High Temperature CuprateSuperconductors

Year Four Theoretical Physics MSci Project

Toby Kingsman

Supervised by Dr Martin Long

Abstract

Our aim is to try and understand the key physics that underpins the experimentally observed resultsof cuprate based high temperature superconductors. We shall consider the effects of doping bothholes and electrons into such materials with a view to creating a theoretical framework that explainsthe asymmetry in their phase diagram. Our method of approach shall be to derive an extendedHamiltonian to model the appropriate interactions within the copper oxide layers. We will showa description for hole-doping that better matches the experimental phase diagram than previouslyproposed models.

School of Physics and AstronomyMarch 2015

Acknowledgements

I would like to take a moment to recognise the impact of several other people on the outcome of thisproject. Particular thanks must go to my supervisor Dr Martin Long for his invaluable help in theproject, the majority of the work being in a similar vein to his previous works in the area over thecourse of his career. He has put up with me with good humour and shown much patience throughoutthe duration of this project despite being a busy man himself.

A special mention must go to my good friend and fellow project partner Ben Horvath who has spent histime this year looking at the same area as myself. This has resulted in many illuminating discussions anddisagreements between us, but having to justify myself to him on many occasions has greatly benefitedmy understanding. It is safe to say that without his company and constant reassurances, I would nothave reached this stage with my sanity entirely intact.

Finally to my friends and family who have taken an interest in my project this year and have beenconstantly asking for a brief or concise explanation into what I have been studying. I now understandthat whilst I have only started to scratch the surface of low temperature physics, I have learnt enoughto provide a reasonable answer to your questions !

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Contents

1 Introduction 41.1 History of High Temperature Superconductors . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Structure of the Cuprates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Crystal Field Splitting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Cuprate Structure 72.1 Introducing Dopants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Parent Compound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.3 Doping the Cuprates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.4 History of Proposed Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3 Second Quantisation 113.1 Defining Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.2 Anti-Commutator Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

4 Deriving the Hamiltonian 124.1 Anderson Lattice Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124.2 Extensions to the Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.3 Transformation of the Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

5 Calculating S 155.1 Part 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155.2 Part 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165.3 Part 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175.4 Part 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175.5 Solving for S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

6 Finding H2 196.1 Solving for (i) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196.2 Solving for (ii) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206.3 Solving H2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

7 H2 for Hole-Doping 247.1 Derivation of Zhang-Rice Singlet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

8 H2 for Electron-Doping 30

9 Superexchange 319.1 The Principle of Superexchange . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319.2 Derivation from the Hubbard Model in Strong-Coupling Limit . . . . . . . . . . . . . . . 31

10 Solving our Hamiltonians 3410.1 Models and Geometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3410.2 Electron Doping - 1D Linear Chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3610.3 Electron Doping - Ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3610.4 Cu3+ Limit - 1D Linear Chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

10.4.1 Term 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4010.4.2 Term 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4110.4.3 Solving the Remainder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

10.5 Cu3+ Limit - Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4310.6 Cu3+ Limit - Square . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4610.7 Connection with Nagaoka’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

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11 Spin Physics 5111.1 Singlet and Triplet Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5111.2 Exchange Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

12 1D Linear Chain - Cu+ Model 5512.1 Oxygen Hopping Part . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5512.2 Singlet Part . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5612.3 Combining the combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5712.4 Illustration of States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

13 Redefined Hamiltonian - Cu+ Model 6113.1 First Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6213.2 Second Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6313.3 Third Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

13.3.1 Denominator of Variational Approximation . . . . . . . . . . . . . . . . . . . . . . 6413.3.2 Numerator of Variational Approximation . . . . . . . . . . . . . . . . . . . . . . . 6513.3.3 Spin Exchange Operator Simplifications . . . . . . . . . . . . . . . . . . . . . . . . 6713.3.4 Results from Takahashi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6813.3.5 Minimising the Variational Approximation . . . . . . . . . . . . . . . . . . . . . . 68

14 Conclusions 70

3

1 Introduction

1.1 History of High Temperature Superconductors

The phenomenon of superconductivity was first discovered in 1911 by Heike Kamerlingh Onnes, who wasstudying the resistivity of mercury at low temperatures. He discovered that the resistivity dropped toessentially zero for temperatures below 4.2K. Further discoveries followed as physicists tried to discoverboth other materials that went superconducting and had higher critical temperatures. It took untilaround seventy years later for the next big discovery to occur, when in the 1980’s cuprate compoundswere found to have critical temperatures higher than the believed theoretical limit of 30K. These havebeen known since as the much studied high temperature cuprate superconductors which are still notfully understood to this day.

1.2 Structure of the Cuprates

The common feature between all the cuprate superconductors as seen below in Figure 1 is that theyall contain layers of copper oxide. These layers have a square planar 2-dimensional structure and areillustrated at the top and bottom of the unit cells of the La2−xSrxCuO4 and HgBa2CuO4+δ cupratesuperconductors. It is this shared layer of copper oxide that is the key to many of the universal propertiesof the cuprates, namely that of superconductivity.

Figure 1: The layered structure of high temperature cuprate superconductors containing the universalCuO2 layers [3].

We will be primarily considering the undoped parent compound known as La2CuO4, in order to build upthe chemistry of the copper oxide layers. If we consider the charges on each of the ions, we can work outthe overall charge on the Cu atoms in the plane. This produces La3+

2 CuθO2−4 , and the knowledge that

the overall charge on the parent compound is zero means that θ = 2. So the copper atoms are actuallyof the form Cu2+.

