Spinon contributions to spin-spin correlations in the Heisenberg XXX chain

Spinon contributions to spin-spin correlations in the Heisenberg

XXX chain

Chelsea Kaandorp July 9, 2015

Verslag van Bachelorproject Natuur- en Sterrenkunde, omvang 15 EC, uitgevoerd tussen 02-03-2015 en 10-07-2015

The Institute for Theoretical Physics Amsterdam

Faculteit der Natuurwetenschappen, Wiskunde en Informatica Begeleider: Jean-S´ebastien Caux

Tweede beoordelaar: Jasper van Wezel Universiteit van Amsterdam

Abstract

In this thesis we considered the Heisenberg XXX chain without a magnetic field. The eigenstates of this system can be analytically calculated by the Bethe Ansatz. An experimentally measurable property of the system is the transverse dynamical structure factor (TDSF). The TDSF is the Fourier transform of the spin-spin correlation function which gives a value to the correlation of spins on different times or on different sites. In this thesis we compute correlations on a state with zero magnetization. The spin-spin correlation function can be written as a sum over states of the system. In this thesis we will look at the contribution of states with an number of downspins equal to M = N

2 − 1. The focus will be on the two and four spinon states. First is calculated how many states are needed to get 99% of the sum in the two and four spinon cases. This number goes approximately with a polynomial of one term with the power 2.0_{± 0.15% respectively} 3.6± 0.18%. Secondly is looked at the relative contribution to the sum of different four spinon states. The states differed from each other by the set of quantum numbers with which they are described. The states with quantum numbers the nearest to zero have the greatest contribution to the sum.

Populair wetenschappelijke samenvatting

Misschien heb je wel eens van elektronen, protonen en neutronen gehoord. Dit was waarschijnlijk bij Scheikunde, maar natuurkundigen kunnen deze kleine deeltjes ook heel boeiend vinden. Alle dingen om ons heen zijn namelijk van deze deeltjes gemaakt. Dit is voor natuurkundigen de reden om meer over deze deeltjes te willen weten. Wat we in dit project hebben onderzocht is een model voor een ketting van deze kleine deeltjes.

Waarom zouden we hiernaar willen kijken? Komen er in de echte wereld ook kettingen van deeltjes voor? Niet van nature, maar we kunnen ze wel maken. We kunnen ze niet alleen in het echt maken, maar we kunnen ook deeltjes erop af schieten en metingen doen. Het resultaat van zo’n meting is dat we meer van die deeltjes en materialen gaan snappen. Zo kunnen we bijvoorbeeld meer leren over waar magnetisatie vandaan komt.

Om magnetisatie beter te begrijpen moeten we eerst een begrip uitleggen. Het begrip ‘spin’. Spin is een eigenschap van kleine deeltjes. Spin zegt iets over hoe een deeltje rond spint. Als een deeltje linksom draait, tekenen we dat als een pijltje omhoog. En als een deeltje rechtsom draait, tekenen we een pijltje omlaag. Onze ketting ziet er nu uit als allemaal pijltjes die omhoog en omlaag staan.

Wat heeft dit nu met magnetisatie te maken? Nou, als alle pijltjes dezelfde kant op staan, dan hebben we een magneet! Als we er dan eentje de andere kant op zetten, wordt de magneet zwakker.

Je kunt je misschien voorstellen dat als je pijltjes gaat omdraaien, dat dat invloed heeft op andere pijltjes op de ketting. In dit project hebben we gekeken hoe het omdraaien van een pijltje invloed heeft op een ander pijltje. Dit hebben we gedaan door te kijken naar spinons. Spinons zijn deeltjes die natuurkundigen hebben bedacht om zo’n ketting beter te kunnen beschrijven. Dat klinkt misschien nep maar het is wel echt iets wat kan bewegen over de ketting. Om te begrijpen wat een spinon is, gaan we eerst beginnen met een ketting die totaal niet magnetisch is. Dat is een ketting waarin de pijltjes om-en-om omhoog en omlaag staan. Deze ketting lijkt op de bovenste ketting op het plaatje. Als we er nu een neutron op schieten, kunnen we een pijltje omdraaien en krijgen we iets zoals de middelste ketting. Wat we willen, is die ketting opknippen in niet-magnetische delen. Dat zijn de rode vlakken die je ziet. Zo’n rood vlak is wat natuurkundigen dan een spinon noemen.

Ik zei net dat natuurkundigen spinons zien als deeltjes die over de ketting kunnen bewegen. Hoe bewegen ze dan? Nou soms gebeurt het dat als een pijltje omhoog naast een pijltje omlaag staat, dat ze dan allebei omflippen. Dan krijg je dus omlaag-omhoog in plaats van omhoog-omlaag. Als dit gebeurt, verplaatsen onze rode vlakken en dus onze spinonen.

Door te kijken naar spinonen kunnen we meer snappen over hoe dat omflippen van pijltjes gaat. Hoe bijvoorbeeld het omflippen van ´e´en pijltje invloed heeft op een ander pijltje. Of

als we dan een pijltje hebben omgeflipt, of hij daarna weer terug wilt flippen.

Dat die rode vlakken spinons heten, is natuurlijk geen toeval. We hadden het immers over pijltjes die we de ‘spin’ van een deeltje noemden. Met die spinons en spins kunnen we dus meer snappen over de dingen om ons heen en magnetisatie.

Contents

1 Introduction 6

2 Theoretical framework of the Bethe Ansatz 7

2.1 Introduction to Heisenberg XXX chain . . . 7

2.2 Introduction to the Bethe Ansatz . . . 8

2.2.1 Two downspins: M = 2 . . . 9

2.2.2 Higher values of M . . . 11

2.3 Spinons . . . 15

3 Results of numerical calculations 19 3.1 Cutoff . . . 20

3.2 Relative contributions of the one-strings tohSj−(t)Sj+0(0)i 4 . . . 23

3.3 Relative contributions of the two-string to_hSj−(t)S_j+0(0)i 4 . . . 24

4 Discussion & conclusion 26

A Transverse dynamical structure factor 28

B Parameters XXZ model 29

C The f-sumrule 30

D Sum rule saturation 31

E Deviation 33

1

Introduction

This thesis is about the isotropic Heisenberg spin-1/2 chain without a magnetic field at zero temperature. This model is important to learn more about quantum magnets. The system where we are looking at is a chain made of N particles at fixed positions j. In this thesis ~ is set equal to one. The hamiltonian of this system is

H = J N X j=1  S_jxS_j+1x + S_jyS_j+1y + S_jzS_j+1z −1 4  ,

where J is called the exchange coupling (Orbach 1958). This is the same as writing

HXXX= J N X j=1  1 2  S_j+S_j+1− + S_j−S_j+1+ + S_jzS_j+1z −1 4  ,

in which Sa with a ∈ {x, y, z, +, −} are spin-1/2 operators. The chain we consider is a closed one, such that j + N _{≡ j.}

An experimentally measurable property of the system is the transverse dynamical structure factor (TDSF): S−+(k, ω) = 1 N N X j,j0₌₁ eik(j−j0) ∞ Z ∞ dteiωt_hSj−(t)S+_j0(0)i .

