R I G O R O U S A S P E C T S O F O R B I T A L M A G N E ... · PDF file1 RESUMÉ...

R I G O R O U S A S P E C T S O F

O R B I T A L M A G N E T I S M I N G R A P H E N E - L I K E

M A T E R I A L S

S P E C I A L E I F Y S I K

M I K K E L H A G G R E N B R Y N I L D S E N

V E J L E D E R E : H O R I A C O R N E A N O G T H O M A S G A R M P E D E R S E N

0 1 - 0 6 - 2 0 1 1

D E P A R T M E N T O F P H Y S I C S A N D N A N O T E C H N O L O G Y

A A L B O R G U N I V E R S I T E T

1

RESUMÉ

Vi udvikler i denne prœsentation en teori for off-diagonal elementet af 2D konduktivitets-

tensoren ←→σ for en todimensionel krystal, påtrykt et statisk, homogent magnetisk felt.

Vi bruger en diskret model, med en Harper-type Hamiltonoperator og Peierls substitu-

tion. Vi viser at σ21 har en asymtotisk ekspansion i den magnetiske feltstyrke b, samt at

σ12(b = 0) er nul, og finder en formel for den første afledte σ′12(b = 0). Denne koefficient

spiller en vigtig rolle i teorien om Faraday ratation.

De anvendte matematiske metoder er: Spektral teori for Harper-type operatorer, Combes-

Thomas lokalisering , magnetisk pertubationsteori og Bloch-Floquet teori.

ABSTRACT

In this presentation we develop a theory for calculating the off-diagonal tensor element

of the 2D conductivity tensor←→σ , for a two dimensional crystal immersed in a constant

homogenous magnetic field, using a Harper-type Hamiltonian and Peierls substituion

and a discrete model of the crystal. Ve show that σ21 has an asymptotic expansion in

the magnetic field strength b, that σ12(b = 0) is zeroand develop a formula for the first

derivative σ′12(b = 0). This coefficient plays a major role in the theory concerning the

Faraday rotation effect.

The main mathematical tolls used here is spectral theory for Harper-type operators,

Combes-Thomas localization, Magnetic pertubation theory and Bloch-Floquet theory.

Draft: August 11, 2011, misc/abstract.tex

3

PREFACE

This thesis was written in the period february 1st to june 1st, 2011, at Aalborg University.

This paper is the result of a 5 year study in Physics and Mathematics.

This presentation requires some basic knowledge of quantum mechanics, real and

complex analysis, and operator theory (bounded operators on Hilbert space).

I would like to thank my main supervisor Horia Cornean for his support.

Aalborg, june 1st, 2011.

Draft: August 11, 2011, misc/forord.tex

til Lars Brynildsen.

CONTENTS

Contents 7

1 Introduction 1

1.1 Physical aspects of Faraday rotation . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Setup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3 Main result of this thesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2 Proof of theorem 1.1(1) 13

2.1 Simple Combes-Thomas theorem . . . . . . . . . . . . . . . . . . . . . . . . 13

2.2 Magnetic pertubation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.3 Magnetic translation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.4 Large N limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3 Proof of theorem 1.1(2) 35

3.1 k-space representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.2 Calculating σ21(0) in reciprocal space . . . . . . . . . . . . . . . . . . . . . 41

3.3 The first derivative of σ21(b) . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

4 Perspectives 49

A Appendices 51

A.1 Selected result concerning bounded operators on a Hilbert space . . . . . 52

Bibliography 55

Draft: August 11, 2011, speciale.tex 7

1

INTRODUCTION

1.1 PHYSICAL ASPECTS OF FARADAY ROTATION

We set the stage by a physical discussion about Faraday rotation. This section is not

“rigorous”, in the mathematical sense.

Graphene in a magnetic field

Graphene is a two-dimensional allotrope of carbon, with the carbon atoms organized in

a “honeycomb” crystal, fig. 1.1. Imagine the following idealization of a experiment: Con-

Figure 1.1: a “honeycomb” pattern

sider a single layer of graphene, placed in a constant magnetic field (“the system”). Let

this layer be subject to a monocromatic (plane-wave) light-beam of known polarization.

We would like to be able to make a quantum mechanical prediction, of how the electrons

of the graphene layer respond to the applied electro-magnetic fields.

Faraday rotation

Faraday rotation is a dispersion effect. It describes the rotation of the polarization of

radiation passing through a substance in the direction of an applied magnetic field. As

in the article by Jian-Ping Peng et.al. [4], we calculate the Faraday rotation, physically a

three-dimensional effect, in a quasi-two-dimensional electron system in a magnetic field

B applied perpendicular to the layer. The electrons are confined within a slab of thickness

d and move freely along the xy plane. Physically the Faraday rotation arises from the

Draft: August 11, 2011, introduction/faradayeffect.tex 1

2 Introduction

difference in the propagation of the two types of circular polarizations into which the

plane polarized beam may be resolved. The Faraday rotation ϑ is defined as

ϑ =ωd(η− − η+)

2c,

where η− and η+ are the indices of refraction of right and left circularly polarized radi-

ation of frequency ω. Expressions for η± can be obtained from Maxwell’s equations for

the electric field E, the electric displacement D, the magnetic induction B, the magnetic

field H and the current density J 1:

∇× E = −1c

∂B

∂t, ∇× H =

4π

cJ +

1c

∂D

∂t

which, adopting a monochromatic plane-wave solution of the form

E = E0ei(ωt−Kz),

combined with the constitutive equations

B = µH, D = εE

and assuming that Ohms law holds:

J =←→σ E,

give

∇(∇ · E)−∇2E = −

(4πµ

c2

)

←→σ ∂E

∂t−(µε

c2

) ∂2E

∂t2 .

For the Faraday effect one takes the propagation along the applies dc magnetic field. This

setup leads to

K2E0 = µε(ω/c)2[←→

1 − (4πi/ωε)←→σ]

E0

(see [2, p.223]) where←→1 is a unit tensor. Assuming a circularly polarized wave E0 =

E0x ± iE0y, then

K2± = µε(ω/c)2

[

1− (4πi/ωε)σ(3D)±

]

.

Where σ(3D)± = σ

(3D)xx ± iσ

(3D)xy . The complex index of refraction (η − iκ) is then obtained

from

(η± − iκ±)2 = µε[

1− (4πi/ωε)σ(3D)±

]

. (1.1)

The diagonal 3D conductivity σzz makes no contribution to the Faraday rotation. The

form of σ(2D)12 is

σ(2D) =

[

σ11 σ12

−σ12 σ22

]

=

[

σxx σxy

−σxy σyy

]

1We use the naming convention from Jackson [2] and in this section we work in gaussian units.

Draft: August 11, 2011, introduction/faradayeffect.tex

Faraday rotation 3

Considering the finite thickness d of the layer, the three-dimensional conductivity tensor

is related to the two-dimensional conductivity tensor by the following expression

σ(3D)xx = σ

(2D)11 /d, ,

and so forth. Working towards isolating η± in formula (1.1), we have

Re[(η± − iκ±)2] = µε

(

1± 4π

ωεσ(3D)xy

)

, Im[(η± − iκ±)2] = −µε

4π

ωεσ(3D)xx .

We are looking for

η± =√

µεRe[

η± − iκ±√µε

]

.

Physically, η± should be non-negative, so we choose the solution,

η±√µε

=

[(

1± 4π

ωεσ(2D)12

)2

+

(4π

ωεσ(2D)11

)2] 1

4

cos

12

arctan

σ(2D)11(

ωε4π ± σ

(2D)12

)

.

Considering σ(2D)11 and σ

(2D)12 as variables u and v, respectively, η± is thus, up to a constant,

given by the function

η±√µε

= f (±u, v) =

[(

1± 4π

ωεu

)2

+

(4π

ωεv

)2] 1

4

cos

(

12

arctan

[

v(

ωε4π ± u

)

])

.

We will show how one can find the asymptotic expansion of σ(2D)12 (b), in powers of the

magnetic field strength b and in particular we will show that in the small b limit, σ12(b)

can be written

σ(2D)12 (b) = bσ

(1)12 +O(b2) ,

using a method that could also be used to find the expansion of σ(2D)11 (b)

σ(2D)11 (b) = σ

(0)11 + bσ

(1)11 +O(b2)

In this presentation we a interested in how ϑ, and therefore (η− − η+) ∝ [ f (−u, v) −f (u, v)], varies with the first power of b, so considering the expansion of f (u, v) in powers

of u we have

f (−u, v)− f (u, v) = −[

2∂ f

∂u(0, v0)

]

u

︸︷︷︸

linear in b

+O(u2)︸︷︷︸

∝b2

where v0 denotes σxx(b = 0). We can omit all but the constant σ(0)11 term in σ

(2D)11 (b). ϑ is

proportional with

b

(ddb

σ(2D)12

)

(0).

The question of calculating the faraday rotation is therefore a question of differentiating

the off-diagonal conductivity σ(2D)12 .

Draft: August 11, 2011, introduction/setup.tex

4 Introduction

1.2 SETUP

One-electron approximation

We work in a one-electron setup, neglecting electron-electron interactions.

The crystal

We settle for a simpler model than graphene. Consider a 2 - dimensional crystal, with

the nuclei fixed at their equilibrium positions

Λ = x ∈ R2 : there is a nucleous at position x.

A point in Λ is denoted a site, and Λ is in this presentation sometimes denoted the ion-

core mesh. The crystal is assumed invariant under translations that transforms functions

defined on R2 according to the rule

(Tγ f )(r) = f (r− γ), γ = ma1 + na2,

where m and n are integers, and the basis vectors a1 and a2 generate the Bravais Lattice,

Γ:

Γ = γ ∈ R2 : γ = ma1 + na2, m, n ∈ Z.

(The two vectors a1 and a2 should not be linearly dependent.) The points of Λ are gen-

erated by appending one unit cell to each point of Γ (making a choice of origo). I denote

the (finite) set of vectors that constitutes the unit cell by Ω:

Ω = xn|Ω|n=1

The choice of Bravais lattice and unit cell should lead to the unique decomposition of all

site position vectors,

x = γ + x, γ ∈ Γ, x ∈ Ω,

for all x ∈ Λ.

When used as a summation index or site label, we will not use the bold notation x, γ

and x, but simply use x, γ and x. That is, |x〉 denotes a localized state at position x, and

the sum ∑γ∈Γ|γ− y| is a sum over all position vectors in Γ. Sometimes we will write ∑γ

for ∑γ∈M, when it is clear from the context, which set, M, γ is to be summed over.

Reciprocal lattice

Define the dual basis b1, b2 by the relations

ai · bj = 2πδij,

where δij is the Kronecker delta. The reciprocal lattice, Γ∗, is the set of all vectors γ∗ ∈ R2

such that

γ∗ = mb1 + nb2, m, n ∈ Z


The unit cell 5

The definition of the reciprocal lattice has the consequence that

γ∗ · γ = n2π, for some n ∈ Z,

for all vectors γ in the Bravais lattice and for all vectors γ∗ in the reciprocal lattice.

We denote by Ω∗ the basic periodic cell for the dual basis (the “Brillouin zone”):

Ω∗ =

t1b1 + t2b2 : −12≤ ti <

12

.

The unit cell

We restrict ourselves to a unit cell with two sites, one at x = (0, 0) and one at x = ( a12 , 0)

(fig. 1.2).

a2 a1

UNIT CELL

AB

Figure 1.2: The ion-core mesh, Λ = x (“A” and “B” type points), is generated by appending a basis oftwo sites, A and B, to the Bravais lattice, Γ = γ (“A” type points), generated by the primitive vectors a1and a2. We choose a simple Bravais lattice where the primitive vectors are orthogonal.

Crystal Hamiltonian, potential term

The potential term, V0 of the crystal Hamiltonian, H0 = T + V0 is in the model given by:

(V0ψ)(γ + x) =

v ψ(γ + x), x = (0, 0)

0, x =( a1

2, 0)

,(1.2)

where the crystal potential strength parameter v is understood to be greater than zero.

Crystal Hamiltonian, kinetic term

We fix the kinetic term kernel t0(x, y) (“hopping matrix elements”) of the Hamiltonian to

be:

t0(x, y) = t0(x1, y1)(1)δx2,y2 + t0(x2, y2)

(2)δx1,y1 , (1.3)

for all sites x = (x1, x2) and y = (y1, y2) in Λ. This models a crystal with no oblique

interactions. We consider a nearest-neighbour model, where

t0(x1, y1)(1) = 0, unless y1 = x1 ±

a1

2,

t0(x2, y2)(2) = 0, unless y2 = x2 ± a2,


6 Introduction

and we fix the magnitude of the non-zero hopping elements to 1, so the crystal Hamilto-

nian has the kernel:

h0(x, y) = δx1,y1±a1/2δx2,y2 + δx2,y2±a2 δx1,y1 + v(x1)δx,y, (1.4)

Finite, but large crystal

Define a central region of the crystal by

ΛN := (γ + x) ∈ Λ : γ = ma1 + na2, |m| ≤ N, |n| ≤ N, x ∈ Ω.

This central region is a region of |ΛN| = (2N + 1)2 unit cells, and |Ω| (2N + 1)2

= 2 (2N + 1)2 sites. The characteristic function, χΛN, of ΛN , is a multiplicative oper-

ator defined by:

(χΛNψ) (x) = χΛN

(x)ψ(x),

where χΛN(x) =

1 , x ∈ ΛN ,

0 , x /∈ ΛN .

The support of ψ ∈ ℓ2(Λ), denoted suppψ, is the collection of all points x ∈ Λ, that

satisfies ψ(x) 6= 0. We will from here on work with the Hamilton operator, subject to

Dirichlet boundary conditions (DBC) in ΛN :

H0,N = χΛNH0 χΛN

, (1.5)

where H0 is an operator defined on ℓ2(Λ). H0,N is a (2N + 1)2|Ω| × (2N + 1)2|Ω|matrix,

so there exists a one-to-one relation between (linear) χΛNA χΛN

operators on ℓ2(Λ), and

(2N + 1)2|Ω| × (2N + 1)2|Ω| matrices. The restricted operator H0,N approximates H0 in

the sense

H0|ψ〉 = χΛNH0 χΛN

|ψ〉, ∗ (1.6)

for |ψ〉 where both suppψ and suppH0ψ is fully enclosed in ΛN .

