Quantum Physics in One Dimension: Basic Training …muellergroup.lassp.cornell.edu/oned.pdfThe Theory of Density Functional Theory: Electronic, Liquid, and Joint (Tomas Arias) Nonlinear

Quantum Physics in One Dimension:

Basic Training in Condensed Matter Physics

Erich Mueller

Lecture Notes: Spring 2018

Preface

About Basic Training

Basic Training in Condensed Matter physics is a modular team taught course

offered by the theorists in the Cornell Physics department. It is designed to

expose our graduate students to a broad range of topics. Each module runs

2-4 weeks, and require a range of preparations. This module, “Greens Function

Approach to Transport,” is designed for students who have completed a standard

one semester graduate solid state physics course. It assumes no prior exposure to

many-body Greens functions, but requires a knowledge of second quantization.

Prior Topics

2006 Random Matrix Theory (Piet Brouwer)

Quantized Hall Effect (Chris Henley)

Disordered Systems, Computational Complexity, and Information Theory

(James Sethna)

Asymptotic Methods (Veit Elser)

2007 Superfluidity in Bose and Fermi Systems (Erich Mueller)

Applications of Many-Body Theory (Tomas Arias)

Rigidity (James Sethna)

Asymptotic Analysis for Differential Equations (Veit Elser)

2008 Constrained Problems (Veit Elser)

Quantum Optics (Erich Mueller)

Quantum Antiferromagnets (Chris Henley)

Luttinger Liquids (Piet Brouwer)

2009 Continuum Theories of Crystal Defects (James Sethna)

Probes of Cold Atoms (Erich Mueller)

Competing Ferroic Orders: the Magnetoelectric Effect (Craig Fennie)

Quantum Criticality (Eun-Ah Kim)

i

2010 Equation of Motion Approach to Many-Body Physics (Erich Mueller)

Dynamics of Infectious Diseases (Chris Myers)

The Theory of Density Functional Theory: Electronic, Liquid, and Joint

(Tomas Arias)

Nonlinear Fits to Data: Sloppiness, Differential Geometry and Algorithms

(James Sethna)

2011 Practical Density Functional Theory (Tomas Arias)

Semiclassical Methods (Erich Mueller)

Ginzburg-Landau Theory of Superconductivity (Chris Henley)

Continuum Quantum Monte Carlo Methods in Chemistry and Physics

(Cyrus Umrigar)

2012 Quantum 14-15 puzzle: mechanics of a single hole in a spin-1/2 lattice

(Veit Elser)

The science of writing Science (Eun-Ah Kim)

Feynman diagrams and ab-initio computation of excited states in con-

densed matter (Tomas Arias)

Probes of Cuprate Superconductors (theorist’s view) (Chris Henley)

Physics of Life (Jane Wang)

2013 Many Body Field Theory (Erich Mueller)

The Theory of Density Functional Theory: Electronic, Liquid, and Joint

(Tomas Arias)

Conformal Symmetry (Andre Leclair)

Nonlinear Fits to Data: Sloppiness, Differential Geometry and Algorithms

(James Sethna)

2014 Rigidity (James Sethna)

Practical Density Functional Theory (Tomas Arias)

Quantum Monte-Carlo (Cyrus Umrigar)

Geometry in Quantum Mechanics (Erich Mueller)

2018 Greens Function approach to Transport (Erich Mueller)

Dynamical Systems that solve hard problems (Viet Elser)

Variational and Projector Monte Carlo Methods for Physics and Chem-

istry (Cyrus Umrigar)

Sloppy Models (Jim Sethna)

ii

Contents

Preface i

Contents iv

1 Introduction – Mar 13, 2019 1

A Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

B What is special about 1D . . . . . . . . . . . . . . . . . . . . . . 1

B.1 Fermi Liquid Theory . . . . . . . . . . . . . . . . . . . . . 1

C From “Coupled Masses” to “Classical Field Theory” . . . . . . . 4

D Quantum Coupled Masses . . . . . . . . . . . . . . . . . . . . . . 6

E HW 1 – Due March 20 . . . . . . . . . . . . . . . . . . . . . . . . 6

2 “Constructive” Bosonization – Mar 15, 2019 11

A Linearizing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

B Density Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

C Klein factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

D Fermion Operators . . . . . . . . . . . . . . . . . . . . . . . . . . 18

E Free Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

F Interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3 Applications – Mar 20, 2019 24

A Occupation Number . . . . . . . . . . . . . . . . . . . . . . . . . 24

A.1 Other Correlation Function . . . . . . . . . . . . . . . . . 26

A.2 Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

B HW 2 – Not Due . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4 Tonks Gas and the Leib-Lininger Model – Mar 27, 2019 33

.1 Permutations . . . . . . . . . . . . . . . . . . . . . . . . . 33

.2 1D Hard Core Gas . . . . . . . . . . . . . . . . . . . . . . 34

A Spin Chains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

iii

5 Bethe Ansatz – Mar 29, 2019 38

A Two-Body Problem . . . . . . . . . . . . . . . . . . . . . . . . . . 38

B Three Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

C Thermodynamic Limit . . . . . . . . . . . . . . . . . . . . . . . . 46

iv

Chapter 1

Introduction – Mar 13,

2019

A. Reading

To learn more about Fermi liquid theory, and how it breaks down in 1D, see

Giamarchi, “Quantum Physics in One Dimension”, chapter 1. To see some

cartoon reasoning about “quasiparticles,” see Mattuck,“A Guide to Feynman

Diagrams in the Many Body Problem.”

B. What is special about 1D

This section takes some of the highlights from Giamarchi’s “Quantum Physics

in One Dimension,” chapter 1. Note, my treatment will be quite a bit simpler

than Giamarchi’s: Though if you have the background, Giamarchi tells a very

nice story.

There are at least two arguments for the importance of studying 1D systems.

The first, which is emphasized in Giamarchi’s chapter, is that 1D is special. The

traditional approaches from 3D break down in 1D. Understanding the break-

down sheds light on the original problem. The unusual physics of 1D also acts as

a template for other exotic phenomena. A second important part of 1D systems

is that we can actually calculate things in 1D. The approaches are technically

beautiful.

B.1. Fermi Liquid Theory

To appreciate the special properties of 1D systems, it is useful to learn a bit

about 3D systems. Giamarchi assumes that the reader already knows about

1

2

them, and does a fairly technical review of their properties. I will just briefly

go over the highlights.

The theory of typical 3D electron systems is “Fermi Liquid Theory.” This

approach was first put forth phenomenologically, and tested experimentally.

Later its theoretical foundations were established. These theoretical foundations

are one of the crowning achievements of many-body theory.

The central premise of Fermi Liquid Theory is that the excitations of a 3D

electron gas are in 1-to-1 correspondence with those of a non-interacting gas,

but these quasiparticles

• have a different effective mass

• have a finite lifetime (the lifetime scales as τ ∝ 1/E2, so low energy

excitations are long-lived)

• are weakly interacting (even if the interactions between the bare electrons

are strong.

The mental picture is that for a Fermi gas, the excitations are:

Imagine starting with these states and adiabatically turning on the interactions.

The resulting state is a ”quasiparticle” or ”quasihole.” One imagines that the

electron (or hole) becomes surrounded by a screening cloud.

3

Multiple representations are always useful, and one concrete picture, is to

consider the operator ck which annihilates a quasiparticle with momentum k.

This operator can be expressed in terms of the original fermion operators ak as

ck =√zak +

∑pq

fkpqak+p−1a†paq + · · · (1.1)

That is, the quasiparticle operator includes a linear combination of all terms

which have total momentum k and total particle number 1. The square of

the coefficient of the single-particle term, z, is known as the “quasi-particle

renormalization factor,” and physically corresponds to the square overlap of the

single particle state and the quasiparticle state. The central premise of Fermi

liquid theory is that z is non-zero.

Another useful picture to have in mind is the “single particle spectral den-

sity.” This is the quantity which is measured in photo-emission, or tunneling.

Imagine you try to inject an electron with momentum k and energy ω. In an

ideal gas, you will only succeed if ω = k2/2m. In an interacting gas, you in-

stead expect that your spectrum (the probability of success) will be bimodal (or

multimodal). You can add the particle, leading to a final state which contains

a single quasiparticle: this will give a Lorentzian contribution to the spectrum.

The width of the Lorentzian is related to the lifetime of the quasiparticle. The

area of the peak is related to the quasiparticle renormalization factor. Alter-

natively you can inject the electron and create a state with some number of

particle-hole excitations. This will give a broad background. Pictorially, the

graph on the left represents the spectral density of a noninteracting Fermi gas,

while the one on the right is a Fermi liquid:

The other story you will commonly see in the textbooks is the distribution

function nk, which represents the number of fermions with momentum k. In

a zero temperature non-interacting gas, nk = 1 for k < kf , and nk = 0 for

k > kf . In particular, there is a discontinuity at k = kf . In a Fermi liquid, the

discontinuity remains, and nk+f− nk−f = z:

4

In 1D we will find that z = 0, but there will still be a singularity at kf :

the occupations will have a kink there: nk ∼ n0 − (k − kf )α for k > kf , and

nk ∼ n0 + (kf − k)α for k < kf :

C. From “Coupled Masses” to “Classical Field Theory”

One of the most dramatic results that we will derive is that the excitations of a

1D fermi gas are exhausted by “phonons.” To understand what this means, it

is useful to remind ourselves where phonons come from. We will begin with a

1D model of masses connected by springs:

If we want to be able to take the continuum limit, then the masses and the

spring constants are related to the equilibrium separation by:

λ =M

d(1.2)

κ = kd. (1.3)

Here λ is the linear mass density, and κ is the stiffness.

