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K±β (x, x , ω) = [K±β (x, x , ω)]—
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’

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