The TDSF is written in terms of momentum and energy (k respectively ω) (Caux et al. 2008). The angular brackets in the TDSF denote the ground state expectation value at zero-temperature with the time denoted by t. The ground state will be denoted by |GSi. The notation conventions and theoretical background of this thesis are from the lecture notes Caux (2014) and the master thesis Mossel (2008).

The factor in the integral can be written in a Lehmann representation such that

hSj−(t)Sj+0(0)i =

X

{α}

| hGS| Sj−|αi |2e−it(Eα−EGS),

where we sum over all states _{|αi of the system (see appendix A). The aim of this study} was to know which states have the biggest contribution to the summation in the Lehmann representation. We particularly focus on the contribution of_{| hGS| Sj}−_{|αi |}2 _{to the sum in the}

equation above.

This system can be solved analytically with the Bethe Ansatz. A method that uses the Bethe Ansatz to calculate properties of the system is the ABACUS method. This method is developed by J.-S. Caux. In this thesis we used ABACUS and made numerical calculations.

2

Theoretical framework of the Bethe Ansatz

In this thesis, we considered the XXX Heisenberg model. We concentrated on a chain with N spin-1/2 particles without a magnetic field. A more general version of the XXX Heisenberg model is the XXZ Heisenberg model. The hamiltonian of the XXZ Heisenberg model is

HXXZ = J N X j=1  1 2  S_j+S_j+1− + S_j−S_j+1+ + ∆(S_jzS_j+1z −1 4)  , (1)

where ∆_{∈ R. Notice that coupling in the x- and y- direction are equal whereas the coupling} in the z-direction is different. Therefore the XXZ Heisenberg model is called anisotropic. The XXX Heisenberg model however is called isotropic, because the coupling in the z-direction is the same as in the x- and y-direction, as ∆ is equal to one in the XXX Heisenberg model. In the following section we will not fix ∆ in order to give the most general story.

A chain with spin-1/2 particles contains particles with spin up or down. The number of down spins is labeled by M and is related to the magnetization of the system in the following way M _≡ N 2 − * _N X j=1 S_jz + . (2)

J can take positive and negative values. In case of J > 0, the ground state of the system has M = N₂ and is called the antiferromagnetic state. Another special state is the one with all spins pointing up along the z-direction. This state will be called the reference state and can be written as

|0i = ⊗Nj=1|+ij. (3)

It is efficient only to consider the states with M ≤ N2. Because for M ≥ N

2 the whole

chain could be turned upside down, which results in the same states for M _≤ N₂.

With this reference state, the states can be written down in terms of down spins only in the following way

|j1, j2, ..., jMi ≡ M

Y

i=1

S_i−|0i , (4)

where ji labels the i’th downspin which is situated at site j.

An eigenstate can now be written as

|ΨMi =

X

j1≤...≤jM

The dimension of the whole Hilbertspace is 2N _{because each site has two options; up and}

down. The Hilbertspace can be devided in fixed magnetization subspaces. This can be done because the total magnetization commutes with the Hamiltonian.

The dimension of a fixed magnetization subspace is given by  N

M 

.

Most of the time we are interested in the energies of the system. So we want to diagonalize this matrix. The Bethe Ansatz gives us a tool to find the eigenfunctions and to calculate the energies of the states.

2.2 Introduction to the Bethe Ansatz

In 1931 Hans Bethe obtains the correct form of the wave functions of the Heisenberg model (Bethe 1931). When periodicity conditions are imposed on the wave functions, we find the Bethe equations. The solution of the Bethe equations gives a set of quantum numbers with which states of the system can be described.

In order to understand where the Bethe Ansatz comes from it is convenient to consider the downspins on the chain as pseudo-particles which each a momentum. These particles are called magnons. The number of particles is equal to M . When two particles scatter, their values of momentum can be interchanged by conservation of momentum. So in a system with M particles, the momenta can be distributed in M ! ways by scattering. The possible permutations of distribution of momenta will be denoted with _{P. When particles scatter,} their wave functions can pick up a phase shift, also called a scattering phase. The idea of the Bethe Ansatz is to solve the Schr¨odinger equation with a wave function in the form of a product of free particle wave functions ψ1 accompanied by an amplitude AP. AP is related

to the scattering phase. This product has to be summed over P, which gives us the wave function ψM(j1, ..., jM) = X P AP M Y i=1 ψ1(ji, kPi). (6)

In order to get familiar with the Bethe Ansatz, we first look at a state where we have two downspins followed by a state with M downspins.

2.2.1 Two downspins: M = 2

Getting a wave function In order to calculate the wave function, we start with the eigenstate from equation (5) with M = 2:

|Ψ2i =

X

j1<j2

Ψ2(j1, j2)|j1, j2i . (7)

Projecting the Schr¨odinger equation onto the bra _hj1, j2| yields the conditions

J 2[Ψ2(j1− 1, j2) + Ψ2(j1, j2− 1) + Ψ2(j1+ 1, j2) + Ψ2(j1, j2+ 1)] = (E2+ 2J∆)Ψ2(j1, j2), 2 < j1+ 1 < j2< N, (8) J 2[Ψ2(j1, j2+ 1) + Ψ2(j1− 1, j2)] = (E2+ J∆)Ψ2(j1, j2), 2 < j1+ 1 = j2 < N. (9)

To solve these equations, we are going to use the Bethe Ansatz. As described in the previous subsection in equation (6), this ansatz is a sum over products of free wave equations and contains amplitudes A. In the case of M = 2 this means

Ψ2(j1, j2) = A12eik1j1+ik2j2 + A21eik2j1+ik1j2, 1≤ j1< j2 ≤ N. (10)

This wave function is an eigenfunction of the anisotropic Hamiltonian and gives us the energy

E2= J(cos k1+ cos k2− 2∆). (11)

Nevertheless, this eigenfunction is not yet fully determined and does not yet meet all the physical characteristics of the system.