The dynamical variables of this set of masses are the displacements from

5

equilibrium, si, and the momenta pi. The equations of motion read:

M∂2sj∂2t

= k(sj+1 + sj−1 − 2sj). (1.4)

This is solved by making the ansatz sj = ei(qdj−ωt, where xj = dj is the equilib-

rium position of the j′th mass. Substituting this into our equations of motion

gives

ω =

√k

M2 sin(qd/2) (1.5)

→ q

√kd

M/d= q

√κ

λ. (1.6)

We teach that argument in PHYS 2218 at Cornell, and it should feel pretty

familiar. One interesting way to think about things is to introduce the continu-

ous function s(x), where sj = s(xj). In the continuum limit this is a well-defined

procedure, and if you want to be pedantic there are formal ways of defining s(x).

We can relate s(x) to the local mass density. The separation between two neigh-

boring masses is

∆j = d+ sj + 1− sj = d

(1 +

∂s

∂x

). (1.7)

That is if s is uniform, then the separation is d. If ∂s/∂x > 0 then the separation

is greater than d. If you know something about elasticity, you would recognize

∂s/∂x as the 1D version of the strain tensor. Regardless, the mass density is

ρ =M

∆=M

d

1

1 + ∂s∂x

(1.8)

≈ λ

(1− ∂s

∂x

). (1.9)

Thus, we can identify −∂s/∂x as the fractional change in the density. Remem-

ber this – we will see this structure again. Note that s is independent of the

discretization.

The classical Lagrangian associated with this system is

L =∑j

M

2s2j −

k

2(sj+1 − sj)2 (1.10)

=∑j

dM

2ds2j − d

kd

2

(sj+1 − sj

d

)2

(1.11)

=

∫dx

[λ

2

(∂s

∂t

)2

− κ

2

(∂s

∂x

)2]. (1.12)

6

The conjugate variable to s(x) is the momentum density

π(x) =δL

δs(x)= λs(x). (1.13)

It has units of momentum per unit length. That is, the momentum of the j’th

mass is pj = dπ(xj).

The classical Hamiltonian is

H =

∫dx

1

2λπ2 +

κ

2(∂xs)

2. (1.14)

D. Quantum Coupled Masses

The quantum version of this problem is essentially identical to the classical prob-

lem. It is convenient to think about the operators sj and pj , with commutation

relations

[si, pj ] = i~δij . (1.15)

We can use our classical definitions of the fields

sj = s(xj) (1.16)

pj = dπ(xj). (1.17)

It is straightforward to argue that

[s(x), π(y)] = i~δ(x− y). (1.18)

We note that the density operator is ρ(x) = λ(1− s′(x)).

E. HW 1 – Due March 20

Problem 1.1. Number Phase Uncertainty: This problem makes concrete

some of the algebra that we will encounter in the lectures.

In your linear algebra class you may have learned about the polar-decomposition:

It is the statement that any matrix can be written as the product of a Hermitian

and Unitary matrix. The name comes from noting that a unitary matrix can

be written as the exponential of a Hermitian matrix, so it looks like the polar

decomposition of a complex number. In other words, the decomposition is:

M = eiθR (1.19)

where

R† = R (1.20)

θ† = θ. (1.21)

7

1.1.1. Two-level system: To make this very concrete, thing about a system

with two states, labeled |0〉 and |1〉. Typically we order these as |1〉, |0〉, but for

our purposes it will be better to write them as |0〉, |1〉. In this basis, with this

order, the “lowering” operator is

a =

(0 1

0 0

). (1.22)

Show that

{a, a†} = 1, (1.23)

and hence a can be thought of as a Fermion lowering operator (where we inter-

pret the two states as “full” and “empty”).

1.1.2. Show that

a =

(0 1

1 0

)(0 0

0 1

), (1.24)

and that the first matrix is Unitary, and the second Hermitian. Thus we write

eiθ =

(0 1

1 0

)(1.25)

R =

(0 0

0 1

). (1.26)

1.1.3. Calculate the anticommutator {eiθ, R†R}.

1.1.4. Find the matrix θ. Hint: Mathematica has a function MatrixLog. Alter-

natively you can write θ as a linear combination of the identity matrix and the

Pauli matrices, and solve for the coefficients. Note, because log is multivalued,

the answer is not unique.

Problem 1.2. Number-phase uncertainty for a bosonic mode:

The bosonic problem is a bit more complicated because the Hilbert space is

infinite. One strategy is to work in a truncated space, where you limit the

occupation to be 0, 1, 2, · · ·n, so the matrices are all finite. We can then take

n→∞ at the end.

1.2.1. Polar decomposition: In this finite space, the annihilation operator is

a =

0√

1 0

0 0√

2...

. . .

0 0√n− 1

0 . . . 0

. (1.27)

8

Show that

a =

0 1 0 0 · · ·0 0 1 0 · · ·...

0 . . . 1

1 . . . 0

0 0 0 0 · · ·0 1 0 0 · · ·0 0

√2 0 · · ·

...

0 . . .√n

. (1.28)

and that the first matrix is Unitary, and the second Hermitian.

1.2.2. Phase Matrix: The matrix θ is quite ugly here – so I won’t ask you to

calculate it. However it is straightforward to calculate

[eiθ, R†R]. (1.29)

A way to make the arithmetic simpler is to write

eiθ =

n∑j=1

|j − 1〉〈j|+ |n〉〈0| (1.30)

R†R =∑

j = 0n|j〉j〈j|. (1.31)

You will find

[eiθ, R†R] = eiθ +X. (1.32)

What is the matrix X?

1.2.3. In the limit n→∞, the matrix X involves a state with infinitely many

particles – and we typically argue that it can be neglected. If we do so, then

[eiθ, R†R] ≈ eiθ. Under this approximation, infer

[θ,R†R]. (1.33)

Problem 1.3. Coherent States

Consider the following two harmonic oscillator Hamiltonians, where we have

taken dimensionless units (~,m, ω all are unity):

H1 =p2

2+

1

2x2 (1.34)

H2 =p2

2+

1

2(x− x0)2. (1.35)

Let |ψ1〉 and |ψ2〉 be the ground states. Clearly

|ψ2〉 = eipx0 |ψ1〉, (1.36)

9

where exp(ipx0) is a translation operator.

We introduce the ladder operators:

a =1√2

(x+ ip) (1.37)

a† =1√2

(x− ip) . (1.38)

Show that |ψ2〉 is an eigenstate of a. Find the eigenvalue.

Hint: What is a|ψ1〉? What is [a, eipx0 ]?

Problem 1.4. Quantum Fluctuations in a 1D harmonic Chain

Consider the 1D chain described by Hamiltonian:

H =∑j

p2j

2m+

1

2k(xj+1 − xj)2, (1.39)

where [xi, pj ] = i~δij , and where j runs from 1 to N . For concreteness we will

use periodic boundary conditions (so xn+1 is identified with x1.

We can diagonalize this Hamiltonian by the transformation:

xj =∑q

eiqjuq(aq + a†−q) (1.40)

pj =∑q

e−iqjvqaq − a†−q

i.

The sums run over q = 2πn/N , with n running from −N/2 to N/2. It turns out

that we can choose uq and vq real. Furthermore, since xj and pj are Hermitian,

we need u−q = uq and v−q = vq. We want the operators aq to obey

[aq, a†q] = δq,q (1.41)

1.4.1. Show that if Eq. (1.41) is satisfied, then

[aq ± a†−q, aq′ ± a†−q′ ] = 0 (1.42)

for any q, q′. Hence the requirements [xi, xj ] = 0 and [pi, pj ] = 0 is automatically

satisfied by this ansatz

1.4.2. Show that if Eq. (1.41) is satisfied, then

[aq + a†−q,aq′ − a†−q′

i] = 2iδq,q′ . (1.43)

for any q, q′.

10

1.4.3. Use that last result to show that [xi, pj ] = i~δij if and only if

uq =

√~

2Nδq (1.44)

vq =

√~

2N

1

δq,

where δq is aribtrary (with δ−q = δq).

1.4.4. Substitute the ansatz, Eq. (1.40) into the Hamiltonian, using Eq. (1.44)

to write the Hamiltonian as a function of the raising/lowering operators and δq.

The resulting expression should be of the form

H =∑q

(aqa−q + a†−qa†q)~2

[Aqδ

2q −

Bqδ2q

](1.45)

+(aqa†q + a†−qa−q)

~2

[Aqδ

2q −

Bqδ2q

].

Find Bq and Aq. We then set

δ4q =

BqAq

, (1.46)

so that

H =∑q

(a†qaq +

1

2

)~ωq. (1.47)

Find ωq.

1.4.5. The ground state of this system will have aq|GS〉 = 0 for all aq. Use this

to write

〈(xi − xj)2〉 =∑q

u2q 4 sin2

(q(i− j)

2

)(1.48)

→ N

2π

∫dq u2

q 4 sin2

(q(i− j)

2

), (1.49)

where we converted the sum to an integral by noting that the separation between

q’s is 2π/N .

If |i − j| � 1, the sin function in Eq. (1.48) is rapidly oscillating, and one

might naively expect that one could replace it with 1/2. Show that if you do, the

integral is divergent. (Hint, the problem happens as q → 0). [You don’t need

to do it, but If you work a bit harder, you can even argue that the divergence

should be 〈(xi − xj)2〉 ∼ log |i− j|.]Regardless, his means that 〈(xi−xj)2〉 → ∞ as |i− j| → ∞. Physically, this

means that the quantum fluctuations disorder the chain. If you did the same

calculation in 3D, you would instead find the integral is convergent.

Chapter 2

“Constructive”

Bosonization – Mar 15,

2019

In this lecture we derive a quite remarkable result: The excitations of an in-

teracting 1D Fermi system can be put in a 1-to-1 correspondence with those

of a harmonic chain. I note that it is an obvious statement that among the

excitations of the electrons are “plasmons” – corresponding to density waves in

the electron fluid. This is true in any dimension. The special thing about 1D is

that all excitations can be written in terms of these phonons.

We will also first neglect spin, then add it later. Our treatment is similar

to Giamarchi chapter 2 – but maybe even closer to Delft and Schoeller’s article

”Bosonization for Beginners...” which is on the arXiv.

