Integrable Quantum Spin Chains

Usman Kayani

March 28, 2012Candidate number: P04349Supervisor: Dr B DoyonKing’s College LondonMSci in Mathematics

Abstract

Overview of Integrable Quantum Spin Chains focusing in particular on theXXX Heisenberg spin chain with topics covering the coordinate Bethe

Ansatz, the algebraic Bethe Ansatz, integrability, higher conserved chargesand correlation functions. I have adapted and filled in some of the proofswhich are commonly left out in the literature. Material for students at a

masters level.

Contents

1 Introduction 31.1 Integrability . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 What is Spin? . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Spin Chains? . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4 Notation, Conventions and Concepts from QM . . . . . . . . 6

2 Heisenberg Spin Chain 92.1 Basis for observables . . . . . . . . . . . . . . . . . . . . . . . 92.2 Derivation of Hamiltonian . . . . . . . . . . . . . . . . . . . . 10

2.2.1 One-particle wave function N = 1 . . . . . . . . . . . 112.2.2 Two-particle wave function N = 2 . . . . . . . . . . . 122.2.3 N Spin chain . . . . . . . . . . . . . . . . . . . . . . . 14

2.3 Energy spectrum of H . . . . . . . . . . . . . . . . . . . . . . 17

3 Bethe Ansatz 193.1 Coordinate BA . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.1.1 Symmetries of Heisenberg Model . . . . . . . . . . . . 203.1.2 Eigenstates of H . . . . . . . . . . . . . . . . . . . . . 22

3.2 BAE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323.3 Algebraic BA . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.3.1 Integrability . . . . . . . . . . . . . . . . . . . . . . . . 343.3.2 Yang-Baxter equation . . . . . . . . . . . . . . . . . . 343.3.3 Spectrum of H . . . . . . . . . . . . . . . . . . . . . . 423.3.4 Bethe Ansatz equations . . . . . . . . . . . . . . . . . 45

3.4 Thermodynamic limit N →∞ - String Hypothesis . . . . . . 483.4.1 XXZ Model . . . . . . . . . . . . . . . . . . . . . . . . 54

4 Higher Conserved Charges of the Heisenberg Hamiltonian 574.1 Logarithmic derivative of the transfer matrix . . . . . . . . . 58

5 Correlation functions 625.1 EFP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

5.1.1 Integral representation of EFP . . . . . . . . . . . . . 645.1.2 Sato’s formula . . . . . . . . . . . . . . . . . . . . . . 66

1

6 Conclusion 68

References 70

2

Chapter 1

Introduction

“The career of a young theoretical physicist consists of treatingthe harmonic oscillator in ever-increasing levels of abstraction.”– Sidney Coleman

H =1

2mp̂2 +

1

2ω2q2

We will begin by briefly talking about the quantum harmonic oscillator - amodel which is easy to understand, familiar amongst most graduate studentsand will help us understand the idea behind a quantum integrable system.The quantum harmonic oscillator is one of the most important models inquantum mechanics, simply because it is one of the few systems for whichan exact, analytic solution is known. Though solving this system directly israther straightforward as you essentially have to solve a differential equation,it can get quite tedious since we will have to deal with Hermite polynomialswhich get increasingly complicated1 for energy eigenstates above the groundstate. Fortunately there exists another method to extract the energy eigen-values without solving the differential equation known as the ladder operatormethod. The familiar terminology “creation” and “annihilation” operatorthat appears in quantum field theory and other area’s of physics comes fromgeneralizing this simple and elegant concept. It will also be essential inthe Algebraic Bethe Ansatz, where the action of the “creation” operator isthought of as a excitation in the chain with respect to the reference ferro-magnetic ground state. The quantum harmonic oscillator is one example ofan integrable system.

1.1 Integrability

The concept of integrability hails from our old friend classical mechanicsand we will later give a more precise definition of this in terms of conserved

1Increase in the polynomial order

3

quantities. For now, we will think of integrability as simply meaning wecan solve the system in question exactly i.e. given the Hamiltonian, whichencodes the dynamics of the system we can find the entire energy spectrumand other physical quantities such as correlation functions. Other examplesof integrable systems include the hydrogen atom, which we will briefly meetin the next chapter, the sine-Gordon model and of course the model of focusfor the project, the Heisenberg spin chain. The concepts that are importantin Quantum integrable models are the Coordinate Bethe Ansatz and theQuantum Inverse Scattering Method also known as the Algebraic BetheAnsatz due to the strong algebraic framework. The CBA was originallydeveloped by Bethe in 1931 and it will be the first method we will use tofind the energy spectrum, after which we will use the ABA and prove itsintegrability.

1.2 What is Spin?

In classical mechanics, we have the notion of orbital angular momentumassociated with the motion of the center of mass. In quantum mechanics, theorbital angular momentum has a simular structure as in classical mechanics.There is however another type of angular momentum which arises known asspin. A fundamental difference between the two types that the spin can haveboth integer and half integer values while only integer values of the orbitalangular momentum are allowed. Spin is a characteristic property of ele-mentary particles (fermions and bosons), composite particles (hadrons) andatomic nuclei. The direction of spin can change but an elementary particlecannot be made to spin faster or slower. For our consideration we will belooking at Spin-1/2 particles i.e. the electron. Particles (like the electron)with Spin posses a magnetic dipole moment, just like a rotating electricallycharged body in classical electrodynamics. As a result, the ‘movement’ ofindividual electrons in their atomic nuclei will generate a weak magneticfield.

Conventionally the direction chosen for the quantization axis is the z-axis, but this choice is arbitrary and chosen for convenience as we shall seeshortly. The spin projection operator Sz affects a measurement of the spinin the z direction and the eigenstates [1, 0]t and [0, 1]t form a complete basisfor the Hilbert space describing a spin-1/2 particle, which we can call ”up”

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and ”down”. This means that we can essentially write any eigenvector bytaking linear combinations of the above eigenstates and hence describe allpossible states of the spin. It is important to note that by the postulates ofquantum mechanics, an experiment designed to measure the electron spinon the x, y or z axis can only yield an eigenvalue of the corresponding spinoperator (Sx, Sy or Sz) on that axis. A measurement of the z-componentof spin destroys any information about the x and y components that mightpreviously have been obtained.

1.3 Spin Chains?

A quantum spin chain is a particular example of an integrable system usu-ally in 1+1 space-time dimensions for the purposes of integrability. Picturea ring of atoms each which posses a quantum degree of freedom (spin), whichwith our convention can point in two directions, up or down and which forma complete basis for the Hilbert space describing the spin-1/2 particle. Thestudy of spin chains naturally arises in the study of magnetism (particularlyferromagnetism). Ferromagnetism is the basic mechanism which certain ma-terials (such as Iron) form permanent magnets or are attracted to magnets.An example of ferromagnetism is a refrigerator magnet used the hold noteson the refrigerator door. It is the only type that creates forces strong enoughto be felt and is responsible for the common phenomena of magnetism en-countered in everyday life. Though the study of spin chains arose to solveparticular problems in condensed matter physics, its application has alsoreached String theory particularly in the AdS/CFT correspondence and inhigh-energy QCD scattering.

In ferrous materials such as Iron, the individual electrons that generatetheir weak fields will ‘combine’ to generate a strong magnetic field and thuswhat we experience as ferromagnetism. At absolute zero temperature, aferromagnet reaches the state of lowest energy (ground state energy) inwhich all the atomic spins align. As the temperature increases more andmore spins deviate from the common direction which can be thought ofexcitations from the ground state caused by a magnon.2 Thus there is anincrease in the internal energy and reducing the net magnetization. Thus

2The concept of a magnon was introduced in 1930 by Felix Bloch in order to explainthe reduction of spontaneous magnetization in a ferromagnet. A magnon can be thoughtof as a quantized spin wave. We will introduce the concept of a magnon in chapter 3 interms of the low-lying excitations from the ground state.

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magnetism arises because the state in which all atomic spins align has lessenergy than disaligned spins. We will begin by considering the diatomicHydrogen Molecule H2 which contains two Hydrogen atoms3. What wefind is that the total internal energy of the system is actually greater thanthe sum of the individual atomic energies. This ‘extra’ energy comes fromthe Coloumb interaction between the protons and electrons. By invokingthe Pauli exclusion principle for the total wave function and making a fewarguments, we will then derive the main system of study in this project -the XXX Heisenberg Hamiltonian. This Hamiltonian describes the nearest-neighbour interaction between electron spins of a linear chain of N atoms.

1.4 Notation, Conventions and Concepts from QM

Unless stated otherwise we will use natural units i.e. set ~ = 1.For Spin-1/2 particles, we can represent the Spin operators in terms of

Pauli matrices:

Sα =1

2σα, α ∈ {x, y, z}

σx =

(0 11 0

), σy =

(0 −ii 0

), σz =

(1 00 −1

)[σa, σb] = 2iεabcσc, (εabc = 1)

{σa, σb} = 2δabI

σaσb = δabI + iεabcσc

trσα = 0

It is convenient to specify the eigenkets of ~S2 and Sz, given a state withspin s and z-component m of the spin:

~S2|s,m〉 = s(s+ 1)|s,m〉

Sz|s,m〉 = m|s,m〉

The possible values for s and m are:

s = 0,1

2, 1,

3

2, · · · , m = −s,−s+ 1, · · · , s− 1, s

In many applications there is no confusion about the spin s of the particlesand therefore the index is dropped. For a spin-1/2 particle s = 1/2, we have

3Each with one electron and one proton

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two possible states |1/2〉 and | − 1/2〉 which is commonly denoted as | ↑〉and | ↓〉.

Tensor productLet A and B be two n × n matrices with elements Aij , B

kl . The upper

index denotes a row and the lower index a column. The tensor productA⊗B is a n2 × n2 matrix elements:

(A⊗B)ikjl = AijBkl

Where i, j are the block indices and k, l the intrinsic indices. For exampleif A and B are 2× 2 matrices with

A =

(a11 a12

a21 a22

), B =

(b11 b12

b21 b22

)Then:

A⊗B =

(a11B a12B

a21B a22B

)=

a11b11 a11b12 a12b11 a12b12

a11b21 a11b22 a12b21 a12b22

a21b11 a21b12 a22b11 a22b12

a21b21 a21b22 a22b21 a22b22

Properties:

det (A⊗B) = (det (A))2(det (B))2

tr (A⊗B) = tr (A) tr (B)

(A⊗B)† = A† ⊗B†

(A+ C)⊗B = A⊗B + C ⊗B

A⊗ (B + C) = A⊗B +A⊗ C

(A⊗B)(C ⊗D) = AC ⊗BD

We also have the notion for a vector tensor product, given x, y ∈ V = C2:

x⊗ y =

(x1

x2

)⊗(y1

y2

)=

(x1y

x2y

)=

x1y1

x1y2

x2y1

x2y2

which is an element of the tensor product space V ⊗ V = C2 ⊗ C2

Let V1 ⊗ V2 ⊗ · · · ⊗ VN be a dN - dimensional vector space. We caninterchange Vj with Vk by means of the permutation matrix Pj,k:

Pj,k(Vj ⊗ Vk) = (Vk ⊗ Vj)

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with property P2j,k = 1. When Vj are a 2-dimensional space e.g. C2 as will

be the case for the discussion in this project, the permutation matrix canbe represented by:

P =1

2(I⊗ I +

∑α

σα ⊗ σα)

which is a permutation in C2 ⊗ C2 i.e. Given x, y ∈ C2 we have:

P (x⊗ y) = y ⊗ x

Now if we use a basis for C2 ⊗ C2 which we will define in the next section,we can see that the permutation operator P can be written explicitly as:

P =

1 0 0 00 0 1 00 1 0 00 0 0 1

Note: Pi,j permutes the spin at site i and j. We also have the followingproperties:

tra Pa,b = trb Pa,b = I

Pa,b = Pb,a

Pn,pPn,q = Pp,qPn,p = Pn,pPp,q

P ∗ = P P 2 = I

The notation for any operator A ∈ End(· · · ), where · · · is the tensorproduct of many vector spaces, means that the operator acts as A in thetensor product space · · · and trivially in any other space.

It will be useful to define the Riemann zeta function ζ(s) since it will beused for the correlation functions in Chapter 5. The Riemann zeta functionis defined as:

ζ(s) =∞∑n=1

1

ns, <(s) > 1

Alternative the Riemann zeta function can be expressed in terms of thealternating zeta series ζa as:

ζ(s) =1

1− 21−s ζa(s), s 6= 1,

where

ζa(s) =∑n>0

(−1)n−1

ns.

The alternating zeta function is well defined for s = 1 and has value ζa(1) =ln 2 unlike the Riemann zeta function which has a pole at s = 1.

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Chapter 2

Heisenberg Spin Chain

In this chapter, we shall derive the Hamiltonian for the Heisenberg SpinChain by considering the Coloumb Hamiltonian. First we will state the setof observables and the vector space on which the operators act. The materialin this chapter is adapted from [1],[2],[3] and [4]. An observable is, roughlyspeaking, any measurable property of a physical system such as: position,spin, energy, momentum or angular momentum. In Quantum Mechanics,each observable is represented by a maximally Hermitian linear operatoracting on the state space. Each eigenstate of an observable corresponds toan eigenvector of the operator, and the associated eigenvalue corresponds tothe value of the observable in that eigenstate. The observable in questionfor the next part is given by the spin operators. For example, for a spin-1/2,Sz affects the measurement in the z direction and has eigenvalues ±~/2,where if we set ~ = 1 simply gives ±1/2. These correspond the eigenstatespointing “up” or “down”, as we shall see shortly.

2.1 Basis for observables

For N = 1, the case where there is one particle in the chain is trivial in thecontext of the Spin Chain. It is still important to state the set of observablesas we will use this to construct the set of observables for N > 1.

The basis for observables for N = 1 are given by the identity I and thefirmilar Pauli Matrices ~σ, that is: {I, ~σ} = {I, σx, σy, σz}

These operators act on a two-dimensional complex vector space V ∼= C2,for our consideration these will operate on the space spanned by the “up”and “down” Spinors:

| ↑〉 =

(10

), | ↓〉 =

(01

)For N = 2 we need to construct our basic observables by using the notion

of a tensor product. The basic observables at each site are: ~σ1 ≡ ~σ ⊗ I and

9

~σ2 ≡ I⊗~σ. The basis for the observables at each site are therefore given by:{I⊗2, ~σi} for i = 1, 2. where I⊗2 = I⊗ I.

These act on the tensor product space V1 ⊗ V2∼= C2 ⊗C2, again for our

consideration these correspond to the space spanned by the vectors:

| ↑↑〉 = | ↑〉 ⊗ | ↑〉 , | ↑↓〉 = | ↑〉 ⊗ | ↓〉 , | ↓↑〉 = | ↓〉 ⊗ | ↑〉 , | ↓↓〉 = | ↓〉 ⊗ | ↓〉

For general N the basic observables ~σn, n = 1 , 2 , . . . , N are defined by:

~σn =

1↓I ⊗ · · · ⊗ I⊗

n↓~σ ⊗I⊗ · · ·⊗

N↓I ,

which acts nontrivially on the nth space and trivially on the rest. A basisfor the observables at each site are therefore given by: {I⊗N , ~σi} for i =1, 2, · · · , N . These are operators which act on the Hilbert (tensor product)space:

HN =

1↓V1 ⊗ · · ·⊗

n↓Vn ⊗ · · ·⊗

N↓VN∼= C2⊗N ,

with dimension dim(HN ) = 2N . This corresponds to the space spanned bythe 2N elements:

| ↑↑ · · · ↑〉, | ↓↑ · · · ↑〉, | ↓↓ · · · ↑〉, | ↑↓ · · · ↑〉, · · · , | ↓↓ · · · ↓〉,

which consists of the complex linear superpositions of all the different pos-sible states.