Cu2+ = 1s22s22p63s23p63d9

O2− = 1s22s22p6

As a result we have that every copper has a single hole located in its 3d orbital, whereas the oxygensform completely closed 2p orbitals [5].

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Figure 2: Every copper site has a single hole and every oxygen site is unoccupied by holes.

1.3 Crystal Field Splitting

We know that if each copper atom were in an isolated region away from other atoms then the 3d orbitalitself would be completely degenerate. However we need to identify the crystal field effect on the Cu2+

ions, as each copper is enclosed by six oxygens forming an octahedron around it as seen in Figure 3.

Figure 3: Octahedron enclosing the central Cu ion [4].

The first splitting of the energy levels of the orbitals is due to the fact that the eg (dx2−y2 and d

3z2−r)

orbitals point directly towards oxygen ions, whereas the t2g orbitals (dxy, dzx and dyz) point towards agap between oxygen ions. As a result the eg orbitals are raised to a higher level due to the larger relativecoulombic repulsion they experience.

Figure 4: Schematics of the Cu 3d9 orbitals.

There is an additional splitting of the eg orbitals due to the tetragonality of the system. As the apical

5

oxygen atoms are further from the central copper atom than the in-plane oxygen atoms, this causes thedx2−y2 orbital to split off from the d

3z2−r orbital. The half occupied orbital then becomes the d

x2−y2

orbital, as seen in Figure 5.

Figure 5: The splitting of a copper 3d level with 9 electrons [18].

Applying the same ideas to the oxygen 2p orbitals themselves we find that the two key orbitals we needto consider are the 2px and 2py orbitals. As seen in Figure 6 and the overall square planar structure,there are cases where both of these orbitals point directly at the copper 3d

x2−y2 orbitals and so are

raised to a higher energy level when compared to pz.

Figure 6: Schematics of O 2p6 orbitals.

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2 Cuprate Structure

2.1 Introducing Dopants

We have seen that due to the chemistry of the problem we are able to simplify the essential physics downto that of a three band model, where the copper oxide layer is compromised of 3d

x2−y2 , 2px and 2py

orbitals. As we have discussed the undoped state has every copper site occupied by a single hole, 3d9,and every oxygen oxygen site empty of holes, 2p6.

Figure 7: Atomic structure of CuO2 plane. Each copper atom has a 3dx2−y2 orbital which overlaps with

the appropriate 2px and 2py oxygen orbitals [3].

The question then becomes one of do the doped holes and electrons choose to reside when they areintroduced to the copper oxide layer through doping? We know that additional electrons entering thesystem would like to form full bands, so they reside on the copper sites to form, 3d10 orbitals.

We would also expect that extra holes released into the system would also like to fully occupy anorbital, so they should end up on the copper orbitals to form, 3d8. However EELS (Electron EnergyLoss Spectroscopy) experiments [14] have determined that this is not what actually happens. It turns outthat the doped holes actually reside on the oxygen orbitals to form, 2p5 orbitals. Essentially it is morepreferably to form O− rather than Cu3+, because of the size of Coulomb energy for double occupation.

Figure 8: U, is the energy needed to doubly occupy a copper site, ∆ is the energy gained by singlyoccupying a copper site [18].

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2.2 Parent Compound

We shall start by considering the undoped parent compound La2CuO4 once again. As we explainedearlier a key orbitals is the 3d

x2−y2 copper orbital which contains a single electron. As this is half-filled,

traditional band theory dictates that this material must be a conductor. This is not what happensphysically though, as it has been observed that the parent compound is actually an insulator [2]. Thereason why usual band theory has apparently broken down is due to the fact that the energy required todoubly occupy a site U, is much bigger than that of the kinetic energy for a single hop between sites,t.As a result the compound prefers to keep the electrons localised. These type of insulators are known asMott insulators.

The logical extension to the Mott insulating nature of the undoped cuprates is that the parent compounddisplays perfect antiferromagnetism. Although the electrons are prevented from doubly occupying sites,they are able to virtually hop between sites. The most energy is gained back from this interaction whenadjacent copper spins are anti-parallel, so that Pauli exclusion principle is obeyed if both were to vir-tually hop onto the same oxygen site. This virtual hopping causes the minimum energy solution to bethat of an antiferromagnetic alignment of spins.

Figure 9: Due to the Pauli principle, the moments induced by the holes on two neighboring copper atomshave to be opposite in order that both holes can meet on the oxygen site [9].

2.3 Doping the Cuprates

The magnetic nature of the cuprate superconductors themselves depends strongly on factors such asthe temperature, amount of doping and the type of dopant used. The procedure of hole doping itselfis performed by replacing some of the La3+ ions for Sr2+ ions, which introduces x additional holes percopper to the copper oxide layer. In the electron doping case Nd ions are replaced by Ce to introduceadditional electrons to the copper oxide layer. As we vary the concentration of dopant x, experimentalistsare able to determine the phase of diagram of both types of dopants as seen below.

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Figure 10: Phase diagram of the cuprate superconductors. Electron-doping is displayed on the left andhole-doping is displayed on the right [10].

Considering electron-doped cuprates such as Nd2−xCexCuO4 we se there is a region of antiferromagneticphase which is preserved until about x=0.14, where the Neel temperature of the system vanishes. Thena short lasting region of superconductivity emerges until x=0.18.