A. Linearizing

The first step will be to linearize about the Fermi surface. It turns out some

of our results do not depend on this approximation, but I think it is best to do

the simplest treatment first. The idea is that we imagine a setting where the

interactions are weak. In that case the ground state of the interacting system

will look a lot like the ground state of the non-interacting system, with the

exception of structure near the Fermi points. Thus it suffices to model physics

11

CHAPTER 2. “CONSTRUCTIVE” BOSONIZATION – MAR 15, 201912

near the Fermi points. In other words we write the kinetic energy as

H = HR +HL +H ′ (2.1)

HR =∑

knear kf

vf (k − kf )c†kck (2.2)

HL =∑

knear−kf

vf (kf − k)c†kck (2.3)

H ′ =∑

other k

εkc†kck. (2.4)

We will come up with a model of HR and HL. We won’t worry about H ′, as it

will not be important to the physics we are after. Clearly if the interactions are

strong, this argument will break down and we will need to include H ′.

It makes sense to shift the origin of momenta, and write

ark = ck−kf (2.5)

a`k = ck+kf . (2.6)

when there is no possibility of confusion, we will leave off the r, ` indices. We

refer to these two “species” of particles as “right movers” and “left movers”.

I have been a little vague about what range of k we are summing over. In

principle it doesn’t matter – if all the action is happening at the Fermi surface,

I can include as many states as I want.

B. Density Operator

In this section we will see that the Fourier transform of the density acts as a

ladder operator. There are a couple tricky things here. The first is just keeping

track of L’s and 2π’s.

The best mnemonic that I have for the factors is to start with the electron

field operator – which is something like

ψ(x) =1√L

∑k

ckeikx, (2.7)

where k = 2πn/L. We can verify that I have the factors right by looking at the

density

ρ(x) = ψ†(x)ψ(x) =1

L

∑kq

ei(k−q)xc†qck. (2.8)

In particular, if we have a homogeneous gas 〈c†qck〉 is zero unless k = q and

k < kf , and we get

ρ(x) =N

L. (2.9)


The Fourier transform of the density operator is

ρp =∑q

a†q+paq. (2.10)

That is

ρ(x) =1

L

∑p

eipxρp. (2.11)

Perhaps I should define ρp with a factor of 1/√L in it. In the end it doesn’t

matter – as long as you are consistent. This convention makes the following

arithmetic a bit easier.

Clearly if p� kf , we can write

ρp = ρRp + ρLp + ρotherp , (2.12)

where

ρRp =∑q

c†r,q+pcrq, (2.13)

and a similar expression for ρL. We will analyze the properties of ρRp .

Now we come to our first profound observation. Suppose |ψ〉 is an eigenstate

of HR:

H|ψ〉 = E|ψ〉. (2.14)

(Note I have left off the R.) Applying ρRp to |ψ〉 will give a superposition of terms

where various right-moving particle have had their momentum increased by p.

But the energy of a given particle is proportional to p – so we have increased

the energy by vfp. In other words

Hρp|ψ〉 = (E + vfp)ρp|ψ〉. (2.15)

Putting these two expressions together, we have Hρp|ψ〉 = ρp(H+vfp)|ψ〉. This

statement is true for all eignenstates, and hence

[ρp, H] = −vfpρp. (2.16)

This is exactly the requirement for a ladder operator. Actually – as you see

ρp appears on both the left and right hand side of that expression, so this is

actually the requirement to be a constant times a ladder operator. Below, we

will show that the correct normalization is (for p > 0),

b†p =2π

L

1

pρp (2.17)

bp =2π

L

1

pρ−p, (2.18)


where these obey the standard relations [bp, b†q] = δp−q.

To establish this result, we need to understand what these operators do to

states. In particular we want to look at how ρpρq|ψ〉 differs from ρqρp|ψ〉. To

make the notation simpler, let us write momenta in units of 2π/~, which is the

spacing between momentum states.

Lets start with a Fermi sea |ψ〉 = | • • • • • : ◦ ◦ ◦ ◦ ◦〉. Applying ρ3 to this

gives three terms

ρ3| • • • • • : ◦ ◦ ◦ ◦ ◦〉 = | • • • • ◦ : ◦ ◦ •◦ ◦◦〉 − | • • • ◦ • : ◦ •◦ ◦ ◦ ◦〉+ | • • ◦ • • : •◦ ◦ ◦ ◦◦〉

Following this with ρ2 gives a total of 8 terms:

ρ2ρ3| • • • • • : ◦ ◦ ◦ ◦ ◦〉 = −| • • ◦ • •◦ : ◦ ◦ •◦ ◦◦〉+ | • • • ◦ ◦ : •◦ ◦ •◦ ◦◦〉+ | • • • • ◦ : ◦ ◦ ◦ ◦ •◦〉+| • ◦ • •◦ • : ◦ •◦ ◦ ◦ ◦〉 − | • • • ◦ • : ◦ ◦ ◦ •◦ ◦〉−| ◦ • •◦ • • : •◦ ◦ ◦ ◦◦〉 − | • • ◦ • ◦ : •◦ •◦ ◦ ◦ ◦〉+ | • • ◦ • • : ◦ ◦ •◦ ◦◦〉

For what it is worth, two of these cancel, leaving

ρ2ρ3| • • • • • : ◦ ◦ ◦ ◦ ◦〉 = +| • • • ◦ ◦ : •◦ ◦ •◦ ◦◦〉+ | • • • • ◦ : ◦ ◦ ◦ ◦ •◦〉+| • ◦ • •◦ • : ◦ •◦ ◦ ◦ ◦〉 − | • • • ◦ • : ◦ ◦ ◦ •◦ ◦〉−| ◦ • •◦ • • : •◦ ◦ ◦ ◦◦〉 − | • • ◦ • ◦ : •◦ •◦ ◦ ◦ ◦〉

Remarkably, if you do these two operators in the opposite order ,you get

exactly the same terms, with exactly the same signs. Even more remarkably,

the same thing happens on any initial state. I will prove this below, but it is

worth playing with this to convince yourself. You will find, however, that if

p = −q, then they don’t commute.

ρ−3ρ3| • • • • • : ◦ ◦ ◦ ◦ ◦〉 = ρ−3 [| • • • • ◦ : ◦ ◦ •◦ ◦◦〉 − | • • • ◦ • : ◦ •◦ ◦ ◦ ◦〉+ | • • ◦ • • : •◦ ◦ ◦ ◦◦〉]= | • • • • •◦ : ◦ ◦ ◦ ◦◦〉+ | • • • •◦ • : ◦ ◦ ◦ ◦ ◦〉+ | • • •◦ • • : ◦ ◦ ◦ ◦◦〉= 3| • • • • • : ◦ ◦ ◦ ◦ ◦〉

but

ρ3ρ−3| • • • • • : ◦ ◦ ◦ ◦ ◦〉 = 0.

Having established plausibility, we now prove the general result [ρp, ρq] =

qδp+q – or in real units [ρp, ρq] = q(L/2π)δp+q. We start with

[ρp, ρq] =∑k,k′

[a†k+pak, a†k′+qa

′k]. (2.19)


This commutator is zero unless k′+q = k or k+p = k′. We need to be somewhat

careful about the bounds on the sum, but for the moment lets not worry about

it,

[ρp, ρq] =∑k

a†k+pak−p −∑k′

a′k+qak′−p. (2.20)

We shift indices on the sum, and everything cancels – except the boundary

terms. In the case p 6= −q, we actually don’t need to worry about the boundary

terms, they are things like a†k+pak−p for k well inside the Fermi surface. Any

states we act on will have both those states occupied, and hence such terms do

nothing. When p = −q these expressions are a†k−qak−p, which is 1 if we are well

inside the Fermi surface, but −1 if we are well outside. A little thought should

convince you that [ρ−q, ρq] = q (or actually [ρ−q, ρq] = qL/2π). [When I have

a chance I’ll write it in more detail, but I already went through it in class, and

am running out of time in typing up the notes.]

We can therefore define ladder operators (for p¿0),

bp =

√2π

Lpρ−p (2.21)

b†p =

√2π

Lpρp. (2.22)

We further have

[bp, b†p] = δpp′ (2.23)

[bp, HR] = vfpbp

For left movers we would instead get (for p < 0)

bp,` =

√2π

L|p|ρ−p,` (2.24)

b†p,` =

√2π

L|p|ρp,`. (2.25)

We further have

[bp,`, b†p,`] = δpp′ (2.26)

[bp,`, HL] = vf |p|bp,`

Thus the Hamiltonian for the non-interacting system reads

H =∑p>0

vfp[b†pr bpr + b†p−`b−p`

]+ ∆ (2.27)


where ∆ commutes with all of the b operators.

Claim: The only operators which commute with all of the b’s and conserve

the number of left and right movers are NL and NR – the total number of left

and right movers – and functions of these.

In other words, given any state, say

|ψ〉 = | • • ◦ • ◦ ◦ • ◦ ◦ ◦〉. (2.28)

The claim is that |ψ〉 can be expressed as a linear superposition of b’s acting on

|ψ0〉 = | • • • • ◦ ◦ ◦ ◦ ◦ ◦〉. (2.29)

This claim is equivalent to the statement that the number of ways of making

orthogonal states of energy E by acting on the ground state with b’s is the

same as the number of ways of making such a state by adding an equal number

of particles and holes. Both constructions make orthogonal states – hence the

Hilbert spaces will have the same dimension – and must be identical.

It is fun to do the counting for the first few states. For example, it is clear

that there is a unique state with energy 1. There are two states with energy 2:

| • • • ◦ : ◦ • ◦◦〉 and | • • ◦ • : • ◦ ◦◦〉. These are linear combinations of b†1b†1|ψ0〉

and b†2|ψ0〉.In general, the fact that these numbers are always the same is equivalent to

the “pentagonal number theorem” – which is at the core of the Jacobi triple

product formula, which can also be considered a Jacobi Theta function identity

(relating it to Wierstrauss functions). I won’t go through the proof of the Jacobi

identiy, but will show how to reduce this physics problem to that result.