2.2 Derivation of Hamiltonian

In this section, we will derive the Hamiltonian for the Heisenberg Spin Chainbased on considerations of the Coulomb interaction Hamiltonian betweenidealized charged bodies. There are many variants of this derivation (See[1],[2],[3] for a formal treatment) that can be found in numerous papersand textbooks and as such I will give quick overview based on my ownunderstanding. Since we are talking about spin-1/2 particles, it is useful toconsider the electron. We will particularly be interested in how the totalenergy is affected by the electron dynamics. By definition, atoms have nooverall electrical charge and so must contain an equal number of protons andelectrons.1 For this reason, when we study electron systems in atoms andsolids we must also take into account the various electrostatic interactionsbetween the electrons and protons, we will see how this comes into playwhen we deal with two-particle wave function (atoms with one un-pairedelectron next to each other). Well-separated atoms are described by atomic

1An atom can gain or lose electrons, becoming an ion which is nothing more than anelectrically charged atom. Adding or removing electrons from an atom does not changethe element, just its net charge.

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wave functions, but in a molecule or solid the wave functions overlap. Theamount they overlap will depend on the spin orientation of the electrons,if they posses the same spin then by the Pauli exclusion principle the twoelectrons cannot be in the same quantum state and so their wave functionswill not remix as much. If the spin is different between the electrons, theirwave functions will overlap more and the electrons have a greater probabilityof being near to each other, this means that electrons with opposite spinsposses a higher potential energy. The more the wave functions overlap, thegreater the Coloumb repulsion between them - this is why the ground statefor a ferromagnetic tends to have all parallel spins as they tend to align.

2.2.1 One-particle wave function N = 1

Firstly, let us consider a one-electron system (i.e. the Hydrogen atom). Thisis a famous integrable model in physics for which the solution can be foundexactly. The Hamiltonian will be given by the kinetic energy of the electronand the attractive Coloumb interaction between the proton and the electron.We will neglect the kinetic energy associated with the motion of the protons,assuming that they are relatively stationary because of their large mass.

Figure 1: The Hydrogen atom

We will introduce some labelling which will be useful later on. For eachelectron we will associate a number (n ∈ E = {1, 2, · · · }) and for eachproton we will associate a letter (α ∈ P = {A,B, · · · }). We will denotethe position of each electron by rn ∈ R3, and the position of each protonby rα. We will be particularly working with the magnitude of the distancebetween the particles i, j, this will be denoted by rij = |rij| = |ri−rj| where

i, j ∈ E,P . The kinetic energy for each electron will be given by − h2

2m∇2n

while the protons will be assumed to be stationary (relative to the electrons),the coloumb interaction term between two such particles will be given by±k e2rij where the sign depends on whether it is repulsive or attractive. For

example, for the one-particle wave function we have one-electron and one-proton with positions r1 and rA, with relative magnitude r1A = |r1 − rA|,in this case since there are no other particle labels to consider we can simplycall this r = r1A for convenience. See Figure 1. The Hamiltonian for this

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system will be given by:

H0(r) = − h2

2m∇2 − ke

2

r

Where k = 1/4πε0. The wave function 2 describing this system will be givenas a solution to H0ψ = E0ψ and ψ(r) ∈ L2(R3).

The total wave function Ψtot for the particle is an element of the Hilbertspace given by the tensor product of the spatial and spin wave functionspaces, i.e. the Hilbert Space is: Hn = L2(R3) ⊗ Vn. Here Vn is a vectorspace spanned by | ↑〉 and | ↓〉 as in the previous section.

2.2.2 Two-particle wave function N = 2

Figure 2: The Helium atom

Now we wish to consider a two-electron system. One such system isthe Helium atom, which contains a doubly positively charged nucleus. SeeFigure 2.

Figure 3: The HydrogenMolecule (H2), we will assume~RAB to be fixed.

For our discussion, since we would like to study the nearest neighbourinteraction between a linear chain of atoms which will give rise to magnetism,we will consider the more general diatomic Hydrogen molecule (H2). It isthe smallest molecule that consists of two Hydrogen atoms. Like Helium, aHydrogen molecule also has two electrons, and so the intermolecular forcesare going to be small - but not as small as Helium. In the Hydrogen molecule,you have two atoms that you can distribute the charge over.

The Hamiltonian will be given by the sum of the Hamiltonians for theone-electron system (for each of the two Hydrogen atoms) and an additional

2The wave function can be determined by the time-independent Schrdinger equation,which is the eigenvalue equation for the Hamiltonian

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term known as the interaction or exchange Hamiltonian, which will encodethe electrostatic Coloumb interaction between the electrons and protons, andeach proton with each corresponding neighbouring electron. See Figure 3.

The Hamiltonian is:

H = H0(r1A) +H0(r2B) +Hexch

where

H0(rnα) = − h2

2m∇2n − k

e2

rnα

and

Hexch = k(− e2

r1B− e2

r2A+e2

r12+

e2

rAB)

and the unperturbed hydrogen atom wave functions ψ1 and ψ2 satisfythe equations:

H0(rnα)ψn(rn) = E0ψn(rn)

We can use these as building blocks to construct symmetric and an-tisymmetric wave functions using the Heitler-London approximation3 (see[3]):

ΨS(r1, r2) =1√2

(ψ1(r1)ψ2(r2) + ψ2(r1)ψ1(r2))

ΨA(r1, r2) =1√2

(ψ1(r1)ψ2(r2)− ψ2(r1)ψ1(r2))

or in compacted notation:

Ψ±(r1, r2) =1√2

(ψ1(r1)ψ2(r2)± ψ2(r1)ψ1(r2))

where ψ1 and ψ2 are wave functions in L2(R3) determined by H0. We cancalculate this energy by using:

E± =〈Ψ± |HΨ± 〉〈Ψ± |Ψ± 〉

We can calculate this by plugging in Hamiltonian and the energy for theunperturbed wave functions. Firstly, calculating the normalization integralin the dominator of E±:

⟨Ψ±

∣∣Ψ± ⟩ =

∫∫Ψ∗±(r1, r2)Ψ±(r1, r2) d3r1 d3r2 = 1± α2

3Based on the Calculation made by Walter Heitler and Fritz London on the hydrogenmolecule (H2) in 1927, it plays an important role in the context of quantum chemistry.

13

Where we have used: ∫|ψi(r)|d3r = 1 i = 1, 2

and α is defined to be the overlap integral:

α ≡∫ψ∗1(r)ψ2(r) d3r

The numerator in expression for E± can be found in a simular manner,after some manipulation:⟨Ψ±

∣∣HΨ±⟩

=

∫∫Ψ∗±(r1, r2)HΨ±(r1, r2) d3r1 d3r2 = 2E0(1±α2) +V ±U

E0 is the one- particle energy defined as before. V and U are called the“Coulomb integral” and “exchange integral”, respectively, defined by:

V ≡∫Hexch|ψ1(r1)|2|ψ2(r2)|2 d3r1 d3r2

U ≡∫ψ∗1(r2)ψ∗2(r1)Hexchψ1(r1)ψ2(r2) d3r1 d3r2

The energy is therefore given by:

E± = 2E0 +V ± U1± α2

Note that α, V and U need to be calculated. This result shows thatthe total energy of the system is in fact greater than the individual atomicenergies.

2.2.3 N Spin chain

Since we are only considering the nearest neighbour interaction, it suffices toconsider the interaction for N = 2, then to generalise for any N by summingup the “interaction” for all possible nearest neighbours. The Hilbert spacefor the total wave function of the Spin Chain will be given by the tensorproduct of all the individual Hilbert spaces for each particle, i.e. for N = 2the total wave function Ψtot is an element of Htot = H1 ⊗H2 = hspatial ⊗hspin, where hspatial ∼= L2(R3) ⊗ L2(R3) is the spatial component of Htot,and hspin ∼= V1 ⊗ V2

∼= C4 is the spin component of Htot.The Pauli exclusion principle states that the total wave function for two

identical fermions is anti-symmetric with respect to the exchange of theparticles.4 This means when we interchange H1 with H2 the total wavefunction is replaced by its negative. The total wave function Ψtot is an

4Definition taken from wiki

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eigenstate of the total Hamiltonian Htot and therefore we can express itas the product Ψspatial ⊗ χspin, where Ψspatial is clearly an eigenstate ofHtot in hspatial and χspin a vector in hspin. We will have symmetric andantisymmetric wave functions for both the spatial and spin components,we have already introduced the symmetric and antisymmetric spatial wavefunctions from the Heitler-London approximation.

The symmetric and antisymmetric spin wave functions are given by:

χsymspin = χS(s1, s2) = | ↑↑〉, 1√2

(| ↑↓〉+ | ↓↑〉), | ↓↓〉

χantispin = χA(s1, s2) =1√2

(| ↑↓〉 − | ↓↑〉)

These are also called the triplet and singlet states. Note these spin wavefunctions are eigenstates of the total spin operator ~S2, with eigenvalues 2and 0 respectively.

It is clear from the Pauli exclusion principle that the symmetric spatialwave function will be associated to the antisymmetric spin wave function andvice versa, in order for the total wave function to always be antisymmetricwith respect to the exchange. This will allow us to associate each spin wavefunction with an energy (calculated from the corresponding spatial wavefunction) and hence construct the spin Hamiltonian Hspin. This will beformally equivalent to the Coloumb Hamiltonian H and will give the sameenergy spectrum.

We can calculate the energy of the spatial wave functions, this will giveE+ for the symmetric and E− for the antisymmetric spatial wave function aswe have shown. Note these are also the energies of the total wave function,since H acts only on the spatial wave functions. We can define the exchangeenergy by J = E+ − E−. We can now define the spin Hamiltonian to be:

Hspin = E+ +1

2(E− − E+)~S2 = E+ − J

2~S2

Where

J = −2V α2 − U1− α4

Let us comment on Hspin, it will have energy E+ for the antisymmetric spinwave function and E− for the symmetric spin wave function. As required,this Hamiltonian will produce the same energy spectrum as H but is inde-pendent of the spatial wave function and hence of the coordinate system forthe spatial component. We have formally replaced a Hamiltonian that de-pends only on the spatial component by a Hamiltonian which depends onlyon the spin component - by virtue of the Pauli exclusion principle with theColoumb potential. We can expand the operator ~S2 = (S1 +S2) · (S1 +S2),

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using the properties of the spin operator and the fact that we have s = 1/2for our consideration we obtain:

Hspin =1

4(E+ + 3E−)− JS1 · S2

Since we measure energy differences (energies are relative), we can ignorethe first term that appears in the Hamiltonian as it is a constant.

Hspin = −J ~S1 · ~S2

Now we consider the Spin chain. In a magnetically ordered solid withno external magnetic field, we will sum this Hamiltonian over all pairs of Nparticles. We thus obtain the famous Heisenberg spin Hamiltonian:

H = −∑i,j

Jij ~Si · ~Sj

The Heisenberg model works in higher dimensions as well such as in a latticeof spins but we restrict our attention to 1 + 1 space-time dimensions forthe purposes of integrability. More generally, a spin chain is a chain oflocal degrees of freedom with a local Hilbert space of low dimension, witha Hamiltonian that represents the interaction between a few neighbourseverywhere along the chain. The Heisenberg spin chain is a special case ofthis where we consider the interaction between nearest neighbouring spins.One can thus simplify the Hamiltonian by assuming the exchange integralJij decreases rapidly with increasing distance between the two electrons,essentially being zero except for nearest neighbours. We will also assumethat the exchange integral will be equal for all nearest neighbour pairs ofelectrons. Note we can expand ~Sj · ~Sj+1 = Sxj S

xj+1 + Syj S

yj+1 + SzjS

zj+1 =

Sαj Sαj+1, giving:

H = −JN∑j=0

~Sj ·~Sj+1 = −JN∑j=0

(Sxj Sxj+1+Syj S

yj+1+SzjS

zj+1) = −J

N∑j=0

Sαj Sαj+1

This is called the XXX Heisenberg Spin chain for N particles. Note, onecan also write this in terms of the Pauli matrices as Sαn = σαn/2 for α = x, y, zif one wishes, The Hamiltonian then becomes

H = −J4

N∑j=0

σαnσαn+1

We will however stick to writing the Hamiltonian in terms of the spin op-erator Sαn in what follows. The sign of the exchange integral J indicateswhether parallel or antiparallel spins are energetically preferred, that is ifJ > 0 means we have a parallel (ferromagnetic) spin coupling and for J < 0

16

means we have a antiparallel (antiferromagnetic) coupling. The ground statefor a ferromagnetic will be given by all spins being parallel (i.e. all spinspointing in the same common direction e.g. up or down, or a linear com-bination) and the ground state for an antiferromagnet will be the highestexcited state of the corresponding ferromagnetic. One can generalise theHamiltonian to an anisotropic arrangement, where the magnetization is dif-ferent in all directions. This gives the generalized anisotropic HeisenbergHamiltonian:

H = −N∑j=0

(JxSxj S

xj+1 + JyS

yj S

yj+1 + JzS

zjS

zj+1)

Which is also known as the XY Z Heisenberg Spin Chain. When there isone preferred direction, we align this direction with the z axis (Jx = Jy) andwe obtain the XXZ Heisenberg Spin Chain. The XXX Spin Chain is anisotropic special case of this Hamiltonian (Jx = Jy = Jz), in this report I willmainly be discussing the XXX Spin Chain and briefly stating results forthe XXZ that can be obtained in a simular manner to the XXX spin chainvia the Algebraic Bethe Ansatz. For my project, I will be using the aboverepresentation of the Hamiltonian for the XXX spin chain. It is sometimescommon however in the literature to subtract a constant proportional to theidentity matrix, this will not change the dynamics of the system and merelybrings shifts the energy, in particular the energy of the ground state to 0.

H = −JN∑j=0

(Sxj Sxj+1 + Syj S

yj+1 + SzjS

zj+1 − 1)

Also, when we talk about a closed chain we will also impose the periodicboundary condition SαN+1 = Sα1 .

2.3 Energy spectrum of H

Now that we have derived the XXX Heisenberg Hamiltonian, we can worktowards the problem of finding the energies of the given system. That is,we wish to solve:

H|φ〉 = E|φ〉

This can of course be done by direct diagonalization of the matrix H,but the matrix will become large very rapidly for increasing N . The Hilbertspace will be of dimension 2N and the matrix will be of size 2N × 2N . Moreover, direct diagonalization is not possible when N →∞ when only analyticmethods will work.

17

We seek a more elegant method than direct diagonalization, one suchmethod is called the Bethe Ansatz which was formulated by Bethe in 1931(See [5]).

18

Chapter 3

Bethe Ansatz

In this chapter we will examine the problem of diagnolozing the Hamiltonianby using the Coordinate Bethe Ansatz (CBA) and the using the AlgebraicBethe Ansatz (ABA). The material on the first part of the chapter is basedon [5],[6] and [7].

Let us introduce the spin flip operators S±n ≡ Sxn ± iSyn, these do exactly

what it says on the tin. S+n flips thenth down spin to up (assuming it is

down in the first place), and S−n does the opposite. As a matrix acting onC (i.e. N = 1):

S+ =

(0 10 0

), S− =

(0 01 0

)We get S±n by applying the appropriate tensor product, as defined in Chapter2. We will write the Hamiltonian in terms of the spin flip operators:

H = −JN∑n=1

SαnSαn+1 = −J

N∑n=1

(1

2

(S+n S−n+1 + S−n S

+n+1

)+ SznS

zn+1

)We will keep the sign of J general unless specifically stated otherwise, thoughour considerations will be made for the ferromagnet J > 0 and our “vacuum”(ground state) will be appropriately defined with this in mind. The physicalproperties of the anti-ferromagnet are very different from the ferromagnet,but for the purposes of calculating the eigenstates and energy eigenvaluesare interchangeable (see comments below).

| ↑〉n | ↓〉nS+n 0 | ↑〉nS−n | ↑〉n 0Szn

12 | ↑〉n −1

2 | ↑〉n

Table 3.1: Action of spin operators on the basis vectors |σ1 . . . σN 〉 with σn =↑, ↓.| ↑〉n = | . . . ↑ . . .〉, | ↓〉n = | . . . ↓ . . .〉.