First, the amplitudes need to be determined. This can be done by substituting equation (11) into equation (9) and using the equality j1+ 1 = j2. We then get

A12

A21

=₋1 + e

i(k1+k2)_{− 2∆e}ik1

1 + ei(k1+k2)−2∆eik2 ≡ −e

−iφ(k1,k2)_{, (12)}

where φ(k1, k2) is defined as the ‘scattering phase shift function’ and is given by

φ(k1, k2) = 2 arctan ∆ sink1−k2 2 cosk1+k2 2 − ∆ cos k1−k2 2 . (13)

This function has a branch point at cosk1+k2

2 −∆ cos k1−k2

2 = 0. The branch can be chosen

in a way such that φ(0, 0) = 0 and φ(k1, k2) =−φ(k2, k1). Implementing the found amplitudes

in equation (12), gives the non-normalized wave function

Ψ2(j1, j2) = eik1j2+ik2j2−

i

2φ(k1,k2)− eik2j1+ik1j2+ i

In order to have a cleaner notation of the scattering phase shift function, new parameters are introduced. These parameters differ for different values of ∆. We will now give the parameters for the case ∆ = 1, so the XXX Heisenberg model. In appendix B the same sort of parameters are given for the case|∆| > 1. The following will hold some math, but hold on, the results will be very useful.

We first introduce a parameter called rapidity and notated by λ. The momentum can be written as function of rapidity:

kXXX(λ) = π− 2 arctan (2λ), (15) = π− θ1, XXX(λ), (16) where θn, XXX(λ) = 2 arctan( 2λ n). (17)

Note that λXXX(k) = 1₂cot(k₂). The function θn, XXX(λ) is not going to be used in this

section, be will be needed for rewriting the Bethe equations later in this section.

The scattering phase shift function can now be written in a more elegant way. The scattering phase shift function in terms of λ for ∆ = 1 is

φXXX(λ) = λ +

i

2. (18)

For other values of ∆ we can also define a k(λ), φ(λ) and θn(λ). So for M = 2, the

wavefunction for the XXZ Heisenberg model is

Ψ2(j1, j2) = eik(λ1)j2+ik(λ2)j2−

i

2φ(λ1,λ2)− eik(λ2)j1+ik(λ1)j2+ i

2φ(λ1,λ2). (19)

This is not yet the most general wave function possible. Note that we defined Ψ2(j1, j2)

in equation (7) with j1 < j2. A more general form of the wave function is one without this

constriction. This can be done by turning equation (19) into

Ψ2(j1, j2)

= sgn(j2− j1)

h

eik(λ1)j1+ik(λ2)j2−₂isgn(j2−j1)φ(λ1,λ2)_{− e}ik(λ2)j1+ik(λ1)j2+₂isgn(j2−j1)φ(λ1,λ2)

i . (20) Now we can make it look more like the ansatz we started with in equation (6) in the following way Ψ2(j1, j2|λP1, λP2) = sgn(j2− j1)sgn(λ2− λ1) X P (−1)[P]_eik(λ_P1)j1+ik(λ_P2)j2−₂isgn(j2−j1)φ(λ_P1,λ_{P2). (21)}

We now have an non-normalized eigenfunction of the XXZ Heisenberg hamiltonian in the case M = 2 in all her glory. It is written in terms of ji and λ. A state is determined by a

set of λ’s. In the following sections we will reconsider these sets of λ’s and illustrate which values λ can take. Note that this function vanishes when λP1 = λP2. This is a manifestation

of an underlying Pauli-like principle.

Getting the Bethe equations Like we said in the previous paragraph, we now are going to further investigate the set of λ’s. Let us start with looking at the periodicity conditions. The chain is made of N particles and we stated that site j is the same site as site j + N . When we take the wave function written as in equation (19) , ie j1 < j2, we have the following

periodicity conditions

Ψ2(j2, j1+ N ) = Ψ2(j1, j2), Ψ2(j2− N, j1) = Ψ2(j1, j2). (22)

Putting together equations (19) and (22) yields the Bethe equations

eik(λ1)N _=−e−iφ(k(λ1),k(λ2))_{, e}ik(λ2)N _=−e+iφ(k(λ1),k(λ2))_{. (23)}

Not only can this result be seen as a product of periodicity conditions. This result can also be seen as a product of scattering factors. In this case we start with particle 1 on spot j1 and particle 2 on spot j2. In order to get particle 1 at spot j1+ N , particle 1 has to pass

particle 2. In this way they will scatter. Scattering gives an extra phase to the wave function. The Bethe equations can also be written down in a logarithmic form

N k(λ1) + φ(k(λ1), k(λ2)) = 2πI1, (24)

N k(λ2)− φ(k(λ1), k(λ2)) = 2πI2, (25)

where Ii are half odd integers.

Note that these equations are coppled equations. When I1and I2are known, the equations

can be solved for k(λ1) and k(λ2). So instead of describing the state with a set of λi, the state

can also be described by a set of integers Ii. These Ii will be called the quantum numbers of

the system.

2.2.2 Higher values of M

We now understand that in the case of M = 2 the Bethe Ansatz comes down to an ansatz for a wave function. The periodicity conditions of the system give us a set of coupled equations with which we can compute all states. We can do the same for a general value of M . The

eigenfunctions of the XXZ Heisenberg model are of the following form: ΨM(j1, ..., jM) = Y M ≥a>b≥1 sgn(ja− jb)× X PM (₋₁₎PeiPMa=1kPaja+₂iPM ≥a>b≥1sgn(ja−jb)φ(kPa,kPb)_{. (26)}

The eigenstates of these eigenfunctions are characterized by a set of rapitidites λj, with

j = 1, ..., M .

Each eigenfunction has a solution to the time-independent Schr¨odinger equation with as eigenvalues the energies

EM = J M

X

a=1

cos ka− ∆. (27)

Similar to the case for M = 2, the chain with M downspins is a periodic one. The periodicity conditions are

ψM(j1, ..., jM) = ψM(j2, ..., jM, j1+ N ). (28)

Like in the previous paragraph, the periodicity conditions give us M coupled Bethe equa-tions with the scattering phase shift function

eikaN _{= (−1)}M −1_e−iPb6=aφ(ka,kb)_{, a = 1, ..., M. (29)}

The Bethe equations can also be written in a logarithmic form in the following way

ka+ 1 N X b6=a φ(ka, kb) = 2π Ia N, (mod2π) a = 1, ..., M, (30)

where Ia are half-odd integers if M is even, and integers if M is odd.