First, the number of ways of using the boson operators to generate a state

of energy m is given by the number of partitions of the integer m. That

is the number of ways of choosing integers which add up to n. For exam-

ple, 2=1+1 or 2=2, has two ways. We write p(2) = 2. On the other hand

4=3+1=2+2+2+1+1=1+1+1+1 has 5 ways. We will define the generating

function

f(x) =∑m

p(n)xm. (2.30)

Standard generating function tricks give

f(x) =∏k

1

1− xk. (2.31)

[I’ll go through these in class.] This is a famous function in number theory.

Lets denote the number of ways of distributing an equal number of particles

and holes to produce energy m as s(m). We define

g(x) =∑m

s(m)xm. (2.32)


The claim is that f(x) = g(x). Calculating g(x) is hard, so we introduce s(m,n)

as the number of ways of distributing particles and holes to produce energy m,

where the number of particles minus the number of holes is n. We define

G(x, y) =∑nm

s(m,n)xmyn. (2.33)

Clearly g(x) is the y0 term in the series. Formally we can write G as the product

of generating functions for each mode

G(x, y) =[(1 + xy)(1 + x2y)(1 + x3y) · · ·

]×[(1 + y−1)(1 + xy−1)(1 + x2y−1) · · ·

].

(2.34)

We then use the “Jacobi triple product” identity, which says

G(x, y) = f(x)∑n

xn(n−1)/2yn. (2.35)

Taking the y = 0 term of both side establishes the proof.

C. Klein factors

There is one other set of operators we will need, namely the ”Klein factors”

F †r and F †` , which are unitary operators which just shift the number of left and

right movers by 1. They are defined as the unique operators which commute

with the b’s, and shift the Fermi sea:

F †r | • • • • : ◦ ◦ ◦◦〉R = (−1)NL | • • • • : • ◦ ◦◦〉R (2.36)

Fr| • • • • : ◦ ◦ ◦◦〉R = (−1)NL | • • • ◦ : ◦ ◦ ◦◦〉R (2.37)

F †` | • • • • : ◦ ◦ ◦◦〉L = | • • • • : • ◦ ◦◦〉L (2.38)

F`| • • • • : ◦ ◦ ◦◦〉L = | • • • ◦ : ◦ ◦ ◦◦〉L (2.39)

On other states, it similarly shifts everything by one:

F †r | • • ◦ • : ◦ • ◦◦〉R = (−1)NL | • • • ◦ : • ◦ •◦〉R. (2.40)

It should be clear that they do indeed commute with the b’s, and

F †rFr = FrF†r = F †` F` = F`F

†` = 1 (2.41)

{Fr, F`} = 0. (2.42)

The factor of (−1)NL was chosen so that this last anticommutation relationship

would hold.


D. Fermion Operators

We can now write the fermion operators in terms of the phonons. In particular

ψ(x) =1√L

∑k

ckeikx (2.43)

=eikfx√L

∑q

arqeiqx +

e−ikfx√L

∑q

a`qeiqx + stuff away from Fermi surface.

In the obvious notation, we call the first two terms ψr and ψ`.

We will now construct ψr in terms of the b’s, by looking at the commutation

relations. Recall

ρr(p) =∑k

c†r k+pcr k (2.44)

Straightforward arithmetic gives

[ρ(p), cq] = −cq−p, (2.45)

and hence

[ρr(p), ψr(x)] = − 1√L

∑q

cr q−peiqx (2.46)

= −eipxψr(x). (2.47)

We now use that

[ρr(p), ρr(−p)] = −p L2π, (2.48)

to conclude that

ψr(x) = (· · · ) exp

(2π

Lpρr(−p)eipx

), (2.49)

where the term in parentheses commutes with ρr(p). The only expression con-

sistent with this (for all p), and which does the right thing to the number of

particles is

ψr(x) = AFr exp∑p

2π

Lpeipxρr(−p) (2.50)

= AFr exp∑p>0

√2π

Lp

[eipxbp − e−ipxb†p

], (2.51)

where A is a constant.

It is traditional to define

ϕr(x) = i2π

L

∑p>0

eipx

pρr(−p) (2.52)

= i

√2π

L

∑p>0

eipx√pbp, (2.53)


and

φr(x) = ϕr(x) + ϕ†r(x) (2.54)

= i2π

L

∑p

eipx

pρr(−p). (2.55)

Then

ψr(x) = AFre−iφr(x) (2.56)

In addition to being a cute way of writing things, we can note an interesting

property of the field φr(x), in particular

∂xφr(x) = −2π

L

∑p

eipxρr(−p) (2.57)

= −2πρr(x). (2.58)

Thus φr has exactly the same meaning as the displacement field that we intro-

duced for the mass chain.

Now Giamarchi makes life more complicated by introducing two other fields

via:

φr(x) = φ(x)− θ(x) (2.59)

φ`(x) = −φ(x)− θ(x). (2.60)

The reason for doing this is that the physical density is the sum of the right-

moving and left-moving densities (plus stuff from other modes). Thus ∂xφ has

the meaning of the total density. The other field θ involves the mis-match of the

left-moving and right-moving densities. It is clearly related to the momentum

density, and will find

ρ(x) = − 1

π∂xφ(x) (2.61)

π(x) =1

π∂xθ(x). (2.62)

Indeed, it is a straightforward exercise, left as a homework exercise, that these

fields obey the right commutation relations.

E. Free Hamiltonian

Lets now consider how the Hamiltonian looks in terms of these operators. First,

as already argued, up to a constant, the free particle Hamiltonian is

H0 =∑p>0

vfp[b†pr bpr + b†−p`b−p`

]+ vf

2π

L(N2

R + N2L)− µ(NR + NL), (2.63)


where we have added a chemical potential. This, and the field operators, are

more conveniently described by

bp>0 = brp (2.64)

bp<0 = b`p. (2.65)

The first term in Eq. (2.63) is then

Hb0 =

∑p

vfpb†pbp, (2.66)

where the superscript b reminds us that we have left off the N terms. Similarly

this makes the expression for θ and φ simpler,

φ(r) = −i2πL

∑p

(L|p|2π

)1/2e−ipx

p(b†p + b−p) (2.67)

θ(r) = −i2πL

∑p

(L|p|2π

)1/2e−ipx

|p|(b†p − b−p). (2.68)

For those who did Problem 1.4, you should recognize this as exactly the expres-

sions for

ρ(x) = − 1

π∂xφ(x) (2.69)

Π(x) =1

π∂xθ(x). (2.70)

In particular, straightforward arithmetic gives

Hb0 =

vf2π

∫dx[(∂xθ)

2 + (∂xφ)2]. (2.71)

F. Interactions

We now consider the role of interactions. A typical model for interactions is

Hint =

∫dx

∫dx′ V (x− x′)ψ†(x)ψ†(x′)ψ(x′)ψ(x). (2.72)

For electrons V could be a Coulomb potential, while for cold atoms it is some

sort of short range interaction. Note, that because of the anti-commutation

relations, one could also write this as

Hint =

∫dx

∫dx′ V (x− x′)ψ†(x)ψ(x)ψ†(x′)ψ(x′)− V (0)

∫dxψ†(x)ψ(x).

(2.73)


The latter term just shifts the chemical potential – possibly by an infinite

amount, depending on how V behaves at short distances. Thus some treat-

ments use

Hint =

∫dx

∫dx′ V (x− x′)ψ†(x)ψ(x)ψ†(x′)ψ(x′). (2.74)

You should keep in mind that the strictly correct expression is Eq. (2.72) or

(2.73), but as long as you don’t worry about a few infinities, it is fine to work

with Eq. (2.74).

We write Eq. (2.72) in momentum space as

Hint =1

L3

∑kpq

Vq ψ†k+qψ

†p−qψpψk. (2.75)

We think about this as a scattering process – where you have incoming particles

with momentum k and p, which exchange momentum q. We only care about

the cases where all of the momenta are near the Fermi surface. There are six

possibilities:

1. k, p, k + q, p− q are all near kf

2. k, p, k + q, p− q are all near −kf

3. k, k + p are all near kf while q and q − p are near −kf


4. k, k + p are all near −kf while q and q − p are near kf

5. k, q − p are near kf while q, k + p are near −kf

6. k, q − p are near −kf while q, k + p are near kf

The first two are referred to as “g4” processes – they involve just scattering

amongst right movers or amongst left movers. For these q ≈ 0, and one often

just replaces Vq with V0 for all these terms. The second two are referred to

as “g2” processes, and involve scattering between right and left movers. Again

q ≈ 0, and one often just replaces Vq with V0. Finally, the last terms are “g1”


processes, which involve scattering from one Fermi surface to the other. Here

q ≈ 2kf , and one often just replaces Vq with V2kf .

If we make this approximation (replacing Vq with V0 in the g4 and g2 pro-

cesses, and replacing it with V2kf in the g1 processes – then both the g1 and g2

terms are of the form

H1/2 = (· · · )∫dx ρR(x)ρL(x). (2.76)

The g4 terms are of the form

H4 = (· · · )∫dx ρR(x)ρR(x) + ρL(x)ρL(x). (2.77)

Following Giamarchi, we write the interaction Hamiltonian as

Hint =

∫dx

g4

2(ρR(x)ρR(x) + ρL(x)ρL(x)) + g2ρR(x)ρL(x) (2.78)

=1

2π

∫dx

g4 − g2

2π(∂xθ)

2 +g4 + g2

2π(∂xθ)

2. (2.79)

Remarkably this is of exactly the same form as the free Hamiltonian, allowing

us to write

H =1

2π

∫dxu

[K(∂xθ)

2 +1

K(∂xφ)

], (2.80)

which is readily diagonalized to produce

H =∑k

ukb†k bk + const, (2.81)

and the ground state has bk|ψ〉 = 0.