19

The spin operators also satisfy the following su(2) spin algebra commu-tation relations:

[San, Sbm] = iδn,mε

abcScm, [Szm, S±n ] = ±δn,mS±m, [S+

m, S−n ] = 2δn,mS

zm

with a, b, c = x, y, z. For the ferromagnet J > 0, the ground state of thesystem will be given by all the spins being aligned (See Chapter 1), we willchoose this to be |0〉 = | ↑↑ . . . ↑〉 = ⊗Nn=1| ↑〉n. We will define the basisvectors for r down spins to be given by:

|m1,m2, . . . ,mr〉 = | ↑〉1 ⊗ · · · | ↓〉m1 ⊗ · · · | ↓〉mr ⊗ · · · | ↑〉N= S−m1

S−m2. . . S−mr

|0〉 = S−M |0〉

Where M = {m1,m2, . . . ,mr} and mi ≤ N ∀i = 1, . . . , r denotes theposition of the down spin with respect to the ground state.

A general eigenstate |ψ〉 will be constructed as a linear combination ofthese basis vectors, we will later see this more concretely and in particularthe exact form of the general eigenstate in accordance with the Bethe Ansatz.It can also be shown (see [7]) that the eigenvalues of H are at least −JN/4which will infact turn out to be the ground state energy of H for J > 0.

Though the sign of the coupling constant J will not affect the diagonal-ization procedure in what follows, it is important to make a few commentsbefore we begin. For the anti-ferromagnet J < 0, |0〉 is not a true groundstate (vacuum) of the system i.e. a state with the lowest energy, it is actuallythe state of highest energy due to the change of sign. Instead it is a knownas a pseudo-vacuum and it will nevertheless be useful as a reference state forthe true vacuum. The true ground state is non-trivial in terms of excitationsin the pseudovacuum and requires some work to be identified. All the eigen-vectors remain the same for both the ferromagnet and anti-ferromagnet, butthe energy eigenvalues have opposite sign.

3.1 Coordinate BA

We will start by using the Coordinate Bethe Ansatz which was originallydeveloped by Bethe in 1931 (See [6]), which will essentially allow us to writethe general eigenstate of H in an elegant form. We will investigate the solu-tions to the Heisenberg Hamiltonian, by considering the spin deviation fromthe ground state. For completeness, I will go through a simular approachtaken in [6], first by considering the low-lying states (r = 1, 2) then leadingup the general case of r down spins. I will use the same notation as in [5]to discuss these low-lying states in terms of the spin flip operator.

3.1.1 Symmetries of Heisenberg Model

We will begin by examining the symmetries of the Heisenberg Hamiltonian,which will be essential in the application of the Bethe Ansatz. The model has

20

a full rotational SU(2) symmetry, though we will only require the rotationalsymmetry about the z-axis in spin space, which will take as the quantizationaxis.

Claim: The conserved quantity which arises from this symmetry is thez-component of the total spin Sz = SzT =

∑Nn=1 S

zn with quantum number

SzT = N/2− r

Proof. We need to consider the action of Sz on the basis vectors |m1,m2, . . . ,mr〉,by using the rules of Table 1.1:

Sz|m1,m2, . . . ,mr〉 =N∑n=1

Szn|m1,m2, . . . ,mr〉

=N∑n=1

SznS−M |0〉

SznS−M |0〉 =

{−S−M |0〉/2 n ∈MS−M |0〉/2 n /∈M

Note for n = 1, . . . , N , it is evident that n ∈M r times since we have rdown spins by definition of the basis vector and n /∈M N − r times.

N∑n=1

SznS−M |0〉 =

(N − r

2− r

2

)S−M |0〉

=

(N

2− r)|m1,m2, . . . ,mr〉

It is now easy enough to show [H,Sα] = 0, for which α = z gives the result:

[H,Sα] = −JN∑

n,m=1

[SβnSβn+1, S

αm] = −J

N∑n,m=1

([Sβn , S

αm]Sβn+1 + Sβn [Sβn+1, S

αm])

= −iJN∑

n,m=1

(δn,mεαβγSβnS

γn+1 − δn+1,mε

αβγSβnSγn+1) = 0

21

One can also verify this by a direct computation, by using what wehave shown above for the quantum number of Sz. Given an eigenstate |ψ〉constructed from the basis vectors, such that H|ψ〉 = E|ψ〉 and Sz|ψ〉 =(N2 − r

)|ψ〉:

[H,Sz]|ψ〉 = (HSz − SzH) |ψ〉

=

(H

(N

2− r)− SzE

)|ψ〉

=

(E

(N

2− r)− E

(N

2− r))|ψ〉

= 0

Since |ψ〉 is an arbitrary eigenstate, we conclude that [H,Sz] = 0 as required.

The model also has a discrete translational symmetry i.e. the invarianceof H with respect to discrete translation by any number of lattice spacings.This can be realised by considering the translation operator T which isdefined by:

T |m1,m2, . . . ,mr〉 = |m1 − 1,m2 − 1, . . . ,mr − 1〉 = |m2,m3, . . . ,mr,m1〉

Which basically shifts the spin chain by one position. Any eigenvector |ψ〉 ofT will satisfy T |ψ〉 = eip|ψ〉 for some momentum p. In fact the interpretationof p as momentum follows from the eigenvalue of T . It is a conserved quantityand hence commutes with H, i.e. [H,T ] = 0. These are the importantsymmetries for the application of the Bethe Ansatz. The model also has aReflection symmetry on the lattice.

3.1.2 Eigenstates of H

Let us first discuss the case where r = 0, this will be given by the (pseudo)-vacuum of H which defined to be |0〉 = | ↑↑ . . . ↑〉 at the beginning of thechapter. Note this is a arbitrary choice and we could have just as well chosenall spins down which will give the same energy. This can be understood interms of the SU(2) symmetry of the model, since rotating the system willleave the system invariant.

Claim: It is an eigenstate H|0〉 = E0|0〉, with energy E0 = −JN/4.

22

Proof. Using Table 1.1 the result is evident since we immediately have:

H|0〉 = −JN∑n=1

(1

2

(S+n S−n+1 + S−n S

+n+1

)+ SznS

zn+1

)|0〉

= −JN∑n=1

SznSzn+1|0〉

= −JN∑n=1

1

4|0〉 = −JN

4|0〉

An excitation in the chain will be thought of as a spin deviation fromthe ground state, i.e. r = 1 corresponds to an excited state with one spindown with respect to the ground state. The N basis vectors in the invariantsubspace with one down spin are labelled by the position of the flipped spin:

|n〉 = S−n |0〉 n = 1, . . . , N

Where S−n is the spin flip operator that changes a spin up to a spin downat position n. These are clearly not eigenstates of H, but one can constructa translationally invariant eigenvector from the |n〉 by writing:

|ψ〉 =N∑n=1

eipn|n〉

With momentum p = 2πm/N,m = 0, 1, . . . , N − 1. This can also beunderstood in terms of the discrete fourier transform of the “position” of thespin down (in analogy with the continuum |p〉 =

∫dx eipx|x〉 from quantum

mechanics).Claim: These vectors are eigenvectors of the translation operator T

with eigenvalue eip and eigenvectors of H with eigenvalue (energy) E =E0 + J(1− cos k).

23

Proof. First we will show the eigenvalue of T :

T |ψ〉 = T

N∑n=1

eipn|n〉

=

N∑n=1

eipnT |n〉

=

N∑n=1

eipn|n− 1〉

=

N∑n=1

eip(n+1)|n〉

= eipN∑n=1

eipn|n〉

= eip|ψ〉

As required, |ψ〉 is an eigenstate of T with eigenvalue eip. Now we will showthe eigenvalue of H. Again, we will show this by applying the rules of Table1.1 in a systematic manner:

H|ψ〉 = −JN∑n=1

(1

2

(S+n S−n+1 + S−n S

+n+1

)+ SznS

zn+1

)|ψ〉

= −JN∑m=1

eipmN∑n=1

(1

2

(S+n S−n+1 + S−n S

+n+1

)+ SznS

zn+1

)|m〉

= −JN∑m=1

eipm(

1

2|m+ 1〉+

1

2|m− 1〉+

(N

4− 1

)|m〉)

= −JN∑m=1

(1

2eip(m−1) +

1

2eip(m+1) +

(N

4− 1

)eipm

)|m〉

= −JN∑m=1

(1

2e−ip +

1

2eip +

(N

4− 1

))eipm|m〉

= −J(

1

2e−ip +

1

2eip +

(N

4− 1

)) N∑m=1

eipm|m〉

= (E0 + J(1− cos p))

N∑m=1

eipm|m〉

= (E0 + J(1− cos p)|ψ〉 = E|ψ〉

By using the identity cos p = (e−ip + eip)/2 and E0 = −JN/4. Note, wehave also used the periodicity condition which allows us to make shifts of

24

the summation variables, which in this case is simply m to bring all thestates to the uniform expression |m〉.

Hence we have the result:

E − E0 = J(1− cos p)

Since we have shown that |ψ〉 is an eigenstate of T with eigenvalue eip, bydefinition it follows that the momentum of the ‘excitation’ can be interpretedas p. The expression above for the eigenvalue (energy) of H is also known asthe dispersion relation between energy E and momentum p. An excitationof the spin chain around the (pseudo)-vacuum |0〉 carrying momentum p iscalled a magnon, which we touched upon briefly in the introduction. Thusa magnon can be thought of a “particle” with momentum p = 2πm/N,m =0, 1, . . . , N − 1 and energy as given above.

For r > 1 this technique of finding the translationally invariant eigenvec-tor by exploiting the symmetries will fail as it will not completely diagonalizethe Hamiltonian. For this reason we will need to use the Bethe Ansatz as apowerful alternative.

For a general eigenstate with r spins down, assuming that the particleswith numbers m1,m2, . . . ,mr corresponds to fixing which atoms point down[5]:

|Ψ〉 =∑

m1,m2,...,mr

a(m1,m2, . . . ,mr)|m1,m2, . . . ,mr〉

where we can write |m1,m2, . . . ,mr〉 = S−m1S−m2

. . . S−mr|0〉. The following

claim was included in the original Bethe paper [6], but the proof is missingfrom the literature in general.

Claim: If we apply the Hamiltonian on this general eigenstate, we willobtain the following consistency equations for the coefficients a(m1,m2, . . . ,mr):

2(E − E0)a(m1, . . . ,mr) = J∑

m′1,...,m′r

[a(m1, . . . ,mr)− a(m′1, . . . ,m′r)] (*),

where particles with numbers m1,m2, . . . ,mr corresponds to fixing whichatoms point down.

Proof. Let us introduce the short hand notation:

|Ψ〉 =∑M

a(M)|M〉 =∑M

a(M)S−M |0〉,

where M = {m1,m2, · · · ,mr}. This will be useful since considering the gen-eral case can get rather tedious and it simplifies the notation, this which will

25

become apparent during the proof. Let us begin straight away by applyingthe Hamiltonian on this general eigenstate:t

H|Ψ〉 = −JN∑n=1

(1

2

(S+n S−n+1 + S−n S

+n+1

)+ SznS

zn+1

)∑M

a(M)|M〉

= −J∑M

a(M)N∑n=1

(1

2

(S+n S−n+1 + S−n S

+n+1

)+ SznS

zn+1

)|M〉

To compute this we need to consider the action of the spin operators on|M〉 = S−M |0〉:∑

n

S+n S−n+1|M〉 =

∑n

S+n S−n+1S

−M |0〉 =

∑M̃

S−M̃|0〉 =

∑M̃

|M̃〉

Where M̃ = {m̃1, · · · , m̃r} is obtained by displacing a ↓ toward the rightin M . We can see this by considering the object Γ1(M) = S+

n S−n+1S

−M |0〉

for different values of n. We see that Γ1(M) vanishes for all n /∈ M orn + 1 ∈ M . The only terms that contribute are terms with neighbouringopposite spins that look like ↓↑ in the spin chain i.e. n ∈M and n+ 1 /∈M ,exchange if mi+1 6= mi + 1 for some i = 1, · · · , r − 1, and m̃i = m1 + 1.Let Λ+(M) be the set of all {m̃1, · · · , m̃r} gotten by displacing a ↓ to in Mtoward the right. i.e. ↓↑ 7→↑↓ and M̃ ∈ Λ+(M).

Similarly, we can consider:∑n

S−n S+n+1|M〉 =

∑M ′′

S−M ′′ |0〉 =∑M ′′

|M ′′〉

Where M ′′ = {m′′1, · · · ,m′′r} is obtained by displacing a ↓ toward the leftin M . We can see this in a simular way as the above, and we see thatthe object in the sum only contributes for neighbouring opposite spins thatlook like ↑↓ in the spin chain i.e. n /∈ M and n + 1 ∈ M , exchange ifmi 6= mi−1 + 1 for some i and m′′i = mi − 1. Let Λ−(M̃) be the set of all{m′′1, · · · ,m′′r} gotten by displacing a ↓ in M toward the left i.e. ↑↓ 7→↓↑ andM ′′ ∈ Λ−(M̃) = {M : M̃ = Λ+(M)}. Combining these results we obtain:

N∑n=1

(S+n S−n+1 + S−n S

+n+1

)|M〉 =

∑M̃

|M̃〉+∑M ′′

|M ′′〉

=∑M ′

|M ′〉

Where M ′ = {m′1, · · · ,m′r} can be obtained by M by exchanging oppo-site neighbouring spins i.e. ↓↑ 7→↑↓ and ↑↓ 7→↓↑. Now we wish to consider:

F (M)|M〉 =N∑n=1

SznSzn+1|M〉 =

N∑n=1

SznSzn+1S

−M |0〉

26

We can denote f(n,M) by:

f(n.M) =

{1 if n ∈M ;−1 if n /∈M.

Then we can write

SznS−M |0〉 =

1

2f(n,M)S−M |0〉

and

SznSzn+1S

−M |0〉 =

1

4f(n,M)f(n+ 1,M)S−M |0〉

then

F (M)|M〉 =N∑n=1

SznSzn+1|M〉 =

1

4

N∑n=1

f(n,M)f(n+ 1,M)|M〉

Now we can compute H|Ψ〉 as above:

H|Ψ〉 = −J∑M

a(M)

(1

2

∑M ′

|M ′〉+ F (M)|M〉

)

Recall in the r = 1 case, we had to apply the periodicity condition insidethe sum which allowed us to make shifts of the summation variables. In thiscase, we will essentially do the same in order to bring all states to the uniformexpression |M〉. Since we are shifting the variables M ′ = {m′1, · · · ,m′r} toM in |M ′〉, we need to compensate by also shifting a(M) appropriately (bythe periodicity condition). When we represent a shift of a ↓ to the right in|M ′〉 (i.e. by M̃), we change a(M) to represent a shift of a ↓ to the left (i.e.by M ′′) and vice versa. Since M ′ is composed of M̃ and M ′′, it is clear thata shift in M ′ in the sum would be compensated by a shifting a(M) to M ′

with M ∈ Λ±(M ′), hence:∑M

a(M)∑M ′

|M ′〉 =∑M

∑M ′

a(M ′)|M〉

Let us consider F (M):

F (M) =1

4

N∑n=1

f(n,M)f(n+ 1,M) =1

4(#parallel−#anti-parallel)

Since N = #parallel + #anti-parallel, this becomes:

F (M) =1

4(N − 2#anti-parallel)

Note:#anti-parallel =

∑M ′

1

27

and therefore

F (M) =N

4− 1

2

∑M ′

1

Putting the above together we obtain:

H|Ψ〉 = −J∑M

(1

2

∑M ′

a(M ′) + a(M)F (M)

)|M〉

= E∑M

a(M)|M〉

This now gives the relation between the coefficients a(M) as:

−J

(1

2

∑M ′

a(M ′) + a(M)F (M)

)= Ea(M)

substituting our F (M):

−J2

∑M ′

a(M ′)− Ja(M)(N

4− 1

2

∑M ′

1) = Ea(M)

Rearranging as in the form of (∗), and substituting E0 = −JN/4, we obtainthe result:

2(E − E0)a(M) = J∑M ′

[a(M)− a(M ′)]

As required.

