p2 dp dˆ = mn p d dˆ (6.142)

p p

we obtain for the inelastic di¬erential cross section for neutron scattering o¬ the

substance

d2 σ m2 L6 p 2π (f) (i)

ˆ

| p, ES |a δ(ˆn ’ RN )|p , ES |2

n

= r

3p

dˆ d (2π )

p

N

(f) (i)

— δ(ES ’ ES ’ ω) (6.143)

which we can express as

∞

d(t ’ t ) ’i(t’t )ω

d2 σ m2 a2 p 2π

dx dx e’

i

n (x’x )·(p’p )

= e

(2π )3 p

dˆ d 2π

p

’∞

(f) (i) (i) (f)

— ES |n(x, t)|ES ES |n(x , t )|ES , (6.144)

where n(x, t) is the density operator for the nuclei of the material in the Heisenberg

ˆ

picture with respect to the substance Hamiltonian HS .

Exercise 6.3. Show that, for scattering o¬ a single heavy nucleus, M mn , we

have for the total cross section

∞

d2 σ mn a 2

σ= dˆ d = 4π . (6.145)

p

2π 2

dˆ d

p

4π 0

176 6. Linear response theory

In the scattering experiment we know only the probability distribution for the

initial state of the material, which we shall assume to be the thermal equilibrium

state

|ES (») P (ES (»)) ES (»)| ,

ρS = (6.146)

»

where

e’ES (»)/kT

HS |ES (») = ES (») |ES (») .

P (ES (»)) = , (6.147)

ZS

For the transition rate weighted over the thermal mixture of initial states of the

substance we have

(i)

≡

“fp P (ES (»)) “¬

»

∞

d(t ’ t ) ’i(t’t )ω

m2 2

a p 2π

n

= P (ES (»)) dx dx e

(2π )3 p 2π

» ’∞

e’

i (f) (f)

— ES |n(x, t)|ES (») ES (»)|n(x , t )|ES

(x’x )·(p’p )

(6.148)

and we obtain for the weighted di¬erential cross section (we use the same notation)

d2 σ d2 σ

(i)

= P (ES (»))

dˆ d dˆ d

p p

»

∞

d(t ’ t ) ’i(t’t )ω

m2 a2 p 2π

n

= P (ES (»)) dx dx e

(2π )3 p 2π

» ’∞

— e’

i (f) (f)

ES |n(x, t)|ES (») ES (»)|n(x , t )|ES

(x’x )·(p’p )

. (6.149)

Furthermore, in the experiment the ¬nal state of the substance is not measured,

and we must sum over all possible ¬nal states of the substance, and we obtain ¬nally

for the observed di¬erential cross section (we use the same notation)

∞

d(t ’ t ) i(t’t )ω

d2 σ m2 a2 p 2π

dx e’ (x’x )·(p ’p)

i

n

= dx e

(2π )3 p

dˆ d 2π

p

’∞

— n(x, t) n(x , t ) , (6.150)

where the bracket denotes the weighted trace with respect to the state of the sub-

stance

≡

n(x, t) n(x , t ) trS (ρS n(x, t) n(x , t )) . (6.151)

6.7. Scattering and correlation functions 177

We thus obtain the formula

d2 σ p m 2 a2

n

= V S(q, ω) (6.152)

p (2π )3 2

dˆ d

p

where S(q, ω) is the Fourier transform of the space-time density correlation function

S(x, t; x , t ) ≡ n(x, t) n(x , t ) , and q ≡ p ’ p and ω is the momentum and en-

ergy transfer from the neutron to the substance. We note that S(’q, ’ω) = S(q, ω).

This correlation function is often referred to as the dynamic structure factor.21 The

dynamic structure factor gives the number of density excitations of the system with

a given energy and momentum. A scattering experiment is thus a measurement of

the density correlation function.

Exercise 6.4. Show that, for a target consisting of a single nucleus of mass M in

the thermal state, the dynamic structure factor is given by

q2

1 2πM ’ 2kM q 2 ω’