Parametrization for ∆ = 1 We now have the Bethe equations in the normal and in the logarithmic form. Remember that φ(ka, kb) depends on the value of ∆. In the case of ∆ = 1

we can rewrite the Bethe equations with our rapidities  λa+ i/2 λa− i/2 N =Y b6=a λa− λb+ i λa− λb− i . (31)

The Bethe equations can be written in a logarithmic form. In order to do this, the we use the identity 1 2ilog  1 + ix 1− ix  = arctan (x). (32)

The logarithmic Bethe equations for the XXX Heisenberg model are given by 2 arctan 2λa− 2 N M X b=1 arctan (λa− λb) = 2π NIa, (33) θ1, XXX(λa)− 1 N M X b=1 θ2, XXX(λa− λb) = 2π NIa, (34)

where we have used our previous definition of θn, XXX as written in equation (17).

The Bethe equations for_{|∆| > 1 can also be written differently then equation (30). θ}n,XXZ

can be defined in such a way that for the XXZ Heisenberg model, the logarithmic Bethe equations are given by

θ1(λa)− 1 N M X b=1 θ2(λa− λb) = 2π N Ia. (35)

Now the set of rapidities with which we can describe the wave function can be determined. When the values of the rapidities of a state are known, the energy of this state can be calculated. With ∆ = 1, the energies are

EXXX=−J M X a=1 2 4λ2 a+ 1 . (36)

Note that the lowest energy is given by a state with rapidities the closest to zero.

To summarise the previous sections: we find eigenfunctions of the XXZ hamiltonian and we can describe states by a set of λi or by a set of Ii. The next section will describe more

about which values can be taken for λi and Ii.

Which values can we take for λ? At this stage, we did not specify the values of λ yet. Is an imaginary or a real number? Can it take an infinite value? In this paragraph we will discuss possible values of λ and the way they influence the Bethe equations for the XXZ Heisenberg model.

Inspection of the Bethe equation shows that if two rapidities are the same, they also have the same quantum number and represent the same state

For λ an imaginary number, a set of λ’s exists where all the λ’s in the set have the same real parts, but different imaginar parts. This set of λ’s with the same real part is called a n-string, where n stands for the number of λ’s in the set. If a certain value of λ is in the set, its self-conjugate is also an element of the set. The real part which they share is notated by λn_α. It is possible to have more than one strings with the same n. This is why the label α is used.

Before we said that for an eigenstate with M down spins, the number of rapidities was equal to M . This is still the case, however the rapidities can now also be a part of a n-string. If the number of strings of length n is notated by Mn. We have

∞

X

n=1

nMn= M. (37)

The total number of strings for a value of M is given by

∞

X

n=1

Mn≡ Ns. (38)

The Bethe equations have to be modified for these string solutions. They are called the Bethe-Gaudin-Takahashi equations: θn(λnα)− 1 N Ns X m=1 Mm X β=1 Θnm(λnα− λmβ) = 2π N I n α. (39)

The string-string scattering phase shift is given by

Θjk(λ) = (1− δjk)θ|j−k|(λ)+ 2θ|j−k|+2(λ) + ... + 2θj+k−2(λ)+ θj+k(λ). (40)

For λ a real number, we make a distinction between finite and infinite λ.

The Bethe equations give a set of quantum numbers associated to different sets of rapidities. A feature of the Bethe equations is that if λn > λm, In > Im1. The Bethe equations allow

taking the limit of λ to infinity. This rapidity will be associated with the quantum number I∞_{. What will happen with the logarithmic Bethe equations if the limit of λ→ ∞ is taken?}

First, notice that the following is true

lim

λ→∞θn(λ) = π, (41)

lim

λ→∞Θjk(λ) = (2min(j, k)− δjk)π. (42)

So taking the limit of λ→ ∞ of equation (39) will give

Ij,∞ = 1 2 N + 2j− 1 − Ns X m=1 (2min(n, m)− δnm)Mm ! . (43)

For example, in the case of a state with only one-strings we have

I_j1,∞ = N− M1+ 1

2 . (44)

1

There are cases for which this condition does not hold. This condition is true for the states which we consider in this thesis.

Where M1 is written instead of M because we only have one-strings.

The values of λ and I are connected. If I∞ is known, the maximal value of the quantum numbers is also known. This will give us a method to calculate the finite possible values of λ. In the case of the one strings the maximum value of the quantum numbers is given by

I1,max= I1,∞_{− 1.} (45)

More generally, the maximum quantum number of an n-string is given by

In,max= In,∞− n. (46)

All quantum numbers obey the inequality |In

j| < In,∞. This is why the total amount of

available quantum numbers for an n-strings is

2· In,max+ 1. (47)

We now know more about the different values that λ can take. It can be a real of an imaginary number. Finite or infinite. Taking the limit of λ to infinity, gives us information about the number of quantum numbers available. Taking a imaginary value for λ gives us n-strings. In the following section we work with a few cases with different n-strings. The cases are specific for the case M = N_{2 − 1 because this is the case we are going to work with} eventually. We are going to see that if we change the number of strings, we can get different states with all the same number of downspins. The property that will distinguish the different states will have something to do with the so called ’spinons’.

2.3 Spinons

With the Bethe ansatz, we can describe a state with a set of quantum numbers. First we are able to calculate the total number of quantum numbers available. Namely 2Imax_{+ 1. It is}

also known how many quantum numbers are needed because of M is fixed. The number of coupled Bethe equations is equal to M , so a state can be described with M different quantum numbers. It is possible to have more quantum numbers available than needed for describing a state. When we think of quantum numbers as sites that can be filled on the number line and we do not need to fill all sites, we have holes on the sites which are not filled. Having two holes is called a two spinon state. Having four holes is called a four spinon state. We will now look into two situations to clearify this subject.

Case 1: Only one-strings If we consider a state with only one-strings. This means that M1 is non-zero and the Mn for n > 1 are equal to zero. In order to get equation (37) right,

This is the state which we get if we begin with the antiferromagnetic state and flip one up-spin down.

The corresponding Bethe equation is

θ1(λa)− 1 N M1 X b6=a θ2(λa− λb) = 2π NI 1 a. (48)

Firstly, we need to find out how many quantum numbers we have. This is done by first looking for I1,∞. We already calculated I1,∞ in equation (44). This was done by taking the limit of λa to infinity in the Bethe equation. In this case M1 = N₂ − 1, so I1,∞ = N₄ + 1.

We now have Imax

1,j = I1,j∞ − 1 = N4 and the number of available quantum numbers is equal to

2Imax

1,j + 1 = N2 + 1. We then have N2 + 1 spots available to fill on the number line, but we

only need N_{2 − 1 spots. This gives us two holes and is therefore called a two spinon state. For} example, take N = 12, we then have 5 quantum numbers taken and still 2 spots left. This will give us different states. As example two different states are

-3 -2 -1 0 1 2 3

⊗ ⊗ ⊗ ⊗ ⊗

⊗ ⊗ ⊗ ⊗ ⊗

.