This useful since to check that an eigenstate satisfies H|Ψ〉 = E|Ψ〉, weneed the coefficients to satisfy (∗). We also have the periodicity conditionfor the coefficients as a(m1, · · · ,mi, · · · ,mr) = a(m1, · · · ,mi +N, · · · ,mr).

Now we will introduce the Bethe Ansatz, which is the “assumed” generalform of the coefficients a(m1,m2, . . . ,mr) in terms of r quasi-momenta kj :

a(m1,m2, . . . ,mr) =∑π∈Sr

Aπ exp

i r∑j=1

kπ(j)mj

The sum π ∈ Sr is over all r! permutations of the labels {1, 2, · · · , r}. Thecoefficients Aπ ∈ C depending on the permutation π need to be determinedand will be given in terms of the scattering phase. We will work throughthe examples r = 1 and r = 2 to verify the Bethe Ansatz and to give a moreexplicit form for Aπ.

For r = 1, we obtain:

|Ψ〉 =N∑m=1

a(m)|m〉 a(m) = Aeikm

28

If we let A = 1, then we get the same eigenstate as we discussed before. Onecan verify this by (∗) since we get:

2(E − E0)a(m) = J [2a(m)− a(m− 1)− a(m+ 1)]

for n = 1, · · · , N and a(n+N) = a(n). This immediately implies as before:

a(m) = eikm, k =2πn

Nn = 0, · · · , N − 1

The energy of the one-magnon state is therefore:

E − E0 = J(1− cos k)

which agree’s with the Bethe Ansatz.For r = 2, we obtain:

|Ψ〉 =N∑m=1

a(m1,m2)|m1,m2〉

a(m1,m2) = Aei(k1m1+k2m2) +A′ei(k2m1+k1m2)

We wish to determine the conditions on A,A′ given that this Ansatz satisfies(∗). We have two cases that we need to consider, either the two down spinsare separate from each other or nearest neighbours. Using (∗) we get thefollowing condition for the former:

2(E − E0)a(m1,m2) = J [4a(m1,m2)− a(m1 − 1,m2)− a(m1 + 1,m2)

− a(m1,m2 − 1)− a(n1,m2 + 1)], m2 6= m1 + 1

And for the latter:

2(E − E0)a(m1,m2) = J [2a(m1,m2)− a(m1 − 1,m2)− a(m1,m2 + 1)],

m2 = m1 + 1

The first group of equations is automatically satisfied for arbitraryA,A′, k1, k2.The second group can be satisfied by choosing A and A′ such that we havethe following eigenstate conditions:

2a(m1,m1 + 1) = a(m1,m1) + a(m1 + 1,m1 + 1) (**)

which is obtained by subtracting the first group by the second for m2 =

29

m1 +1. To obtain the energy, we can substitute the Ansatz in (∗) to obtain:

2(E − E0)(Aei(k1m1+k2m2) +A′ei(k2m1+k1m2)

)= J

[4(Aei(k1m1+k2m2) +A′ei(k2m1+k1m2)

)−(Aei(k1m1+k2m2)eik1 +A′ei(k2m1+k1m2)eik2

)−(Aei(k1m1+k2m2)e−ik1 +A′ei(k2m1+k1m2)e−ik2

)−(Aei(k1m1+k2m2)eik2 +A′ei(k2m1+k1m2)eik1

)−(Aei(k1m1+k2m2)e−ik2 +A′ei(k2m1+k1m2)e−ik1

)]We see that the dependence on A and A′ cancel out completely as we expectand we get the following equation for the energy:

E − E0 = J2∑j=1

(1− cos kj) = J(2− cos k1 − cos k2)

We observe that the energy of the two-magnon state appears to be equalto the sum of the energies of the one-magnon state. This observation isquite remarkable as it appears that the energy is additive. This shows thatmagnons essentially behave themselves as free particles in a box. We willnow use the condition (∗∗) to determine the restrictions placed on A and A′

Labelling n1 = n and substituting our Ansatz into this condition we get:

Aei(k1+k2)n +A′ei(k1+k2)n +Aei(k1+k2)(n+1) +A′ei(k1+k2)(n+1)

− 2(Aei(k1n+k2(n+1)) +A′ei(k2n+k1(n+1))

)= 0

This allows us to determine the ratio in terms of a phase angle θ12 = −θ21 ≡θ as:

A

A′≡ eiθ = −e

i(k1+k2) + 1− 2eik1

ei(k1+k2) + 1− 2eik2(***)

i.e. by setting A = eiθ12/2 and A′ = eiθ21/2. This requirement is incorporatedinto our Ansatz as phase factors:

a(m1,m2) = ei(k1m1+k2m2+ 12θ12) + ei(k2m1+k1m2+ 1

2θ21)

= ei(k1m1+k2m2+ 12θ) + ei(k2m1+k1m2− 1

2θ)

Where the phase angle θ depends on the undetermined the as yet undeter-mined k1, k2 via:

2 cotθ12

2= 2 cot

θ

2= cot

k1

2− cot

k2

2

30

The periodicity condition requires:

a(m1,m2) = a(m2,m+ 1 +N)

Inserting this into our Ansatz we obtain:

ei(k1m1+k2m2+ 12θ) + ei(k2m1+k1m2− 1

2θ)

= ei(k2(m1+N)+k1m2+ 12θ) + ei(k1(m1+N)+k2m2− 1

2θ)

Since this is true for all m1 and m2 we have:

eik1N = eiθ, eik2N = e−iθ

If we take the logarithm of these we obtain:

Nk1 = 2πI1 + θ, Nk2 = 2πI2 − θ

Where the integers Ij ∈ {0, 1, · · · , N − 1} are called the Bethe quantumnumbers. These numbers are useful to distinguish eigenstates with differentphysical properties. The total momentum is given by the sum of all quasi-momenta:

p = k1 + k2 =2π

N(I1 + I2)

p is a true integration constant of the problem. This is the same momentumassociated to the translation operator T as we saw earlier, since displacinga down spin by one place on the lattice does not change the physics of thesystem. Now that we have discussed the case for r = 2 i.e. two downspins, we can use the concepts that we introduced to general r down spinsin a natural way. The generalization of (∗ ∗ ∗) for the Aπ in terms of thescattering phase θjl = −θlj for each pair (kj , kl) is given by:

Aπ = exp

i

2

∑l<j

θπ(l)π(j)

i.e. the Bethe Ansatz for the coefficients a(m1, · · · ,mr) can be written as:

a(m1,m2, . . . ,mr) =∑π∈Sr

exp

i

2

∑l<j

θπ(l)π(j)

exp

i r∑j=1

kπ(j)mj

=∑π∈Sr

exp

i r∑j=1

kπ(j)mj +i

2

∑l<j

θπ(l)π(j)

If we assume that these coefficients satisfy the consistency equations (∗)then the energy depends on k1, k2, · · · , kr given as a straightforward gener-alization of the two-particle case:

E − E0 = J

r∑j=1

(1− cos kj)

31

The eigenstate condition analogous to (∗∗) is given by:

2a(m1, · · · ,mj ,mj + 1, · · · ,mr) = a(m1, · · · ,mj ,mj , · · · ,mr)

+ a(m1, · · · ,mj + 1,mj + 1, · · · ,mr)

These conditions relate every phase angle θjl to the (as yet undetermined)kj as a generalization of (∗ ∗ ∗):

eiθjl = −ei(kj+kl) + 1− 2eikk

ei(kj+kl) + 1− 2eikl

or in real form is given as

2 cotθjl2

= cotkj2− cot

kl2, j, l = 1, · · · , r

3.2 BAE

Here we will derive the Bethe Ansatz Equations by considering the period-icity condition on the Ansatz given for the coefficients a(m1,m2, · · · ,mr).By the periodicity condition (closed chain)

a(m1,m2, · · · ,mr) = a(m2, · · · ,mr,m1 +N)

Which from the general eigenstate (Bethe Ansatz) gives:

r∑j=1

kπ(j)mj +1

2

∑l<j

θπ(l)π(j)

=1

2

∑l<j

θπ′(l)π′(j) − 2πIπ′(r) +r∑j=2

kπ′(j−1)mj + kπ′(r)(m1 +N)

The permutations π′(j) and π(j) on the left and right are related by:

π′(j − 1) = π′(j), j = 2, · · · , r

π′(r) = π(1)

The terms not involving the index π′(r) = π(1) cancel and we are thus leftwith the following set of r equations:

Nkj = 2πIj +∑l 6=j

θjl, j = 1, · · · , r (+)

Claim: If we introduce the rapidities λj to parametrize the momentakj :

λj =1

2cot

kj2

32

then we can rewrite (+) as the Bethe Ansatz equation (BAE):(λj + i/2

λj − i/2

)N=

r∏l=1,l 6=j

λj − λl + i

λj − λl − i

Proof. Before we begin the proof, we will need the following simple trigono-metric identity:

eiφ(cot (φ/2)− i) = (cot (φ/2) + i)

We can show this by using Euler’s identity with:

eiφ =eiφ/2

e−iφ/2=

cos (φ/2) + i sin (φ/2)

cos (φ/2)− i sin (φ/2)=

cot (φ/2) + i

cot (φ/2)− i

and hence the result. Now if we take multiply both sides of (+) by i andtake the exponential, we get:

eiNkj = ei∑

l 6=j θjl

Since ei2πIj = 1, as Ij is an integer. Now, applying the trigonometric identityto eikj and substituting the rapidity λj :

eikj =2λj + i

2λj − i=λj + i/2

λj − i/2

and so the LHS becomes:

(eikj )N =

(λj + i/2

λj − i/2

)NWe can also rewrite the RHS as:

ei∑

l 6=j θjl =r∏

l=1,l 6=jeiθjl

Again, by applying the trigonometric identity on eiθjl and substituting λjwe obtain:

eθjl =λj − λl + i

λj − λl − iWhich then gives using the above:(

λj + i/2

λj − i/2

)N=

r∏l=1,l 6=j

λj − λl + i

λj − λl − i

As required.

33

3.3 Algebraic BA

Here we will solve the XXX Heisenberg Spin chain using a variant of theBethe Ansatz known as the Algebraic Bethe Ansatz (ABA). The ABA ismore powerful since it will allow us to derive properties of the system suchas integrability, it also allows us to more easily generalise the XXZ spinchain in which we state results for at the end of the chapter. Before webegin by applying the ABA we will first introduce some concepts which willbe essential in its application. The material in this section will be adaptedfrom [7],[8] and [9].

3.3.1 Integrability

Let us define what it means for a Quantum system to be integrable.A classical system with N degrees of freedom, described by a Hamilto-

nian H is Liouvelle integrable if there exists N conserved charges Qi withzero poisson bracket in involution i.e. {Qi, Qj} = 0 and the HamiltonianH is one of the charges. Each of the conserved charges yields a conserva-tion law that can be solved (integrated) to fix all the independent degreesof freedom. This notion of integrability can be carried over to Quantumsystems.

A Quantum system withN degrees of freedom defined by a Hamiltonian-operatorH, will be called integrable if there existsN local conserved chargesQi that commute i.e. [Qi, Qj ] = 0 and the Hamiltonian H is one of them.Therefore, all the charges can be diagonlalized simultaneously and the set ofcomplete eigenstates with corresponding eigenvalues can be found exactly.1We will later see how we can compute these conserved charges.

3.3.2 Yang-Baxter equation

The Yang-Baxter equation is an essential tool in showing the integrabilityof the Heisenberg model. There are different conventions for the form of theequation, we will briefly look at the different conventions and finally stickwith the most common. The definition involves the nth local quantum spaceVn = C2 and the auxiliary space A, which we will also take to be C2 toallow us to make sense of the permutation operator as defined above actingon both spaces. Given λ, µ, χ ∈ C, and u, v ∈ {λ, µ, χ}, the invertible Rmatrix Ra1,a2(λ1, λ2) ∈ End(A1 ⊗A2) satisfies:

Ra1,a2(λ, µ)Ra1,a3(λ, χ)Ra2,a3(µ, χ) = Ra2,a3(µ, χ)Ra1,a3(λ, χ)Ra1,a2(λ, µ)

1This definition of quantum integrability is not entirely correct and it is still a subjectof research on how to make a better definition. For example, the XXX model actuallyhas 2N degrees of freedom since each lattice site can be occupied by a spin up or a spindown electr on. But in this definition we only consider the lattice site of the XXX modelof having one degree of freedom.

34

On End(A1⊗A2⊗A3). The index (a1, a2) indicates that it acts nontriviallyon two copies of the auxiliary spce. Assume Rai,aj (u, v) = Rai,aj (u − v),then the Yang-Baxter equation can be written as:

Ra1,a2(λ−µ)Ra1,a3(λ−χ)Ra2,a3(µ−χ) = Ra2,a3(µ−χ)Ra1,a3(λ−χ)Ra1,a2(λ−µ)

If we let χ = 0 or by relabelling u− χ→ u, we obtain:

Ra1,a2(λ− µ)Ra1,a3(λ)Ra2,a3(µ) = Ra2,a3(µ)Ra1,a3(λ)Ra1,a2(λ− µ)

The Lax operator La,n(λ) ∈ End(A ⊗ Vn) satisfies the FCR (Interwiningequation):

Ra1,a2(λ− µ)La1,n(λ)La2,n(µ) = La2,n(µ)La1,n(λ)Ra1,a2(λ− µ)

On End(A1 ⊗A2 ⊗ Vn). The index (a, n) indicates that it acts nontriviallyon the auxiliary space and the nth local quantum space. We also havethe commutativity of the Lax operator with completely different indices(Ultralocality condition):

[La1,n(λ), La2,m(µ)] = 0

On End(A1 ⊗A2 ⊗ Vn ⊗ Vm). This is obvious since Lai,n acts trivially withLaj ,m given both auxiliary spaces are distinct i.e. i 6= j and both localquantum spaces are distinct i.e. n 6= m.

For the XXX Heisenberg model in particular, the Lax operator is givenby:

La,n(λ) = λIa ⊗ In + i∑α

Sαn ⊗ σαa

Explicitly as a 2× 2 matrix in the auxiliary space:

La,n(λ) =

(λ+ iSzn iS−niS+n λ− iSzn

)Which is a matrix acting in the auxiliary space A, with entries being oper-ators on the Quantum space Vn. Given that the operator:

P =1

2(I⊗ I +

∑α

σα ⊗ σα)

is a permutation in C2 ⊗ C2, we can rewrite the Lax operator in terms ofPa,n:

La,n(λ) = (λ− i

2)Ia,n + iPa,n

In order for this to satisfy the FCR, we must have:

Ra1,a2(λ) = λIa1,a2 + iPa1,a2 ,

35

which we will refer to as our R-operator (or matrix). It is easy to see thatthe Lax operator and the R-matrix are in a simular form, indeed La,n(λ) =Ra,n(λ− i/2).