The hole-doped side of the spectrum provides us with an altogether more exotic range of behavioursfor the system. In this case the antiferromagnetic phase is completely destroyed at around 2% dopingand then a strange spin-glass behaviour is exhibited until a level of 5% doping. This spin-glass behaviouris thought of as where the spin fluctuations of the compound start slowing down. Continuing to increasethe levels of doping yields us with the superconducting state which is preserved until x=0.25. This re-gion has the classic dome shape of all high temperature superconductors, with an optimal doping level of16% which corresponds to a critical temperature, Tc, of 40K for La2−xSrxCuO4. Doping past the super-conductive state we eventually return to the normal state which is well explained by Fermi-Liquid physics.

Finally there is also the presence of a pseudo-gap region, which is not well understood. Its namesoriginates from the fact that there exists a gap in the spin excitations. It is believed that this is theregion where the superconductivity and antiferromagnetism directly compete and also exhibits otherstrange properties such as having resistivity that is linear in temperature.

It is also worth noting that the families of hole-doped cuprates tend to have critical temperatures that areon average 2-3 times larger than that of electron-doped cuprates. For example the highest known criticaltemperature belongs to Hg-1223, a member of the hole-doped (HBCO) family, and is 130K, whereas theelectron-doped (LSCO) family only reaches around 40K. Empirical observations have suggested that thecritical temperature is proportional to the number of CuO2 planes [6], for example Tl2Ca2Ba2Cu3O10

has three CuO2 planes and a critical temperature of 125K.

The key point of interest for us is that the antiferromagnetic phase is disappears at a much lowerlevel of doping for electrons than for holes, a key asymmetry in the phase diagram. This is suggestive oftwo different mechanisms at work, each being responsible for one type of doping.

It shall be shown later on that the traditionally used models cannot predict this experimentally observedasymmetry in the phase diagram and therefore must lack a mechanism for breaking the symmetry be-tween hole doping and electron doping. Our main objective is to see if we can propose a model thatcan explain this asymmetry better than previous models and produce predictions for the ground state

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energy.

2.4 History of Proposed Models

We previously stated that the inherent physics of the cuprate superconductors can be explained by theuse of a full 3 band model. A Hamiltonian to describe this was first put forward by Emery and Reiter [16]and retains the essential character of the problem. The issue was however that it was a very complicatedmodel containing many parameters which meant it was to all intense purposes intractable. As suchpeople wished to try and map the problem onto a simpler one by reducing the number of bands needingto be considered.

The culmination of the efforts to simplify the problem down to a one band model came in 1988 throughthe work of Zhang and Rice [22]. They made several simplifications in their work that allowed themto create the now well known t-J model. This became widely accepted for its apparent simplicity andthe fact that it could be unified with the already mainstream Hubbard model in the appropriate strong-coupling limit [6]. Due to the t-J model’s success little effort was given to developing alternatives to itdespite the objections of Reiter, who continued to argue that it could not describe the essential physicsas it disregarded the oxygen orbitals.

However in more recent years it has been shown that the t-J is indeed an oversimplification of theproblem after all. As we will highlight, by ignoring the superexchange antiferromagnetic interaction(J = 0) the resulting t model predicts that for square loops we will obtain a ferromagnetic ground state,which disagrees with experiment.

We have decided to consider a form of the Anderson Lattice Model as our starting point, which isbased on the three band model, since this is now commonly agreed to be the best starting point whenconsidering the cuprate superconductors.

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3 Second Quantisation

Second quantisation is a technique that allows us to simplify the formulation of many body problems inquantum mechanical and condensed matter systems. The main idea is that we can encode the results ofquantum mechanics and statistics within the method by focusing on the occupation of particular states,rather than by utilising a quantum-mechanical wavefunction.

As we are considering fermionic particles such as electrons and electronic holes, the correct approach isto employ anticommutation relations. These are commutation relations of the form, {A,B} =AB+BA={B,A} , for two operators A and B. By using anti-commutators we automatically ensure that the Pauliexclusion principle is maintained as well as the fact that fermions remain antisymmetric under exchange.

This is the language that we will be using to describe the mathematical model of the cuprate layersin subsequent chapters. We shall briefly discuss the main points that need to be considered in ourapplication of the technique to conclude this section.

3.1 Defining Operators

In our model there is not only a single creation or annihilation operator, but an entire set of them whereeach individual operator refers to a particular state. In general as we are concerned with copper d-shelland oxygen p-shell holes, we will use the following representation:

• Creation of a p-shell hole at state j with a spin σ is: p†jσ

• Annihilation of a p-shell hole at state j with a spin σ is: pjσ

• Creation of a d-shell hole at state i with a spin σ is: d†iσ

• Annihilation of a d-shell hole at state i with a spin σ is: diσ

We will generally use the convention that i will be for copper sites and j for oxygen sites. Note that σ isdefined to always be the opposite spin to σ. So if σ =↑, then σ =↓ and vice versa.