A consequence of the tight binding model is, that functions ψ with suppH0ψ * ΛN

but suppψ ⊂ ΛN can only be functions ψ(x) with non-zero values near the edge of

ΛN . Conversely, the transformed vector H0ψ is not different from the vector H0,Nψ for

wavefunctions ψ with their support localized in the “inner part” 2 of the central region,

that is for ψ with the support “situated far enough from” the edge of ΛN . This line of

thought will be formalized and expanded later , as a key part of the proof of the large N

limit.

We use as a notation for an operator O with the DBC the subscript N, ON

Operators with the subscript “N” are from here on always subject to DBC.

2How close to the edge this “inner” part can come, is of course linked to the number of “neighbours”of a given site one allow to have non-zero hopping matrix elements in to the model. A second-nearest tb.model will have a smaller inner part than a nearest-neighbour tb. model, for otherwise equal models (fixedN).


Peierls substitution 7

Peierls substitution

We include a static magnetic field into the model, it is thought of as having always existed

(before the light pertubation was turned on). This is done by the use of Peierls substitution.

In the Peierls substitution, the Hamiltonian matrix element in a constant magnetic field

can be obtained through multiplication of the crystal Hamiltonian matrix element h0 by

a phase factor [7],

hb(x, y) = eibϕ(x,y)h0(x, y).

The magnetic field is assumed perpendicular to the crystal, directed in the z+ direction,

having constant magnitude b, . The phase factor is given by the flux of a +z directed field

of strength unity through a triangle defined by origo, the site x and the site y3. We set

ϕ(x, y) :=12(y1x2 − x1y2). (1.7)

With a slight abuse of notation (using 3D cross-product), ϕ(x, y) given by formula (1.7)

can be calculated by:

ϕ(x, y) =12[(y× x)]z ,

where in this 3D mindset, x→ (x1, x2, 0), y→ (y1, y2, 0) and [v]z denotes the z-component

of the 3D vector v. Note that ϕ(x, y) is not bound by a constant multiplying ‖x− y‖! but

we can use the estimate

|ϕ(x, y)| ≤ min‖x‖, ‖y‖‖x− y‖.

Note also that ϕ is antisymmetric, that is

ϕ(x, y) = −ϕ(y, x).

We denote by fl(x, x′, x′′) the magnetic flux of a field of unity field-strength through the

triangle defined by origo, the site x− x′ and the site x′ − x′′. We will need the equality,

where x, x′ and y′ are vectors in the xy-plane extended of R3,

12(x′ × x) +

12(x′′ × x′) =

12(x′′ × x) +

12

[(x′ − x′′)× (x− x′)

],

Equating z-components, this gives

ϕ(x, x′) + ϕ(x′, x′′) = ϕ(x, x′′) + fl(x, x′, x′′),

with the definition

fl(x, x′, x′′) =12

[(x′ − x′′)× (x− x′)

]

z.

The fact that ϕ(x, x) = 0 means that we only need to consider the effect of the phase

factor on the kinetic term of the Hamiltonian, since the potential term has the factor

δ(x, y) in the kernel:

eibϕ(x,y)(

δx1,y1±a1/2δx2,y2 + δx2,y2±a2 δx1,y1 + v(y1)δx,y

)

= eibϕ(x,y)(

δx1,y1±a1/2δx2,y2 + δx2,y2±a2 δx1,y1

)

+ v(y1)δx,y,

3See R. Saito et al. section 6.2, expecially around formula (6.14).


8 Introduction

that is

Hb = Tb + V0,

with the “magnetic” kinetic operator, Tb, defined as the operator having the kernel

tb := eibϕ(x,y)(

δx1,y1±a1/2δx2,y2 + δx2,y2±a2 δx1,y1

)

.

Units

To simplify the notation we work in a system of units, where

h = 2melectron = e = 1

Here h is Plancks reduced constant and e denotes the charge of an proton.

Position operator

We define the position operators X1 and X2, by how they transform the basis elements

δx, x ∈ Λ,

X1δx := x1δx, ∀x ∈ Λ,

X2δx := x2δx, ∀x ∈ Λ,

(remember x = (x1, x2)). For a general ψ in ℓ2(Λ), and fixed x ∈ Λ, we have

(X1ψ) (x) = ∑y∈Λ

X1(x, y)ψ(y) = x1ψ(x),

so the kernel of Xν, in the δx representation, must be Xν(x, y) = δx,yxν for ν ∈ 1, 2.Since these operators do nothing else than pick up the spatial label of the basis states, we

could denote it the label operator, but we will use it as the representation of the position

operator in the δx basis. Note that the position operator is not bounded in the oper-

atornorm, and that even though ψ ∈ ℓ2(Λ) it is not always true that Xνψ ∈ ℓ2(Λ). It

holds, though, that Xν,Nψ ∈ ℓ2(Λ) for Xνψ ∈ ℓ2(Λ).

The field of the light beam

We introduce the light-beam incident on the crystal as a monochromatic, plane-wave

EM-field, with the electric field parallel with the x-axis

E(t) = (Ex, Ey) := (Eeiωt, 0), E ≥ 0,

with a negative imaginary part of the frequency; Imω = −η < 0.

Note on notation: We write cartesian vectors component-wise the same way as we

write points in the plane; v = (vx, vy), but nonetheless a vector (vx, vy) is to be thought

of as a column vector when multiplying matrices.

(vx, vy)→[

vx

vy

]

.


Dynamic equation for the density operator 9

We incorperate the light beam into the model as an external time-dependent potential

term VE(t). The Hamilton operator is now given by

HE(t) = Hb + VE(t). (1.8)

The potential operator representing the light beam, as introduced above, can be calcu-

lated by use of the first component position operator,

VE(t) = EeiωtX1. (1.9)

Dynamic equation for the density operator

We denote by µ the chemical potential, and β := 1kBτ is the inverse temperature, so the

Fermi-Dirac distribution can be written

fFD(z) =1

eβ(z−µ) + 1.

We set as a condition for the system, that the density operator in the limit t → −∞ is

given by the Fermi-Dirac distribution function :

N(t→ −∞) = 0,N = fFD(Hb,N).

An example: if |ψj〉 is a basis of energy-eigenvectors for the self-adjoint matrix Hb,N ,

we have Hb,N |ψj〉 = λj|ψj〉. In this case for the infinite past, the state of system should be

defined by the density operator:

∑j

fFD(λj)|ψj〉〈ψj|,

where |ψj〉 is an energy eigenvector basis for wavefunctions on the central region ΛN

of the site-mesh.

We will sometimes just use “density” for “density operator”. To find the density at

an arbitrary time t, we need to solve the dynamic (Liouville) equation for E,N(t). This

is:

i ˙ E,N(t) = [HE,N , E,N(t)],

E,N(t→ −∞) = fD(Hb,N),(1.10)

note that the negative imaginary part of ω has the effect that VE(t)→ 0 for t→ −∞, that

is HE,N(t)→ Hb,N as t→ −∞.

Current density

The current operator is defined:

jν,b,N := i[Hb,N , Xν,N] ν = 1, 2

with (DBC), and

jν,b := i[Hb , Xν] ν = 1, 2


10 Introduction

on the full site-mesh. The fact that [X1, X2] = 0 implies that

j2,E,N(E) = j2,b,N + [EeiωtX1, X2] = j2,b,N.

The expectation value, over the central region, of the current density J2,E,N = |ΛN |−1j2,E,N

in the y-direction is 〈 J2,E,N(E) 〉, at t = 0, we denote J2,E,N(E). J2,E,N(E) is given by the

trace of the operator product of ρE,N(0) and J2,E,N.:

J2,E,N(E) =1|ΛN |

TrNρE,N(t = 0)j2,E,N (1.11)

where TrN· is a notational shorhand for a trace over the central region ΛN . We will see

that J2,N,E(E) has an analytic expansion in E:

J2,b(E) = J2,b(E = 0) + Eσ21 +O(E2). (1.12)

The linear term in E, in formula (1.12), is just the off-diagonal conduction element that

we seek to identify, because the light-field is polarized in the x-direction.

1.3 MAIN RESULT OF THIS THESIS

We shall see that σ21 has the form:

σ21(b, N) =1

2πω|ΛN |∮

C

dz fFD(z)(

TrN

(HN,b − z + ω)−1 j1,b,N(HN,b − z)−1j2,b,N

+ TrN

(HN,b − z)−1j1,b,N (HN,b − z−ω)−1 j2,b,N

)

,

(1.13)

see equation (2.26) and figure (1.3) for details and definitions.

Define the operators in ℓ2(Λ):

Db,+(z) := (Hb − z + ω)−1 j1,b(Hb − z)−1j2,b

Db,−(z) := (Hb − z)−1j1,b (Hb − z−ω)−1 j2,b,

where jν,b := i[Hb, Xν], ν = 1, 2. We are now ready to state the main result of this paper.

Theorem 1.1 (Main Result). The following results hold true:

1.

σ21(b) := limN→∞

σ21(b, N)

=1

2|Ω|πω

∮

C

dz fFD(z) ∑x∈Ω

(Db,+(x, x; z) + Db,−(x, x; z)) ,(1.14)

where the smooth path C enclose, but has no points in common with, the (real, bounded)

spectrum of Hb, and C is chosen such that ω lies outside C . C must furthermore be chosen

so “close” to the real line, that fFD(z) has no singularities inside C , see fig. 1.3.

2. The function σ21(·) is smooth and has an asymptotic expansion in b around 0, in particular,

σ21(0) = 0. All the derivatives of σ21 can be written only in terms of the fiber operators

associated to Bloch-decomposition. The Faraday rotation ϑ is proportional with σ′21(0),

which is given by formula (3.19).

Draft: August 11, 2011, introduction/introduction.tex

1.3. Main result of this thesis 11

Re(z)

Im(z)

C

ω

σ(H0)

Figure 1.3: The path C should enclose σ(Hb), and ω should be outside C .

Draft: August 11, 2011, introduction/introduction.tex

2

PROOF OF THEOREM 1.1(1)

2.1 SIMPLE COMBES-THOMAS THEOREM

In this section we discuss general operators defined on ℓ2(Λ). It should be understood

that the results 2.1, 2.2, 2.3, 2.4, 2.6 and 2.5 also hold if Λ is replaced with ΛN , uniformly

in N.

Exponentially almost diagonal operators

Definition 2.1 (Schur-Holmgreen bound). For a linear operator A : ℓ2(Λ) → ℓ2(Λ) with

the kernel a(x, x′), x, x′ ∈ Λ, we define the Schur-Holmgreen bound, ‖A‖1 of A by:

‖A‖1 := max

supx∈Λ

∑x′∈Λ

|a(x, x′)|, supx′∈Λ

∑x∈Λ

|a(x, x′)|

(2.1)

If an operator has ‖A‖1 < ∞ it is said to be Schur-Holmgreen bounded.

Lemma 2.2. Let A be defined as in definition 2.1. If ‖A‖1 < ∞, then ‖A‖ ≤ ‖A‖1, where ‖A‖is the usual operator norm.

‖A‖ ≤(

supx∈Λ

∑x′∈Λ

|a(x, x′)|)1/2(

supx′∈Λ

∑x∈Λ

|a(x, x′)|)1/2

≤ ‖A‖1 (2.2)

Proof. Let ψ ∈ ℓ2(Λ), with ‖ψ‖ = 1, assume that ψ has compact support. Let x be an

arbitrary point in Λ and consider

|(Aψ) (x)| ≤ ∑y∈Λ

|a(x, y)ψ(y)| (2.3a)

≤ ∑y∈Λ

(|a(x, y)|)1/2 (|a(x, y)|)1/2 |ψ(y)| (2.3b)

≤

∑y∈Λ

|a(x, y)|

1/2

∑y∈Λ

|a(x, y)||ψ(y)|2

1/2

. (2.3c)

Draft: August 11, 2011, proofOne/combesThomas.tex 13

14 Proof of theorem 1.1(1)

Inequalities (2.3) implies that

‖Aψ‖2 = ∑x∈Λ

|(Aψ) (x)|2

≤ ∑x∈Λ

∑y∈Λ

|a(x, y)|

∑y∈Λ

|a(x, y)||ψ(y)|2

≤(

supx′

∑y|a(x′, y)|

)

∑x

(

∑y|a(x, y)| |ψ(y)|2

)

≤(

supx′

∑y|a(x′, y)|

)(

supy

∑x|a(x, y)|

)

∑y|ψ(y)|2

︸︷︷︸

=1

,

which shows (2.2). This result can be extended to all ψ ∈ ℓ2(Λ) by the extension-by-

continuity principle.

Definition 2.3 (Exponentially Almost Diagonal Operator). Let H be ℓ2(Λ). We say that

an operator A : H → H is exponentially almost diagonal, if there exists constants C1, C2,

both strictly positive, so that the kernel of A satisfies

|a(x, y)| ≤ C1e−C2‖x−y‖ (2.4)

for all x, y in Λ.

Theorem 2.4. An exponentially almost diagonal operator is Schur-Holmgreen bounded.

Proof. Consider an exponentially almost diagonal operator A. For a real number x, the

geometric series ∑∞k=1 xk converges iff |x| < 1, so the series ∑

∞k=1 e−ck converge iff c>0.

For a exponentially almost diagonal operator, positivity of the norm shows that the sums

(2.1) converge iff C2 > 0, which is always true for an exponentially almost diagonal

operator.

CT-property

We now show a property for exponentially almost diagonal operators, which is a simple

version of Combes-Thomas theorem [5]. We start with a prepatory lemma:

Lemma 2.5. Let H ⊂ ρ(A) be a compact set, and suppose σ(A) is bounded, then the distance

between H and σ(A), defined by

dist(H, σ(A)) := infz∈Hdist(z, σ(A) ) ,

is actually a minimum, and strictly larger than zero.

Proof. H and σ(A) are disjoint sets, so by positivity of the complex norm, for all z ∈ H

and s ∈ σ(A) we must have

‖z− s‖ > 0. (2.5)

Draft: August 11, 2011, proofOne/combesThomas.tex

CT-property 15

The distance function ‖· − ·‖ : C×C → R,

‖z− s‖ z, s ∈ C,

is a continous mapping. Since σ(A) is compact, implying that H × σ(A) is a compact

subset of C2, the value

inf|z− s| : z ∈ H, s ∈ σ(A)

must be attained for at least one pair (z0, s0) ∈ H × σ(A), by the extreme-value theorem

for real-valued continous functions. Therefore the infimum is actually an minimum, and

by inequality (2.5) strictly larger than zero.

Theorem 2.6 (CT-property). Let z be an complex number in the resolvent set ρ(A) of an self-

adjoint exponentially almost diagonal operator A operating on the Hilbert Space H = ℓ2(Λ) for

some lattice Λ. Let constants C1 and C2 be defined as in definition (2.4).