The effects of interactions are parameterized by the coefficients u and K. In

particular, repulsive interactions correspond to K < 1, and attractive to K > 1.

The relationship between θ, φ and the b’s is the same as the relationship with

the b’s, except for a scaling. In particular, if we write θ =√Kθ and φ = φ/

√K,

these scaled fields have the same commutation relations as the original,

[φ(x), ∂xθ(x)] = [φ(x), ∂xθ(x)] = iπδ(x− y). (2.82)

But in these scaled variables, H looks like the free Hamiltonian,

H =1

2π

∫dxu

[(∂xθ)

2 + (∂xφ)], (2.83)

and hence

φ =φ√K

=

(−iπL

)∑p

(L

2π

)1/2e−ipx√|p| sign p

(b†p + b−p

)(2.84)

θ =√Kθ =

(iπ

L

)∑p

(L

2π

)1/2e−ipx√|p|

(b†p − b−p

). (2.85)

Chapter 3

Applications – Mar 20, 2019

In this lecture we will use the results from the previous lecture to calculate

various physical quantities. We will need the following facts. First, for an

eigenstate of any quadratic Hamiltonian

〈b1b2b3b4〉 = 〈b1b2〉〈b3b4〉+ 〈b1b3〉〈b2b4〉+ 〈b1b4〉〈b2b3〉, (3.1)

where each b is either a bq or a b†q. This fact, known as Wick’s theorem, is proven

in any standard book on many-body theory. It is a straightforward argument

which we skip here. A related fact is that for an eigenstate of a quadratic

Hamiltonian, and any operator X with 〈X〉 = 0, one has

〈e〈X〉〉 = e〈X2〉/2. (3.2)

Finally, we need that for the ground state

〈b1b2〉 = 0 (3.3)

unless there is some q such that b2 = b†q and b1 = bq

A. Occupation Number

The first thing that we want to calculate is the occupation number 〈nk〉 = 〈c†kck〉.If we choose k near kf , we can just consider

nkf+q = c†rqcrq (3.4)

=1

L

∫dx

∫dx′ ψ†r(x)ψr(x

′)eiq(x−x′). (3.5)

Taking the expectation value we have

〈nkf+q〉 =1

L

∫dx

∫dx′ 〈ψ†r(x)ψr(x

′)〉eiq(x−x′). (3.6)

24

CHAPTER 3. APPLICATIONS – MAR 20, 2019 25

We then note that the expectation value should be translationally invariant, so

〈nkf+q〉 =

∫dx 〈ψ†r(x)ψr(0)〉eiqx. (3.7)

We now use our bosonized expression

ψr(x) = AFre−i(φ−θ), (3.8)

which gives

〈nkf+q〉 = A2

∫dx 〈ei(φ(x)−θ(x))e−i(φ(0)−θ(0))〉eiqx. (3.9)

We can combine the two terms using “Baker-Campbell-Hausdorff,” which says

that if [A,B] = C, and C is a constant, then

eAeB = eA+B+(1/2)[A,B]. (3.10)

In the present case

[φ(x), θ(0)] =iπ

2sign(x), (3.11)

[θ(x), φ(0)] =iπ

2sign(x), (3.12)

and hence

〈nkf+q〉 = A2

∫dx 〈ei(φ(x)−φ(0))−i(θ(x)−θ(0))+iπ sign(x)e−i(−θ(0))〉eiqx(3.13)

= A2

∫dx 〈ei(φ(x)−φ(0))−i(θ(x)−θ(0))〉eiqx (3.14)

At this point we use Eq. 3.2, to get

〈nkf+q〉 = A2

∫dx e−

12 〈[φ(x)−φ(0))−i(θ(x)−θ(0)]2〉eiqx. (3.15)

Thus we see that calculating the momentum distribution reduces to calculating

the correlators of the fields φ and θ.

We start with

〈[φ(x)− φ(x′)]2〉 = K

(−iπL

)(L

2π

)∑pq

(e−ipx − e−ipx′√|p| sign(p)

)(e−iqx − e−iqx′√|q| sign(q)

)×〈(b†p + b−p)(b

†q + b−q)〉. (3.16)

The operator expectation value is 1 if q = −p and otherwise is 0, and hence

〈[φ(x)− φ(x′)]2〉 = K

π

L

∑p

sin2 p(x−x′)2

|p|. (3.17)


In the thermodynamic limit we replace the sum with an integral, and,

〈[φ(x)− φ(x′)]2〉 = K

∫ ∞0

dpsin2 p(x−x′)

2

p. (3.18)

Unfortunately that integral is infinite. In particular, the sin2 is rapidly oscillat-

ing, and (at least in the tails of the integral) can be replaced by 1/2. We then

get the area under the 1/p curve, which is infinite.

The standard argument to save us is to say that we are only considering p

near the Fermi surface, and hence we can introduce a cutoff. A better argument

is to use a finite range interaction to cut things off. Nonetheless, we introduce

a cutoff Λ, and replace the sin2 with 1/2 – this latter substitution only works if

k(x− x′) > 1, and hence we also introduce a lower cutoff 1/(x− x′). Thus

〈[φ(x)− φ(x′)]2〉 = K log[Λ(x− x′)]. (3.19)

Similarly we get

〈[θ(x)− θ(x′)]2〉 =1

Klog[Λ(x− x′)], (3.20)

and

〈[φ(x)− φ(x′)] [θ(x)− θ(x′)]〉 = 0. (3.21)

Plugging this into our expression for the occupation numbers yields

nkf+q ∼∫dx

eiqx

|x|(K+K−1)/2(3.22)

∼ q(K+K−1)/2−1. (3.23)

And hence, the occupation number has a power law singularity at the Fermi

surface.

In the non-interacting case, K = 1, instead we have that nk is flat, and we

have a discontinuity at the Fermi surface.

A.1. Other Correlation Function

We can do the same arithmetic with the density-density correlation function.

In particular,

ρ(x) = ρr + ρl + e2ikfxψ†rψl + e−2ikfxψ†l ψr (3.24)

= − 1

π∂xφ+A2e2ikfxe2iφ +A2e−2ikfxe−2iφ. (3.25)


The first part is smooth, while the latter show oscillations on the scale of 2kf .

Turning the crank yields

〈ρ(x)ρ(0)〉 = ρ2 − K

(2π)2

1

x2+ (· · · ) cos(2kfx)

(1

x

)2K

. (3.26)

We see that the larger K is, the faster the 2kf oscillations decay: Attractive

interactions make these oscillations decay faster, while repulsive makes them

decay slower. One way to interpret this is that repulsive interactions lead to a

fluctuating charge density wave.

A complementary variable is the “pairing operator,”

∆(x) = π†(x)ψ†(x+ a), (3.27)

where a is some small number. If a was zero, the ∆ involves taking two spinless

fermions from the same location in space, and hence ∆ would vanish. For small,

but finite a, ∆ is related to how likely it is that two fermions are close together.

Being sloppy about the arguments,

∆(x) ≈ e2ikfxψ†Rψ†R + e−2ikfxψ†Lψ

†L + ψ†Rψ

†L + ψ†Lψ

†R. (3.28)

The first two terms are typically small, and

∆(x) ∼ e2iθ + e−2iθ, (3.29)

is obviously complementary to ρ – which involved φ’s. Clearly

〈∆(x)∆†(0)〉 ∼ 1

x1/2K. (3.30)

Attractive interactions make the pair correlations stronger.

A.2. Spin

I just want to very quickly sketch out what happens when one adds spin. In the

absense of interactions the ↑ and ↓ fermions are independent, so everything we

did goes through with a φ↓, φ↑, θ↓ and θ↑. The Free Hamiltonian is

H0 =1

2π

∫dx vf

[(∂xθ↑)

2 + (∂xθ↓)2 + (∂xφ↑)

2 + (∂xφ↓)2]. (3.31)

The same interaction terms as before occur, but now we should be careful about

the spin. The g4 processes


can occur with both spins, and gives terms

H4 ∝∫dx [ρ2

r + ρ2l ] (3.32)

∝∫dx [∂x(φ↑ + θ↑ + φ↓ + θ↓)]

2 + [∂x(φ↑ − θ↑ + φ↓ − θ↓)]2

The g2 processes

give

H2 ∝∫dx [ρrρl] (3.33)

∝∫dx ∂x(φ↑ + θ↑ + φ↓ + θ↓)∂x(φ↑ − θ↑ + φ↓ − θ↓)

Finally, the g1 processes

separate into sets of terms. First if the spins are all the same we get

Ha1 ∝

∫dx ρr↑ρl↑ + ρr↓ρl↓ (3.34)

∝∫dx ∂x(φ↑ + θ↑)∂x(φ↑ − θ↑) + ∂x(φ↓ + θ↓)∂x(φ↓ − θ↓).


The terms where the spins are different have another structure

Hb1 ∝

∫dxψ†r↑ψ

†l↓ψr↓ψl↑ + ψ†r↓ψ

†l↑ψr↑ψl↓ (3.35)

∝∫dx cos(2i[φ↑ − φ↓]). (3.36)

These terms can be combined to give a Hamiltonian which separately involves

spin and charge degrees of freedom. In particular we define

φρ =1√2

(φ↑ + φ↓) (3.37)

φσ =1√2

(φ↑ − φ↓) (3.38)

θρ =1√2

(θ↑ + θ↓) (3.39)

θσ =1√2

(θ↑ − θ↓), (3.40)

in terms of which

H = Hρ +Hσ (3.41)

where

Hρ =

∫dx(· · · )(∂xθρ)2 + (· · · )(∂xφρ)2 (3.42)

Hσ =

∫dx(· · · )(∂xθσ)2 + (· · · )(∂xφσ)2 + (· · · ) cos(2

√2φσ). (3.43)

The charge degrees of freedom can be analyzed using the techniques we have

already discussed. The spin degrees of freedom are a bit more complicated: It

turns out there is a phase transition between a phase where φσ is frozen (which

it turns out corresponds to pairing) and one where it fluctuates. The phase

transition can be mapped onto the classical Kosterlitz-Thouless transition of a

2D XY model.