Note that the first row stands for the state with the lowest energy. This is because this is the state with quantum numbers the closest to zero. A state with these quantum numbers is equivalent with a state with a set of rapidities the closest to zero, and by that with the lowest energy by equation (36).

The possible ways of distributing the quantum numbers on the number line is equal to N 2 + 1 N 2 − 1  . (49)

So the total number of two spinon states is

N2 =

1

8(N + 2)N. (50)

It can be proven that the state with only one-strings gives all the two spinon state for M = N_{2 − 1.}

Case 2: One strings and just one two-string The second case which we are going to describe is a state with one-strings and only one two-string. So we take M2 equal to one. In

First we look at the Bethe-Gaudin-Takahashi equations for the one-strings: θ1(λ1α)− 1 N M1=N₂−3 X β6=α θ2(λ1α− λ1β) − 1 NΘ21(λ 1 α− λ2β) = 2π NI 1 α. (51)

Also in this case we want to know the number of available quantum numbers. We proceed as usual. First I1,∞ is needed and is calculated by taking the limit of lambda to infinity of equation (51) (remember limλ→∞θn(λ) = π and limλ→∞Θjk(λ) = (2min(j, k)− δjk)π)

lim λ∞ α→∞ θ1(λ1α)− 1 N M1=N₂−3 X β6=α θ2(λ1α− λ1β) − 1 NΘ21(λ 1 α− λ2β)≡ 2π NI 1,∞ α . (52) We thus find 2π NI (1),∞ _{= π−} π N( N 2 − 4) − 2π N, I1,∞ = N 4 + 1, (53)

from which follows that

I1,max = N

4 , (54)

2I1,max+ 1 = N

2 + 1. (55)

The needed number quantum numbers was M1 = N₂ − 3. So there are 4 more available

quantum numbers than needed. This means that there are 4 holes. This is called a four spinon state.

To total number of possibilities of chosing a set of quantum numbers is now equal to N 2 + 1 N 2 − 3  = ( N 2 + 1) N 2( N 2 − 1)( N 2 − 2) 4! , = 1 16 (N + 2)N (N − 2)(N − 4) 4! , = (N + 2)N (N − 2)(N − 4) 384 .

This was the situation for only the one-strings of this configuration. The configuration is named after the state of the one-string. So in this case, we will call the whole configuation a four spinon state.

Nevertheless, it is also interesting to calculate the quantum numbers of the two-string. Equation (39) gives us θ2(λ2α)− 1 N N 2−3 X β Θ21(λ2α− λ1β) = 2π NI 2 α.

Taking the limit for λ to infinity gives us an expression for I2,∞ lim λ2 α→∞ θ2(λ2α)− 1 N N 2−3 X β Θ21(λ2α− λ1β)≡ 2π NI 2,∞ α . (56) We thus find 2π N I 2,∞ ₌ 1 2(N− ( N 2 − 3) · 2), (57) I2,∞ = 3, (58)

from wich follows that

I2,max= 1, (59)

2I(2),max+ 1 = 3. (60)

So there are three available quantum numbers.

The total number of states we can get for having one two-string and one-strings is now the product of number of possibilities to choose the quantum numbers:

3×(N + 2)N (N − 2)(N − 4)

384 . (61)

The total number of four spinon states is therefore equal to

N4 =

(N + 2)N (N − 2)(N − 4)

128 . (62)

To summarise, each state can be described by a set of quantum numbers. Sometimes more quantum numbers are available then needed. In this case we talk about holes and spinons. The number of two spinon states scales with a power of N2 _{and the number of four spinon}

3

Results of numerical calculations

General introduction We have looked to several systems varying the number of particles, the number of spinons or the value of momentum. The data files for each system were pro-duced by ABACUS and contained information about the states, such as energy, momentum, | hGS| Sj−|αi |2 and places of the holes on the numberline. All files were sorted on the values

of| hGS| Sj−|αi |2.

As told in the introduction, a way to calculate the transverse dynamical structure factor is to first calculate the spin-spin correlation functionhSj−(t)Sj+0(0)i. The spin-spin correlation

function can be written as a sum over all states in the following way

hSj−(t)Sj+0(0)i =

X

{α}

| hGS| Sj−|αi |2e−it(Eα−EGS). (63)

For this thesis we have only looked at the following sum (see appendix C for explanation)

X

{α}

| hGS| S−j |αi |2. (64)

In the following programs the contribution of just a fraction of states to the sum in equation (64) will play a significant role. We mainly look at the relative contributions of different states and conclude which states contribute the most. The sum over a fraction of states will be notated by

hSj−(t)Sj+0(0)i

k

i , (65)

where k stands for the value of the momentum and i the spinon state. If no k or i are notated, the values are not fixed. The number of states of a certain fraction of states will be notated by

Nik. (66)

For this thesis we made the choice only to look at the two and four spinon states. We did this because their contributions to hS−j (t)Sj+0(0)i is significant (for further explanation, see

appendix D). For fixed momentum, we have chosen to look at the values k = π and k = π₂. The contribution per state is of the form| hGS| Sj−|αi |2. The value of| hGS| S

−

j |αi |2gives

an idea of the overlap between a state _{|αi and the ground state. The overlap is the greatest} when these states are alike. As said in the previous section, a state with the holes located near the boundaries has the lowest energy and therefore is likely to have a bigger overlap with the ground state. Our expectation is that this nature will show in the results.

3.1 Cutoff

Description Inspection of the data files showed that the first states in both the two and four spinon cases had a significant higher value of| hGS| Sj−(t)|αi |2than the rest of the states.

The first question was, how many states are needed to get 99% of_hSj−(t)S_j+0(0)i

k

i. The number

of states needed to get 99% of the integral is named the ‘cutoff’ in this thesis and will be notated by_N(99)_{. The choice of 99% is motivated by how precise measurements are taken in}

current research.

The second question was how the cutoff is related to the number of particles, N .

In this section we have looked at six different kind of states. The used state was either a two spinon state or a four spinon state. For each spinon state we have looked at three different situations, namely k unfixed, k = π or k = π₂.

For each kind of state,_hSj−(t)S+ j0(0)i

k

i could be calculated by ABACUS. So 99% of this value

was easily calculated. In order to know how many states are needed to come down to this value we wrote a program. This program summed over the first values of_{| hGS| Sj}−(t)_{|αi |}2

of the sorted file until the 99% was reached and calculated the number of terms.