Then the resolvent, (A− z)−1 is also exponentially almost diagonal. That is, there exists two

positive constants, C3 and C4 so that the kernel of (A− z)−1 fullfills

|(A− z)−1(x, y)| ≤ C3e−C4‖x−y‖.

If z is restricted to belong to a closed subset H ⊂ ρ(A), then we have the following lower bound

on C3, and upper bound on C4:

C3 = 2 maxz∈H

1

dist(H, σ(A) )

C4 = min

dist(H, σ(A) )

2C,

C2

4

> 0,

C := maxa1, a216 C1 ∑x∈Λ

e−C22 ‖x‖,

where a1 and a2 are the length of the orthogonal vectors generating the Bravais lattice.

Proof. For α > 0, and an arbitrarily fixed lattice point x0 ∈ Λ, define the operator Aα :

H →H by

Aα := eα‖· −x0‖Ae−α‖· −x0‖

(Aαψ)(x) = eα‖x−x0‖ ∑y∈Λ

a(x, y)e−α‖y−x0‖ψ(y) = ∑y∈Λ

aα(x, y)ψ(y),

, where eα|· −x0| and e−α|· −x0| are multiplicative operators, i.e.

(eα‖· −x0‖ψ)(x) = eα‖x−x0‖ψ(x),

for all x ∈ Λ. Similarly for e−α|· −x0|.

If α is small enough, Aα is bounded, since

|aα(x, y)| =∣∣eα‖x−x0‖a(x, y)e−α‖y−x0‖∣∣ =

∣∣eα‖x−y+y−x0‖a(x, y)e−α‖y−x0‖∣∣

≤∣∣a(x, y)eα‖x−y‖∣∣ ≤

∣∣C1e−(C2−α)‖x−y‖∣∣,



where we have used the triangle inequality in the second-last inequality. This shows that,

as long as C2− α is positive, Aα is exponentially almost diagonal, and therefore bounded.

We will now show that if α is small enough, Aα − z is invertible. Using a geometric

series argument, like in the proof of theorem (A.3) from Peter D. Lax’s book [3], we see

that

(Aα − z) = (1− (A−Aα)(A− z)−1)(A− z)

is invertible if ‖(A− Aα)(A− z)−1‖ ≤ ‖(A− Aα)‖‖(A− z)−1‖ < 1. For a fixed z this

can be achieved if

‖A−Aα‖α→ 0

GGGGGGGGGGGA 0. (2.6)

This is in fact true, which we can see by evaluating the Schur-Holmgreen norm of Aα−A

‖Aα −A‖1 = max

supx

∑y|(Aα −A)(x, y)|, sup

y∑x|(Aα −A)(x, y)|

,

we evaluate supx ∑y|(Aα −A)(x, y)|:

supx

∑y|(Aα −A)(x, y)| = sup

x∑y

∣∣∣eα‖x−x0‖a(x, y)e−α‖y−x0‖ − a(x, y)

∣∣∣

= supx

∑y|a(x, y)|

(

eα(‖x−x0‖−‖y−x0‖) − 1)

≤ supx

∑y|a(x, y)|

(

eα‖x−y‖ − 1)

≤ supx

∑y

C1e−C2‖x−y‖α‖x− y‖eα‖x−y‖., (2.7)

where we in have used that ex − 1 − exx ≤ 0 for x ∈ [0, ∞[ (it is true in x = 0, and

ex − 1 − exx is a decreasing function of x on [0, ∞[). Using an inequality of the type

∑n≥0 e−2nnN ≤ ∑n≥0 e−nN!, it is possible after some tedious rewriting and evaluation, to

recast inequality (2.7) in the form

supx

∑y|(Aα −A)(x, y)| ≤ αC1 sup

x∑y

e(α−C22 )‖x−y‖, C1 := max(a1, a2) · 16 C1

And if α < C2/4 the series is convergent. Evaluating supy ∑x|(Aα − A)(x, y)| is done

similarly, leading to the same bound, and we have

‖Aα −A‖1 ≤ α · C,

C := supx∈Λ

∑y∈Λ

C1e−C24 ‖x−y‖ = C1 ∑

y∈Λ

e−C24 ‖y‖.

which implies (2.6), and therefore implies that (Aα − z) is invertible. A sufficient condi-

tion for the limit (2.6), on α, is therefore

α <C2

4. (2.8)


CT-property 17

The equality

e−α|· −x0|(Aα − z) = (A− z)e−α|· −x0|

implies that

(A− z)−1e−α|· −x0| = e−α|· −x0|(Aα − z)−1.

This implies that for arbitrarily chosen φ ∈ ℓ2(Λ) we have

(A− z)−1e−α‖· −x0‖φ = e−α‖· −x0‖(Aα − z)−1φ, (2.9)

and therefore we can multiply the left side of (2.9) with eα‖· −x0‖ to get a formula for

(Aα − z)−1:

eα‖· −x0‖(A− z)−1e−α‖· −x0‖ = (Aα − z)−1.

If α, for a given z ∈ ρ(A), is chosen to satisfy (in addition to being less than C2/4)

α(z) ≤ 12C‖(A− z)−1‖ , (2.10)

so

‖(A−Aα)(A− z)−1‖ ≤ ‖A−Aα‖‖(A− z)−1‖ ≤ α(z)C‖(A− z)−1‖ ≤ 12

we have∥∥∥eα‖· −x0‖(A− z)−1e−α‖· −x0‖

∥∥∥ = ‖(Aα − z)−1‖

≤ ‖(A− z)−1‖ 11− ‖(A−Aα)(A− z)−1‖

≤ 2‖(A− z)−1‖,

and therefore

supx0∈Λ

∥∥∥eα‖· −x0‖(A− z)−1e−α‖· −x0‖

∥∥∥ ≤ 2‖(A− z)−1‖ = C(z) if 0 ≤ α ≤ α(z).

We examine the kernel of (Aα − z)−1:

⟨δx , (Aα − z)−1δy

⟩=⟨

δx , eα‖· −x0‖(A− z)−1e−α‖· −x0‖δy

⟩

= eα‖x−x0‖(A− z)−1(x, y)e−α‖y−x0‖,

where we have used the properties of the delta-function. If we choose x0 = y we have

⟨δx , (Aα − z)−1δx0

⟩= eα‖x−x0‖(A− z)−1(x, x0),

that is, using the Cauchy-Schwartz inequality, we have the final result

eα‖x−x0‖∣∣∣

(

(A− z)−1)

(x, x0)∣∣∣ = |

⟨δx , (Aα − z)−1δx0

⟩|

≤ ‖δx‖‖(Aα − z)−1δx0‖= ‖(Aα − z)−1‖≤ C(z),



which shows that (A− z)−1 is exponentially almost diagonal.

If we restrict z to belong to a compact subset H of ρ(A), then because of the equality

‖(A− z)−1‖ = 1dist(z, σ(A) )

,

and lemma 2.5, we can evaluate

minz∈H

1

‖(A− z)−1‖

= minz∈Hdist(z, σ(A) ) = dist(H, σ(A) ).

This, used on the bound (2.10), leads to the z-independent bound on α:

α0 ≤dist(H, σ(A) )

2C,

such that for 0 ≤ α ≤ α0 , we know that

|(A− z)−1(x, y)| ≤ 2 supz∈H

‖(A− z)−1‖e−α‖x−y‖

= 2 supz∈H

1dist(z, σ(A) )

e−α‖x−y‖.

2.2 MAGNETIC PERTUBATION

the S operator

We define the operator Sb(z), for z ∈ ρ(H0) as the operator having the matrix element

sb(x, y; z) := eibϕ(x,y)(

(H0 − z)−1)

(x, y)︸︷︷︸

s0(x,y;z)

.

Now we examine the matrix element (HbSb) (x, x′):

(HbSb) (x, x′) = ∑y∈Λ

hb(x, y)sb(y, x′; z)

= ∑y

eib(ϕ(x,y)+ϕ(y,x′))h0(x, y)s0(y, x′; z)

= eibϕ(x,x′) ∑y

(

1 + eib fl(x,y,x′) − 1)

h0(x, y)s0(y, x′; z). (2.11)

If we introduce the operator Kb with the kernel

kb(x, x′; z) := eibϕ(x,x′) ∑y

(

eib fl(x,y,x′) − 1)

h0(x, y)s0(y, x′; z),

and notice that

eibϕ(x,x′) ∑y

h0(x, y)s0(y, x′; z) = eibϕ(x,x′)(H0S0)(x, x′),

Draft: August 11, 2011, proofOne/magneticPertubation.tex

the S operator 19

then (2.11) has the form

(HbSb) (x, x′) = eibϕ(x,x′)(H0S0)(x, x′) + kb(x, x′; z). (2.12)

If we examine (H0S0)(x, x′), we see that

(H0S0)(x, x′) = ((H0 − z1)S0) (x, x′)︸︷︷︸

δ(x,x′)

+z s0(x, x′)

which gives that (2.12) is equivalent with

(HbSb) (x, x′) = eibϕ(x,x′)δ(x, x′) + z eibϕ(x,x′)S0(x, x′)︸︷︷︸

sb(x,x′)

+kb(x, x′),

or

((Hb − z)Sb) (x, x′) = δ(x, x′) + kb(x, x′),

that is

(Hb − z)Sb = 1 + Kb. (2.13)

Using the evaluation

eix − 1 = ei x2

(

ei x2 − e−i x

2

)

= 2iei x2 sin(

x

2),

|eix − 1| ≤ 2|sin(x

2)| ≤ |x|,

for real x ≥ 0, we evaluate the Schur-Holmgren bound for the operator norm of Kb,

consider

supx′

∑x|kb(x, x′; z)| ≤ sup

x′∑x

∑y

∣∣∣eib fl(x,y,x′) − 1

∣∣∣|h0(x, y)||s0(y, x′; z)|

≤ supx′

∑x

∑y

b fl(x, y, x′)C1e−C2‖x−y‖C3(z)e−C4(z)‖y−x′‖

≤ b supx′

∑x

∑y|x− y||y− x′|C1e−C2‖x−y‖C3(z)e

−C4(z)‖y−x′‖

≤ b C1C3(z) supx′

∑x

∑y|x− y|e−C2‖x−y‖︸︷︷︸

≤C5e−C2/2‖x−y‖

|y− x′|e−C4(z)‖y−x′‖︸︷︷︸

≤C6e−C4(z)/2‖y−x′‖

.

(2.14)

That is, there exists a z-dependent constant, C(z)

C(z) = C1C3(z)C5C6 supx′

∑x

∑y

e−C22 ‖x−y‖− C4(z)

2 ‖y−x′‖.

This constant can be chosen uniformly on compact sets M ⊂ ρ(H0):

supz∈M

supx′

∑x|kb(x, x′; z)|

≤ bC. (2.15)

Similarly we can show an estimate for supx ∑x′ |kb(x, x′; z)|, and this implies, using the

Schur-Holmgreen bound, that

supz∈M

‖Kb(z)‖ ≤ bC. (2.16)



Stability of ρ(H) under Peierls substitution.

Theorem 2.7. Let M ⊂ ρ(H0) be compact. Then there exists bM > 0, sufficiently small, such

that M ⊂ ρ(Hb) for all 0 ≤ b ≤ bM.

Proof. We can evaluate the norm of Sb by using the Schur-Holmgreen norm again, equiv-

alently to the above calculations for Kb. Doing this, we show that there exists a constant

CS only dependent on M, such that

‖Sb‖ ≤ ‖Sb‖1 ≤ CS, for all z ∈ M.

Hb is a self-adjoint operator with a symmetric kernel, so the spectrum of Hb must belong

to the real line. Assume that M contains a real λ 6= 0, λ ∈ M ∩R. Note that, because the

spectrum of Hb is real, then if ε 6= 0, the operator Hb − (λ + iε)1 is invertible.

For all z ∈ [λ − i, λ + i] ⊂ ρ(H0), the estimate (2.16) holds, uniformly in λ ∈ M.

Therefore we can choose b so small that (1 + Kb) is invertible.

Substitute z = λ + iε into identity (2.13), and multiply on the right by (1 + Kb)−1 and

on the left by (Hb − λ− iε)−1 to obtain the formula for (Hb − λ− iε)−1:

Sb(λ + iε)(1 + Kb)−1 = (Hb − λ− iε)−1

By the use of corollary A.4 and (2.16) we can evaluate the norm of (Hb − λ− iε)−1:

∥∥(Hb − λ− iε)−1

∥∥ ≤ ‖Sb(λ + iε)‖ 1

1− ‖bC‖

≤ CS1

1− ‖bC‖ = Cst(M),

for b < bM and some M dependent number Cst(M). The now familiar expansion

(Hb − λ) =(

1 + iε(Hb − λ− iε)−1)

(Hb − λ− iε)

shows that Hb− λ is invertible; choose small enough ε, then(1 + iε(Hb − λ− iε)−1

)and

Hb − λ− iε are both invertible.

Let z ∈ ρ(Hb). We use the identity

(1 + Kb)−1 = ∑

n≥0(−1)nKn

b = 1 + ∑n≥1

(−1)nKnb

= 1 + ∑n≥0

(−1)n+1Kn+1b = 1−

(

∑n≥0

(−1)nKnb

)

Kb

= 1− (1 + Kb)−1Kb

to expand (Hb − z)−1:

(Hb − z)−1 = Sb(z)(1 + Kb)−1

= Sb(z)(

1− (1 + Kb)−1Kb

)

= Sb(z)− (Hb − z)−1Kb.


2.3. Magnetic translation 21

And we can iterate this one more time to obtain

(Hb − z)−1 = Sb(z)− Sb(z)Kb +Rb(z), (2.17)

with the definition

Rb(z) := (Hb − z)−1K2b,

which we will give the name “the remainder term”. We prove that the remainder term is

an O(b2) function, in the sense of theorem 2.8.

Theorem 2.8 (Remainder term O(b2)). Let M ⊂ ρ(H0) be closed. The remainder term is

exponentially almost diagonal, and there exists constants C5 ≥ 0, C6 > 0, such that

supz∈M

∣∣Rb(x, y; z)

∣∣ ≤ b2 C5e−C6‖x−y‖, for all z ∈ M.