The R-operator given explicitly is a solution of the Yang-Baxter equa-tion. One way to see this is explicitly, one can express Ra1,a2(λ) as a matrixin an appropriate basis and perform a tedious task in matrix multiplication.

Let us define the Monodromy matrix Ta(λ) ∈ End(A⊗HN ) by:

Ta(λ) = Ta,N (λ) = La,N (λ) · · ·La,1(λ) =N∏n=1

La,N−(n−1)

Claim: This matrix also has a FCR which is in the same form as the localFCR i.e.

Ra1,a2(λ− µ)Ta1(λ)Ta2(µ) = Ta2(µ)Ta1(λ)Ra1,a2(λ− µ)

On End(A1 ⊗A2 ⊗HN ). We shall call this the Global FCR.

Proof. We will start with the left hand side of the equation, which we willwrite in terms of the L-operators:

Ra1,a2(λ− µ)La1,N (λ) · · ·La1,1(λ)La2,N (µ) · · ·La2,1(µ)

= Ra1,a2(λ− µ)La1,N (λ)La2,N (µ) · · ·La1,1(λ)La2,1(µ)

Where the last line by the application of the ultralocality condition repeat-edly. Next we will use the local FCR for the L-operators in order to transferthe R-operator from the left to the right of the product i.e.:

· · · = La2,N (µ)La1,N (λ)Ra1,a2(λ− µ) · · ·La1,1(λ)La2,1(µ)

= La2,N (µ)La1,N (λ) · · ·La2,1(µ)La1,1(λ)Ra1,a2(λ− µ)

= La2,N (µ) · · ·La2,1(µ)La1,N (λ) · · ·La1,1(λ)Ra1,a2(λ− µ)

= Ta2(µ)Ta1(λ)Ra1,a2(λ− µ)

Where in the third line we have used the ultralocality condition again inorder to make the monodromy matrix appear.

Note: La,n(λ) is linear in λ i.e. ddλLa,n = Ia,n ∀λ. Since Ta(λ) is a

product of N of these, this implies that the power expansion for Ta(λ) isfinite

Ta(λ) =

∞∑n=1

an(λ− λ0)n =

N∑n=1

an(λ− λ0)n = λN +

N−1∑n=1

an(λ− λ0)n

where

an =1

n!

dn

dλnTa(λ)

∣∣∣λ=λ0

=T

(n)a (λ0)

n!

36

and λ0 = i/2 is the usual parametrization, but it doesn’t matter whichparametrization we use as the expansion is finite. In other words, it isa polynomial of order N with leading coefficient (of λN ) equal to unity.This is obvious by looking at the explicit expression for Ta(λ) but can also

be realised by computing aN and using dN

dλNTa(λ) = N ! Ia,N ∀λ. Let us

compute the coefficient of the next to highest degree aN−1:

aN−1 =1

(N − 1)!

dN−1

dλN−1Ta(λ)

∣∣∣λ=λ0

Which is the coefficient of λN−1, here we will take λ0 = 0 for convenience.If fn(x) is linear i.e.

d

dxfi(x) = 1 ∀i→ fn(x) = x+ rn

dN−1

dxN−1

(N∏n=1

fn(x)

)= (N − 1)!

N∑n=1

fn(x)

From this we can compute:

dN−1

dλN−1Ta(λ)

∣∣∣λ=0

=dN−1

dλN−1(La,N · · ·La,1)

∣∣∣λ=0

= (N − 1)!

N∑n=1

La,n(λ)∣∣∣λ=0

= (N − 1)!

N∑n=1

La,n(0)

La,n(0) = i∑α

(Sαn ⊗ σαa )

The total spin is defined as:

Sα = SαT =

N∑n=1

Sαn

Therefore we have:

aN−1 = i∑α

N∑n=1

(Sαn ⊗ σα)

= i∑α

(Sα ⊗ σα)

Ta(λ) = λN + iλN−1∑α

(Sα ⊗ σα) + · · ·

37

This operator can also be represented by the following 2 × 2 matrix inthe auxiliary space A, with entries being operators on the Quantum spaceVn:

Ta(λ) =

(A(λ) B(λ)C(λ) D(λ)

)The transfer matrix is defined by tracing over the auxiliary space:

t(λ) = tra Ta(λ) = A(λ) +D(λ)

Claim: The transfer matrix consists of a one-parameter family of com-muting operators:

[t(λ), t(µ)] = 0

Proof.

t(λ)t(µ) = (tra1 Ta1(λ))(tra2 Ta2(µ))

= tra1,a2 [Ta1(λ)Ta2(µ)]

= tra1,a2 [Ra1,a2(λ− µ)−1Ra1,a2(λ− µ)Ta1(λ)Ta2(µ)]

= tra1,a2 [Ra1,a2(λ− µ)−1Ta2(µ)Ta1(λ)Ra1,a2(λ− µ)]

= tra1,a2 [Ta2(µ)Ta1(λ)]

= (tra2 Ta2(µ))(tra1 Ta1(λ))

= t(µ)t(λ)

Where we have taken the double trace over both auxiliary spaces and haveused the cyclic property of the trace.

Note since this holds for ∀λ, µ ∈ C, we can take the series expansiont(λ) =

∑i λ

iQi and by the commutation relation:∑i,j

λi µj [Qi, Qj ] = 0

This automatically tells us that we have

[Qi, Qj ] = 0 ∀i, j

Let us compute the trivial terms in the power expansion for the transfer ma-trix t(λ) = aλN +bλN−1+· · · , the transfer matrix will be a finite polynomial

38

of order N , the coefficients are given by:

a =1

N !

dN

dλNt(λ)

=1

N !tra(

dN

dλNTa(λ))

=1

N !tra (N ! Ia,N )

=1

N !tra (N ! Ia ⊗ IN )

= tra(Ia) tra(In) = 2IN

This is true for all λ, we shall now compute the λN−1 term, we will find thatthe λN−1 term vanishes since the Pauli matrices σα are traceless:

b =1

(N − 1)!

dN−1

dλN−1t(λ)

∣∣∣λ=0

=1

(N − 1)!tra

(dN−1

dλN−1Ta(λ)

) ∣∣∣λ=0

=1

(N − 1)!tra

(i(N − 1)!

∑α

Sα ⊗ σαa

)

= i tra

(∑α

Sα ⊗ σαa

)= i∑α

tra(Sα) tra(σ

αa )

= i∑α

Sα tra(σαa ) = 0

Where we have used the identity tr(A⊗B) = trA trB.This tells us that the non-trivial expansion begins with power λN−2:

t(λ) = 2λN +

N−2∑n=0

Qnλn

and produces N − 1 commuting operators Qi (conserved charges). As weshall see in Chapter 4, these charges will turn out to be intractable and non-local, it will therefore be useful to consider the expansion of the logarithmof the transfer matrix since it has been shown by Lusher [17] to producelocal charges which will be algebraically related to the coefficients Qn. TheHamiltonian will be related to the first term of the expansion (the firstderivative) which we can write in terms of t(λ) as:

d

dλln t(λ) = t′(λ)t(λ)−1

39

Claim: The Hamiltonian H for the XXX model can be generated by thecoefficients Qn as:

H = −J(i

2Q−1

0 Q1 −N

4

)Proof. Let us consider the point λ = i/2:

La,n(i/2) = iPa,n

we also have for any λ:d

dλLa,n(λ) = Ia,n

This makes it easy to control the expansion of t(λ) in the neighborhood ofλ = i/2. We have:

Q0 = t(i/2) = tra (La,N (i/2) · · ·La,1(i/2))

= tra ((iPa,N ) · · · (iPa,1))

= iN tra (Pa,N · · ·Pa,1)

These string of permutations inside the trace can be transformed by theidentities from Chapter 1 to isolate a single permutation carrying the indexof the auxiliary subspace:

Pa,N · · ·Pa,1 = PN,N−1PN,N−2 · · ·PN,1Pa,N= P1,2P2,3 · · ·PN−1,NPa,N

Recall that tra Pa,N = IN , therefore we have:

t(i/2) = iNP1,2P2,3 · · ·PN−1,N

We can define the shift (translation) operator T as:

T = i−N t(i/2) = P1,2P2,3 · · ·PN−1,N

The operator T is unitary T ∗T = TT ∗ = I because the permutations havethe properties P ∗ = P and P 2 = I, it generates a shift along the chain as:

T−1XnT = Xn−1

By definition an operator of the infinitesimal shift is the momentum on thelattice defined as

T |ψ〉 = eip|ψ〉

where p is the momentum as we used in the CBA for the translation operator.

40

Now let us differentiate the logarithm of the transfer matrix:

Q1 = t′(i/2) =d

dλt(λ)

∣∣∣λ=i/2

=d

dλ(tra [La,N (λ) · · ·La,1(λ)])

∣∣∣λ=i/2

= trad

dλ([La,N (λ) · · ·La,1(λ)])

∣∣∣λ=i/2

Let us compute the object inside the trace:

d

dλ([La,N (λ) · · ·La,1(λ)])

∣∣∣λ=i/2

= iN−1∑n

(Pa,N · · · P̂a,n · · ·Pa,1)

Thus we get:

t′(i/2) = iN−1∑n

tra(Pa,N · · · P̂a,n · · ·Pa,1)

= iN−1∑n

(P1,2 · · ·Pn−1,n+1 · · ·PN−1,N )

Putting the above together, we are now in a position to compute the loga-rithmic derivative:

d

dλln t(λ)

∣∣∣λ=i/2

=

(d

dλt(λ)

)t(λ)−1

∣∣∣λ=i/2

= i−NQ1(PN,N−1PN−1,N−2 · · ·P2,1)

=1

i

N∑n=1

Pn,n+1

Recall that:

H = −J4

N∑n=1

σαnσαn+1 = −J

N∑n=1

SαnSαn+1 = −J

(1

2

N∑n=1

Pn,n+1 −N

4

)

Hence we can write the Hamiltonian in terms of the transfer matrix as:

H = −J(i

2Q−1

0 Q1 −N

4

)= −J

(i

2

d

dλln t(λ)

∣∣∣λ=i/2

− N

4

)

Therefore, H belongs to a family of N − 1 commuting operators. If weadd the total spin operator SzT to this family we get N commuting operatorsand hence the XXX-model is integrable.

41

3.3.3 Spectrum of H

Now that we have proven the integrability of the model, we need to solve iti.e. find the spectrum of the Hamiltonian H. It is clear that [H, t(λ)] = 0,this implies that eigenstates of H will also be eigenstates of t(λ), hence inorder to solve the model it suffices to diagonalize t(λ). Given | ↑〉n ∈ Vnand v ∈ a, the action of the Lax operator La,n(λ) : A ⊗ Vn → A ⊗ Vn onarbitrary element of v ⊗ | ↑〉n ∈ A⊗ Vn:

La,n(λ)(v ⊗ | ↑〉n) =

((λ+ iSzn)| ↑〉n iS−n | ↑〉niS+n | ↑〉n (λ− iSzn)| ↑〉n

)v

=

((λ+ i

2)| ↑〉n i| ↓〉n0 (λ− i

2)| ↑〉n

)v

The action on v is that of matrix multiplication on Va, in what follows wecan omit it. Let us now compute the action of La,n(λ) on the ground state|0〉 = | ↑↑ · · · ↑〉 = ⊗Nn=1| ↑〉n ∈ HN :

La,n(λ)|0〉 =

(λ+ iSzn iS−niS+n λ− iSzn

)|0〉

=

((λ+ iSzn)|0〉 iS−n |0〉iS+n |0〉 (λ− iSzn)|0〉

)Note (from the table at the beginning of the section):

Szn|0〉 =1

2|0〉 S+

n |0〉 = 0

The action of S−n on the ground state merely changes the nth spin in thechain to a spin down, the explicit form will be unimportant in what followssince we will be considering the action of the trace of the Monodromy matrix(i.e. the transfer matrix) on the ground state. Denoting any unimportantitems with a ∗ i.e., we get from the above:

La,n(λ)|0〉 =

((λ+ i

2)|0〉 ∗0 (λ− i

2)|0〉

)=

((λ+ i

2) ∗0 (λ− i

2)

)|0〉

Using this, we can now compute the action of the Monodromy matrix

42

Ta(λ) on |0〉 by the repeated application of the above:

Ta(λ)|0〉 = La,N (λ) · · ·La,1(λ)|0〉

=

(λ+ iSzN iS−NiS+N λ− iSzN

)· · ·(λ+ iSz1 iS−1iS+

1 λ− iSz1

)|0〉

=

(λ+ iSzN iS−NiS+N λ− iSzN

)· · ·(

(λ+ i2)|0〉 ∗

0 (λ− i2)|0〉

)=

((λ+ i

2)N |0〉 ∗0 (λ− i

2)N |0〉

)=

((λ+ i

2)N ∗0 (λ− i

2)N

)|0〉 =

(A(λ) B(λ)C(λ) D(λ)

)|0〉

More importantly, what this tells us is that the ground state |0〉 is aneigenstate of A(λ) and D(λ) simultaneously i.e.:

A(λ)|0〉 = a(λ)|0〉 D(λ)|0〉 = d(λ)|0〉

Where a(λ) = (λ + i2)N and d(λ) = (λ − i

2)N . Consequently, it is also aneigenstate of the transfer matrix t(λ) = A(λ) +D(λ). This is not surprisingsince we know that [H, t(λ)] = 0 and we have also proved that |0〉 is aneigenstate of H earlier in the section, this immediately implies that it is alsoan eigenstate of t(λ). It is also clear from the above that |0〉 is annihilatedby C(λ) i.e.:

C(λ)|0〉 = 0

The algebraic Bethe Ansatz (ABA) is the ’“assumption” that C(λ) andB(λ) are annihilation and creation operators. A general state with r downspins will be given by choosing λ such that:

|λ1, · · · , λr〉 = B(λ1) · · ·B(λr)|0〉 =r∏

n=1

B(λn)|0〉

The conditions on λ for the above to be an eigenstate of t(λ) will producethe BAE’s, which we have also derived using the CBA. The λi, the so-calledBethe roots are comparable to the ki (momenta) from the CBA and weshall see how they are related to each other. To calculate the eigenvalue forthe above eigenstate will require us to establish the commutation relationsbetween [A(λ), B(µ)], [B(λ), B(µ)] and [D(λ), B(µ)]. These will result fromthe RTT relations (the Global FCR).