As shown in equations (50) and (62), the number of spinon states goes polynomial with the orders

N2(N )∼ N2, (67)

N4(N )∼ N4. (68)

This is true for general value of k. A reason to look at fixed values of k, is that the number of states will be a factor N less than with unfixed value of k. This is because the number of degrees of freedom is one less.

So we then have Nk=π, π 2 2 (N )∼ N, (69) Nk=π, π 2 4 (N )∼ N3. (70)

A reason to look at a fixed value of k is that less states are needed to be calculated by ABACUS. Which makes it easier to look at system with higher value of N .

Because_Nk=π,π2(N ) goes with a factor less, we expect the order ofN(99),k=π, π

2(N ) to lower

with one value in comparison to when k is not fixed.

We expected the cutoff to be a polynomial function of the number of particles. Looking at the number of degrees of freedom of a state can provide us an argument for expecting

a certain order for the polynomials. A state can be described by k and ω, so there are two degrees of freedom. A two spinon state is also described by two parameters, namely the places of the two holes. A direct mapping between the states described by k and ω and two spinon states described by the places of holes is then expected. The number of two spinon states goes with a power of two. Therefore we also expect the cutoff of the two spinon state to go with a power of two. In the case of the four spinon states, we do not have any arguments to estimate the dependence of _N4(99) on N .

Results In the three figures below, the calculated cutoff for six kind of states are plotted against the number of particles used. Also is chosen to fit a polynomial function through the calculated points. The error on the fit is only the error given by the fitting program.

Fit cutoff: General k

1.0· 103 2.0· 103 3.0· 103 4.0· 103 5.0· 103 0 50 100 150 200 250 N (99) N 2 spinon states 0 0.5· 106 1· 106 1.5· 106 2· 106 0 50 100 150 200 250 N 4 spinon states

Figure 2: The dots are the calculated values of the cutoff. The dotted line is a fitted function. The used function is a_{· N}m _{with parameter values a = 0.1± 1.6% and m = 2.0 ± 0.15% for}

the two spinon case and with parameter values a = 5.1· 10−3_{± 3.6% and m = 3.6 ± 0.18%}

Fit cutoff: k = π 0 10 20 30 40 50 60 70 80 0 100 200 300 400 500 N (99) ,k = π N 2 spinon states 1· 104 2· 104 3· 104 4· 104 5· 104 0 100 200 300 400 500 N 4 spinon states

Figure 3: The dots are the calculated values of the cutoff. The dotted line is a fitted function. The used function is a_{· N}m with parameter values a = 0.21_{± 3.3% and m = 0.93 ± 0.60% for} the two spinon case and with parameter values a = 6.4_{· 10}−3_{± 2.6% and m = 2.5 ± 0.17%}

for the four spinon case

Fit cutoff: k =π 2 10 20 30 40 0 100 200 300 400 500 N (99) ,k = π 2 N 2 spinon states 2· 104 4· 104 6· 104 8· 104 10· 104 0 100 200 300 400 500 N 4 spinon states

Figure 4: The dots are the calculated values of the cutoff. The dotted line is a fitted function. The used function is a· Nm _{with parameter values a = 0.12± 4.9% and m = 0.96 ± 0.85% for}

the two spinon case and with parameter values a = 3.0_{· 10}−3_{± 2.3% and m = 2.8 ± 0.14%} for the four spinon case

The following three subsection are written about calculations with the four spinon states.

3.2 Relative contributions of the one-strings to _hSj−(t)S+ j0(0)i

4

Description This program is about the one-strings of the four spinon state. We looked at the correlation between the placement of the holes and the contribution to_hS−_j (t)S_j+0(0)i

4. For

every hole is looked at the relative contribution to_hSj−(t)S_j+0(0)i

4 of the terms| hGS| S − j |αi |2

for a fixed site of one hole. So for example, we fixed the place of the first hole on site 1 and summed over the values of _{| hGS| Sj}−_{|αi |}2 _{of all states with the first hole fixed on site one.}

The ‘ratio’ is calculated by taking that sum and dividing it byhSj−(t)Sj+0(0)i

4.

Results This program is used to calculate the relative contributions for different values of N . Namely N _{∈ {32, 64, 96, 128, 160, 192, 224}. The results for different values of N were} quite similar. In figures (5) and (6) the results for N = 32 and N = 224 are shown.

What we see in the small window within the graph is that the system is symmetric. The states with holes located on the borders contribute the most tohSj−(t)S+j0(0)i

4. However the

contribution of the states with holes in the middle can not be neglected. In this program we did not look at the relation between the holes.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 −8 −6 −4 −2 0 2 4 6 8 Ratio Site

Relative contribution per site, N = 32

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 −8 −6 −4 −2 0 2 4 6 8 hole 1 hole 2 hole 3 hole 4

Figure 5: For every hole the relative contribution tohSj−(t)Sj+0(0)i

4is considered. So for

exam-ple, we fixed the place of the first hole on site 1 and summed over the values of_{| hGS| Sj}−_{|αi |}2

of all states with the first hole fixed on site one. The ratio is calculated by taking that sum and dividing it byhSj−(t)Sj+0(0)i

4. In the small window, the ratio for all the holes are plotted.

Notice that the result is symmetric. This is why only the ratios of the first two holes are plotted in the bigger window.

0 0.1 0.2 0.3 0.4 0.5 0.6 −50 −40 −30 −20 −10 0 10 20 30 40 50 Ratio Site

Relative contribution per site, N = 224

0 0.1 0.2 0.3 0.4 0.5 0.6 −50−40−30−20−10 0 10 20 30 40 50 hole 1 hole 2 hole 3 hole 4

Figure 6: For every hole the relative contribution to_hSj−(t)S_j+0(0)i

4is considered. So for

exam-ple, we fixed the place of the first hole on site 1 and summed over the values of| hGS| Sj−|αi |2

of all states with the first hole fixed on site one. The ratio is calculated by taking that sum and dividing it by_hSj−(t)S+

j0(0)i

4. In the small window, the ratio for all the holes are plotted.

Notice that the result is symmetric. This is why only the ratios of the first two holes are plotted in the bigger window.