Proof. For x, y ∈ Λ and z ∈ M, we have the bound

supz∈M

∣∣(Hb − z)−1(x, y)

∣∣ ≤ C1e−C2‖x−y‖.

so if we for a small α > 0 and z ∈ M ⊂ ρH0 (M closed), examine the matrix element

〈δx, eα‖· −y‖Rb(z)e−α‖· −y‖δy〉:

〈δx, eα‖· −y‖Rb(z)e−α‖· −y‖δy〉

= eα‖x−y‖ ∑x′,x′′

[(Hb − z)−1](x, x′)[Kb(z)](x′, x′′)[Kb(z)](x′′, y)

≤ eα‖x−y‖ ∑x′,x′′

C1e−C2‖x−x′‖bC3(M)e−C4(M)‖x′−x′′‖bC3(M)e−C4(M)‖x′′−y‖

≤ eα‖x−y‖b2C1C3(M)2 ∑x′

e−C2‖x−x′‖∑x′′

e−C4(M)‖x′−x′′‖e−C4(M)‖x′′−y‖

︸︷︷︸

≤C5 exp(−C4/2 ‖x′−y‖)

≤ eα‖x−y‖b2C1C3(M)2C5C6 exp(

−min

C2

2,

C4

4

‖x− y‖)

.

That is, by restricting α ≤ min

C24 , C4

8

, the kernel of the remainder term is seen to be

exponentially bounded, with the evaluation

∣∣∣Rb(x, y; z)

∣∣∣ ≤ b2C7(M)e

−min

C24 , C4

8

‖x−y‖,

uniformly in z ∈ M.

2.3 MAGNETIC TRANSLATION

We define the magnetic translation operator as the operator that transforms ψ ∈ ℓ2(Λ)

according to the rule:

(Tb,γψ

)(x) := eibϕ(x,γ)ψ(x− γ), for all γ ∈ Γ, x ∈ Λ.

The magnetic translations has some properties we prove here:

Draft: August 11, 2011, proofOne/magneticTranslations.tex


Theorem 2.9. The Hamilton operator Hb commutes with Tb,γ for all γ ∈ Γ.

Proof. H0 commutes with lattice translation, by construction:

h0(x + γ, y + γ) = h0(x, y) for all γ ∈ Γ, x, y ∈ Λ,

Consider how Tb,γHb transforms an arbitrarly chosen ψ ∈ ℓ2(Λ):

(Tb,γHbψ

)(x) = eibϕ(x,γ) (Hbψ) (x− γ)

= eibϕ(x,γ) ∑y

Hb(x− γ, y)ψ(y)

= eibϕ(x,γ) ∑y

eibϕ(x−γ,y)H0(x− γ, y)ψ(y)

= ∑y

eib(ϕ(x,γ)+ϕ(x,y)−ϕ(γ,y)H0(γ, y + γ)ψ(y)

= ∑y

eibϕ(x,y+γ)H0(γ, y + γ)eib(ϕ(y,γ)ψ(y)

and if we change the summation index by equating x′ = y + γ, then we have

(Tb,γHbψ

)(x) = ∑

x′eibϕ(x,x′)H0(γ, x′)eibϕ(x′−γ,γ)ψ(x′ − γ) (2.18)

Remembering that ϕ(·, ·) has basically the same algebraic properties as a crossproduct, it

is easy to see that

ϕ(x′ − γ, γ) = ϕ(x′, γ),

Therefore equation (2.18) reduces to

(Tb,γHbψ

)(x) = ∑

x′eibϕ(x,x′)H0(γ, x′)eibϕ(x′,γ)ψ(x′ − γ)

= ∑x′

hb(x, x′)(Tb,γψ

)(x′)

=(HbTb,γψ

)(x),

which proves the theorem.

Lemma 2.10. The inverse of the magnetic translation operator Tb,γ (where γ ∈ Γ) is given by

Tb,−γ.

T −1b,γ = Tb,−γ

Proof. We prove this simply by calculating how Tb,−γTb,γ transform an arbitrarily chosen

ψ ∈ ℓ2(Λ), using the identity ϕ(x,−y) = −ϕ(x, y)):

(Tb,−γTb,γψ

)(x) = e−ibϕ(x,γ)

(Tb,γψ

)(x + γ)

= e−ibϕ(x,γ)eibϕ(x+γ,γ)ψ(x + γ− γ)

= ψ(x).

This proves the lemma.

Draft: August 11, 2011, proofOne/magneticTranslations.tex

Current operator commutes with magnetic translations 23

Since Hb commutes with magnetic translations, according to theorem 2.9, and using

lemma 2.10, we have the identities:

Hb = Tb,γHbTb,−γ

and

Hb − z = Tb,γ(Hb − z)Tb,−γ. (2.19)

Inverting the operator (2.19), we have a result for the resolvent of Hb.

Theorem 2.11. For z ∈ ρHb we have

(Hb − z)−1 = Tb,γ(Hb − z)−1Tb,−γ,

and therefore that the resolvent also commutes with magnetic translations.

Current operator commutes with magnetic translations

Consider the matrix element of the current operator

j1,b(x, y) = ihb(x, y)(y1 − x1)

= eibϕ(x,y)j1,0(x, y),

where we denote

j1,0(x, y) = ih0(x, y)(y1 − x1).

From this we deduce, that, since h0(x + γ, y + γ) = h0(x, y) (for all γ ∈ Γ), j1,0 must have

the same property:

j1,0(x + γ, y + γ) = j1,0(x, y) for all γ ∈ Γ.

The current operator commutes with magnetic translations, a property we state as a

lemma

Lemma 2.12. For γ ∈ Γ we have that

Tb,γj1,b = j1,bTb,γ

Proof. The proof of this is in every way similarly to the proof of theorem 2.9, so we omit

it here.

2.4 LARGE N LIMIT

We return to the off-diagonal element of the conductivity tensor, σ21,N(t = 0).

In the Interaction picture the y-current operator is given by:

j2N(t) = exp (itHb,N) j2,N exp (−itHb,N) ,

Draft: August 11, 2011, proofOne/largeNLimit.tex


we denote other operators in the interaction picture by a “tilde”, for instance j1,N, and

Xν,b,N, ν = 1, 2.

In the interaction picture, an power series expansion of ˜ N,E(t = 0) in powers of E can

be found iteratively, (under the condition that the interaction term of the Hamiltonian is

wellbehaved), since ˜ N,E(t) is a solution of:

ddt

˜ N,E(t) = −i[VN,E(t), ˜ N,E(t)]

We can transform the differential Liouville equation for ˜ N,E(0) (1.10) into an integral

equation:

˜ N,E(0) = ˜ N,E(t0) +∫ 0

t0

dsdds

˜ N,E(s)

= ˜ N,E(t0)− i∫ 0

t0

ds[VN,E(s), ˜ N,E(s)],

where t0 should thought of as a negative real number of large absolute value (we will

take the limit t0 → −∞ where we require the density operator to be fFD(HN,b), since for

all t0 we have exp(iHN,bt0) fFD(HN,b) exp(−iHN,bt0) = fFD(HN,b)). We assume a unique

solution exists that can be written as power series in E. We find the term linear in E by

inspecting

˜ N,E(0) = ˜ N,E(t0)− i∫ 0

t0

ds1

[

VN,E(s1) , ˜ N,E(t0)− i∫ s1

t0

ds2[VN,E(s2), ˜ N,E(s2)]

]

,

= ˜ N,E(t0)− i∫ 0

t0

ds1

[

VN,E(s1) , ˜ N,E(t0)]

−∫ 0

t0

ds1

∫ s1

t0

ds2

[

VN,E(s1) , [VN,E(s2), ˜ N,E(s2)]]

= ˜ N,E(t0)− i∫ 0

t0

ds1

[

VN,E(s1) , ˜ N,E(t0)]

+O(E2),

and in the limit t0 → −∞ this becomes

˜ N,E(0) = fFD(HN,b)− i∫ 0

−∞ds[VN,E(s) , fFD(HN,b)

]+O(E2) .

Therefore, in the linear response approximation we set

N,E,lin(0) = ˜ N,E,,lin(0) := fFD(HN,b)− iE∫ 0

−∞ds eiωs

[X1N,E(s) , fFD(HN,b)

], (2.20)

andwe focus our attention on calculating

J2,b,N,,lin(E, t = 0) = TrN

N,E,,lin(0)j2,b,N

|ΛN |

.

By examining formulae (1.11), (1.12) and (2.20), we can single out the conduction term,

and see that

σN,21(t = 0) = − i

|ΛN |∫ 0

−∞ds eiωs TrN

[X1,b,N(s) , fFD(HN,b)

]j2,b,N, (2.21)


Some commutators 25

where we have interchanged the order of the trace and the integral (this doesn’t change

the sum or integral by Fubini’s theorem).

The zero-field term, J2(0) is given by:

J2,b,N(0) ∝ TrN fFD(HN,b)[HN,b, X2,N]= TrN[ fFD(HN,b), HN,b]X2,N= 0,

which makes sense; no light gives no (oblique) current.

Some commutators

With the choice of hopping integrals as described in section 1.2, the first component of

the current operator i[Tb,N , X1,N] commutes with the second component of the position

operator:

([j1,b,N, X2,N]ψ) (x) = ([i[Tb,,N , X1,N] , X2,N]ψ) (x)

= ([i(Tb,NX1,N)− i(X1,NTb,N) , X2,N]ψ) (x)

= ((i(Tb,NX1,NX2,N)− i(X1,NTb,NX2,N)− i(X2,NTb,NX1,N) + i(X2,NX1,NTb,N))ψ) (x)

= iχΛN(x)∑

y

eibϕ(x,y)t0(x, y)(y1y2 − x1y2 − x2y1 + x2x1)χΛN(y)ψ(y)

= iχΛN(x)∑

y

eibϕ(x,y)t0(x, y)(x1 − y1)(x2 − y2)χΛN(y)ψ(y)

= i ∑y

χΛN(x)eibϕ(x,y)

(δx1,y1±a1/2δx2,y2 + δx2,y2±a2 δx1,y1

)(x1 − y1)(x2 − y2)χΛN

(y)ψ(y)

= 0, for all x ∈ ΛN .

This also hold on the full site-mesh. Doing equivalent calculations, one can see, that

([j1,b,N, X2,N]ψ) (x) = 0 for all x ∈ ΛN .

Using the trace-commutator rule (A.2), we see that expression (2.21) can be re-expressed

as:

σ21,N(t = 0) =−i

|ΛN |∫ 0

−∞ds eiωs TrN

fFD(HN,b) [j2,b,N , eisHN,b X1,Ne−isHN,b ]

=−i

|ΛN |∫ 0

−∞ds eiωs TrN

[j2,b,N , eisHN,b X1,Ne−isHN,b ] fFD(HN,0)

(2.22)

Now use the identity

eiωs =1

iω

dds

eiωs

and the partial integration rule for a function G(s)

∫ 0

−∞

dds

(1

iωeiωs

)

G(s)ds =

[1

iωeiωsG(s)

]0

−∞

−∫ 0

−∞

1iω

eiωsG′(s)ds



to re-express (2.22):

σ21,N(t = 0) =−i

|ΛN |

[1

iωeiωs TrN

[j2,b,N, eisHN,b X1,Ne−isHN,b ] fFD(HN,b)]0

−∞

+i

|ΛN |∫ 0

−∞ds

1iω

eiωs dds

(

TrN

[j2,b,N, eisHN,b X1,Ne−isHN,b ] fFD(HN,b))

.

(2.23)

The first term of (2.23) is zero, because:

• The s = 0 term; the exponential factors vanish, since s = 0, furthermore, j2,b,N and

X1,N commute.

• The s → −∞ term; the choice of negative imaginary part of ω ensures, that eiωs

tends to zero as s tends to minus infinity and the trace is performed over a finite set

of sites, so it must be bounded. i · (−i|Imω|) · t = |Imω|t→ −∞ as t→ −∞.

The differentiating in the second term of (2.23) is done by considering that j2,b has no

explicit time dependence, noticing that

ddt

(

eitHN,b X1,Ne−itHN,b

)

= eitHN,b i(HN,bX1,N − X1,NHN,b)e−itHN,b

= eitHN,b i[HN,b, X1,N]e−itHN,b = eitHN,b j1,b,Ne−itHN,b

by the definition of j1,b. We thus have

σ21,N(t = 0) =1

ω|ΛN |∫ 0

−∞ds eiωs TrN

[j2,b,N , eisHN,b j1,b,Ne−isHN,b ] fFD(HN,b)

=1

ω|ΛN |∫ 0

−∞ds eiωs TrN

eisHN,b j1,b,Ne−isHN,b [ fFD(HN,b) , j2,b,N]

,

where we have used the trace-commutator rule in the last equation. The problem is

therefore to evaluate

σ21,N(t = 0) =1

ω|ΛN |∫ 0

−∞

(

TrN

eis(ω+HN,b)j1,b,Ne−isHN,b fFD(HN,b)j2,b,N

−TrN

eis(ω+HN,b) j1,b,N e−isHN,b j2,b,N fFD(HN,b))

ds.(2.24)

By using the trace-permutation rule on the second term, and by noticing that fFD(HN,b)

and eis(ω+HN,b) commute, we have

σ21,N(t = 0) =1

ω|ΛN |∫ 0

−∞

(

TrN

eisHN,b)j1,b,Ne−is(HN,b−ω) fFD(HN,b)j2,b,N

−Tr

eis(ω+HN,b) fFD(HN,b)j1,b,NeisHN,b j2,b,N))

ds

We can use the Cauchy integral formula to express the operator eis(ω+HN,b) fFD(HN,b) by

a curve integral in the complex plane, involving the resolvent of HN,b:

eis(ω+HN,b) fFD(HN,b) =i

2π

∮

C

eis(z+ω) fFD(z)(HN,b − z)−1dz,


Some commutators 27

where the path C enclose, but has no points in common with, the (real, bounded) spec-

trum, σ(H0,N), of H0,N. We claim that it is possible, given ω, β and µ, to choose such a

curve such that ω lies outside C , and so “close” (small imaginary parts of z ∈ C ) to the

real line, that

fFD(z) =1

eβ(z−µ) + 1

has no singularities inside C . (µ ∈ R, β ∈ R, if Rez = µ, then we can make sure that

β · Imz cannot atain the value π, i.e. choose that all z ∈ C fullfills |Imz| < πβ , also ω is

assumed to have a non-zero, (negative) imaginary part.) This leads to

σ21,N(t = 0)

=1

ω|ΛN |∫ 0

−∞ds

TrN

eisHN,b j1,b,N

(i

2π

∮

C

dz e−is(z−ω) fFD(z)(HN,b − z)−1)

︸︷︷︸

=e−is(HN,b−ω)

fFD(HN,b)

j2,b,N

− TrN

(i

2π

∮

C

dz eis(z+ω) fFD(z)(HN,b − z)−1)

︸︷︷︸

=eis(HN,b+ω)

fFD(HN,b)

j1,b,N e−isHN,b j2,b,N

(2.25)

. The two integrals,∮

Cdz· · · and

∫ 0−∞

ds· · · in equation (2.25) are both absolutely conver-

gent, therefore we can exchange integration order (Fubini). Furthermore, as e±is(z±ω) is

nothing but a complex scalar, we can freely place this factor in the operator product,

σ21,N(t = 0)

=1

ω|ΛN |∮

C

dz∫ 0

−∞ds

(

TrN

eis(HN,b−z+ω)j1,b,Ni

2πfFD(z)(HN,b − z)−1j2,b,N

− TrN

i

2πfFD(z)(HN,b − z)−1j1,b,Neis(z+ω−HN,b)j2,b,N

)

.