3.2 Anti-Commutator Relations

The only anti-commutator relations that do not commute in general are,

{diσ, d†lσ} = δilδσσ = {d†lσ, diσ} (1)

{pjσ, p†mσ} = δjmδσσ = {p†mσ, pjσ} (2)

While the ones that always commute are,

{diσ, dlσ} = 0 = {pjσ, pmσ} (3)

{d†iσ, d†lσ} = 0 = {p†jσ, p

†mσ} (4)

{diσ, pjσ} = 0 = {pjσ, diσ} (5)

{d†iσ, pjσ} = 0 = {pjσ, d†iσ} (6)

{diσ, p†jσ} = 0 = {p†jσ, diσ} (7)

{d†iσ, p†jσ} = 0 = {p†jσ, d

†iσ} (8)

We will see in Chapter 4, that we will need to calculate out lengthy commutators involving multipleterms in terms of their constituent fermionic anti-commutator relations. This will be done by employingidentities similar to the following, where A,B,C and D are operators,

[AB,CD] = C[AB,D] + [AB,C]D

= C(A{B,D} − {A,D}B) + (A{B,C} − {A,C}B)D

= CA{B,D} − C{A,D}B +A{B,C}D − {A,C}BD

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4 Deriving the Hamiltonian

4.1 Anderson Lattice Model

It has been proposed that a ground state Hamiltonian for a CuO2 square plane might be the followingHamiltonian.

H0 = εd∑iσ

d†iσdiσ + εp∑jσ

p†jσpjσ + U∑i

d†iσdiσd†iσdiσ (9)

The first term in this Hamiltonian corresponds to awarding an energy cost εd to every copper hole, whilstthe second awards an energy εp for every oxygen hole. The third term adds in an energy cost U if aparticular copper site is doubly occupied with holes, noting that we have forbidden the violation of thePauli exclusion principle on these sites.

We can make further simplifications to this model by setting the energy of the oxygen holes, εp, equal tozero. We have already seen that the energy gap between having a single hole on the copper and addinga single hole to the oxygen is ∆ in Figure 8. Hence εp − εd = ∆, which allows us to replace εd with −∆.

H0 = −∆∑iσ

d†iσdiσ + U∑i

d†iσdiσd†iσdiσ (10)

We also need to allow for the possible hopping of copper holes to the oxygen and oxygen holes to thecopper. So we now introduce a hopping term that allows the possibility of this, with V representingthe hybridisation energy of the hopping. We use the <ij> to represent the fact that i and j must all beadjacent to each other.

H1 = V∑<ij>σ

(d†iσpjσ + p†jσdiσ) (11)

The appropriate Hamiltonian to use to model the system is as follows and is also known as the AndersonLattice Model [19],

H = H0 +H1 = −∆∑iσ

d†iσdiσ + U∑i

d†iσdiσd†iσdiσ + V

∑<ij>σ

(d†iσpjσ + p†jσdiσ) (12)

Figure 11: Schematic describing the CuO2 layer for an arbitrary site i. l is a nearest neighbour coppersite to i, j is one of the nearest neighbour oxygen sites m to i.

It would be of interest to know the values of the parameters described in our model, however they cannotbe determined directly from experiment. Only quantities such as the optical gap ∆op [21], the bandwidth

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W [7], superexchange J and cyclical Jc coupling can be measured experimentally. It is then necessaryto calculate the values of these experimental parameters in terms of ∆, U and V , so that we can get anapproximate estimate for the values we are interested in. Long and Styles [11] found a good match with∆ ≈ 2eV, U ≈ 10eV and V ≈ 0.75eV.

4.2 Extensions to the Hamiltonian

It is important to notice that our proposed model of the copper oxide layer is a degenerate system.This is due to the fact that there are lots of different ways of forming the ground state of the system.For example in the parent compound, we have a single hole on every copper and every oxygen emptyof holes. If we were to dope just a single hole it could reside on any of the multitude of oxygen sitesin the layer causing a massive degeneracy in the system. There is also a large contribution due to thespin degeneracy as well, since ever hole has a spin. As a result we need to ensure that we maintain thisdegenerate ground state when we calculate extensions to the Hamiltonian.

We now need to consider the possibility of hopping between copper and oxygen orbitals in our model asdescribed by H1. This extension to the Hamiltonian provides a hybridisation between the copper andoxygen orbitals, but actually needs to be excluded from our ground state. For instance in the undopedcase, H1 would hop a hole from a copper site to an oxygen site causing us to exit the ground state. Inorder to preserve the ground state we need to consider even numbers of hops, the first of which cor-responds to H2. With this we could hop a doped hole from an oxygen to a copper and then back toanother oxygen orbital maintaining the ground state.

4.3 Transformation of the Hamiltonian

So the aim will now be to try and obtain an extension to the Hamiltonian which corresponds to the caseof two hops between orbitals, whilst eliminating our first order hopping Hamiltonian. The method weshall employ was first proposed by Schrieffer as a canonical transformation of the following form [17],

f(λ) = eλS(H0 +H1)e−λS (13)

Here f(1) = eS(H0 + H1)e−S is the change of basis from our original Hamiltonian and S is of orderV. We will now try to represent f(λ) as a Taylor expansion, so we can consider orders of V. We cancalculate the first derivative of f with respect to λ in the following way,

df

dλ= eλS .(H0 +H1).− Se−λS + SeλS .(H0 +H1)e−λS

= eλS(S.(H0 +H1)− (H0 +H1).S)e−λS

= eλS [S, (H0 +H1)] e−λS

(14)

It is merely a matter of algebraic manipulation to again achieve the result for the second derivative of fin terms of λ,

d2f

dλ2 = eλS [S, [S, (H0 +H1)] ] e−λS (15)

Now we can use these differentials to apply Taylor’s theorem to expand f(λ) out, noting that S is oforder V.

f(1) = (H0 +H1) + [S, (H0 +H1)] +1

2[S, [S, (H0 +H1)] ] + . . .