B. HW 2 – Not Due

These problems are just to help understanding. You neither need to do them

nor to hand them in.

Problem 3.1. Squeezed States

We can “solve” the Harmonic oscillator

H0 =p2

2+x2

2, (3.44)


with [x, p] = i, by the transformation:

a =1√2

(x+ ip) (3.45)

a† =1√2

(x− ip) , (3.46)

in terms of which

H0 = a†a+1

2. (3.47)

The operators a, a† obey the standard ladder operator commutators, and are

ladder operators for this Hamiltonian. The ground state is defined by

a|ψ0〉 = 0. (3.48)

Consider now the Hamiltonian

H = α

(λ2 p

2

2+

1

λ2

x2

2

), (3.49)

3.1.1. Show that x = x/λ and p = λp obey the same commutation relations as

x and p.

3.1.2. Use that result to argue that

H = α

(b†b+

1

2

), (3.50)

where b and b† obey the standard commutation relations. Express b and b† in

terms of x and p.

3.1.3. Express a and a† in terms of b and b†.

3.1.4. Consider the state |ψ〉 defined by

b|ψ〉 = 0. (3.51)

This is a “squeezed state” in the original variables. Find

〈ψ|a†a|ψ〉 (3.52)

〈ψ|aa†|ψ〉 (3.53)

〈ψ|aa|ψ〉 (3.54)

〈ψ|a†a†|ψ〉. (3.55)

The relevance to the Luttinger liquid is that the ground state of the interacting

system is exactly this sort of squeezed state.


3.1.5. A more tedious approach to the same problem is to consider the Hamil-

tonian

H = A(a†a+ aa†) +B(aa+ a†a†). (3.56)

Show that Eq. (3.56) is of the same form as Eq. (3.49), and relate α and λ to

A and B.

I am not going to make you go through it, but the “Bogoliubov” approach to

diagonalizing Eq. (3.56) is to make the general transformation:

a = ub+ vb† (3.57)

a† = ub† + vb, (3.58)

with u2 − v2 = 1. One can easily verify that the b’s obey the standard ladder

operator commutation relations. Substituting into Eq. (3.56) yields

H =[A(u2 + v2) +B(2uv)

](b†b+ bb†) +

[B(u2 + v2) +A(2uv)

](b†b† + bb).(3.59)

To diagonalize the Hamiltonian, one chooses u and v to obey

u2 − v2 = 1 (3.60)

B(u2 + v2) +A(2uv) = 0. (3.61)

One traditionally writes u = cosh(ξ) and v = sinh(ξ), in which case

u2 + v2 = cosh(2ξ) (3.62)

2uv = sinh(2ξ), (3.63)

and

tanh(2ξ) = −B/A. (3.64)

Clearly we require |A| ≥ |B|. Simple “trig” then gives

u2 + v2 =A√

A2 −B2(3.65)

2uv = − B√A2 −B2

, (3.66)

and

H =√A2 −B2(b†b+ bb†) (3.67)

which should agree with the less tedious “scaling” method.

Problem 3.2. This is mostly an exercise in book-keeping. Feel free to skip it if

it seems too dull. Nonetheless going through this once is useful – and if you end


up doing more 1D stuff, getting fast at it is useful, as you need to keep doing

this to find all your π’s and − signs.

3.2.1. In the main text we defined:

φr(x) = i2π

L

∑p

eipx

pρr(−p) (3.68)

φ`(x) = i2π

L

∑p

eipx

pρ`(−p), (3.69)

and we have

bp = θ(p)

√2π

L|p|ρr(−p) + θ(−p)

√2π

L|p|ρ`(p) (3.70)

or for p > 0

ρr(−p) =

√Lp

2πbp (3.71)

ρ`(p) =

√Lp

2πb−p (3.72)

ρr(p) =

√Lp

2πb†p (3.73)

ρ`(−p) =

√Lp

2πb†−p. (3.74)

Write φr and φ` in terms of the b’s.

3.2.2. Write φ = (φr − φ`)/2 and θ = −(φr + φ`)/2 in terms of the b’s.

3.2.3. As argued in the main text ρ = ∂xφ/π, and hence φ has the physical

meaning of a displacement field. Show that

[φ(x), ∂yθ(y)] = iπδ(x− y), (3.75)

and hence we can interpret Π(x) = ∂xθ(x)/π as the conjugate momentum to θ.

3.2.4. The Hamiltonian is

Hb0 =

∑p

vfpb†pbp. (3.76)

Show that

Hb0 =

vf2π

∫dx[(∂xθ)

2 + (∂xφ)2]. (3.77)

Chapter 4

Tonks Gas and the

Leib-Lininger Model – Mar

27, 2019

In the previous chapters we explored the “bosonization” techniques, largely in

the context of spin-free fermions. The remarkable result there was that the low

energy physics was describable in terms of a 1D elastic medium, and that phys-

ical quantities, such as the momentum occupation space occupation numbers,

could be expressed in terms of corelators of the fields describing this elastic

medium. Both a positive and negative feature of this bosonization approach is

that it is constructed in terms of operators. In the next two chapters we con-

sider a complementary point of view, and explicitly construct the wavefunction

of a 1D system. This is clearly more concrete, but it means that calculating

physical quantities is a little more difficult.

The approach, widely known as the “Bethe Ansatz” is applicable to a number

of systems with “point interactions.” It has been applied to bosons (the Lieb-

Lininger model), spin-1/2 fermions (the Gaudin-Yang model), spin-1/2 lattice

fermions (the Hubbard model), and spin chains (the XXZ model). We will

explicitly discuss only the simplest of these, namely the boson system.

.1. Permutations

Permutations play an important role in our arguments, so it is good to briefly

introduce some notation that we will use. Consider the permutations of n

objects. These form a group, Sn, consisting of n! elements, and known as the

“symmetric group”. We think about the permutations as functions which map

33

CHAPTER 4. TONKS GAS AND THE LEIB-LININGER MODEL – MAR 27, 201934

the integers 1, 2, · · · , n onto themselves. For example, with n = 5, we might

have σ(1) = 3, σ(2) = 2, σ(3) = 4, σ(4) = 1, σ(5) = 5. The composition of

two permutations is a permutation, and we just use standard “multiplication”

notation for composition. So if τ(1) = 5, τ(2) = 2, τ(3) = 1, τ(4) = 3, τ(5) = 4,

then we can define their product λ = στ by λ(1) = σ(τ(1)) = σ(5) = 5, and so

on. A permutation is uniquely defined by how it scrambles the ordered set, for

example 1, 2, 3, 4, 5. Thus we can write σ = [3, 2, 4, 1, 5] or τ = [5, 2, 1, 3, 4].

Another useful notation for permutations is to decompose them into disjoint

cycles. For example σ = (134)(2)(5), meaning that σ(1) = 3 and σ(3) = 4

and σ(4) = 1. Up to cyclical permutations of each cycle, this decomposition is

unique. We can safely ignore the 1-cycles (or unicycles as Veit suggested) as

they don’t do anything. Hence σ = (1, 2, 3) and τ = (1, 5, 4, 3).

A further refinement is that each cycle can be written as a product of

non-disjoint 2-cycles, also known as transpositions. For example, (1, 2, 3) =

(1, 2)(2, 3). The decomposition into transpositions is not unique, but the parity

of the number of transpositions is. The permutations consisting of even numbers

of transpositions forms a subgroup known as the “alternating group,” An.

Clearly the 2-cycles generate Sn. It turns out that a minimal set of generators

is just the “neighboring” transpositions (1, 2), (2, 3), · · · (n− 1, n). For example

(1, 3) = (1, 2)(2, 3)(1, 2). This decomposition is not unique.

.2. 1D Hard Core Gas

Let us now think about bosons with hard-core interactions. This is a bit of a

warm-up, that will introduce some of the basic concepts. The hard core Bose

gas is often known as the Tonks-Girardeau model, and sometimes it seems that

keeping track of the names of the various models is half the work. [For those

who care, Tonks actually studied the classical model, but nonetheless his name

is still associated with this system.]

As we know from solving the “particle in an infinite well,” infinitely strong

potentials can be replaced by a boundary condition on the wavefunction. By

the same argument, the hard core Bose gas can be described by a free-particle

Schrodinger equation,

−1

2

∑j

∂2jψ(x1, · · · , xn) = Eψ(x1, · · · , xn), (4.1)

with the boundary condition that ψ = 0 if any of the two x’s are equal. We

should also probably add the Bose symmetry requirement that ψ is unchanged

by permuting any two of the particles. Finally, if we want a finite particle

density, we should put this in a box of length L. Lets use periodic boundary


conditions, so ψ(x1 + L) = ψ(x1), where I have held all the other coordinates

fixed.

There is a wonderful trick for solving this problem. Namely we introduce

a function A(x1, x2, · · ·xn) which is always either 1 or −1. It equals 1 if x1 <

x2 < · · · < xN , or if the ordering of the particles is an even permutation. It is

−1 if it is an odd permutation. Given A, and a bosonic wavefunction ψB that

obeys our hard core boundary conditions, we can always generate a fermionic

wavefunction ψF = AψB . Conversely, given an arbitrary fermionic wavefunction

ψF we can generate a hard-core Bose wavefunction vie ψB = AψF . Clearly ψBobey EQ. (4.1) if and only if ψF also does. Thus we have established a one-to-

one mapping between hard core Bose wavefunctions, and non-interacting Fermi

wavefunctions in one dimension.

In particular, the fermionic wavefunction can always be expressed as a Slater

determinant

ψF =

∣∣∣∣∣∣∣∣eik1x1 eik1x2 · · · eik1xn

eik2x1 eik2x2 · · ·...