We now evaluate the (time) integration:

∫ 0

−∞ds eis(z+ω−HN,b) =

[

(i((z + ω)−HN,b))−1 eis(z+ω−HN,b)

]0

−∞,

= (i((z + ω)−HN,b))−1

= i (HN,b − z−ω)−1

and similar for∫ 0−∞

ddds eis(HN,b−(z−ω)) = (i(HN,b − z + ω))−1. Where we note that for



z ∈ C , z±ω cannot be in the spectrum for HN,b. This lead to the formula for σxy(0)

σ21,N(t = 0) =

=1

ω|ΛN |∮

C

dz

(

TrN

(−i) (HN,b − z + ω)−1 j1,b,Ni

2πfFD(z)(HN,b − z)−1j2,b,N

− TrN

i

2πfFD(z)(HN,b − z)−1j1,b,N i (HN,b − z−ω)−1 j2,b,N

)

=1

2πω|ΛN |∮

C

dz fFD(z)(

TrN

(HN,b − z + ω)−1 j1,b,N(HN,b − z)−1j2,b,N

+ TrN

(HN,b − z)−1j1,b,N (HN,b − z−ω)−1 j2,b,N

) (2.26)

Dealing with Dirichlet

Now we would like to get rid of the Dirichlet boundary condition. Let ε be a number

between 0 and 1, 0 < ε < 1. We divide the central region, ΛN , up into a Nε unit cells

wide rim, ˜ΛN„ and a core part, a square, 2(N − Nε) + 1, unit cells wide, figure 2.11.

˜ΛN ΛN

Figure 2.1: We divide the central region, ΛN , up into a rim ˜ΛN , Nε unit cells wide, and a core part ΛN ,2(N − Nε) + 1, unit cells wide. 0 < ε < 1

The different parts of ΛN relate to each other by the formula:

˜ΛN = ΛN \ ΛN .

The number of sites in ˜ΛN is dominated by N2,

| ˜ΛN |N2 → 0, N → ∞,

1Of course, it makes no sense to take a non-integer number of unit cells, so in practice, take the floorof the number Nε. In the large N regime, which is the interesting case for this presentation, this makes nodifference in the limits we are to examine.


Some auxilliary operators 29

as a consequence of the choice of ε, whereas the number of sites in ΛN converge to 4N2:

(2(N − Nε) + 1)2 = 4N2(

1− 1N1−ε

+1

2N

)2

→ 4N2, N → ∞.

Some auxilliary operators

To simplify notation and maximize readability, we introduce a shorthand for the charac-

teristic functions for ΛN , ΛN and ˜ΛN :

χN := χΛN,

χN := χΛN,

˜χN := χ ˜ΛN.

We now introduce a new auxilliary operator by the definition:

Ab,N(z) := χN (Hb − z)−1 χN + ˜χN (Hb,N − z)−1 ˜χN.

We can express the resolvent (Hb,N − z)−1 in terms of (Hb − z)−1 and different charac-

teristic functions, as a step towards taking the large N limit.

If we multiply Ab,N(z) on the left by (Hb,N − z), we have

(Hb,N − z)Ab,N(z) = (Hb,N − z) χN (Hb − z)−1 χN + (Hb,N − z) ˜χN (Hb,N − z)−1 ˜χN.

(2.27)

Note that for large N, the nn-interaction property of Hb,N implies that (Hb,N − z) χN =

(Hb − z) χN. Therefore, if we use this in the first term of formula (2.27), and then com-

mute (Hb − z) and χN, we have:

(Hb,N − z) χN (Hb − z)−1 χN = [Hb, χN] (Hb − z)−1 χN + χN. (2.28)

Similarly we commute Hb,N and ˜χN in the second term of formula (2.27) to get:

(Hb,N − z) ˜χN (Hb,N − z)−1 ˜χN = [Hb,N, ˜χN] (Hb,N − z)−1 ˜χN + ˜χN.. (2.29)

Combining, and noticing that χN + ˜χN = χN, we have the expression for (2.27):

(Hb,N − z)Ab,N(z) = χN + [Hb, χN] (Hb − z)−1 χN + [Hb,N, ˜χN] (Hb,N − z)−1 ˜χN,

which is equivalent with (we off course choose z ∈ ρ(Hb), so (Hb,N − z) is invertible...):

(Hb,N − z)−1 χN = (Hb,N − z)−1 = Ab,N(z)− (Hb,N − z)−1 Bb,N(z), (2.30)

where we have collected the terms with commutators in the definition

Bb,N(z) := [Hb, χN] (Hb − z)−1 χN + [Hb,N, ˜χN] (Hb,N − z)−1 ˜χN.



Since we in the trace in formula (2.26) we only sum over ΛN , in this trace we omit χN on

the left side of equation (2.30). If we therefore substitute

(Hb,N − z)−1 = Ab,N(z)− (Hb,N − z)−1 Bb,N(z),

into the formula for the off-diagonal part of the conductivity, formula (2.26), for z = z

and z = z±ω, we then get a lot off different terms. We claim that only one of these terms

contribute in the large N limit, that is the term:

12πω|ΛN |

∮

C

dz fFD(z)(

TrN

χN (Hb − z + ω)−1 χNjN,1χN(Hb − z)−1χNjN,2

+ TrN

χN(Hb − z)−1χNjN,1χN (Hb − z−ω)−1 χNjN,2

)

.

The other terms all have factors of type ˜χN (Hb,N − z)−1, [Hb, χN] or [Hb,N, ˜χN]. In the

large N limit terms having these factors vanish, which we explain in the following.

First three kind of junk terms

The ˜χN (Hb,N − z)−1 type factors:

By the trace permutation rule, it is enough to examine terms of the type

1|ΛN |

TrN

˜χN (Hb,N − z)−1 · · ·

∝1|ΛN | ∑

x∈ ˜ΛN

· · · (x, x).

One example is

1|ΛN |

TrN

(Hb,N − z + ω)−1 ˜χNj1,b,NχN(Hb − z)−1χNj2,b,N

. (2.31)

By the definition of j1,b, we have the kernel element for the first current factor given as

j1,b(x, y) = Hb(x, y)(y1 − x1)), where the zero-field (nn) Hamiltonian, Hb is known to

be bounded. Similarly for j2,b, j2,b(x, y) = Hb(x, y)(y2 − x2)). Now, we also know that

the kernel elements of the resolvent (Hb,N − z)−1 decays exponentially and are bounded.

The sum defining the trace (2.31) is therefore absolutely bounded by the sum (by omitting

some characteristic functions, we can only add to the sum of non-negative numbers.)

∑x∈ ˜ΛN

x′′′

,x′′

,x′∈ΛN

∣∣∣∣∣

[

(HN,N − z + ω)−1]

(x, x′)j1,b,N(x′, x′′)[

(HN − z)−1]

(x′′, x′′′)j2,b,N(x

′′′, x)

∣∣∣∣∣

≤ ∑x∈ ˜ΛN

x′′′

,x′′

,x′∈Λ

C1e−C2‖x−x′‖Hb(x′, x′′)|x′′1 − x

′1|C3e−C4‖x

′′−x′′′‖Hb(x′′′

, x)|x2 − x′′′2 |.

For N-independent constants Ci, i = 1, 2, 3, 4. By using evaluations of the type

Hb(x′, x′′)|x′′1 − x

′1| ≤ C5 exp(−c6‖x′ − x

′′‖)‖x′ − x′′‖,

for constants C5 and C6, and the sum rule ∑ e−2Cnn ≤ ∑ e−Cn, we see that all terms are

exponentially located, so each sub-sum for a fixed x ∈ ΛN is convergent and bounded,


First three kind of junk terms 31

uniformly in N. This is supposing z and ω is chosen appropriately - according to the

restrictions mentioned earlier. For each N we have finitely many sites in ˜ΛN to sum over,

so the sum converges, and even if the number of sites inside ˜ΛN grow as N · Nε, this

number is dominated by the factor |ΛN |−1∝ N−2, the crucial piece of information being

ε < 1:∣∣∣

1|ΛN |

TrN

˜χN (Hb,N − z)−1 · · ·∣∣∣ ≤ 1|ΛN |

(| ˜ΛN | ·Cst.)→ 0, N → ∞.

Therefore all these types of terms goes to zero in the large N limit.

The [Hb, χN] type factors:

Again, we can use the trace-permutation rule and need only look at terms of the type

1|ΛN |

TrN [Hb, χN] · · · .

We introduce another subset of the central region, a band centered on the edge of ΛN of

width 10 unit cells as sketched in figure 2.2, we denote this subset ∂ΛN . Now we claim

10

∂ΛN

ΛN

Nε

Figure 2.2: ∂ΛN is a 10 unit cell wide band centered on the edge of ΛN .

that for [Hb, χN]-type factors, it suffices to sum over ∂ΛN in the trace. This is due to the

equality

[Hb, χN](x, y) = Hb(x, y)(χN(y)− χN(x)).

The factor (χN(y)− χN(x)) is zero unless one of x or y falls in ΛN and the other in ˜ΛN .

The nearest neighbour form of the Hamiltonian makes the matrix element Hb(x, y) be

zero unless both ‖x− y‖ is less than, say 2a (depending on the choice of hopping matrix

elements in the tb-model). This proves the claim.

Again the subsum of the trace-sum for each fixed x in ∂ΛN is absolutely bounded,

uniformely in N. (since we sum over kernel elements of an operator that is the product



of exponentially almost diagonal operators). The number of sites in ∂ΛN grows as N

whereas we divide by a factor proportional to N2, so terms of this type must also vanish

in the large N limit.

The [Hb,N, ˜χN] type factors:

The reason why [Hb,N, ˜χN]-type factorsvanish in the large N limit is the same as for

[Hb, χN] type factors (above), only with the band ∂ ˜ΛN (see figure 2.3) instead of ∂ΛN .

10

10

∂ ˜ΛN

Nε

ΛN

Figure 2.3: ∂ ˜ΛN is a 10 unit cell wide band centered on the inner and outher edges of ˜ΛN .

Removing more remainder terms

If a term has more than one of these first three types of “junk” terms, then it will vanish

even more rapidly as N increases. We are left with one term, given by formula (2.33). We

rewrite

χN = 1− (1− χN),

and inserting this into (2.33), we have that the first trace of (2.33) is given by:

1|ΛN |

Tr

χN (Hb − z + ω)−1 χN j1,b,N χN︸︷︷︸

=χNj1,bχN

(Hb − z)−1 χN j2,b,N︸︷︷︸

=χN j2,b χN


Reducing the problem to the unit cell 33

We notice that

χNj1,b,NχN = χNj1,bχN,

by same arguments as above, so we consider the trace

1|ΛN |

Tr

χN (Hb − z + ω)−1 (1− (1− χN)) j1,b × · · ·

× (1− (1− χN)) (Hb − z)−1 (1− (1− χN)) j2,bχN

,(2.32)

this trace consists of the main term

1|ΛN |

Tr

χN (Hb − z + ω)−1 j1,b(Hb − z)−1j2,b

,

and remainder terms

− 1|ΛN |

Tr

χN (Hb − z + ω)−1 (1− χN)j1,b(Hb − z)−1j2,b

+ · · ·

All the remainder terms have at least one factor of the type

1− χN = 1− χN + ˜χN

where 1 − χN is the characteristic function of the complementary set of ΛN . The terms

arising from ˜χN part can be brought to the form

− 1|ΛN |

Tr

˜χN (Hb − z + ω)−1 × · · ·

which is a ˜χN (Hb,N − z)−1 type factor, and vanishes in the large N limit.

The term arising from the 1 − χN part also goes to zero, because resolvent kernel

elements sanwiched between 1− χN and χN is evaluated at pairs of sitepoints seperated

by at least Nε unit cells, and therefore the sum terms are absolutely bounded by terms

with a factor of e−CNεwhich dominates the number of sites summed over( ∝ N2), so

terms of this type also goes to zero as N → ∞.

Reducing the problem to the unit cell

According to the above and formula (2.33), the only non-vanishing contribution to the

off-diagonal conduction element in the large N-limit isgiven by

12πω|ΛN |

∮

C

dz fFD(z)(

Tr

χN (Hb − z + ω)−1 j1,b(Hb − z)−1j2,bχN

+ Tr

χN(Hb − z)−1j1,b (Hb − z−ω)−1 j2,bχN

)

.(2.33)

We now introduce further notational simplifications

Db,+(z) := (Hb − z + ω)−1 j1,b(Hb − z)−1j2,b (2.34a)

Db,−(z) := (Hb − z)−1j1,b (Hb − z−ω)−1 j2,b (2.34b)



So, using this notation, we wish to calculate the traces

1|ΛN | ∑

x∈ΛN

Db,+(x, x; z) +1|ΛN | ∑

x∈ΛN

Db,−(x, x; z) (2.35)

By theorem and theorem, Db,±(z) is a operator product of operators that all commute

with magnetic translations, so Db,±(z) must also commute with magnetic translations.

This leads to the important lemma

Lemma 2.13. For x ∈ Ω, and γ ∈ Γ, we have that

Db,±(x + γ, x + γ; z) = Db,±(x, x; z).