= H0 + (H1 + [S,H0] ) + ([S,H1] +1

2[S, [S,H0] ] ) + . . .

(16)

The next step is required by the definition of the ground state. As we want to remain in the groundstate, we want to eliminate any order V dependence since this corresponds to a single hop, yet allow

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order V2 dependence, which is two hops. We can see that the order V dependence is now completelycontained within the second term, so we require the following to be true,

H1 + [S,H0] = 0⇒ H1 = −[S,H0] = [H0, S] (17)

We can then further substitute this requirement back into f(λ) to let us calculate the new Hamiltionian.

f(1) = H0 + (H1 + [S,H0] ) + ([S,H1] +1

2[S, [S,H0] ] ) + . . .

= H0 + 0 + [S,H1] +1

2[S,−H1] + . . .

= H0 + [S,H1] − 1

2[S,H1] + . . .

= H0 +1

2[S,H1] + . . .

(18)

So we obtain the new Hamiltonian which has the small perturbation to it H2 as,

H = H0 +1

2[S,H1] = H0 +H2 (19)

The aim for the next part of the project is to solve equations (17) and (19) to obtain a form for S andthe new Hamiltonian H2.

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5 Calculating S

As we wish to calculate a correct form of S we need to evaluate out the following commutator to obtaina form for S. This is a long and tedious process that will take up the rest of this chapter.

H1 = −[S,H0] = [H0, S] (20)

We already have the correct form for H which is given in (12), but we shall guess the following form forS, where the constants α and β need to be determined,

S = αS0 + βS1

S0 =∑<ij>

(d†iσpjσ − p†jσdiσ)

S1 =∑<ij>

(d†iσpjσ − p†jσdiσ)d†iσdiσ

(21)

Hence we now want to evaluate the following commutator,

H1 = [−∆∑<iσ>

d†iσdiσ + U∑i

d†iσdiσd†iσdiσ, αS0 + βS1] (22)

Once we substitute the values for S0 and S1 into the above equation, we then need to utilise dummylabels as below,

H1 = [−∆∑lτ

d†lτdlτ , α∑<ij>

(d†iσpjσ − p†jσdiσ)]

+ [−∆∑lτ

d†lτdlτ , β∑<ij>

(d†iσpjσ − p†jσdiσ)d†iσdiσ]

+ [U∑l

d†lτdlτd†lτdlτ , α

∑<ij>

(d†iσpjσ − p†jσdiσ)]

+ [U∑l

d†lτdlτd†lτdlτ , β

∑<ij>

(d†iσpjσ − p†jσdiσ)d†iσdiσ]

(23)

From now onwards we shall refer to the first commutator in (23) as (a), the second as (b), the third as(c) and the fourth as (d).

5.1 Part 1

Now we can calculate the first one of these commutators explicitly:

(a) = −∆α∑lτ

∑<ij>

[d†lτdlτ , d†iσpjσ − p

†jσdiσ]

= −∆α∑lτ

∑<ij>

([d†lτdlτ , d†iσpjσ] − [d†lτdlτ , p

†jσdiσ] )

= −∆α∑lτ

∑<ij>

(d†iσ[d†lτdlτ , pjσ] + [d†lτdlτ , d†iσ] pjσ − p

†jσ[d†lτdlτ , diσ] − [d†lτdlτ , p

†jσ] diσ)

(24)

Let us evaluate out the required commutators,

[d†lτdlτ , pjσ] = d†lτ{dlτ , pjσ} − {d†lτ , pjσ} dlτ = d†lτ .0− 0.dlτ = 0

[d†lτdlτ , d†iσ] = d†lτ{dlτ , d

†iσ} − {d

†lτ , d

†iσ} dlτ = d†lτδilδστ − 0.dlτ = d†lτδilδστ

[d†lτdlτ , diσ] = d†lτ{dlτ , diσ} − {d†lτ , diσ} dlτ = d†lτ .0− δilδστdlτ = −δilδστdlτ

[d†lτdlτ , p†jσ] = d†lτ{dlτ , p

†jσ} − {d

†lτ , p

†jσ} dlτ = d†lτ .0− 0.dlτ = 0

15

Now substituting these results into (24),

(a) = −∆α∑lτ

∑<ij>

(d†lτδilδστpjσ + p†jσδilδστdlτ )

(a) = −∆α∑<ij>σ

(d†iσpjσ + p†jσdiσ)(25)

5.2 Part 2

(c) = αU∑l

∑<ij>

[d†lτdlτd†lτdlτ , (d

†iσpjσ − p

†jσdiσ)]

= αU∑l

∑<ij>

([d†lτdlτd†lτdlτ , d

†iσpjσ] − [d†lτdlτd

†lτdlτ , p

†jσdiσ] )

= αU∑l

∑<ij>

d†lτdlτ [d†lτdlτ , d†iσpjσ] + [d†lτdlτ , d

†iσpjσ] d†lτdlτ − d

†lτdlτ [d†lτdlτ , p

†jσdiσ] − [d†lτdlτ , p

†jσdiσ] d†lτdlτ

(26)