∣∣∣∣∣∣∣∣ (4.2)

=∑σ∈Sn

sig(σ)∏j

eikσ(j)xj (4.3)

where k1, k2, · · · kn are distinct wave-vectors, satisfying kj = nj(2π/L), and njare distinct integers. The energy is simply E =

∑j k

2j/2, and the ground state

consists of choosing all of the nj ’s to have as small as possible magnitudes – ie

forming a Fermi surface.

There is a minor subtlety worth mentioning. Namely, if there are an even

number of particles, and we want the bosonic wavefunction to have periodic

boundary conditions, then the fermions need to have antiperiodic boundary

conditions, ie kj = (nj + 1/2)(2π/L).

It is also useful to note that one can explicitly write down the ground state

wavefunction by using the fact that for numbers z1, z2, · · · zn, the Vandermond

determinant can be evaluated as∣∣∣∣∣∣∣∣∣∣1 1 1 · · ·z1 z2 z3 · · ·z2

1 z22 z2

3 · · ·...

∣∣∣∣∣∣∣∣∣∣=∏i<j

(zi − zj). (4.4)

This can be proven by noting that both sides vanish when zi = zj , and both


are n’th order homogeneous polynomials. We then write

ψf = e−inπL

∑j xj

∣∣∣∣∣∣∣∣∣∣1 1 · · · 1

e2πix1/L e2πix2/L · · · e2πixn/L

e4πix1/L e44πix2/L · · · e4πixn/L

...

∣∣∣∣∣∣∣∣∣∣(4.5)

= e−inπL

∑j xj∏i<j

(e2πixi/L − e2πixj/L( (4.6)

∝∏i<j

sinπ

L(xi − xj) (4.7)

The bosonic wavefunction is then

ψb =∏i<j

sinπ

L|xi − xj |. (4.8)

The remarkable thing is that all thermodynamic properties of the hard core

bosons and the free fermions are identical. Also the density, or the density-

density correlations functions should be identical. I should caution that other

properties are not the same. For example, the k’s in the determinant, referred to

as rapidities, do not correspond to the momenta of the bosons. The momentum

distribution is

nk =∝∫dx′1

∫dx1 dx2 · · · dxne−ik(x−x′)ψ∗(x′1, x2, · · ·xn)ψ(x1, x2, · · ·xn).

(4.9)

At small k, nk ∼ k−1/2, while for large k, nk ∼ k−4.

A. Spin Chains

To further illustrate the principle of mapping onto a Fermi system, we can

consider a chain of spin-1/2 objects,

H =∑j

Jx(Sxj Sxj+1 + Syj S

yj+1) + Jz(S

zj S

zj+1). (4.10)

This is referred to as the XXZ model, and represents some sort of anisotropic

magnetic interactions. The isotropic case Jx = Jz is the Heisenberg model,

while the extremes case Jx = 0 and Jz = 0 are the Ising model and XY models.

The operators Sx, Sy, Sz are Pauli operators. It is convenient to introduce

ladder operators S+ and S−, in terms of which

H =∑j

Jx2

(S+j S−j+1 + S−j S

+j+1) + Jz(S

zj S

zj+1). (4.11)


We map this onto a Fermi gas by identifying a down-spin site with an empty

site, and an up-spin site with an occupied site. One naturally expects something

like

S+j = f†j (4.12)

S−j = fj (4.13)

Szj = 2f†j fj − 1. (4.14)

One can readily verify that the {fj , f†j } = 1. Unfortunately the fj ’s on different

sites commute. We can fix this with what is known as a “Jordan-Wigner”

transform

a†j = (−1)∑k<j f

†kfkf†j (4.15)

aj = (−1)∑k<j f

†kfkfj , (4.16)

which gives

H =Jx2

∑i

(a†iai+1 + a†i+1ai

)+ Jz

∑i

(2ni − 1)(2ni+1 − 1). (4.17)

This is the Hubbard model. Note the special case Jz = 0 corresponds to free

Fermions, and we can trivially solve that case.

Chapter 5

Bethe Ansatz – Mar 29,

2019

The Bethe ansatz is a generalization of the approach in Sec. .2 – and is best

illustrated using a generalization of the model discussed their, namely a gas of

1D bosons with point interactions,

−1

2

∑j

∂2jψ + c

∑i<j

δ(xi − xj)ψ = Eψ. (5.1)

As before, we impose bosonic symmetry, and periodic boundary conditions in a

box of size L. The limit c→∞ corresponds to the Tonks gas, which we solved

by mapping onto free fermions. The limit c→ 0 corresponds to the free bosons,

with ψ = 1.

We will approach this problem, by first solving the two-body problem, then

the three-body problem, then generalizing to arbitrary particle number.

A. Two-Body Problem

The two-body problem is a classic undergraduate quantum mechanics problem.

In lecture I jumped to the answer, but in these notes I will go through all of the

details (since I never remember them, and have to do it from scratch anyways.

We begin with

−∂21 + ∂2

2

2ψ(x1, x2) + cδ(x1 − x2)ψ(x1, x2) = Eψ(x2, x2). (5.2)

38

CHAPTER 5. BETHE ANSATZ – MAR 29, 2019 39

We change variables to center-of-mass and relative:

y = x1 − x2 (5.3)

s =x1 + x2

2. (5.4)

We then use the chain rule to find:(∂ψ

∂x1

)x2

=

(∂ψ

∂y

)s

(y

∂x1

)x2

+

(∂ψ

∂s

)y

(s

∂x1

)x2

(5.5)

=

(∂ψ

∂y

)s

+1

2

(∂ψ

∂s

)y

. (5.6)

We then repeat to get

∂21 = ∂2

y + ∂y∂s +1

4∂2s . (5.7)

To get ∂22 , we just flip the sign of y, so

∂22 = ∂2

y − ∂y∂s +1

4∂2s . (5.8)

Hence, we can rewrite Eq. (5.9) as

−∂2yψ −

1

4∂2sψ + cδ(y)ψ = Eψ. (5.9)

We then integrate over y around y = 0 to yields

∂yψ|y=0+

y=0− = c ψ|y=0 . (5.10)

We now convert this back to our original variables to get

(∂1 − ∂2)ψ|x1=x2+εx1=x2−ε = 2c ψ|x1=x2

. (5.11)

The only tricky thing with that whole argument is getting the factors of 2 right.

We can think of Eq. (5.9) as a free Schrodinger equation with the boundary

condition in Eq. (5.11).

Lets now think about this as a scattering problem in the x1 − x2 plane. In

our elementary classes, we solve scattering problems by imagining an incoming

plane wave, and an outgoing plane wave, then choosing amplitudes and phases

so that the boundary conditions are satisfied. Thus we imagine an incoming

wave, ei(k1x1+k2x2), in the regime x1 > x2:


By Bose symmetry, one must also have an incoming wave ei(k1x2+k2x1) with

x1 < x2. One will also end up with reflected and transmitted rays from each

of these two incoming waves. The reflected ray from the x1 > x2 wave has the

same wave-vector as the transmitted ray from the x1 < x2 wave, so we only

need one coefficient R to describe both these processes. And hence we draw the

various rays as:

To be explicit, the wavefunction for x1 > x2 is

ψ12(x1, x2) = eik1x1+ik2x2 +Reik2x1+ik1x2 , (5.12)

and the wavefunction for x2 > x1 is

ψ21(x1, x2) = eik1x2+ik2x1 +Reik2x2+ik1x1 . (5.13)


We now implement the boundary condition in Eq. (5.11), and calculate

(∂1 − ∂2)ψ|x1=x2+ε = (∂1 − ∂2)ψ12|x1=x2(5.14)

= i(k1 − k2)(1−R)ei(k1+k2)x (5.15)

(∂1 − ∂2)ψ|x1=x2−ε = (∂1 − ∂2)ψ21|x1=x2(5.16)

= −i(k1 − k2)(1−R)ei(k1+k2)x (5.17)

ψx1=x2= (1 +R)ei(k1+k2)x. (5.18)

For the boundary condition to be satisfied, we need

i(1−R)(k1 − k2) = c(1 +R) (5.19)

which can be manipulated into

R = −eiφ (5.20)

= −(c− i(k1 − k2)

c+ i(k1 − k2)

), (5.21)

which defines the angle

φ = 2 arctank2 − k1

c. (5.22)

The first observation is that it makes sense for φ to simply be a phase: this

is consistent with conservation of probability. One may be curious about why I

included the minus sign in R = −eiφ. The reason is that with this definition, in

the hard core case φ = 0. It turns out that the expansion about this hard core

limit is better behaved than the expansion about the non-interacting limit, so

it is convenient to set up the notation in anticipation.

A second observation is that the boundary condition can be satisfied by any

k1 and k2. This is analogous to the observation in section .2, that the hard core

constraints and Bose symmetry could be satisfied by any Slater determinant. In

both that case, and this one, the quantization of the k’s comes from imposing

periodic boundary conditions:

ψ(x1 + L, x2) = ψ(x1, x2). (5.23)

If we take x1 < x2, this requires

ψ12(x1 + L, x2) = ψ21(x1, x2) (5.24)

which means

eik1Leik1x1+ik2x2 +Reik2Leik2x1+ik1x2 = eik1x2+ik2x1 +Reik2x2+ik1x1 .(5.25)


This must be true for all x1 and x2, so we need

Reik2L = 1 (5.26)

eik1L = R. (5.27)

If I multiply these two equations I find

ei(k1+k2)L = 1. (5.28)

This is just quantization of the center of mass momentum. If I divide the

equations, I get

ei(k1−k2) = R2 = e2iφ (5.29)

Taking the logarithm, and defining δ = k1 − k2, we have

δ = 2 logc− iδc+ iδ

+ 2πn (5.30)

= 2πn− 4 arctanδk

c. (5.31)

We can solve this graphically. The left hand side is a straight line, while the

right hand side is a series of steps. For smallish c, it looks like

The arctangent lines are labeled by n. The n = 0 case gives δ = 0, which

gives R = −1, and ψ = 0. Thus we should neglect the n = 0 line, which I

colored green. As c → 0, the remaining solutions approach δ = 2πn/L, n =

±0,±1,±2, · · · . They come in plus-minus pairs because permuting k1 and k2

yields the same wavefunction. For large c, the curves become flat:


Again as c → ∞ we have δ = 2πn/L, but this time n = ±1,±2, · · · . [Recall,

the green curve should be ignored.]