Proof. Let γ be a fixed vector in Γ. We calculate how the operator Db,±Tb,γ and Tb,γDb,±transforms an arbitrarily chosen vector ψ in ℓ2(Λ) (nowing that the two calculations

should be equal since Db,± commutes with Tb,γ ).(

Db,±Tb,γψ)

(x) = ∑y

Db,±(x, y; z)eibϕ(y,γ)ψ(y− γ) (2.36)

with the definition x′ := y− γ this is equal to(

Db,±Tb,γψ)

(x) = ∑x′

Db,±(x, x′ + γ; z)eibϕ(x′+γ,γ)ψ(x′) (2.37)

Calculating Tb,γDb,± gives(

Tb,γDb,±ψ)

(x) = eibϕ(x,γ) (Db,±ψ) (x− γ)

= ∑x′

eibϕ(x,γ)Db,±(x− γ, x′; z)ψ(x′)

These equations must be true for all ψ, so taking ψ = δx−γ for a fixed x ∈ Ω, we have

Db,±(x, x; z)eibϕ(x,γ) = eibϕ(x,γ)Db,±(x− γ, x− γ; z)

and taking x = x we have the desired result.

Now it is easy to see, that we only need to take the trace over the unit cell, in the

trace (2.35):

1|ΛN | ∑

x∈ΛN

Db,+(x, x; z) +1|ΛN | ∑

x∈ΛN

Db,−(x, x; z)

=the number of unit cells in ΛN

|ΛN | ∑x∈Ω

(Db,+(x, x; z) + Db,−(x, x; z))

and the fraction in this is

the number of unit cells in ΛN

|ΛN |=

(2(N − Nε) + 1)2

(2N + 1)2|Ω|) → 1|Ω| , N → ∞

so in the large N limit the off-diagonal conduction tensor element is given by:

σ21(b) =1

2|Ω|πω

∮

C

dz fFD(z) ∑x∈Ω

(Db,+(x, x; z, b) + Db,−(x, x; z, b)) (2.38)

Draft: August 11, 2011, proofOne/proofOne.tex

3

PROOF OF THEOREM 1.1(2)

3.1 K-SPACE REPRESENTATION

The crystal potential is periodic in Γ,

(V0ψ)(γ + x) = (V0ψ)(γ + β + x). for all β ∈ Γ.

We define H ′ := ℓ2(Ω) and introduce 1

HF :=∫ ⊕

Ω∗H′d2k.

The linear space HF can be shown to be a Hilbert space [6], with the inner product∫

Ω∗d2k ∑

x∈Ω∗f (x, k)g(x, k).

(subscript “F” for “Floquet”.)

We define the operator taking vectors ψ in ℓ2(Λ), ψ having compact support, into

HF, U : ℓ2(Λ)→HF, as the operation that satisfies

(Uψ)(k, x) =1

√

|Ω∗| ∑γ∈Γ

exp(ik · γ)ψ(x− γ), k ∈ Ω∗, x ∈ Ω, (3.1)

We will now show that ℓ2(Λ) and HF are isomorphic, because U has an unitary exten-

sion. We denote by ℓ2c(Λ) the vectors ψ in ℓ2(Λ) that has compact support.

Theorem 3.1. U defined by formula (3.1) has an unitary extension.

Proof. It can be shown that ℓ2c(Λ) is a dense subset of ℓ2(Λ). For an arbitrarily ψ ∈ ℓ2

c(Λ),

we examine ‖Uψ‖F:

‖Uψ‖2F =

∫

k∈Ω∗d2k ∑

x∈Ω

(

1√

|Ω∗| ∑γ∈Γ

e−ik·γψ(x− γ)

)(

1√

|Ω∗| ∑β∈Γ

eik·βψ(x− γ)

)

.

(3.2)

1 The ⊕ marks a “constant fiber direct integral”.

Draft: August 11, 2011, proofTwo/kSpaceDecomp.tex 35


The integral∫

k∈Ω∗ d2k e−ik·γeik·β will give zero unless for γ = β where it gives |Ω∗|, we

thus have

‖Uψ‖2F = ∑

x∈Ω

∑γ∈Γ

ψ(x− γ)ψ(x− γ) = ‖ψ‖2ℓ2(Λ),

Therefore U has a unique extension by continuity to ℓ2(Λ), and we denote this extension

also by U.

The adjoint operator U∗ satisfies, for ψ ∈ ℓ2(Λ) and f ∈ HF .

〈 f , Uψ〉F = 〈U∗ f , ψ〉ℓ2(Λ) . (3.3)

Now we want a explicit formula for U∗ f , f being sufficiently smooth and Ω∗ periodic in

the k variables.

If, in addition, ψ ∈ ℓ2c(Λ), then the computation

〈 f , Uψ〉F =∫

Ω∗d2k ∑

x∈Ω

f (k, x)(Uψ)(k, x)

=∫

Ω∗d2k ∑

x∈Ω

f (k, x)1

√

|Ω∗| ∑γ∈Γ

exp(ik · γ)ψ(x− γ)

= ∑γ∈Γ

∑x∈Ω

(

(|Ω∗|)−1/2∫

Ω∗d2k f (k, x) exp(−ik · γ)

)

ψ(x− γ)

and the equality (3.3), indicates that the adjoint of U is given by

(U∗ f ) (x− γ) =1

√

|Ω∗|

∫

Ω∗d2k f (k, x) exp(−ik · γ).

A computation similar to formula (3.2) shows that for ψ ∈ ℓ2c(Λ) and φ ∈ ℓ2

c(Λ) it is true

that

〈Uψ, Uφ〉F = 〈ψ, U∗Uφ〉ℓ2(Λ)

= 〈ψ, φ〉ℓ2(Λ).

We therefore have, for all ψ in ℓ2(Λ), that

〈ψ, U∗Uφ−φ〉ℓ2(Λ) = 0,

which indicate that U∗Uφ = φ for all φ in ℓ2c(Λ), and therefore the extensions obey

U∗U = 1. (identity in ℓ2(Λ) ) (3.4)

For f ∈ HF, f being sufficiently smooth and Ω∗ periodic in the k variables, consider

UU∗ f :

(UU∗ f ) (k, x) =1

√

|Ω∗| ∑γ∈Γ

eik·γ(U∗ f )(x− γ)

=1

√

|Ω∗| ∑γ∈Γ

eik·γ 1√

|Ω∗|

∫

Ω∗d2k f (k, x)e−ik·γ,

Draft: August 11, 2011, proofTwo/kSpaceDecomp.tex

3.1. k-space representation 37

which is nothing but the Fourier series decomposition of f (k, x) 2 . If we require f to be

sufficiently smooth, so that the Fourier inversion theorem holds, we therefore have

(UU∗ f ) (k, x) = f (k, x),

and from this it follows that

UU∗ = 1.

with the restrictions mentioned above.

The functions in HF satisfying smoothness and Ω∗ periodicity are dense in HF.

The fact that U has a unitary extension on ℓ2(Λ) and U∗ an extension on HF, satisfy-

ing all of the above, can then be shown using continuity arguments.

K-space representation of operators allows us to work with finite matrices when cal-

culating energy-levels and expectation values of observables. For instance, let H = T + V0

be an (hermitian adjoint. exponentially almost diagonal 3 ) hamiltonian operator on Λ,

that commutes with lattice translations Tγ, that is TγH = HTγ. The translation operator

transforms according to the rule (Tγψ)(x) = ψ(x− γ). We can find the k-space represen-

tative of this operator as it must be UHU∗ : HF → HF4. We examine how this operator

transforms an sufficiently smooth, Ω∗-periodic f ∈ HF

(UHU∗ f ) (k, x) = ∑γ∈Γ

1√

|Ω∗|eik·γ (HU∗ f ) (x− γ)

= ∑γ∈Γ

1√

|Ω∗|eik·γ (TγHU∗ f ) (0 + x)

= ∑γ∈Γ

1√

|Ω∗|eik·γ (HTγU∗ f ) (x)

= ∑γ∈Γ

1√

|Ω∗|eik·γ ∑

y∈Λ

(

t(x, y)(U∗ f )(y− γ)︸︷︷︸

kinetic term

+ v0(y)(U∗ f )(y− x)

︸︷︷︸

potential term

)

.

Examine the kinetic term first.

∑γ∈Γ

1√

|Ω∗|eik·γ ∑

β∈Γ

∑y∈Ω

t(x, β + y)(U∗ f )(β + y− γ)

= ∑β∈Γ

∑y∈Ω

t(x, β + y) ∑γ∈Γ

eik·(γ+β−β) 1|Ω∗|

∫

Ω∗d2k′ f (k′, y)e−ik

′·(γ−β)

= ∑β∈Γ

∑y∈Ω

t(x, β + y)eik·β ∑γ∈Γ

eik·(γ−β) 1|Ω∗|

∫

Ω∗d2k′ f (k′, y)e−ik

′·(γ−β)

= ∑y∈Ω

∑β∈Γ

t(x, β + y)eik·β f (k, y), (3.5)

2For f ∈ HF, sufficiently smooth, and x ∈ Ω, k ∈ Ω∗ the fourier series of f (k, x)is f (k, x) = ∑γ∈Γ f (γ, x) exp(ik · γ), with the Fourier coefficient, f (γ, x), given by f (γ, x) =

|Ω∗|−1 ∫

Ω∗ d2k f (k, x) exp(−ik · γ).3We want to be able to use Fubini’s theorem over and over - swapping integration and summation

orders.4The operators UHU∗ : HF → HF and H are said to be unitarily equivalent if U maps the domain of H

bijectively onto the domain of UHU∗ : HF → HF.



where we have used the Fourier expansion identity again in the last inequality. If we for

each k define the x, y matrix element, tF(k; x, y), in a |Ω| × |Ω|matrix tF(k), as

tF(x, y; k) := ∑β∈Γ

t(x, β + y)eik·β,

then identity (3.5) kan be seen as a matrix product of the tF(k) matrix and a |Ω|-tuple

( f (k, y1), f (k, y

2), · · · , f (k, y|Ω|)):

(tF(k) f (k, ·)) (x) = ∑y∈Ω

tF(x, y; k) f (k, y).

Note on notation: In order to lessen the amount of subscripts, we use the same op-

erater name in both ℓ2 and k-space, if there occurs a k in the variables of an kernel, for

instance a(x, y; k), then it is meant that this is a fiber in k-space, representing an operator

in ℓ2 with the kernel a(x, y) .

The k-space representation potential operator, V0, is a little more simple to find, be-

cause of the periodicity of v0:

(UV0U∗ f )(k, x) = ∑γ∈Γ

eik·γ(V0U∗ f )(x− γ)

= ∑γ∈Γ

eik·γv0(x− γ)(U∗ f )(x− γ)

= ∑γ∈Γ

eik·γv0(0 + x)(U∗ f )(x− γ)

= v0(x)(UU∗ f )(k, x)

= v0(x) f (k, x),

where v0 : Ω→ C is the values v0 can take on the unit cell. The k-space representation of

the potential operator is thus a k-independent multiplication operator. The full Hamilton

operator for a given k ∈ Ω∗ (the fiber) can therefore be expressed by the kernel

hF(x, y; k) := ∑γ∈Γ

eik·γt(x− γ, y) + v0(x)δ(x, y). (3.6)

The common notation is

UHU∗ =∫ ⊕

Ω∗hF(x, y; k)dk.

We have used the calculation rule for the kinetic term of the Hamiltonian:

t(x, y + γ) = t(x− γ, y).

One can easily see, by inspecting formula (3.6), that the matrix hF(k) inherits the property

of being self-adjoint from the operator T:

t(x− γ, y) = t(y, x− γ) ⇒ hF(x, y; k) = hF(y, x; k).


Energy bands 39

Energy bands

The spectral theorem [1] shows that for each k, hF(k) has a diagonal matrix with respect

to some orthonormal basis of ℓ2(Ω) and thus have up to |Ω| distinct eigenvalues. These

energy eigenvalues defines the energy bands. We define the eigenvalue with the lowest

value to be ε1(k), the eigenvalue with the second lowest value to be ε2(k), and so on:

ε1(k) ≤ ε2(k) ≤ · · · ≤ ε |Ω|(k).

The set of energies defined by

E ∈ R : there exists a k ∈ Ω∗ such that ε i(k) = E

is denoted the ith energy band for i ∈ 1, 2, · · · , |Ω|. If the ith energy band and the

union of other energy bands is disjoint, then we say that the ith energy band is non-

degenerate. If this is not the case then we say that the ith energy band is degenerate.

A slight modification, more about operators periodic in Γ

Later we would like to find the k-space operator representing the discrete curret opera-

tors j1 and j2, these operators will be defined later. To this end, we modify the transform

operator U defined by formula (3.1) slightly, introducing the new operator UF:

UF : ℓ2(Λ)→HF, (UFψ) (k, x) := ∑γ∈Γ

e−ik·(x−γ)ψ(x− γ), (3.7)

for all ψ ∈ ℓ2c(Λ)). The connection betwen the vector component (Uψ) (k, x) and the vec-

tor component (UFψ) (k, x) is thus a factor e−ik·x, that is UF is the operator product of U

and the multiplicative operator e−k·(·). The inverse operator to e−ik·(·) : ℓ2(Ω) → ℓ2(Ω)

is eik·(·), and the fact that this operator is unitary combined with theorem 3.1 implies that

UF has an unitary extension as well.

Theorem 3.2. The operator UF defined by formula (3.7) has an unitary extension.

Assuming a Γ-translation invariant Hamilton operator H. Doing calculations much

alike the ones leading up to expression (3.5), it can be shown that for the transformed

Hamiltonian having the definition

UFH0U∗F : HF →HF,

the fibers has the matrix element

h0(x, y; k) = ∑γ∈Γ

h(x, y + γ)e−ik·(x−γ−y),

which has the alternate form

h0(x, y; k) = ∑γ∈Γ

h0(x + γ, y)e−ik·(x+γ−y).



We denote the fiber matrices, in the model of this presentation being 2× 2 matrices, by

h0(k) =

[

h0(x1, x1; k) h0(x1, x2; k)

h0(x2, x1; k) h0(x2, x2; k),

]

where x1 = (0, 0) and x2 = (a1/2, 0).

If we now differentiate the fiber with respect to the first component of k, k1, we have

∂

∂k1h(x, y; k) = − ∑

γ∈Γ

e−ik·(x+γ−y)

i(x1 + γ1 − y1)h(x + γ, y). (3.8)

The right side of equation (3.8) is the fiber of the transform of the operator

−i[X1, H] = i[H, X1],

that is the commutator of the Hamilton operator (first component) and the position op-

erator - which is the current operator (x-direction). This shows that a way to work with

current operators, is basically to shift to k-space and differentiate a (finite) matrix. We

will use this technique later.

decomposition of the product of two Γ-periodic operators

Suppose we have two operators on ℓ2(Λ), A and B, both Γ-periodic, and exponentially

almost diagonal. How does the product AB behave in the k-space representation?