3.3 Relative contributions of the two-string to _hSj−(t)S_j+0(0)i

4

Description In case 2 in section (2.3) the two strings states were labeled with the quantum numbers I _{∈ {−1, 0, 1}. In this section the aim was to determine whether or not there is a} difference in the choice of I for the contribution to hSj−(t)Sj+0(0)i

4. Note thathS −

j (t)Sj+0(0)i

4

should be the same for I =_{−1 and I = 1 because of symmetry.}

The state with I = 0 resembles the most with the ground state because the holes are located near the boundaries. Following this we can expect this choice for I will have the greatest contribution. We do not have any reasons to assume that in the thermodynamic limit, I = 0 will be the only state to contribute to hS−j (t)Sj+0(0)i

Results The relative contributions of the different two-string states are pictured below. 0 0.2 0.4 0.6 0.8 1 0 50 100 150 200 Ratio N

Distribution two string

I = 0 I _{∈ {−1, 1}}

Figure 7: Relative contribution of the two string states to _hSj−(t)S_j+0(0)i

4

Discussion & conclusion

Cutoff In section (2) the number of the two and four spinon state where given as follows

N2 = 1 8(N + 2)N = 1 8N 2_{+ O(N ), (71)} N4 = (N + 2)N (N − 2)(N − 4) 128 = 1 128N 4_{+ O(N}3_{). (72)}

In this project we fitted the polynomial functionNi(99)= a· Nm through our results, with

a and m free parameters. The found values of m are presented as ‘power’ in the following tabel.

Value of k Power _N2(99) Power _N4(99) - 2.0_{± 0.15%} 3.6_{± 0.18%} k = π 0.93_{± 0.60%} 2.5_{± 0.17%} k = π₂ 0.96± 0.85% 2.8± 0.14%

In the two spinon case, _N2(99) is proportional to _N2 as was expected. This means that

for high values of N it will not be more efficient to only calculate 99% of the sum instead of calculating the whole sum. In the four spinon case, _N4(99) is not proportional to _N4. So

the computer capacity needed to calculate 99% of hSj−(t)Sj+0(0)i is less than when 100% is

calculated. For high values of N this will save computer capacity.

It is important to note that we fitted a polynomial fit with only one term, althoughN2 and

N4 have more terms. This is done because of the small amount of data points available. The

different polynomial behaviour between_N4(99) and _N4 can be measured more accurate when

more data points are available. In that case polynomial functions which are more similar to N2 andN4 could be fitted through the data points. It could also be interesting to look at the

error and compare that with the expected error of equations (71) and (72).

There were more data points available in the cases of fixed momenta. What we see is that the orders of the cutoff for fixed momenta was one less, as expected. In order to say more about the found powers, more knowledge is needed about the momentum distribution over states.

Relative contributions of the one-strings to hSj−(t)Sj+0(0)i

4 Figures (5) and (6) show

that the relative contributions are symmetric and that the states with the holes on the border have the greatest contribution to the sum.

The first argument deals with the manner in which we excite the system. The configuration is such that we first begin with a chain in an antiferromagnetic state and we then flip one of the up-spins down by scattering this spin with a neutron. This is how we get a M = N_{2 − 1} state. One spin flip creates directly an excitation of a two spinon states. This first excitation will carry the most of the transferred momentum and energy of the neutron. Sometimes more excitations are created, since all the spins on the chain are interacting with each other. These higher order excitations will carry much less of the transferred energy and momentum. This effect is visible in figures (5) and (6). What we see is that two holes have a high contribution only when they are located near the boundary, the other two holes can also have a significant contribution when they are in the middle. Their contribution is still the greatest when they are near one of the boundaries. This shows that only one spinon is likely to move, while the other spinons stay near the boundaries.

The second argument explains why the holes near the boundaries have such a high contri-bution. By calculating _{| hGS| Sj}−_{|αi |}2 _{we take into account the overlap between the ground}

state and state _{|αi. Looking at equation (36) we know that in the XXX Heisenberg model} the state with the lowest energies is described by a set of λ the closest to zero. This is the same as saying that the quantum numbers are the closest to zero. Therefore, the state with the holes near the boundaries have the biggest overlap with the ground state.

For further studies it may be interesting to look at the relation between the energy of states and the value of_{| hGS| Sj}−_{|αi |}2_.

Another proposal for ulterior research is to make a classification of states. We have already identified the state with the first and fourth hole at the boundary as a special one.

A last proposal would be to think about a method to picture the relation between the distribution of holes and other characteristics of the state. The difficulty is that we then would like to picture something with five dimensions. Four dimensions are taken for the distribution of the four holes and the last one can be the value of for example the energy.

Relative contributions of the two-string to _hSj−(t)S_j+0(0)i

4 What we see in figure (7)

is that the contribution of the I = 0 case increases for higher values of N . With the eye it seems that the data points go with an asympotic behaviour to 0.15 respectively 0.85. It was not possible to make an exploration of the data points. This would probably be possible when more data points are available. It could be interesting to do calculations in the thermodynamic limit and then compare these results with the results in the thermodynamic limit.

A

Transverse dynamical structure factor

The transverse dynamical structure factor (TDSF) is given by

S−+(k, ω) = 1 N N X j,j0₌₁ eik(j−j0) ∞ Z ∞ dteiωt_hSj−(t)S+_j0(0)i , (73)

where the angular brackets denote the ground-state expectation value at zero-temperature. In general a state can be written as a time dependent wave function using unitary time evolution and a basis of Bethe states_{ψn} in the following way

|Ψ(t)i = e−i ˆHt|Ψ(0)i =X

n

e−iEnt_c

n|ψni . (74)

The initial state_{|Ψ(0)i is determined by the coefficients c}n=hψn|Ψ(0)i.

We now are going to look at the average in the integral of equation (73). This term is called the spin-spin correlation function. It is a measure of the correlations of spins on different times and different sites. Note that the spin-spin correlation function is a ground-state expectation value:

hSj−(t)Sj+0(0)i = hGS| S_j−(t)S_j+0(0)|GSi . (75)

By using the identity

X

α

|αi hα| = 1 (76)

we can rewrite the spin-spin correlation function

hGS| Sj−(t)Sj+0(0)|GSi = X {α} hGS| Sj−(t)|αi hα| Sj+0 |GSi , (77) =X {α}

hGS| eiHtS_j−e−iHt|αi hα| S+

j0|GSi , (78)

=X

{α}

hGS| Sj−|αi hα| Sj+0|GSi e−it(Eα−EGS), (79)

=X

{α}

| hGS| Sj−|αi |2e−it(Eα−EGS), (80)

where|αi are intermediate states and where we have used the unitary matrix. So _hSj−(t)S_j+0(0)i in the Lehman representation is given by

hSj−(t)Sj+0(0)i =

X

{α}

| hGS| Sj−|αi |2e−it(Eα−EGS). (81)

The ground state is a zero-magnetization state, so the number of downspins is M = N₂. Note that_{hGS| Sj}−_{|αi is only nonzero if |αi is a state with M =} N_{2 − 1.}