If c < 0 we also have complex solutions, which correspond to bound states.

We will not consider these here.

B. Three Particles

We now solve the three-particle problem. We simply follow the same approach

as the two-particle problem, and the only challenge is book-keeping. Instead

of dividing space into two sectors we divide it into six sectors: x1 < x2 < x3,

x2 < x1 < x3,· · · . I denote the wavefunction when x1 < x2 < x3 as ψ123, and

when x2 < x1 < x3 as ψ213, · · · . More formally, for a permutation σ we define

ψσ to describe the sector where xσ(1) < xσ(2) < xσ(3). We then postulate that

in each sector the wavefunction is a sum of plane waves, which we can think of

as all the rays which happen when one of the particles scatter off one-another,

ψ123(x1, x2, x3) = A123123e

i(k1x1+k2x2+k3x3) +A123213e

i(k2x1+k1x2+k3x3)(5.32)

+A123132e

i(k1x1+k3x2+k2x3) +A123231e

i(k2x1+k3x2+k1x3)

+A123312e

i(k3x1+k1x2+k2x3) +A123321e

i(k3x1+k2x2+k1x3).

That is A123τ is the coefficient of ei(kτ(1)x1+kτ(2)x2+kτ(3)x3). In other sectors one

would define

ψ213(x1, x2, x3) = A213123e

i(k1x2+k2x1+k3x3) +A213213e

i(k2x2+k1x1+k3x3)(5.33)

+A213132e

i(k1x2+k3x1+k2x3) +A213231e

i(k2x2+k3x1+k1x3)

+A213312e

i(k3x2+k1x1+k2x3) +A213321e

i(k3x2+k2x1+k1x3).

That is, in the sector where xσ(1) < xσ(2) < xσ(3), the factor Aστ is the coefficient

of ei(kτ(1)xσ(1)+kτ(2)xσ(2)+kτ(3)xσ(3)). The generalization to more particles should

be clear.

We now consider Bose symmetry, which says that ψ123(x, y, z) = ψ213(y, x, z).

Writing things out, we find A123τ = A213

τ . Clearly the generalization is that Aστis independent of σ. This generalization works for all particle numbers.

We now turn to the boundary conditions, for example, lets take x3 to be

bigger than x1 and x2, and consider the condition

(∂1 − ∂2)ψ|x1=x2+εx1=x2−ε = 2c ψ|x1=x2

. (5.34)

This can be expressed as

(∂1−∂2)ψ213(x, x, x3)−(∂1−∂2)ψ123(x, x, x3) = 2cψ213(x, x, x3) = 2cψ123(x, x, x3).

(5.35)


There are three sorts of terms, proportional to ei(k1+k2)x+k3x3 , ei(k2+k3)x+k1x3 ,

and ei(k1+k3)x+k2x3 . One can equate the coefficient of each of these. Lets look

at the terms proportional to ei(k1+k2)x+k3x3 :[i(k2 − k1)A213

123 + i(k1 − k2)A213213

]−[i(k1 − k2)A123

123 + i(k2 − k1)A213123

]= 2c(A123+A213)

(5.36)

We added the superscripts just to make it more obvious where each term came

from. We can then group terms, and find

A213

A123= R12 = −eiθ12 = −c+ i(k1 − k2)

c− i(k1 − k2). (5.37)

We can similarly define

Rij = −eiθij = −c+ i(ki − kj)c− i(ki − kj)

, (5.38)

which relatesAjikAijk

=AkjiAkij

= Rij . (5.39)

That is, because the boundary conditions relate permutations of neighboring

particles, we get relations between A’s where neighboring indices are exchanged.

Because the neighboring transpositions generate all the permutations, we can

find a general Aσ in terms of A123. We can further always choose A123 = 1.

That is

A123 = 1 (5.40)

A213 = R12 (5.41)

A132 = R23 (5.42)

A231 = R31R12 (5.43)

A312 = R13R23 (5.44)

A321 = R12R13R23. (5.45)

These are clearly consistent with one-another: There are two distinct ways to

reach A321 from A123, but they give the same R’s. The generalization to more

particles will also be consistent.

For any set of k’s, our ansatz solves Schrodinger’s equation (as long as no

two k’s are not identical – for otherwise the wavefunction vanishes). Adding

periodic boundary conditions will then restrict the allowed k’s. Take 0 <=

x1 < x2 < x3 < L, then

ψ(x1 + L, x2, x3) = ψ(x1, x2, x3). (5.46)


The wavefuction of the left is clearly ψ231, and the wavefunction on the right

is ψ123. There are several ways to do the arithmetic, but I like to use the Bose

symmetry so that these are both in the 123 sector,

ψ123(x2, x3, x1 + L) = ψ123(x1, x2, x3). (5.47)

Equating terms with the same x dependence yields

eik3LA123 = A312 = R13R23A123 (5.48)

and permutations. The way I like to think of it, is with periodic boundary

conditions you can go from 123 to 312 in two ways: You do two transpositions,

or you simply slip particle number 1 off the left. Equating these gives Eq. (5.48).

We consequently find three equations

R13R23 = eik3L (5.49)

R12R32 = eik2L (5.50)

R21R31 = eik1L (5.51)

Taking logs yields

k3L = (π + θ13) + (π + θ23) + 2πn3 (5.52)

k2L = (π + θ12) + (π + θ32) + 2πn2 (5.53)

k1L = (π + θ21) + (π + θ31) + 2πn1, (5.54)

where the nj are integers, and

θij = −i log

(c+ i(ki − kj)c− i(ki − kj)

)= 2 tan−1 ki − kj

c. (5.55)

If we define θii = 0, then the general many-particle result is clearly

kjL =∑i

θij + 2π(nj + ξ/2) (5.56)

where xi = 0 if there are an odd number of particles, but ξ = 1 for an even.

I claim that Eq. (5.56) has a unique solution once we specify the nj ’s. The

best way that I know to prove this uniqueness is to first show that Eq. (5.56) is

equivalent to extremizing the function

F =∑j

(k2j

2− 2π(nj + ξ/2)

Lkj

)− 1

2

∑ij

Λij (5.57)

with

Λij = − i

L[(c+ (ki − kj)) log (c+ (ki − kj)) + (c− (ki − kj)) log (c− (ki − kj))] .

(5.58)


We then prove that F is concave up. Thus we calculate the second derivatives

∂2F

∂k2i

= 1 +2

L

∑j 6=i

c

c2 + (ki − kj)2(5.59)

∂2F

∂ki∂kj= − 1

L

2c

c2 + (ki − kj)2. (5.60)

We now consider the quadratic form

G =∑ij

∂2F

∂ki∂kjxixj (5.61)

=∑i

x2i +

1

L

∑j

c

c2 + (ki − kj)2(xi − xj)2. (5.62)

Clearly, as the sum of positive quantities, G is positive – hence F is concave up.

C. Thermodynamic Limit

While numerically it is quite efficient to simply solve Eq. (5.56) for fixed N , it

is appealing to understand how to take the thermodynamic limit. To do so, we

consider an odd number of particles, and order the n’s such that n1 < n2 <

· · ·nN . The ground state will have

nj = −(N + 1)/2 + j. (5.63)

We then try to find the spacing between subsequent k’s:

(kj − kj−1)L =∑i

(θij − θij−1) + 2π. (5.64)

In the thermodynamic limit the spacing will be small, and we define the density

of k’s to be

Lf(kj) =1

kj − kj−1. (5.65)

We then divide Eq. (5.64) by (kj − kj−1)L to arrive at

1 =∑i

θij − θi,j−1

L(kj − kj−1)+ 2πf(kj). (5.66)

We then note that in the thermodynamic limit, the term in the sum is a deriva-

tive:

1 =∑i

1

L

∂θij∂kj

+ 2πf(kj). (5.67)

Finally we convert the sum to an integral,

1 =

∫dq f(q)∂kθ(q − k) + 2πf(k), (5.68)


where we have replaced kj with k and ki with q. We have also introduced the

function

θ(q) = −i logc+ iq

c− iq. (5.69)

Taking the derivative, we find the integral equation

2πf(k)− 1 =

∫dq

2c

c2 + (q − k)2f(q). (5.70)

This should be supplemented with the constraint∫f(k)dk =

N

L. (5.71)

The traditional approach to solving this integral equation involves first as-

suming that in the thermodynamic limit there is a maximum k, called k0. This

is certainly true in the hard core case, so it seems plausible. Thus f(k) = 0 for

all |k| > k0. We rescale by this maximum, writing x = k/k0. We also rescale

c, writing λ = c/k0. We also introduce γ = cL/N , which is a dimensionless

measure of the interaction strength. In the thermodynamic limit, one expects

that the interactions only enter in this combination. Finally to avoid confusion

we write g(x = k/k0) = f(k).

Equation (5.71) relates λ to g and the physical quantity γ via

λ = γ

∫ 1

−1

g(x)dx. (5.72)

The other equation becomes

2πg(y) = 1 + 2λ

∫ 1

−1

g(x)

λ2 + (x− y)2dx. (5.73)

One, simple approach to solving these equations is to begin with some trial

g0. one then uses Eq. (5.72) to calculate λ0. One substitutes this into Eq. (5.73)

to get a new g,

2πg1(y) = 1 + 2λ0

∫ 1

−1

g0(x)

λ2 + (x− y)2dx. (5.74)

one then iterates.

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