We now show that there is no difference between taking the operator product in

L(ℓ2(Λ)) or multiplying the finite fiber-matrices representing A and B in L(ℓ2(Ω)). This

is a huge advantage when it comes to calculating products of operators! now we can

work with finite matrices!

Theorem 3.3. For Γ-periodic, exponentially almost diagonal operators A and B, it holds that

(AB)(x, y; k) = ∑x′∈Ω

a(x, x′; k)b(x′, y; k)

.

Proof. This is proven by straight calculation:

(AB)(x, y; k) = ∑γ∈Γ

e−k·(x+γ−y)

(AB)(x + γ, y)

= ∑γ∈Γ

e−k·(x+γ−y)

∑x∈Λ

a(x + γ, x)b(x, y).

If we in the above sum decompose the site points x into Bravais lattice vectors and unit

cell vectors, x = γ + x′, then we have, rearranging the order of summation a little:

(AB)(x, y; k) = ∑x′∈Ω

∑γ,β∈Γ

e−1k·(x+γ−y+β−β+x′−x′)

a(x + γ, β + x′)b(β + x′, y)

= ∑x′∈Ω

∑γ,β∈Γ

e−ik·(x+γ−β−x′)e−ik·(β+x′−y)

a(x + γ− β, x′)b(β + x′, y),


3.2. Calculating σ21(0) in reciprocal space 41

where we have used Γ-periodicity of A to rewrite a(x + γ, β + x′) = a(x + γ − β, x′).

Now in the last sum, two of the factors, e−ik·(β+x′−y) and b(β + x′, y) are independent

of the sumation variable γ. Furthermore in the other two factors eik·(x+γ−β−x′) and

a(x + γ− β, x′), γ only occur in terms of γ− β, and since γ run through all of Γ and, we

can decouple the summing over γ′ = γ− β and β:

(AB)(x, y; k) = ∑x′∈Ω

∑γ′

e−ik·(x+γ′−x′)a(x + γ′, x′)

∑β

e−ik·(β+x′−y)

b(β + x′, y)

.

This is what we need to show the theorem. This theorem is reminiscent of how a con-

volution of two functions transform in to a product in k-space under the Fourier trans-

form.

3.2 CALCULATING σ21(0) IN RECIPROCAL SPACE

the term independent of b

By setting b = 0, all the exponential factors in Db,±(z) defined by formulae (2.34) vanish

and by inserting D0,±(z) in to (2.38), we have the part of σ21 independent of b:

σ21(0) =1

4πω

∮

C

dz fFD(z) ∑x∈Ω

[

(H0 − z + ω)−1j1,0(H0 − z)−1j2,0 + · · · (3.9)

+ (H0 − z)−1j1,0(H0 − z−ω)−1j2,0

]

(x, x). (3.10)

One term in the trace sum consist of to products of four operators, each factor being

periodic in the Bravais lattice Γ. The trace is therefore over operators periodic in Γ. This

allows us to switch to k-space and take the trace over products of 2x2 matrices. We use

the modified Floquet transform. The resolvents like (Hb − z)−1 are transformed into the

inverse of regular 2× 2 matrices:[

(Hb − z)−1]

(k) = (h0(k)− z12)−1

The current operators we find simply by differentiating with respect to the proper com-

ponent of k = (k1, k2):

jν(x, y; k) =∂

∂kν

h(x, y; k),

if we remember to use the modified Floquet transform throughout the calculations!

k-space representation of the crystal Hamiltonian

If we for a little while permit ourselves to working with column vectors, we can write

a matrix element of the crystal Hamilton operator fiber in the following form, using the

modified Floquet transform:

h0(x, y; k) = ∑γ∈Γ

exp

(

−i

[

k1

k2

]

·[

x1 + γ1 − y1

x2 + γ2 − y2

])

h0

([

x1 + γ1

x2 + γ2

]

,

[

y1

y2

])

,

Draft: August 11, 2011, proofTwo/sigmaTwoOneConstant.tex


notice the difference between the dot-product in the exponential term, and the comma in

the variable part of h0(x + γ, y)...

If we fix the hopping integral to having value 1 for nearest neighbours, h0(x, y) has

the form:

h0(x, y) = δx1,y1±a1 δx2,y2+ δx2,y2±a2 δx1,y1

+ v(x1)δx,y, (3.11)

where the potential has the form

v(x1 = γ1 + x1) =

v > 0 , x1 = 0,

0 , x1 =a1

2.

For fixed x and y, both belonging to the origo unit cell Ω, evaluating the expression (3.11)

involves nine non-zero terms of h0(x, y), x ∈ Λ, y ∈ Ω :

h0

([

0

0

]

,

[

0

0

])

= v, h0

([

0

0

]

,

[

± a12

0

])

= 1,

h0

([

± a12

0

]

,

[

0

0

])

= 1, h0

([

0

±a2

]

,

[

0

0

])

= 1,

h0

([a12

±a2

]

,

[

± a12

0

])

= 1

Using the values, we calculate, for instance

h0

([

0

0

]

,

[

0

0

]

;

[

k1

k2

])

= ∑γ∈Γ

e−i(k1γ1+k2γ2)h0

([

γ1

γ2

]

,

[

0

0

])

= h0

([

0

0

]

,

[

0

0

])

+ e−ik2a2 h0

([

0

a2

]

,

[

0

0

])

+ eik2a2 h0

([

0

−a2

]

,

[

0

0

])

= v + 2 cos(k2a2).

Making equivalent calculations for the other three matrix elements of h0(k), we have

h0(k) =

2 cos(k2a2) + v , 2 cos

(k1a1

2

)

2 cos(

k1a12

)

, 2 cos(k2a2)

.

We see that all matrix elements are even, periodic functions of kν, ν ∈ 1, 2. The band

structure is found easily by finding the two eigenvalues of h0(k):

λ±(k1, k2) = 2 cos(k2a2) +v

2± 1

2

√

v2 + 16 cos2

(k1a1

2

)

,

which we will not use much, but they convey some information about the model. Letting

k vary over the first Brilluoin zone Ω∗ we can find the energy bands.

We are now in a position to find the k-space fibers jν,0(k) and s0(z, k).

j1,0(k) =∂

∂k1h0(k)

=

0 ,−a1 sin

(k1a1

2

)

−a1 sin(

k1a12

)

, 0

,

Draft: August 11, 2011, proofTwo/sigmaTwoOneConstant.tex

Constant term of σ21 is zero 43

j2,0(k) =∂

∂k2h0(k)

=

[

−2a2 sin(k2a2) , 0

0 ,−2a2 sin(k2a2)

]

,

s0(z0, k) = (h0(k)− z0)−1, z0 ∈ ρ(H0)

=1

Det(z0, v)

2 cos(k2a2)− z0 ,−2 cos

(k1a1

2

)

−2 cos(

k1a12

)

, 2 cos(k2a2) + v− z0

,

Det(z0, v) = (2 cos (k2a2)− z0) (2 cos (k2a2) + v− z0)− 4 cos2(

k1a1

2

)

.

Notice that the matrix elements of j1,0 only depend on k2 and is uneven in this variable,

likevise, the matrix elements of j2,0 only depend on k1 and is uneven in this variable.

Constant term of σ21 is zero

We now return to the b-independent off-diagonal term of the conduction tensor, and

consider the fiber of formula (3.10):

σ21(0) =1

4πω

∮

C

dz fFD(z) ∑x,x′,x′′,x′′′∈Ω

1|Ω∗|

∫

Ω∗d2k · · ·

(

s0(x, x′; z−ω, k)∂

∂k1h0(x′, x′′; k)s0(x′′, x′′′; z, k)

∂

∂k2h0(x′′′, x; k) + · · ·

+s0(x, x′; z, k)∂

∂k1h0(x′, x′′; k)s0(x′′, x′′′; z + ω, k)

∂

∂k2h0(x′′′, x; k)

)

.

(3.12)

It turns out that σ21(0), given by formula (3.12) is zero. Consider the term:

∫

Ω∗d2ks0(x, x′; z−ω, k)

∂

∂k1h0(x′, x′′; k)s0(x′′, x′′′; z, k)

∂

∂k2h0(x′′′, x; k).

When integrating k1 out, we integrate the product of three even and one uneven func-

tions on the interval (first Brillouin zone), which amounts to integrate an uneven function

over k1 ∈ [− πa1

, πa1]. This must return zero. The other term in formula (3.12) is zero by the

same line of reasoning.

That σ21(b = 0) is zero makes perfect sense, physically.

3.3 THE FIRST DERIVATIVE OF σ21(b)

the term linear in b

Now we seek to isolate the terms in

σ21(b) =1

4πω

∮

C

dz fFD(z) ∑x∈Ω

[Db,− + Db,+] (x, x),

Draft: August 11, 2011, proofTwo/sigmaTwoOneFirstDerivative.tex


containing b to the first power. Using the definitions of Db,±, we have

σ21(b) =1

4πω

∮

C

dz fFD(z) ∑x∈Ω

[

(Hb − z + ω)−1j1,b(Hb − z)−1j2,b + · · ·

+ (Hb − z)−1j1,b(Hb − z−ω)−1j2,b

]

(x, x),

(3.13)

We have two types of factors in Db,±; resolvents like (Hb − z)−1, and current opera-

tors jν,b, ν = 1, 2. We discard the O(b2) -remainder term of equation (2.17) and subtitute

(Hb − z0)−1 ≈ Sb(z0)− Sb(z0)Kb(z0). (3.14)

into formula (3.13), isolating an expression containing the linear terms in b:

σ21,lin(b) =1

4πω

∮

C

dz fFD(z)× · · ·

× ∑x∈Ω

[(Sb(z−ω)− Sb(z−ω)Kb(z−ω)) j1,b (Sb(z)− Sb(z)Kb(z)) j2,b + · · ·

+ ((Sb(z)− Sb(z)Kb(z)) j1,b (Sb(z + ω)− Sb(z + ω)Kb(z + ω)) j2,b] (x, x).

(3.15)

In order to keep track of the big picture when multiplying these operator products out,

we introduce some shorthand notation:

S := Sb(z),

S− := Sb(z−ω),

S+ := Sb(z + ω),

SK := Sb(z)Kb(z),

SK− := Sb(z−ω)Kb(z−ω),

SK+ := Sb(z + ω)Kb(z + ω).

For x ∈ Ω, we need focus on the matrix element, with the new shorthand, suppresing

the integrals and sums a little while:

[ (S− − SK−) j1b (S − SK) j2b + (S − SK) j1b (S+ − SK+) j2b ] (x, x)

= [ S−S − S−SK − SK−S + SK−SK+ SS+ − SSK+ − SKS+ + SKSK+ ] (x, x),

(3.16)

where we have suppressed the sandwiched j1b and the ending j2b factor, common to all

terms.

Consider the factor SK = Sb(z)Kb(z):

[Sb(z)Kb(z)] (x, y) = ∑xI ,xI I∈Λ

eibϕ(x,xI)so(x, xI ; z)eibϕ(xI ,y)(

eibfl(xI ,xI I ,y) − 1)

× · · ·

× h0(xI , xI I)so(xI I , y; z).


The S−S term 45

The fact that the summand contains the factor(


is crucial here, as the

b-independent part of this, using the power series expansion of the exponential function,

ex = ∑∞xn=0 xn/n! = 1 + x +O(x2) is zero, and next term, linear in b is

(


≈ ibfl(xI , xI I , y).

Thereby we see that all terms in (3.16) containing at least one of the factors SK, SK−or SK+ must vanish in the case b = 0. Furthermore, when can discard the two terms

SK−SK and SKSK+, when isolating the terms linear in b, since these contain no term

of lower order than b2. The matrix element [Sb(z)Kb(z)] (x, y) is in the first order ap-

proximation thus given by

[Sb(z)Kb(z)] (x, y) = ∑xI ,xI I∈Λ

eibϕ(x,xI)so(x, xI ; z)eibϕ(xI ,y)ibfl(xI , xI I , y)h0(xI , xI I)so(xI I , y; z),

(3.17)

and we are down to considering 6 terms in the sum (3.16).

The S−S term

The matrix element of considaration here is

[S−S ] (x, x) = [Sb(z−ω)j1,bSb(z)j2,b] (x, x), x ∈ Ω.

Isolating the b-dependence into exponential factors this is equal to

∑xI ,xI I ,xI I I

eibϕ(x,xI)s0(x, xI ; z−ω)eibϕ(xI ,xI I) j1,0(xI , xI I)eibϕ(xI I ,xI I I)s0(xI I , xI I I ; z)eibϕ(xI I I ,x) j2,0(xI I I , x)

We collect all the exponential factors

eibϕ(x,x′)eibϕ(x′,x′′)eibϕ(x′′,x′′′)eibϕ(x′′′ ,x) = eib(FL),

with the definition

(FL) = ϕ(x, x′) + ϕ(x′, x′′) + ϕ(x′′, x′′′) + ϕ(x′′′, x).

The power series expansion of eib(FL) in powers of b is

eib(FL) = 1 + ib(FL) +O(b2). (3.18)

It can be shown that

(FL) = fl(x, x′, x′′) + fl(x, x′′, x′′′)

=12

[(x′1 − x′′1 )(x2 − x′2)− (x′2 − x′′2 )(x1 − x′1)

]+

12

[(x′′1 − x′′′1 )(x2 − x′′2 )− (x′′2 − x′′′2 )(x1 − x′′1 )

].



Inserting this into formula (3.18) discarding all terms not linear in b we will make the

substitution

eib(FL) → ib

2

[(x′1 − x′′1 )(x2 − x′2)− (x′2 − x′′2 )(x1 − x′1)

]+ · · ·

+ib

2

[(x′′1 − x′′′1 )(x2 − x′′2 )− (x′′2 − x′′′2 )(x1 − x′′1 )

]

=ib

2

[(x′1 − x′′1 )(x2 − x′2)− (x′2 − x′′2 )(x1 − x′1)

]+ · · ·

+ib

2

[(x′′1 − x′′′1 )(x2 − x′2) + (x′′1 − x′′′1 )(x′2 − x′′2 )− (x′′2 − x′′′2 )(x1 − x′1)− (x′′2 − x′′′2 )(x′1 − x′′1 )

]

into the formula for [S−S ] (x, x). (In the last sum, we make sure that only term of the

type (xrν − xr+1

ν ) occur in the expression). We have

[S−S ] (x, x) =ib

2 ∑xI ,xI I ,xI I I∈Λ

[(x1 − x′1)s0(x, x′; z−ω) (x′2 − x′′2 )j10(x′, x′′) s0(x′′, x′′′; z)j0(x′′′, x) + · · ·

− (x2 − x′2)s0(x, x′; z−ω) (x′1 − x′′1 )j10(x′, x′′) s0(x′′, x′′′; z)j0(x′′′, x) + · · ·+ (x2 − x′2)s0(x, x; z−ω) j10(x′, x′′) (x′′1 − x′′′1 )s0(x′′, x′′′; z) j20(vt′′′, x) + · · ·− (x1 − x′1)s0(x, x; z−ω) j10(x′, x′′) (x′′2 − x′′′2 )s0(x′′, x′′′; z) j20(vt′′′, x) + · · ·+ s0(x, x′; z−ω) (x′1 − x′′1 )j10(x′, x′′) (x′′2 − x′′′2 ) s0(x′′, x′′′; z) j20(x′′′, x) + · · ·−s0(x, x′; z−ω) (x′2 − x′′2 )j10(x′, x′′) (x′′1 − x′′′1 ) s0(x′′, x′′′; z) j20(x′′′, x)

].