Claim: Applying the RTT relations we get the following:

[B(λ), B(µ)] = 0

A(λ)B(µ) = f(λ− µ)B(µ)A(λ) + g(λ− µ)B(λ)A(µ)

D(λ)B(µ) = h(λ− µ)B(µ)D(λ) + k(λ− µ)B(λ)D(µ)

43

Where:

f(λ) =λ− iλ

, g(λ) =i

λ, h(λ) =

λ+ i

λ, k(λ) = − i

λ

Proof. We will do this by direct computation of the RTT relations andcomparing coefficients. We will use the explicit matrix representation of theFCR in End(A1 ⊗ A2). All objects are 4× 4 matrices in A1 ⊗ A2 with thebasis given by H2. The R-matrix in this representation looks like:

Ra1,a2(λ) = (λ+ i)

1 0 0 0

0 λλ+i

iλ+i 0

0 iλ+i

λλ+i 0

0 0 0 1

The matrices Ta1(λ) and Ta2(µ) in this representation are given by:

Ta1(λ) = Ta(λ)⊗ I

Ta2(µ) = I⊗ Ta(µ)

Explicitly:

Ta1(λ) =

A(λ) B(λ)

A(λ) B(λ)C(λ) D(λ)

C(λ) D(λ)

Ta2(µ) =

A(µ) B(µ)C(µ) D(µ)

A(µ) B(µ)C(µ) D(µ)

Thus:

Ta1(λ)Ta2(µ) =

A(λ)A(µ) A(λ)B(µ) B(λ)A(µ) B(λ)B(µ)A(λ)C(µ) A(λ)D(µ) B(λ)C(µ) B(λ)D(µ)C(λ)A(µ) C(λ)B(µ) D(λ)A(µ) D(λ)B(µ)C(λ)C(µ) C(λ)D(µ) D(λ)C(µ) D(λ)D(µ)

and

Ta2(µ)Ta1(λ) =

A(µ)A(λ) B(µ)A(λ) A(µ)B(λ) B(µ)B(λ)C(µ)A(λ) D(µ)A(λ) C(µ)B(λ) D(µ)B(λ)A(µ)C(λ) B(µ)C(λ) A(µ)D(λ) B(µ)D(λ)C(µ)C(λ) D(µ)C(λ) C(µ)D(λ) D(µ)D(λ)

By direct computation of the RTT relation:

Ra1,a2(λ− µ)Ta1(λ)Ta2(µ)− Ta2(µ)Ta1(λ)Ra1,a2(λ− µ) = 0

By looking at the 1st row and 4th column of the matrix on the LHS andequating it to 0, we obtain the first relation. Doing the same with the 1st rowand 3rd column and interchanging µ with λ we obtain the second relation,the third is obtained by looking at the 3rd row and 4th column.

44

3.3.4 Bethe Ansatz equations

Now we are in a position to derive the BAE from the above considerations,these will follow from the following claim.

Claim: The condition that |λ1, · · · , λr〉 is an eigenstate of the trans-fer matrix t(λ) will lead to a set of algebraic relations on the parametersλ; {λi} = {λ1, · · · , λr}, these will be the BAE.

Proof. To this end, let us compute the action ofA(λ) andD(λ) on |λ1, · · · , λr〉:

A(λ)|λ1, · · · , λr〉 = A(λ)B(λ1) · · ·B(λr)|0〉

= a(λ)

r∏n=1

f(λ− λn)B(λ1) · · ·B(λr)|0〉

+

r∑n=1

Mn(λ, {λj})B(λ)r∏

j=1,j 6=nB(λj)|0〉

Where

Mn(λ, {λj}) = g(λ− λn)a(λn)

r∏j=1,j 6=n

f(λn − λj)

We can see this explicitly by using the commutation relations repeatedlyto move A(λ) to the right, lets compute the coefficient for the nth term. Notewe can write |λ1, · · · , λr〉 in the following way:

|λ1, · · · , λr〉 = B(λn)r∏

j=1,j 6=nB(λj)|0〉 n = 1, · · · , r

since the operators B(λ) commute with each other, using this along withthe second commutation relation, we can move the A(λ) past the B(λn) i.e.:

A(λ)|λ1, · · · , λr〉 = A(λ)B(λn)r∏

j=1,j 6=nB(λj)|0〉

= f(λ− λn)B(λn)A(λ)

r∏j=1,j 6=n

B(λj)|0〉

+ g(λ− λn)B(λ)A(λn)r∏

j=1,j 6=nB(λj)|0〉

From this it is clear that the first term will not contribute to Mn(λ, {λi})since it contains B(λn), only the second term will contribute. If we now moveA(λn) past the B(λj) we see that the only way to avoid the appearance of

45

B(λn) is by using the first term of the A-B commutation relation. Theresulting term will be:

g(λ− λn)a(λn)r∏

j=1,j 6=nf(λn − λj)B(λ)

r∏j=1,j 6=n

B(λj)|0〉

i.e. we have:

Mn(λ, {λi}) = g(λ− λn)a(λn)

r∏j=1,j 6=n

f(λn − λj)

Similarly we have:

D(λ)|λ1, · · · , λr〉 = D(λ)B(λ1) · · ·B(λr)|0〉

= d(λ)r∏

n=1

h(λ− λn)B(λ1) · · ·B(λr)|0〉

+r∑

n=1

NnB(λ)r∏

j=1,j 6=nB(λj)|0〉

Where

Nn = Nn(λ, {λj}) = k(λ− λn)d(λn)r∏

j=1,j 6=nh(λn − λj)

Combining these results we get for the transfer matrix t(λ):

t(λ)|λ1, · · · , λr〉 = (A(λ) +D(λ))|λ1, · · · , λr〉

=

(a(λ)

r∏n=1

f(λ− λn) + d(λ)

r∏n=1

h(λ− λn)

)|λ1, · · · , λr〉

+r∑

n=1

(Mn +Nn)B(λ)r∏

j=1,j 6=nB(λj)|0〉

The first term has a desirable form and if the second term was absentthen we would immediately have that |λ1, · · · , λr〉 is an eigenstate of t(λ) forall λ. Since it is present however, it will produce restrictions on λ which wewill see will lead to the BAE. We can now make the following statement: Thegeneral eigenstate with r spins down, produced by the creation operatorsB(λi), |λ1, · · · , λr〉 is an eigenstate of t(λ) (and consequently of H) witheigenvalue:

Λ(λ) =

(a(λ)

r∏n=1

f(λ− λn) + d(λ)r∏

n=1

h(λ− λn)

)

46

provided that the second term vanishes i.e. Mn +Nn = 0 ∀n. This meansone needs to satisfy the condition:

g(λ− λn)a(λn)r∏

j=1,j 6=nf(λn − λj) = −k(λ− λn)d(λn)

r∏j=1,j 6=n

h(λn − λj)

−g(λ− λn)a(λn)

k(λ− λn)d(λn)=

r∏j=1,j 6=n

h(λn − λj)f(λn − λj)

Substituting the explicit expressions for f(λ), g(λ), h(λ), k(λ) and the eigen-values of the ground state a(λ), d(λ), we obtain the Bethe Ansatz Equa-tion: (

λn + i/2

λn − i/2

)N=

r∏j=1,j 6=n

λn − λj + i

λn − λj − i

As required.

We will now look at the expressions for the eigenvalues of the componentsof spin on |λ1, · · · , λr〉. To this end, we will look at the limit λ → ∞ inthe RTT relation, using the power expansion of Ta(λ) and the R- matrixrepresentation in terms of the permutation matrix P. We get:(

(λ− µ) +i

2[Ia1 ⊗ Ia2 +

∑α

(σαa1 ⊗ σαa2)

)(λN + iλN−1

∑α

(Sα ⊗ σα) + · · ·

)Tb(µ)

= Tb(µ)

(λN + iλN−1

∑α

(Sα ⊗ σα) + · · ·

)((λ− µ) +

i

2[Ia1 ⊗ Ia2 +

∑α

(σαa1 ⊗ σαa2)

)(∑

α

(σαa ⊗ Ib ⊗ Sα)

)(Ia ⊗ Tb(µ)) +

1

2

∑α

(σαa ⊗ σαb ⊗ Ih)(Ia ⊗ Tb(µ))

= (Ia ⊗ Tb(µ))

(∑α

(σαa ⊗ Ib ⊗ Sα)

)+ (Ia ⊗ Tb(µ))

1

2

∑α

(σαa ⊗ σαb ⊗ Ih)

∑α

σαa ⊗ [Tb(µ), Ib ⊗ Sα +1

2(σαb ⊗ Ih)] = 0

[Tb(µ), Ib ⊗ Sα +

1

2(σαb ⊗ Ih)

]=

[(A(µ) B(µ)C(µ) D(µ)

),

(Sα 00 Sα

)+

1

2(σαb ⊗ Ih)

]= 0

Taking α = z:[(A(µ) B(µ)C(µ) D(µ)

),

(Sz 00 Sz

)+

1

2

(I 00 −I

)]= 0

47

The (1, 2) matrix entry of this equation gives us

[Sz, B] = −B

Taking α = x, y we get the relation:

[S+, B] = A−D

For our ground state |0〉 we have:

S+|0〉 = 0 Sz|0〉 =N

2|0〉

Which tells us that |0〉 is a highest weight vector for the spin algebra. Letus look at the eigenstates |λ1, · · · , λr〉. Using the above relations we get:

Sz|λ1, · · · , λr〉 =

(N

2− r)|λ1, · · · , λr〉

which confirms the result we got in the first part of the section. One canshow in a simular fashion to the proof of the BAE that:

S+|λ1, · · · , λr〉 = 0

This shows that the eigenstates obtained from the ABA are also the highestweight vectors of the spin algebra.

We will now calculate the energy for the corresponding eigenstates usingthe logarithmic derivative of the eigenvalue of the transfer matrix. We obtainafter straightforward differentiation:

E = −J(i

2

d

dλln Λ(λ)

∣∣∣λ=i/2

− N

4

)= E0 +

J

2

N∑i=1

1

λ2i + 1/4

If we identify λi as the rapidity as shown in the CBA and use the parame-terization λk = 1

2 cot(kj2 ) we get:

E − E0 = J

N∑i=1

2

cot(kj/2)2 + 1= J

N∑i=1

2 sin2(kj2

) = J

N∑i=1

(1− cos (kj))

3.4 Thermodynamic limit N → ∞ - String Hy-pothesis

We will use our analysis on the CBA and ABA in order to analyse the be-haviour at N →∞. The mathematical and physical features of the N →∞are completely different for the cases the cases J > 0 and J < 0. We willconsider these separately corresponding to the ferromagnetic and antifer-romagnetic phases respectively. We saw earlier that the rapidities λj can

48

have real and complex solutions, so far our discussion has been general butit turns out that the majority of the states are characterized by complexrapidities. In general these have to found numerically but they can be ex-tremely hard to find. We would like to understand the structure of thecomplex rapidities to develop more suitable ways to determine them, butfor finite N this can be intractable. The advantage of analysing N →∞ isthat a simple structure emerge, called the string hypothesis which allows usto analyse these complex rapidities in more detail. We will begin by talkingabout the FM case J > 0, though some of the deductions will allow us toexplore the physics of the AFM J < 0. The material in this section will bebased on [8] ,[11] and [12].

Recall the BAE:(λn + i/2

λn − i/2

)N=

r∏j=1,j 6=n

λn − λj + i

λn − λj − i

Rapidities λj to parameterize the quasi-momenta kj as:

λj = cotkj2

or

kj =1

ilnλj + i

λk − i= π − θ1(λj),

where θn(λ) = 2 arctan λn .

Taking the Log of the BAE or using what we obtained from the CBAwe obtain (kj = k(λj)):

Nk(λj) = 2πIj +∑l 6=j

θjl, j = 1, · · · , r

In the BAE, N enters the exponent of the LHS. For real λ1, · · · , λr, bothsides of the BAE are functions on the circle with the LHS oscillating wildlywhen N is large. For large N and I with r kept fixed, the second term ofthe above is negligible thus we have

Nkj ∼ 2πIj =⇒ kj ∼2πIjN

Which is simply the momentum of a free particle on the chain. When J > 0,the energy of this particle is ε(p) = 1 − cos p. The “correction” caused bythe second term of the RHS encodes the information about the scatteringof these particles.

The BAE also allows for λj ∈ C, which correspond to bound states forJ > 0. Let us look at the case r = 2 (two down spins), the BAE are:(

λ1 + i/2

λ1 − i/2

)N=λ1 − λ2 + i

λ1 − λ2 − i(1)

49

(λ2 + i/2

λ2 − i/2

)N=λ2 − λ1 + i

λ2 − λ1 − i(2)

=⇒(λ1 + i/2

λ1 − i/2

)N (λ2 + i/2

λ2 − i/2

)N= 1

Since we require that the total momentum k(λ1) + k(λ2) be real. ForImλ1 6= 0 the LHS of (1) grows (or decreases) exponentially when N →∞ and thus the RHS must have Im (λ1 − λ2) = ±i. λ1 and λ2 can beinterchanged and hence their form is given by:

λ1 = λ1/2 +i

2, λ2 = λ1/2 −

i

2

The momentum and energy of this state are:

eip1/2(λ) = eip0(λ+i/2)+ip0(λ−i/2) =λ+ i

λ− i

ε1/2(λ) = −Jd p1/2

dλ= ε0(λ+ i/2) + ε0(λ− i/2) =

4J

λ2 + 4

Which gives the dispersion relation:

ε1/2(p) =J

2(1− cos p1/2)

For J > 0 we have ∀p, p′ < 2π:

ε1/2(p) < ε0(p− p′) + ε0(p′)

Bound states are energetically favoured compared to the real solutions inthe ferromagnetic case.

For r > 2 more complex solutions can appear and we can describe themin a simular way. Roots λr are combined with complexes (or strings) of typeM , where M can have half-integer values M = 0, 1/2, 1, · · · , we can havecomplexes of 2M+1 rapidities characterized by the same real value λM , butdifferent imaginary parts.

λM ;m = λM + im −M ≤ m ≤M

Where m is a integer or half-integer together with M . Counting all thecomplexes of length (type) M , we can denote the number of complexes byνM . The set of integers {νM} defines the configuration space of Bethe roots.For a state with a given magnetization we can define the partition:

r =∑M

(2M + 1)νM

50

It is important to stress that this string structure is only value in the ther-modynamic limit N → ∞. The energy and momentum of a M -complex isgiven by:

pM (λM ) =1

ilnλM + i(M + 1/2)

λM − i(M + 1/2)

εM (λM ) =2J(M + 1/2)

λ2M + (M + 1/2)2

=J

2M + 1(1− cos pm)

For a ferromagnetic coupling the completely aligned state |0〉 can betaken as the ground state. Excited states above the ground state can beconstructed in terms of magnon excitations, keeping into consideration thatbound states corresponding to 2M + 1 complexes have lower energies com-pared to states with 2M + 1 real magnons. For the Anti-Ferromagnetic(J < 0) case we will have to do more work in order to determine the phys-ical spectrum as the completely polarized |0〉 used as our reference stateis clearly different from the anti-ferromagnet ground state. In the ferro-magnetic regime, string solutions have lower energy than unbound, purelyreal ones. The opposite is true for the AFM, the string solutions will havehigher energy. From here on we will let J = −1 for the anti-ferromagnetwithout loss of generality. Let us assume that N is even in what follows. Theground state configuration for the AFM will be composed of single particleexcitations:

ν0 =N

2; νM = 0 M ≥ 1

2

For this configuration we have r = N/2, this implies that the spin of thestates vanishes. Excited states over the ground state can be characterisedby κ with:

ν0 =N

2− κ; νM = 0 M ≥ 1

2

We will now proceed to approximate the distribution of the solutions of theBethe equation with their continuous distribution. We will start with theground state, for which the quantum numbers I0,j fill the allowed intervalof vacancies without holes. Let us assume that N/2 is odd (with the evencase requiring a minor modification), so that:

I0,j = −N4

+1

2,−N

4+

3

2, · · · , N

4− 1

2

We can write the Bethe equations as:

θ1(λj) =π

NI0,j +

1

N

∑k

θ2(λj − λk)

Explicitly this is:

arctanλj =π

NI0,k +

1

N

∑k

arctan

(λj − λk

2

)

51

In the limit N → ∞, the variable x = I0,k/N becomes continuous andlimited in the range −1/4 ≤ x ≤ 1/4. The set of roots λj becomes afunction λ(x) and we obtain:

arctanλ(x) = πx+

∫ 1/4

−1/4arctan

(λ(x)− λ(y)

2

)dx

Using: ∫ 1/4

−1/4f(λ(x)) dx =

∫ ∞−∞

f(λ)ρ0(λ) dλ

where the change of variables x→ λ(x) maps interval −1/4 ≤ x ≤ 1/4 intothe real line −∞ < λ <∞ due to the monotonicity of λ(x), i.e. explicitly:

ρ0(λ) =dx

dλ=

1

λ′(x)

∣∣∣x=λ−1(λ)

Now differentiation this wrt to λ, the Bethe equations for the ground statebecome:

ρ0(λ) =1

π

1

1 + λ2− 1

π

∫ ∞−∞

2

(λ− µ)2 + 4ρ0(µ) dµ

In the thermodynamic limit, the energy and momentum can also be ex-pressed in terms of the density function ρ0(λ) as:

EAFM ≡ E0 +N

∫ε0(λ)ρ0(λ) dλ

KAFM ≡ N∫p0(λ)ρ0(λ) dλ

We can solve ρ0(λ) from the above integral equation by using the Fouriertransform, we get (see [11]):

ρ0(λ) =1

4 cosh (πλ2 )

And we thus obtain the anti-ferromagnet (J = −1) ground state energy:

EAFM = N

(1

4− ln 2

)and

KAFM =Nπ

2mod 2π

Now we turn to the state with ν0 = N/2− 1 and νM = 0 for M ≥ 1/2.The state is characterized by two wholes which we place at j1 and j2:

I0,j = j +H(j − j1) +H(j − j2),

52

where H(x) is the usual Heaviside step-function. The integral equation forthe rapidity density of the real roots ρt(λ) (where t stands for triplet) is:

ρt(λ) =1

π

1

1 + λ2− 1

π

∫ ∞−∞

2

(λ− µ)2 + 4ρt(µ) dµ− 1

N[δ(λ− λ1) + δ(λ− λ2)],

where λi, i = 1, 2 are the images of xi = ji/N under the map x → λ(x).Since we are dealing with linear equations, we can write the solution forρt(λ) as:

ρt(λ) = ρ0(λ) +1

N[τ(λ− λ1) + τ(λ− λ2)],

where τ(λ) solve the integral equation:

τ(λ) +1

π

∫ ∞−∞

2

(λ− µ)2 + 4τ(µ) dµ+ δ(λ)

As before, solving this integral equation using the fourier transform as in[11], we obtain the total momentum and energy as:

K = N

∫p0(λ)ρt(λ) dλ = KAFM + κ(λ1) + κ(λ2),

E = N

∫ε0(λ)ρt(λ) dλ = EAFM + ε(λ1) + ε(λ2),

where

κ(λ) ≡ π

2− arctan sinh

πλ

2, ε(λ) ≡ − π

2 cosh πλ2

This is a state with two-particle excitations (spinons) and ε(λ), κ(λ) aretheir dressed energy and momentum. Combing the two, we see that theseexcitations are characterized each by the dispersion relation:

ε(k) = −π2

sin k, −π2≤ k ≤ π

2

Therefore each hole in the quantum numbers generates a quasi-particle ex-citation, which is called a spinon, i.e. a spin- 1/2 excitation.

We see that if we follow the same procedure for the state with ν0 =N/2− 2, ν1/2 = 1 and νM = 0 for M ≥ 1. We can compute rapidity densityfor the real roots ρs(λ) (s is for singlet) by solving its integral equation usinga simular procedure to the above given in [11], after which we obtain for theenergy:

E = EAFM +N

∫ε0(λ)ρS(λ) dλ+ ε1/2(λ1/2)

= EAFM + ε(λ1) + ε(λ2)

We see that the contributions from the string cancel exactly and this statehas the same momentum, energy and (dispersion relation) as the one without

53

complexes given by the state with ν0 = N/2−1 and νM = 0 forM ≥ 1/2. Wesee that the excitations in the anti- ferromagnetic ground state |AFM〉 canbe interpreted as spinon excitations over the spinon vacuum, this point ofview is particularly suitable to desribe the anti-ferromagnetic case only. Forthe ferromagnet case the ground state is a magnon-vacuum and excitationsare interpreted as magnons.

3.4.1 XXZ Model

We will briefly state the results for the XXZ-model that can be obtained bythe ABA through a very simular method to the XXX-model. The resultspresented here will be based on [12] and [13]. The generalized anisotropicHeisenberg Hamiltonian as introduced in Chapter 2 is given by:

H = −N∑j=0

(JxSxj S

xj+1 + JyS

yj S

yj+1 + JzS

zjS

zj+1)

The XXZ model in particular corresponds to Jx = Jy = J and Jz = ∆J ,where ∆ is the measure of anisotropy:

H = −JN∑j=0

(Sxj Sxj+1 + Syj S

yj+1 + ∆SzjS

zj+1),

where ∆ = 1 corresponds to the XXX model. For the XXZ model, the Laxoperator is given by:

La,n(λ) =

(sinh (λ+ ηSzn) i2S−n sinh(η)S+n sinh (η) sinh (λ− ηSzn)

),

where λ is the spectral parameter. The relationship between η and ∆ is givenby ∆ = cosh η. This will satisfy the Yang-Baxter algebra, in particular theFCR relation as given before:

Ra1,a2(λ− µ)La1,n(λ)La2,n(µ) = La2,n(µ)La1,n(λ)Ra1,a2(λ− µ),

on End(A1 ⊗ A2 ⊗ Vn). This will be given as a direct consequence by thefollowing R-matrix (R(λ) acting on C2 ⊗ C2) which is a solution to theYang-Baxter equation as before:

R(λ) =1

sinh (λ+ η)

sinh (λ+ η) 0 0 0

0 sinhλ sinh η 00 sinh η sinhλ 00 0 0 sinh (λ+ η)

.

Indeed we have the R-matrix and the Lax operator in a simular form asLa,n(λ) = Ra,n(λ − η/2) which is simular to what we had in the XXX

54

model. As with the XXX model, we have the Monodromy matrix:

Ta(λ) = La,N (λ) · · ·La,1(λ) =

(A(λ) B(λ)C(λ) D(λ)

)Transfer matrix:

t(λ) = tra Ta(λ) = A(λ) +D(λ)

It follows as from the isotropic case that these transfer matrices generatehigher conserved charges, in particular the Hamiltonian which can be writtenas:

H = (−2J sinh η) t−1(η/2)t′(η/2) + const

= −2J sinh η∂

∂λln t(λ)

∣∣∣λ=η/2

+ const

The eigenvalues of t(λ) are again given as excitations from the pseudo-vacuum |0〉,

|λ1, · · · , λn〉 =r∏i=1

B(λi)|0〉,

where B(λ) and C(λ) are assumed to be creation and annihilation operatorsas before. λi also satisfy the BAE’s which in the XXZ model are given by:(

sinh (λj + η/2)

sinh (λj − η/2)

)N=

r∏l=1,l 6=j

sinh (λj − λl + i)

sinh (λj − λl − i), j = 1, · · · , r

For the XXZ model, the energy is given by:

E = −JN∆

4+ J

r∑j=1

(∆− cos kj),

which for the isotropic limit ∆→ 1 reduces to the energy that we calculatedbefore for the XXX model.

The construction of |0〉AFM = |AFM〉 (J = −1) is credited due Hulthen[23] in the thermodynamic limit N →∞. For −1 < ∆ < 1 the value in thethermodynamic limit for the AFM (J = −1):

EAFM = N∆

4−N sinπν

2π

∫ ∞−∞

dzsinh (1− ν)z

sinh z cosh νz

= N∆

4+N

sinπν

2π

∫ ∞+i/2

−∞+i/2dx

1

sinhx

cosh νx

sinh νx

Where the anisotropy parameter ∆ is parameterized in [23] as ∆ = cosπν.In the XXX limit ∆→ 1, the ground state energy is more simply given by:

EAFM = N

(1

4− ln 2

),

55

which agree’s with the result that we obtained in the previous section in thethermodynamic limit N− > ∞ for the anti-ferromagnet ground state. Itis useful to introduce the ground state energy per site which will simply bedefined as:

e0 = limN→∞

EAFMN

For the above we obtain:

e0 =∆

4+

sinπν

2π

∫ ∞+i/2

−∞+i/2dx

1

sinhx

cosh νx

sinh νx,

and in particular for the XXX spin chain:

e0 =

(1

4− ln 2

)

56

Chapter 4

Higher Conserved Charges ofthe Heisenberg Hamiltonian

In this chapter we will be looking at the Higher Conserved charges of theHeisenberg Hamiltonian which will produce the local conservation laws ofthe XXX-model. The material in this chapter will be based on [14],[15] and[16].

In the previous chapter we saw that the transfer matrix t(λ) produceda mutually commuting set Qn of N − 1 independent conserved quantities,essentially given as the coefficients of the power series of t(λ). We alsosaw that we could generate the Hamiltonian from the transfer matrix, inparticular it was related to second term in the expansion (coefficient of theλ term) via:

H = −J(i

2t(i/2)Q1 −

N

4

)and

Q1 =d

dλt(λ)

∣∣∣λ=i/2

A local conserved charge is a sum over the whole chain of terms that aresupported (i.e. non-trivial) only on a finite number of sites, which makessense as a definition in the large N limit. For finite N , these number of siteswill be sufficiently small compared to N , we will give a more mathematicaldescription in the next section. In order to satisfy the locality conditions, aHamiltonian should be expressible as a sum of terms each one only involvingspins close to each on the chain. The Heisenberg Hamiltonian in particularis a local conservation law, but the other coefficients Qn are in generalnot. These conserved quantities are rather intractable due to their non-locality, we wish to consider only local conservation charges since they arethe characteristic of integrability. If we look at the Hamiltonian, we see thatin the left hand side we notice that t(i/2)Q1 = t(i/2) t′(i/2) corresponds

57

to the first derivative of the logarithm of the Transfer matrix evaluated atλ = i/2. This provides us a clue to what we must do in order to generatelocal conserved quantities which we address in this section.

4.1 Logarithmic derivative of the transfer matrix

A more useful set of operators to consider are given by coefficients fromthe expansion of P (λ) = ln (t(λ)), given essentially as derivatives of P (λ)evaluated at some point λ = λ0. It can be shown that the higher derivativesof this logarithm commute and so form a set of conserved quantities, takingthe logarithm ensures locality but it is very non-trivial to see this explicitly.

We can write P (λ) as a power series in the spectral parameter λ aroundλ0 = i/2 as:

P (λ) = ln (t(λ)) =

∞∑n=1

(λ− λ0)npnn!

Notice that the expansion of P (λ) will provide infinitely many coeffi-cients, but these coefficients will be expressed as functions of the coefficientsof t(λ) (i.e. Qn) - so they are not all algebraically independent, there arenon-linear algebraic relations between them. In general if Q1, · · · , Qn areconserved then so is pi = f(Q1, · · · , Qn), so these coefficients constructedfrom other conserved charges do not produce any new conservation laws.

It is clear from the expansion of t(λ) in the previous section that wecannot have more than N−1 conserved quantities. Since t(λ) is a polynomialof order N , we have the following properties which we have been shown orare inferred, true ∀λ:

t(N−1)(λ) = 0, t(N)(λ) = 2N !, t(i)(λ) = 0 ∀i > N

This controls the expansion of P (λ) in some sense and prevents higherthan N order coefficients from being algebraically independent. The coeffi-cient of P (λ) at order higher than N will therefore not produce new charges,these will be algebraically related to the local conserved charges at lower or-ders and will lose locality since it is focused around a site of extent of orderN .

The Hamiltonian is related to p1 by:

H = ap1 + b

Where a and b are constants, in this case given by a = −J/2i, b = JN/4Since p1 corresponds to the Hamiltonian, the coefficients of the above

expansion will form a commuting set of local conservation laws, the localitywas shown by Luscher [17]. In the previous section when we showed that wecan generate the Hamiltonian from the derivative of the transfer matrix, itwas explicitly clear that the Hamiltonian was local. When we multiplied the

58

derivative of t(λ) by t−1(λ), we cancelled out the most of the permutationmatrices and we were left with identities everywhere except for a small bit ofsize n (∼ Pn,n+1). This is simular to what happens in higher local charges,admitting an expansion as:

pn =d

dλln (t(λ))

∣∣∣λ=λ0

= t−1(λ0)t(n)(λ0) + polynomials,

where the polynomials depend on lower order local charges. It can be shownin [21] that the expectation value of such higher charges is actually equal tothe expectation value of the first term i.e. t−1(λ0)t(n)(λ0).

Higher charges pi for i > 1 correspond to Hamiltonians with ’more neigh-bours interacting’. Since these charges are local operators, they can be putin the form [17]1:

pn =∑

{i1,··· ,in−1}

GTn−1(i1, · · · , in−1)

Where the summation is over the ordered subsets {i1, · · · , in−1} ∈ {1, · · · , N}and GT is a translationally covariant and totally symmetric function, obey-ing the locality property:

GTn (i1, · · · , in) = 0

for |in − i1| ≥ n. For the infinite XXX chain, further properties of theconserved charges, including their completeness, have been proved in [22].

It is very difficult to extract the explicit form of the charges directly bycomputing higher derivatives of P (λ), this is because the size for the transfermatrix grows exponentially with the length of the chain. Fortunately thereexists a shortcut which will involve an alternative method of constructinghigher charges with a boost operator. See [14]. The Boost operator is definedas the first moment of the Hamiltonian i.e.:

B̂ =N∑n=1

n~Sn · ~Sn+1

We use the fact that its commutator with the transfer matrix is equal to thederivative i.e.:

[B̂, t(λ)] =∂

∂λt(λ)

A consequence of this is that up to some constant terms, the boost operatorgenerates conserved quanties recursively i.e.:

pn+1 = [B̂, pn]

1Shown by Luscher for the XYZ model

59

Recall

p1 =d

dλln (t(λ))

∣∣∣λ=i/2

=1

i

N∑n=1

Pn,n+1 = − J

2a

N∑n=1

SαnSαn+1 −

b

a

We can omit the constants in this expression for convenience, since multiply-ing the charge by a constant will not change its conservation or locality and[B̂, k] = 0 for any constant k. This allows us to concentrate mainly on thepart with the spin matrices i.e. p1 ∼

∑Nn=1 S

αnS

αn+1. Using our knowledge

of p1 and the above, we will now proceed to calculate p2 and p3:

p2 = [B̂, p1]

=∑n,m

n[SanSan+1, S

bmS

bm+1]

In order to compute this, we will need the following identities for commuta-tors and the knowledge of the su(2) algebra of spins (see appendix):

[A,BC] = [A,B]C +B[A,C]

[AB,C] = A[B,C] + [A,C]B

[AB,C D] = A[B,C]D + [A,C]BD + C A[B,D] + C[A,D]B

[San, Sbm] = iδn,mε

abcScm

· · · p2 ∼∑n,m

mεabc(δm+1,nSamS

ciS

bi+1 + δm,nS

cnS

am+1S

bn+1

+ δm+1,n+1SbnS

amS

cn+1 + δm,n+1S

bnS

cn+1S

am+1)

=

N∑n=1

εabc[(n− 1)San−1ScnS

bn+1 + nScnS

an+1S

bn+1

+ nSbnSanS

cn+1 + (n+ 1)SbnS

cn+1S

an+2]

Since εabc is antisymmetric, terms with two S’s acting on the same sitevanish, we obtain:

p2 =N∑n=1

εabc[(n− 1)San−1ScnS

bn+1 + (n+ 1)SbnS

cn+1S

an+2]

If we shift the first term by 1 then recombine both terms we get:

p2 =

N∑n=1

εabc[nSanScn+1S

bn+2 + (n+ 1)SbnS

cn+1S

an+2]

=

N∑n=1

(nεacb + (n+ 1)εcab)SanSbn+1S

cn+2

=

N∑n=1

εabcSanSbn+1S

cn+2

60

By the definition of the cross product:

(A×B)i = εijkAjBk

Finally we have:

p2 =

N∑n=1

(~Sn × ~Sn+1) · ~Si+2

Indeed this has a local form that acts on the nearest and next-to- nearestneighbouring sites. In the same way iteratively we will obtain p3:

p3 = [B̂, p2] =∑m,n

mεabc[SαmSαm+1, S

ai S

bi+1S

ci+2]

By repeatedly using the commutator identities, we get:

[AB,C DE] = A[B,C]DE + [A,C]BDE + C A[B,D]E

+ C AD[B,E] + C[A,D]EB + C D[A,E]B

p3 =∑n,m

mεabc(Sαm[Sαm+1, San]Sbn+1S

cn+2 + [Sαm, S

an]Sαm+1S

bn+1S

cn+2

+ SanSαm[Sαm+1, S

bn+1]Scn+2 + SanS

αmS

bn+1[Sαm+1, S

cn+1]

+ San[Sαm, Sbn+1]Scn+2S

αm+1 + SanS

bn+1[Sαm, S

cn+1]Sαm+1)

Inserting the commuting relation for the spin algebra:

p3 ∼∑n,m

mεabc(εαadδm+1,nSαmS

dnS

bn+1S

cn+2 + εαadδm,nS

dnS

αm+1S

bn+1S

cn+2

+ εαbdδm+1,n+1SanS

αmS

dn+1S

cn+2 + εαcdδm+1,n+2S

anS

αmS

bn+1S

dn+2

+ εαbdδm,n+1SanS

dn+1S

cn+2S

αm+1 + εαcdδm,n+2S

anS

bn+1S

dn+2S

αm+1)

Using the same procedure as before and shifting appropriate terms wefinally get:

p3 = 2N∑n=1

[(~Sn × ~Sn+1)× ~Sn+2 · ~Sn+3 + ~Sn · ~Si+2]− 4p1

61

Chapter 5

Correlation functions

The material based here will be from [17], [18] and [19].The matrix element computed by inserting a product of operators be-

tween two states, usually the vacuum states, is called a correlation function.Consider an operator acting on the j-th site, S

ajj ∈ {Sxn, S

yn, Szn}. The aver-

age of this operator with respect to the ground state |0〉 is:

〈0|n∏j=1

Sajj |0〉 = 〈

n∏j=1

Sajj 〉

We will particularly be looking at the thermodynamic limit for the anti-ferromagnet (J < 0) since the correlation functions for the ferromagnet(J > 0) are trivial since the ground state has all the spins aligned. TheXXZ model for the antiferromagnetic phase (J = −1), the energy is givenby:

E =N∆

4+

M∑j=1

(cos kj −∆)

The construction of |0〉AFM (see [19]) is credited to Hulthen [23] in thethermodynamic limit N → ∞, which we also stated in Section 3. For−1 < ∆ < 1 the value in the thermodynamic limit the ground state energyper site is given by:

e0 = limN→∞

EAFMN

=∆

4− sinπν

2π

∫ ∞−∞

dzsinh (1− ν)z

sinh z cosh νz

=∆

4+

sinπν

2π

∫ ∞+i/2

−∞+i/2dx

1

sinhx

cosh νx

sinh νx

Given

H =

N∑j=1

(Sxj Sxj+1 + Syj S

yj+1 + ∆SzjS

zj+1)

62

Where ∆ = 1 corresponds to the XXX model.

e0(∆) =〈0|H|0〉N

= (2a+ b∆)

Where a, b are given by:

a = 〈Sxj Sxj+1〉 = 〈Syj Syj+1〉, b = 〈SzjSzj+1〉

By the Hellmann Feynman theorem:

〈SznSzn+1〉 =d

d∆e0(∆)

∣∣∣∆=1

,

i.e we can compute the nearest-neighbour correlation directly from our ex-pression for the anti-ferromagnetic ground state energy. It it well knownthat the nearest neighbour correlator can be obtained from the ground stateenergy per site by Hulthen [23], what is less obvious is calculating higherneighbouring correlators. I will state a few results that have been obtained:

〈SznSzn+1〉 =1

12− 1

3ln 2

=1

12− 1

3ζa(1) = −0.1477157168 · · ·

which is obtained as above from the ground state energy per site.

〈SznSzn+2〉 =1

12− 4

3ln 2 +

3

4ζ(3)

=1

12− 4

3ζa(1) + ζa(3) = 0.06067976995 · · · ,

which is obtained from the half-filled Hubbard model by Takahashi [25].The next nearest correlators and other type of correlation functions up to8 lattice sites have been calculated in many papers (see [18]) and it hasthe same pattern structure as above, given in terms of the Riemann zetafunctions1. We generally would like to compute 〈SzjSzj+n〉 ∀n.

5.1 EFP

An important class of correlation functions called the Emptiness FormationProbability (EFP) which is the probability to find the formation of a ferro-magnetic string of length n in the anti- ferromagnetic ground state |AFM〉.The EFP is defined as:

P (n) = 〈AFM |n∏j=1

Pj |AFM〉 = 〈n∏j=1

(1

2+ Szj )〉

1This was defined in the introduction

63

Where Pj = 1/2 + Szj is the projector on the state with the spin up in thej-th lattice site. The importance of the EFP was realised in [28]. It wasconjectured in [18] that P (n) is always in terms of ln 2 and Riemann zetafunctions ζ(n) with odd arguments and rational coefficients. This is alsothe case for 〈SzjSzj+n〉 since it was also conjectured that we can write thesecorrelators in terms of P (n). In particular, we can recover the first andsecond-neighbour correlators by the following relations

P (2) = 〈(1

2+ Szj )(

1

2+ Szj+1)〉 = 〈SzjSzj+1〉+

1

4

P (3) = 〈SzjSzj+1〉+1

2〈SzjSzj+2〉+

1

8

Though it is important to note that 〈SzjSzj+3〉 cannot solely be determinedfrom P (4). It was also shown in [26] that P (n) exhibits interesting Gaussiandecay at n→∞. The first four values of the EFP look as follows:

P (1) =1

2= 0.5

P (2) =1

3(1− ln 2) = 0.102284273 · · ·

P (3) =1

4− ln 2 +

3

8ζ(3) = 0.007624158 · · ·

P (4) =1

5− 2 ln 2 +

173

60ζ(3)− 11

6ζ(3) ln 2− 51

80ζ2(3)

− 55

24ζ(5) +

85

24ζ(5) ln 2

The value of P (1) is evident from the symmetry, P (2) can be extractedfrom the explicit expression of the ground state energy from 〈SzjSzj+1〉 as wehave shown above. P (3) can be extracted from the results M. Takahashi in[25] on the calculation of the next to nearest neighbour correlation, as wediscussed above. P (4) can be extracted from the integral representation ofthe EFP as in [26].

5.1.1 Integral representation of EFP

There is an explicit formula for P (n) which was determined in [28] and inthe XXX limit reads:

P (n) =

n∏j=1

∫C

dλj2πi

Un(λ1, · · · , λn)Tn(λ1, · · · , λn)

Where:

Un(λ1, · · · , λn) = πn(n+1)/2

∏1≤k<j≤n sinhπ(λk − λj)∏n

j=1 sinhn πλj

64

Tn(λ1, · · · , λn) =

∏nj=1 λ

j−1j (λj + 1)n−j∏

1≤k<j≤n(λj − λk − i)The contour C goes parallel to the real axis with the imaginary part

confined between 0 and −i for each integral, though in principle it can bechosen arbitrary between the two points.

It can be shown in [29] that the expression for P (2) reduces to the onedimensional integral for arbitrary ν as:

P (2) =1

2+

1

2π2 sinπν

∂

∂ν

(sinπν

∫ ∞−∞

sinh (1− ν)x

sinhx cosh νxdx

),

which leads to a one-dimensional integral representation for 〈SznSzn+1〉, sincewe have 〈SznSzn+1〉 = P (2) − 1/4. This reduces to the same form as theground-state energy per site e0 shown above by the Feynmann Hellmantheorem.

One can compute these type of integrals in general by transforming theintegrand to a canonical form and thereby reducing it in such a way thatthe integral doesn’t change, then one can perform the integral simply usingthe residue theorem. This method of evaluating integrals was introducedby Boos and Korepin (see [26],[27]). I will briefly go through the details forP (3) as in [27].

P (3) =3∏j=1

∫ ∞−i/2−∞−i/2

dλj2πi

U3(λ1, λ2, λ3)T3(λ1, λ2, λ3)

U3(λ1, λ2, λ3) = π6

∏1≤k<j≤3 sinhπ(λk − λj)∏3

j=1 sinh3 πλj

T3(λ1, λ2, λ3) =

∏3j=1 λ

j−1j (λj + 1)3−j∏

1≤k<j≤3(λj − λk − i)

=(λ1 + i)λ2(λ2 + i)λ2

3

(λ2 − λ1 − i)(λ3 − λ1 − i)(λ3 − λ2 − i)By utilizing the anti-symmetry and some analytic properties of U3(λ1, λ2, λ3),we can replace the integrand T3(λ1, λ2, λ3) by a certain canonical formTC3 (λ1, λ2, λ3), the details of the transformation to canonical form are givenin [26],[27], which was derived as:

T3(λ1, λ2, λ3) ∼ TC3 (λ1, λ2, λ3) = P(0)3 +

P(1)3

λ2 − λ1

Where P(0)3 = −2λ2λ

23 and P

(1)3 = 1/3− iλ1− iλ2− iλ3−2λ1λ3. Performing

the integration by the residue theorem, we obtain:

J(0)3 =

3∏j=1

∫ ∞−i/2−∞−i/2

dλj2πi

U3(λ1, λ2, λ3)P(0)3 =

1

4

65

J(1)3 =

3∏j=1

∫ ∞−i/2−∞−i/2

dλj2πi

U3(λ1, λ2, λ3)P

(0)1

λ2 − λ1= − ln 2 +

3

8ζ(3)

and hence the result for P (3) is given by:

P (3) = J(0)3 + J

(1)3 =

1

4− ln 2 +

3

8ζ(3),

which agree’s with the result as above. This result also allows us to com-pute the second-neighbour correlation function since we have 〈SznSzn+2〉 =2(P (3) − P (2)) + 1/4 which is obtained by rearranging the expression forP (3) in terms of 〈SznSzn+2〉 and substituting the expression for P (2).

5.1.2 Sato’s formula

The exact calculation of correlation functions has been a non-trivial task,except in the limit ∆ → 0 for the XX Ising model. In this case an exactformula can be derived in terms by the use of Wicks theorem as in [30],[31]:

〈SznSzn+k〉 = −(1− (−1)k)

2π2k2.

As we have mentioned before, it was conjectured that any correlation func-tions of the XXX spin chain can be represented in terms of ln 2 and Riemannzeta functions ζ(n). This conjecture was actually proved in 2006 by H.boos,M. Jimbo, T. Miwa, F smirmnov and Y.Takeyama [24]. It is reasonabletherefore to ask whether there is an explicit expression for all the rationalcoefficients that enter the expressions for the correlation function, this doesnot exist yet and is an open problem. Fortunately, a recent development byJuw Sato [18]2, who obtained a formula for the linear terms and some of thehigher terms for the coefficients. In order to present the formula let ζa(s)be the alternating Riemann zeta function, we will define:

c0 =1/4− ζa(1)

3

cn =ζa(2n− 1)− ζa(2n+ 1)

3

Then Sato’s formula is defined by:

〈SzjSzj+n〉 =

n∑k=0

(−1)k(n− 1

k

)(2k + 1

k

)ck−24

(n− 1

2

)c1(c0+c1)+higher terms

The “higher terms” represent terms which contribute for n > 7. This for-mula can reproduce the result up to n = 7, though the calculation itself can

2Though this result was not directly obtained in [18], the author has listed the referenceas a private communication, as such no other references exist to this

66

be rather tedious. For the nearest-neighbour correlator, we immediatelyhave according the formula:

〈SznSzn+1〉 =1∑

k=0

(−1)k(

0

k

)(2k + 1

k

)ck = c0 =

1

12− 1

3ζa(1),

which is the same result obtained by the ground state energy and the integralformulation.

67

Chapter 6

Conclusion

We started our discussion by deriving the Heisenberg Hamiltonian throughconsiderations of the Hydrogen Molecule and the Pauli exclusion principle.We then examined some of the symmetries of the model and saw that itpossed full rotational, translational and reflection symmetry. The key be-hind the CBA is through exploiting these known symmetries . We thensolved the model via the CBA and obtained the energy spectrum by makingan Ansatz about the coefficients for the general eigenstate of the model.By considering the periodicity condition on these coefficients led to the for-mulation of the Bethe Ansatz equation (BAE). In the context of the ABAframework, we then introduced the mathematical description of integrabil-ity. We thus showed from that the transfer matrix we can produce theHamiltonian, and generate local conserved quantities that commute withthe Hamiltonian from which we were able to deduce the integrability of themodel. The ’Ansatz’ for the ABA is the assumption that B(λ) and C(λ)through the entries of the transfer matrix are creation and annihilation op-erators respectively, this lead to conditions on the parameter λ which led tosame BAE’s as from the CBA. By using the fact that the (pseudo)-vacuum|0〉 is a highest weight vector for the global spin algebra su(2), we thenshowed that this was the case for any general eigenstate. We then lookedat the thermodynamic limit N →∞ and in particular examined the stringhypothesis. This lead us to calculate the non-trivial anti- ferromagneticground state energy.

We then stated some results for the XXZ-model in the context of theABA and the thermodynamic limit, which immediately reduced to the XXX-model in the limit ∆→ 1. We then examined the higher conserved charges,where we had to take the charges hailing from the expansion of the loga-rithm transfer matrix in order to maintain locality. We thus showed thatwe can produce a family of N − 1 indepedant local charges from the trans-fer matrix for the XXX Hamiltonian, which when by adding the total spinoperator Sz (which is also local) produces N such charges and thus proves

68

the integrability of the model. Finally, we looked at correlation functionsfor the XXX anti-ferromagnetic spin chain, first the nearest neighbouringcorrelator which we saw can be calculated directly from the ground stateenergy per site. Other correlators as we saw are not as obvious to compute,we then stated some known results in and in particular stated modestly waysto compute such correlators.

Some comments have to made on the ABA and CBA: Though the CBAand ABA led to the same spectra, the ABA has the advantage of an algebraicframework easily generalizable to other models such as the XXZ spin chain,it allows us to prove the integrability of the model which the CBA lacks andit also allows us to more readily compute other quantities such as correlationfunctions, using methods such as the quantum inverse scattering method.The CBA does provide a “picture” of the excitations in the spin chain interms of the concept of a magnon (for the ferromagnet magnon-vacuum),with which its description leads natrually to establish the momentum ofthe magnon. We also saw that the energy was additive and the that thetotal momentum of the system was conserved, this description is harder toimagine in the ABA because of its purely algebraic nature.

Though the Heisenberg XXX spin chain is one of the bestl understoodmodels in terms of its integrability, open problems within the area still existsuch as to produce a general formula for any correlation function. As we sawin Chapter 5, these correlators have been proven to be expressible in termsof polynomials of the alternating zeta function, with a regular pattern as canbe directly observed. It is reasonable to assume that such a formula doesexist, perhaps as a generalization of Sato’s formula given in Chapter 5. It isadvantageous to therefore to try to determine the coeffcients either by bruteforce in trying to recognize a pattern by finding higher correlators, or by amore elegant approach utilizing mathematical tools from number theory inthe description of the Riemann zeta function. I will personally look into sucha problem during my spare time as I feel a more mathematically rigoroustreatment is needed in order to realize the full potential of the connectionbetween the Riemann zeta function and the Heisenberg spin chain.

69

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