B

Parameters XXZ model

In this appendix we are going to give an overview of the Bethe Ansatz for the XXZ Heisenberg model. In section 2 we gave a derivation for the isotropic case by starting with the hamiltonian of which we wanted to know the eigenstates. The hamiltonian for the XXZ Heisenberg model is H = J N X j=1  1 2  S_j+S_j+1−+ S− j S + j+1  + ∆(S_jzS_j+1z ₋1 4)  , (82)

of which eigenstates are in the form of

|ΨMi = N

X

j1<j2<..<jM

ΨM(j1, j2, ..., jM)|j1, j2, ..., jMi . (83)

The Bethe Ansatz is a solution of the wave function for a general value of M and is given by ΨM(j1, ..., jM) = Y M ≥a>b≥1 sgn(ja− jb)× X PM (₋₁₎PeiPMa=1kPaja+₂iPM ≥a>b≥1sgn(ja−jb)φ(kPa,kPb)_{. (84)}

Solving the periodicity conditions yields the Bethe equations. The Bethe equations in the logarithmic form are as follows:

ka+ 1 N X b6=a φ(ka, kb) = 2π Ia N, (mod2π) a = 1, ..., M. (85)

It is convenient to write the Bethe equations in terms of rapidities λ. In this way the Bethe equations can be written down more elegantly. The parametrization is different for different classes of ∆. First we need to define some new functions. These functions are given in the table below.

∆ = 1 |∆| < 1 ∆ > 1

ζ - arccos(∆) arccosh(∆)

φn(λ) λ + in₂ sinh λ +inζ₂ sin λ +inζ₂

θn(λ) 2 arctan 2λ_n
2 arctan  tanhλ tannζ₂  2 arctan  λ tanhnζ₂ 

We can now write the momentum and the logaritmic Bethe equations in terms of rapidity as

and θ1(λa)− 1 N M X b=1 θ2(λa− λb) = 2π N Ia. (87)

C

The f-sumrule

A property of the TDSF is that it obeys several sum rules. One of them is the f-sumrule. The f-sumrule is given by

X k ∞ Z −∞ dω 2πS −+ (k, ω) = N 2 (88)

and is only the case where the correlation is evaluated on a zero-magnetization state (Caux et al. 2008).

The f-sumrule can be written differently. First we need to rewrite the TDSF in the following way: S+−(k, ω) = 2πX α | hGS| Sk−|αi | 2_δ(ω − Eα+ EGS). (89)

Now putting equations (88) and (89) together, gives

X k ∞ Z −∞ dω 2π2π X {α} | hGS| Sk−|αi | 2_δ(ω − Eα+ EGS) = N 2, (90) X {α} | hGS| Sk−α−kGS|αi | 2 ₌ N 2. (91)

Note that the delta function drops out when we integrate over ω. A similar thing happens with the energies in the exponent of the spin-spin correlation function. This is why this exponent is not taken into account in our programs.

S_k− can be written in terms of S_j− using the Fourier series

S_k−= √1 N N X j=1 eikjS_j−. (92)

D

Sum rule saturation

In the previous section we described that the TDSF obeys the f-sumrule. The fraction in which a part of the states contributes to the total f-sumrule is called the sum rule saturation (SRSAT). Plotted are the SRSAT’s for different values of k. What can be shown is that for low N the contribution of the two spinon states is high. For a larger number of particles the contribution of four spinon states becomes larger. We expect that at a certain value of N the contribution of the four spinons states will decrease and the contribution of higher spinon states will increase. We expect this because for a higher value of N higher spinon states are possible to be accessed.

SRSAT 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 50 100 150 200 250 300 SRSA T N General k 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 150 200 250 300 350 400 450 500 550 SRSA T N k = π 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 150 200 250 300 350 400 450 500 550 SRSA T N k =π 2 Two spinon Four spinon Two spinon Four spinon Two spinon Four spinon

E

Deviation

The used data files contained numerical calculated values of_{| hGS| Sj}−_{|αi | for every state. The} data files also contained the estimated error of these values called the ‘deviation’ and denoted by δ. In this section we researched if we could make a motivated choice for a maximum deviation. The states with a deviation above this maximum deviation are then not taken into account when the cutoff is calculated. This is because we have stated that their errors are too big.

In order to look if we could make this choice, we wrote a program. It calculates the following sum for different values of a maximum deviation ‘δmax’

X

{α} δα<δmax

| hGS| Sj−|αi |2. (93)

A motivation to choose a certain value of δmax is when there is an interval where the

sum above does not seem to depend on the maximum deviation. In the following figures, we calculated the sum above in the cases N ∈ {32, 96, 160, 224}.

Deviation 0.7 0.75 0.8 0.85 0.9 0.95 1 Ratio N = 32 0.7 0.75 0.8 0.85 0.9 0.95 1 N = 96 0.7 0.75 0.8 0.85 0.9 0.95 1 −10 −8 −6 −4 −2 0 Ratio log(δmax) N = 160 0.7 0.75 0.8 0.85 0.9 0.95 1 −10 −8 −6 −4 −2 0 log(δmax) N = 224

Figure 9: The ratio is calculated by P

{α},δα<δmax | hGS| Sj−|αi |2 / P {α}| hGS| S − j |αi |2

. This ratio is calculated for different values of δmax and different number of particles N .

What we see in the figure above is that there does not seem to exist an interval where the sum does not depend on the maximum deviation for every value of N . We can not make a motivated choice with this result. This is why we have chosen to set the maximum deviation equal to zero.

F

Acknowledgements

This bachelor thesis is made in three months at the Institute for Theoretical Physics Amster-dam (ITFA) at the University of AmsterAmster-dam. Normally, bachelor students can work on their theses in the library of the university, but during my project I got a really special chance. My supervisor Jean-S´ebastien Caux could get me a working place at the ITFA where I could work at the same place as the research group and easily ask my questions to all the people of this group. For answering all my questions with a lot of patience I want to thank in special Rianne van den Berg, Jean-S´ebastien Caux, Rogier Vlijm and Bram Wouters.

For being so kind to be my second assessor I want to thank Jasper van Wezel.

Another advantage of working at the ITFA was that I could have lunch with the research group. This was really a learning experience which I enjoyed a lot.

References

Bethe, H. 1931, Zeit. f¨ur Physik, 71, 205

Caux, J.-S. 2014, Integrability in atomic and condensed matter physics, NORDITA school

Caux, J.-S., Mossel, J., & Castillo, I. P. 2008, Journal of Statistical Mechanics: Theory and Experiment, 08

Mossel, J. 2008, Master’s thesis

Mourigal, M., Enderle, M., Kl¨opperpieper, A., et al. 2013, Nature Physics, 09