Let us once more evaluate the necessary commutators,

[d†lτdlτ , d†iσpjσ] = d†iσ[d†lτdlτ , pjσ] + [d†lτdlτ , d

†iσ] pjσ

= d†iσd†lτ{dlτ , pjσ} − d

†iσ{d

†lτ , pjσ} dlτ + d†lτ{dlτ , d

†iσ} pjσ − {d

†lτ , d

†iσ} dlτpjσ

= d†iσd†lτ .0 + d†iσ.0.dlτ − d

†lτ{dlτ , d

†iσ} pjσ − 0.dlτpjσ = d†lτδilδστpjσ

[d†lτdlτ , d†iσpjσ] = d†iσ[d†lτdlτ , pjσ] + [d†lτdlτ , d

†iσ] pjσ

= d†iσd†lτ{dlτ , pjσ} − d

†iσ{d

†lτ , pjσ} dlτ + d†lτ{dlτ , d

†iσ} pjσ − {d

†lτ , d

†iσ} dlτpjσ

= d†iσd†lτ .0− d

†iσ.0.dlτ + d†lτ{dlτ , d

†iσ} pjσ − 0.dlτpjσ = d†lτδilδστpjσ

[d†lτdlτ , p†jσdiσ] = p†jσ[d†lτdlτ , diσ] + [d†lτdlτ , p

†jσ] diσ

= p†jσd†lτ{dlτ , diσ} − p

†jσ{d

†lτ , diσ} dlτ + d†lτ{dlτ , p

†jσ} diσ − {d

†lτ , p

†jσ} dlτdiσ

= p†jσd†lτ .0− p

†jσ{d

†lτ , diσ} dlτ + d†lτ .0.diσ − 0.dlτdiσ = −p†jσδilδστdlτ

[d†lτdlτ , p†jσdiσ] = p†jσ[d†lτdlτ , diσ] + [d†lτdlτ , p

†jσ] diσ

= p†jσd†lτ{dlτ , diσ} − p

†jσ{d

†lτ , diσ} dlτ + d†lτ{dlτ , p

†jσ} diσ − {d

†lτ , p

†jσ} dlτdiσ

= p†jσd†lτ .0− p

†jσ{d

†lτ , diσ} dlτ + d†lτ .0.diσ − 0.dlτdiσ = −p†jσδilδστdlτ

Hence (26) can be evaluated to obtain,

(c) = αU∑l

∑<ij>

d†lτdlτd†lτδilδστpjσ + d†lτδilδστpjσd

†lτdlτ + d†lτdlτp

†jσδilδστdlτ + p†jσδilδστdlτd

†lτdlτ

= αU∑<ij>

d†iτdiτd†iτδστpjσ + d†iτδστpjσd

†iτdiτ + d†iτdiτp

†jσδστdiτ + p†jσδστdiτd

†iτdiτ

= αU∑<ij>

δστ (d†iτdiτd†iτpjσ + d†iτdiτp

†jσdiτ ) + δστ (d†iτpjσd

†iτdiτ + p†jσdiτd

†iτdiτ )

= αU∑<ij>

δστd†iτdiτ (d†iτpjσ + p†jσdiτ ) + δστ (d†iτpjσ + p†jσdiτ )d†iτdiτ

= αU∑<ij>

d†iτdiτ (d†iτpjτ + p†jτdiτ ) + (d†iτpjτ + p†jτdiτ )d†iτdiτ

= αU∑<ij>

(d†iτpjτ + p†jτdiτ )d†iτdiτ + (d†iτpjτ + p†jτdiτ )d†iτdiτ

= αU∑<ij>τ

(d†iτpjτ + p†jτdiτ )d†iτdiτ

= αU∑<ij>σ

(d†iσpjσ + p†jσdiσ)d†iσdiσ

(27)

16

So we have calculated our solution for (c),

(c) = αU∑<ij>σ

(d†iσpjσ + p†jσdiσ)d†iσdiσ (28)

5.3 Part 3

We can now use a trick from the properties of the anti-commutator relations, which if we recognise willsave us a great deal of algebra. The general equation for (b) is given below,

(b) = −∆β∑lτ

∑<ij>

[d†lτdlτ , (d†iσpjσ − p

†jσdiσ)d†iσdiσ] (29)

The idea is that the number operator niσ = d†iσdiσ commutes with d†iσdiσ and as a result that term is

zero. This means in effect that we can pull the d†iσdiσ outside the commutator for (b).

(b) = −∆β∑lτ

∑<ij>

(d†iσpjσ − p†jσdiσ)[d†lτdlτ , d

†iσdiσ] + [d†lτdlτ , (d

†iσpjσ − p

†jσdiσ)] d†iσdiσ

(b) = −∆β∑lτ

∑<ij>

[d†lτdlτ , (d†iσpjσ − p

†jσdiσ)] d†iσdiσ

(30)

The resulting commutator sum is simply that which we worked out in part (a) but with the additional

factor of d†iσdiσ at the end and a different coefficient at the front. It is therefore trivial to write downthe solution for (b),

(b) = −∆β∑<ij>σ

(d†iσpjσ + p†jσdiσ)d†iσdiσ (31)

5.4 Part 4

We can then employ the same basic trick again to simplify down (d) into the form of (c) as we did tosolve for (b) in terms of (a). The commutator that is represented by (d) is,

(d) = βU∑l

∑<ij>

[d†lτdlτd†lτdlτ , (d

†iσpjσ − p

†jσdiσ)d†iσdiσ] (32)

Evaluating out this commutator in the same manner again and employing the trick yields us with thefollowing,

(d) = βU∑l

∑<ij>

(d†iσpjσ − p†jσdiσ)[d†lτdlτd

†lτdlτ , d

†iσdiσ] + [d†lτdlτd

†lτdlτ , (d

†iσpjσ − p

†jσdiσ)] d†iσdiσ

(d) = βU∑l

∑<ij>

[d†lτdlτd†lτdlτ , (d

†iσpjσ − p

†jσdiσ)] d†iσdiσ

(33)

By inspection with the answer we derived for (c), we can write the solution to (d) as,

(d) = βU∑<ij>σ

(d†iσpjσ + p†jσdiσ)d†iσdiσ (34)

5.5 Solving for S

Now we simply need to put the calculated equations for (a),(b),(c) and (d) back into the original equationthat we had to find S and its coefficients, α and β.