Switching to k-space, we have that multiplying an operator-kernel a(xr, xr+1) with (xrν −

xr+1ν ) transfers into differentiating the fiber with respect to kν, ν ∈ 1, 2.:

i(xrν − xr+1

ν )a(xr, xr+1) → 1|Ω∗|

∫

Ω∗d2k

∂

∂kν

a(xr, xr+1; k).

In k-space, we also again make the switch

∑···x,x′···∈Λ

(· · · )jν,0(x, x′)(· · · )→ 1|Ω∗|

∫

Ω∗d2k (· · · ) ∂

∂kν

h0(x, x′; k)(· · · )


All terms in k-space 47

Using this rule, we have

∑x∈Ω

[S−S ] (x, x) =b

2i|Ω∗| ∑x,xI ,xI I ,xI I I∈Ω

∫

Ω∗d2k · · ·

×[

∂

∂k2s0(x, x′; z−ω, k)

∂

∂k1

∂

∂k1h0(x′, x′′; k)s0(x′′, x′′′; z, k)

∂

∂k2h0(x′′′, x; k) + · · ·

− ∂

∂k1s0(x, x′; z−ω, k)

∂

∂k2

∂

∂k1h0(x′, x′′; k)

︸︷︷︸

=0

s0(x′′, x′′′; z, k)∂

∂k2h0(x′′′, x; k) + · · ·

+∂

∂k2s0(x, x′; z−ω, k)

∂

∂k1h0(x′, x′′; k)

∂

∂k1s0(x′′, x′′′; z, k)

∂

∂k2h0(x′′′, x; k) + · · ·

− ∂

∂k1s0(x, x′; z−ω, k)

∂

∂k1h0(x′, x′′; k)

∂

∂k2s0(x′′, x′′′; z, k)

∂

∂k2h0(x′′′, x; k) + · · ·

+ s0(x, x′; z−ω, k)∂

∂k2

∂

∂k1h0(x′, x′′; k)

︸︷︷︸

=0

∂

∂k1s0(x′′, x′′′; z, k)

∂

∂k2h0(x′′′, x; k)

−s0(x, x′; z−ω, k)∂

∂k1

∂

∂k1h0(x′, x′′; k)

∂

∂k2s0(x′′, x′′′; z, k)

∂

∂k2h0(x′′′, x; k)

]

,

where two terms vanish as a consequence of our choice of model, the mixed second

derivatives of the Hamilton operator fiber h0(k) is the zero matrix. In this way we can

calculate explicitly ∑x∈Ω [S−S ] (x, x) when z, z±ω lies in the resolvent set of H0.

All terms in k-space

Using the method above, we can calculate the traces of S−SK, SK−S , SS+, SSK+ and

SKS+, for a given z, z± ω in the resolvent set of H0, explicitly, by simply inverting and

differentiating known 2× 2-matrices.

Written as matrix product in k-space, suppressing the k-dependence all the terms are

(also stating the S−S term in compressed notation):

trΩ S−S :b

2i|Ω∗|∫

Ω∗d2k · · ·

trΩ

∂2s0(z−ω) ∂1∂1h0 s0(z) ∂2h0 − s0(z−ω) ∂1∂1h0 ∂1s0(z) ∂2h0

+∂2s0(z−ω) ∂1h0 ∂1s0(z) ∂2h0 − ∂1s0(z−ω) ∂1h0 ∂2s0(z) ∂2h0

.

trΩ S−SK :b

2i|Ω∗|∫

Ω∗d2k · · ·

trΩ

s0(z−ω) ∂1h0 s0(z)

[∂2h0 ∂1s0(z) − ∂1h0 ∂2s0(z)

]∂2h0

trΩ SK−S :b

2i|Ω∗|∫

Ω∗d2k · · ·

trΩ

s0(z−ω)

[∂2h0 ∂1s0(z−ω) − ∂1h0 ∂2s0(z−ω)

]∂1h0 s0(z) ∂2h0



trΩ SS+ :b

2i|Ω∗|∫

Ω∗d2k · · ·

trΩ

∂2s0(z) ∂1∂1h0 s0(z + ω) ∂2h0 − s0(z) ∂1∂1h0 ∂1s0(z + ω) ∂2h0

+∂2s0(z) ∂1h0 ∂1s0(z + ω) ∂2h0 − ∂1s0(z) ∂1h0 ∂2s0(z + ω) ∂2h0

.

trΩ SSK+ :b

2i|Ω∗|∫

Ω∗d2k · · ·

trΩ

s0(z) ∂1h0 s0(z + ω)

[∂2h0 ∂1s0(z + ω) − ∂1h0 ∂2s0(z + ω)

]∂2h0

trΩ SKS+ :b

2i|Ω∗|∫

Ω∗d2k · · ·

trΩ

s0(z)

[∂2h0 ∂1s0(z) − ∂1h0 ∂2s0(z)

]∂1h0 s0(z + ω) ∂2h0

,

where we can see some of the “cross product component” f1g2− g f2g1 structure is inher-

ited from the definition of ϕ(x, x′) = (1/2)(x1x′2 − x2x′1). Some terms, where the mixed

second derivative of h0 occur, has been omitted from the S−S and SS+ terms, since these

vanish in our nn. tb. model.

Collecting the terms

Collecting the terms, and inserting into formula (3.15), we have:

σ′21(0) =1

4πω

∮

C

dz fFD(z)1

2i|Ω∗|∫

Ω∗d2k · · ·

trΩ

∂2s0(z−ω) ∂1∂1h0 s0(z) ∂2h0 − s0(z−ω) ∂1∂1h0 ∂1s0(z) ∂2h0

+ ∂2s0(z−ω) ∂1h0 ∂1s0(z) ∂2h0 − ∂1s0(z−ω) ∂1h0 ∂2s0(z) ∂2h0

− s0(z−ω) ∂1h0 s0(z)[∂2h0 ∂1s0(z) − ∂1h0 ∂2s0(z)

]∂2h0

− s0(z−ω)[∂2h0 ∂1s0(z−ω) − ∂1h0 ∂2s0(z−ω)

]∂1h0 s0(z) ∂2h0

+ ∂2s0(z) ∂1∂1h0 s0(z + ω) ∂2h0 − s0(z) ∂1∂1h0 ∂1s0(z + ω) ∂2h0

+ ∂2s0(z) ∂1h0 ∂1s0(z + ω) ∂2h0 − ∂1s0(z) ∂1h0 ∂2s0(z + ω) ∂2h0

− s0(z) ∂1h0 s0(z + ω)[∂2h0 ∂1s0(z + ω) − ∂1h0 ∂2s0(z + ω)

]∂2h0

− s0(z−ω)[∂2h0 ∂1s0(z−ω) − ∂1h0 ∂2s0(z−ω)

]∂1h0 s0(z) ∂2h0

.

(3.19)

Which only contains known matrices and their derivatives.

Draft: August 11, 2011, proofTwo/proofTwo.tex

4

PERSPECTIVES

The method shown in this paper is not restricted to working on the simple model. The

model basically just simplifies the calculations to a minimum, while still not being totally

unphysical. In this last chapter we will outline some questions arising, which could be a

natural next step.

Actually calculating the first derivative of σ12(b) with a result from which the inte-

grals can be calculated (probably using residue integral techniques for the path integral)

should be a first step. Extending the model to a tight binding model encompassing real

graphene (“honeycomb” lattice nn. tb. model) and comparing theoretical predictions

of the Faraday rotation with experimental results the next. The simplest tight binding

model for graphene that has a perpendicular choice of basic lattice vectors a1 and a2,

would be using a unit cell with four sites, fig. 4.1.

a2

a1

unit cell

Figure 4.1: a choice of a perpendicular set of basic lattice vectors for graphene. This model has four sitesper unit cell, |Ω| = 4.

Higher powers

The mehod allows for computation of the second dervative of σ12(b) (relates to the “Cotton-

Mouton” effect) and so on. Laborous calculations should be expected, some mean of

simplifying the trace-expressions would be good.

Draft: August 11, 2011, perspectives/perspectives.tex 49

50 Perspectives

Second quantization

It would be interesting to imcoorperate the quantized fields into the model, ideally both

the light EM field aswell as the static magnetic field.

Self-consistence

Viewing the light EM field aswell as the static magnetic field as a single EM field, working

with only one vector potential, and one scalar potential. This single EM field should for

instance reproduce the experimental setting wherein Faraday rotation is measured.

Draft: August 11, 2011, perspectives/perspectives.tex

A

APPENDICES


52 Appendices

A.1 SELECTED RESULT CONCERNING BOUNDED OPERATORS ON A

HILBERT SPACE

Trace math

There are a few well-known tricks that we use so often, which we state here for com-

pletenes.

Theorem A.1 (Permutation rule for Trace). For operators A,B and C the trace is invariant

under cyclic permutations:

TrABC = TrCAB (A.1)

Theorem A.2 (Trace-Commutator Rule). For operators A,B and C, we have

Tr[A, B]C = TrB[C, A] (A.2)

On resolvents

Presented here are excerpts from Peter D. Lax [3], chapter 17.

Theorem A.3. Let L be a complete space of bounded linear operators on H (a Banach Algebra).

Suppose that K ∈ L(H ) is invertible; then so are all elements of L(H ) of the form L = K−A,

provided that

‖A‖ < 1‖K−1‖ . (A.3)

Proof. We treat first the special case K = 1; we claim that all elements of the form 1− B

are invertible, provided that

‖B‖ < 1.

The inverse of 1− B is given by the geometric series

∞

∑n=0

Bn = S. (A.4)

Since ‖B‖ < 1, the sequence of partial sums is a Cauchy sequence; since L(H ) is com-

plete, the series converges. It follows from the properties of the operator norm on infinite

series that a convergent series can be multiplied termwise: multiplying (A.4) on the left

by B, we get

BS = B∞

∑n=0

Bn =∞

∑k=1

Bk = S− 1,

from which it follows that (1− B)S = 1. Similarly, multiplying (A.4) on the right shows

that S(1− B) = 1. This shows that S is the inverse of 1− B.

We now return to (A.3); we factor

K−A = K(1−K−1A). (A.5)

Draft: August 11, 2011, appendices/boundedOpTheory.tex

On resolvents 53

Set B = K−1A; by submultiplicavity, and by inequality (A.3),

‖B‖ = ‖K−1A‖ = ‖K−1‖‖A‖ < 1.

Using (A.4), we invert (A.5):

(K−A)−1 =(

1−K−1A)−1

K−1 =∞

∑n=0

(

K−1A)n

K−1.

This proves that (K−A) is invertible.

Corollary A.4. Consider the case of the inverse to (1− B). The inverse exists if ‖B‖ < 1, and

we can evaluate the operator norm of (1− B)−1:

‖(1− B)−1‖ =∥∥∥

∞

∑n=0

Bn∥∥∥ ≤

∞

∑n=0‖B‖n =

11− ‖B‖ .

Where the last equality is the known sum of a geometric series.

Definition A.5 (Resolvent set, spectrum). The resolvent set of A in H consists of those

complex numbers λ for which λ1−A is invertible. The spectrum of A consists of those λ

for which λ1 −A is not invertible. The resolvent set of A is denoted ρ(A), its spectrum

by σ(A).

It is an important fact, that the resolvent set for an operator A is an open set.

Theorem A.6 (Resolvent set open). For a bounded operator A, ρ(A) is open.

Proof. Let z0 ∈ ρ(A). For an arbitrarily chosen z ∈ C, the identity

A− z = A− z0 − (z− z0) = (1− (z− z0)(A− z0)−1)(A− z0)

show that if (A− z)−1 exists, it must satisfy

(A− z)−1 = (A− z0)−1(1− (z− z0)(A− z0)

−1)−1.

Denote (z− z0)(A− z0)−1 by N(z), the proof of theorem A.3 shows that if ‖N(z)‖ < 1

then

(1− (z− z0)(A− z0)−1) = (1−N(z))

is invertible. Now set δ = 1/‖(A− z0)−1‖, then

‖z− z0‖ < δ implies ‖N(z)‖ = ‖z− z0‖‖(A− z0)−1‖ < 1, (A.6)

and therefore all z in the open ball ‖z− z0‖ < δ must belong to ρ(A).

Corollary A.7. Denote by dist(z, σ(A)) the distance between z and σ(A) ( dist(z, σ(A)) is the

infimum of ‖s− z‖ over all s ∈ σ(A) ), then the implication (A.6) implies that

dist(z, σ(A)) ≥ 1‖(A− z0)−1‖

Draft: August 11, 2011, appendices/boundedOpTheory.tex

BIBLIOGRAPHY

[1] Sheldon Axler. Linear Algebra done right. Springer publishing, 1997.

[2] John David Jackson. Classical Electrodynamics. John Wiley and Sons, 1962.

[3] Peter D. Lax. Functional Analysis. John Wiley and Sons, 2002.

[4] Jian-Ping Peng, Shi-Xun Zhou, and Xue-Chu Shen. Faraday rotation in quasi-two-

dimensional electron systems in the quantized hall regime. Phys. Rev. B, 44(8):4021–

4023, Aug 1991.

[5] Michael Reed and Barry Simon. Ananlysis of Operators, vol 4. Academic Press, INC,

1972.

[6] Michael Reed and Barry Simon. Functional Analysis, Volume 1. Academic Press, INC,

1972.

[7] R. Saito, G. Dresselhaus, and Dresselhaus M. S. Physical properties of carbon nanotubes.

World scientific publishing, 1998.


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