H1 = [−∆∑<iσ>

d†iσdiσ + U∑i

d†iσdiσd†iσdiσ, αS0 + βS1] (35)

17

Writing out in full we have,

H1 = [−∆∑lτ

d†lτdlτ , α∑<ij>

(d†iσpjσ − p†jσdiσ)]

+ [−∆∑lτ

d†lτdlτ , β∑<ij>

(d†iσpjσ − p†jσdiσ)d†iσdiσ]

+ [U∑l

d†lτdlτd†lτdlτ , α

∑<ij>

(d†iσpjσ − p†jσdiσ)]

+ [U∑l

d†lτdlτd†lτdlτ , β

∑<ij>

(d†iσpjσ − p†jσdiσ)d†iσdiσ]

(36)

We now want to compare our calculated value of H1 with the actual value of H1 in our model so we cancompare coefficients,

V∑<ij>σ

(d†iσpjσ + p†jσdiσ) = −∆α∑<ij>σ

(d†iσpjσ + p†jσdiσ)

+−∆β∑<ij>σ

(d†iσpjσ + p†jσdiσ)d†iσdiσ

+ αU∑<ij>σ

(d†iσpjσ + p†jσdiσ)d†iσdiσ

+ βU∑<ij>σ

(d†iσpjσ + p†jσdiσ)d†iσdiσ

(37)

We can simplify this equation down into two simultaneous equations for the coefficients which containα and β,

V = −∆α

0 = −∆β + αU + βU(38)

The solutions to (38) provide the coefficients of S0 and S1 as follows,

α = −V∆

β =αU

∆− U=

UV

∆(U −∆)

(39)

All that remains is to substitute these values of α and β back into S, and we can produce the correctform of S that eliminates the order V dependence in the Hamiltonian,

S = −V∆

∑<ij>σ

(d†iσpjσ − p†jσdiσ) +

UV

∆(U −∆)

∑<ij>σ

(d†iσpjσ − p†jσdiσ)d†iσdiσ (40)

We have finally determined the correct form of S needed for our derivation of H2 through the use ofvarious types of commutator and anti-commutator relations. The physical interpretation of this term Shas already been discussed in the project proposal and is of no real relevance to us, so our next aim willbe to to use this form of S to actually calculate H2.

18

6 Finding H2

We now need to extend the work in the previous chapter by solving the earlier defined commutator toproduce the actual form of H2 for our model. We will need to use the form of S we just calculated andapply a similar method of calculating out the components of the commutator.

H2 =1

2[S,H1] = −1

2[H1, S] (41)

This requires us to calculate the following commutator,

[S,H1] = [−V∆

∑<ij>

(d†iσpjσ − p†jσdiσ) +

UV

∆(U −∆)

∑<ij>σ

(d†iσpjσ − p†jσdiσ)d†iσdiσ, V

∑<ij>σ

(d†iσpjσ + p†jσdiσ)]

(42)

Upon employing appropriate dummy variables and calculating this commutator the other way aroundwe get,

−[H1, S] = −[V∑

<lm>τ

(d†lτpmτ + p†mτdlτ ),−V∆

∑<ij>

(d†iσpjσ − p†jσdiσ) +

UV

∆(U −∆)

∑<ij>σ

(d†iσpjσ − p†jσdiσ)d†iσdiσ]

−[H1, S] = −[V∑

<lm>τ

(d†lτpmτ + p†mτdlτ ),−V∆

∑<ij>

(d†iσpjσ − p†jσdiσ)]

− [V∑

<lm>τ

(d†lτpmτ + p†mτdlτ ),UV

∆(U −∆)

∑<ij>σ

(d†iσpjσ − p†jσdiσ)d†iσdiσ]

(43)

So expanded out into a nicer form we obtain,

[S,H1] = +V 2

∆

∑<lm>τ

∑<ij>

[(d†lτpmτ + p†mτdlτ ), (d†iσpjσ − p†jσdiσ)]

− UV 2

∆(U −∆)

∑<lm>τ

∑<ij>

[(d†lτpmτ + p†mτdlτ ), (d†iσpjσ − p†jσdiσ)d†iσdiσ]

(44)

From this point onwards we shall call the first term in (44) as (i) and the second term in (44) as (ii).

6.1 Solving for (i)

Our objective is to now try to evaluate out the first commutator in (44) which is known as (i). Basiccommutator relations allow us to do the following,

(i) =V 2

∆

∑<lm>τ

∑<ij>

[(d†lτpmτ + p†mτdlτ ), (d†iσpjσ − p†jσdiσ)]

=V 2

∆

∑<lm>τ

∑<ij>

[d†lτpmτ , d†iσpjσ] − [d†lτpmτ , p

†jσdiσ] + [p†mτdlτ , d

†iσpjσ] − [p†mτdlτ , p

†jσdiσ]

=V 2

∆

∑<lm>τ

∑<ij>

−[d†lτpmτ , p†jσdiσ] + [p†mτdlτ , d

†iσpjσ]

(45)

We know that d†iσpjσ and p†jσdiσ both always commute with themselves, so we have dropped the firstand last commutators in the line above as they equal to zero. Now we will evaluate out the remaining

19

commutators,

−[d†lτpmτ , p†jσdiσ] = −p†jσ[d†lτpmτ , diσ] − [d†lτpmτ , p

†jσ] diσ

= −p†jσd†lτ{pmτ , diσ} + p†jσ{d

†lτ , diσ} pmτ

− d†lτ{pmτ , p†jσ} diσ + {d†lτ , p

†jσ} pmτdiσ

= p†jσ{d†lτ , diσ} pmτ − d

†lτ{pmτ , p

†jσ} diσ

= p†jσδilδστpmτ − d†lτδjmδστdiσ

(46)

[p†mτdlτ , d†iσpjσ] = d†iσ[p†mτdlτ , pjσ] + [p†mτdlτ , d

†iσ] pjσ

= d†iσp†mτ{dlτ , pjσ} − d

†iσ{p

†mτ , pjσ} dlτ

+ p†mτ{dlτ , d†iσ} pjσ − {p

†mτ , d

†iσ} dlτpjσ

= −d†iσ{p†mτ , pjσ} dlτ + p†mτ{dlτ , d

†iσ} pjσ

= −d†iσδjmδστdlτ + p†mτδilδστpjσ

(47)

Let us now substitute these simplified commutators back into (i),

(i) =V 2

∆

∑<lm>τ

∑<ij>

p†jσδilδστpmτ − d†lτδjmδστdiσ − d

†iσδjmδστdlτ + p†mτδilδστpjσ

=V 2

∆

∑<lm>σ

∑<ij>

p†jσδilpmσ − d†lσδjmdiσ − d

†iσδjmdlσ + p†mσδilpjσ

=V 2

∆

∑<lm>σ

∑<ij>

δil(p†jσpmσ + p†mσpjσ)− δjm(d†lσdiσ + d†iσdlσ)

=V 2

∆

∑<lm>σ

∑<ij>

δil(p†jσpmσ + p†mσpjσ)− V 2

∆

∑<lm>σ

∑<ij>

δjm(d†lσdiσ + d†iσdlσ)

=V 2

∆

∑<im>σ

∑<ij>

(p†jσpmσ + p†mσpjσ)− V 2

∆

∑<lj>σ

∑<ij>

(d†lσdiσ + d†iσdlσ)

(48)

So our final form for (i) is,

V 2

∆

∑<im>σ

∑<ij>

(p†jσpmσ + p†mσpjσ)− V 2

∆

∑<lj>σ

∑<ij>

(d†lσdiσ + d†iσdlσ) (49)

6.2 Solving for (ii)

We shall do a similar process to simplify down the second commutator involved in the calculation of (44)which we have labelled as (ii),

(ii) = − UV 2

∆(U −∆)

∑<lm>τ

∑<ij>

[(d†lτpmτ + p†mτdlτ ), (d†iσpjσ − p†jσdiσ)d†iσdiσ] (50)

= − UV 2

∆(U −∆)

∑<lm>τ

∑<ij>

(d†iσpjσ − p†jσdiσ)[(d†lτpmτ + p†mτdlτ ), d†iσdiσ] (51)

− UV 2

∆(U −∆)

∑<lm>τ

∑<ij>

[(d†lτpmτ + p†mτdlτ ), (d†iσpjσ − p†jσdiσ)] d†iσdiσ (52)

20

We can again identify that the second term aka (52) is of a form that is similar to what we have calculatedfor (i). Now we will expand out the commutator involved in line (51),

−[(d†lτpmτ + p†mτdlτ ), d†iσdiσ] = [d†iσdiσ, d†lτpmτ ] + [d†iσdiσ, p

†mτdlτ ]

= d†iσδilδστpmτ − p†mτδilδστdiσ

= δilδστ (d†iσpmτ − p†mτdiσ)

= δilδστ (d†iσpmσ − p†mσdiσ)

(53)

Hence the first term in (ii) from line (51), denoted as (ii1) below, becomes,

(ii1) = − UV 2

∆(U −∆)

∑<lm>τ

∑<ij>

(d†iσpjσ − p†jσdiσ)[(d†lτpmτ + p†mτdlτ ), d†iσdiσ]

=UV 2

∆(U −∆)

∑<lm>τ

∑<ij>

(d†iσpjσ − p†jσdiσ)δilδστ (d†iσpmσ − p

†mσdiσ)

=UV 2

∆(U −∆)

∑<im>σ

∑<ij>

(d†iσpjσ − p†jσdiσ)(d†iσpmσ − p

†mσdiσ)

=UV 2

∆(U −∆)

∑<im>σ

∑<ij>

(d†iσpjσ − p†jσdiσ)(d†iσpmσ − p

†mσdiσ)

(54)

Now we can calculate what the second term in (ii) from line (52), denoted as (ii2) below, becomes,

(ii2) = − UV 2

∆(U −∆)

∑<lm>τ

∑<ij>

[(d†lτpmτ + p†mτdlτ ), (d†iσpjσ − p†jσdiσ)] d†iσdiσ

= − UV 2

∆(U −∆)

∑<lm>τ

∑<ij>

δστ (δil(p†jσpmσ + p†mσpjσ)− δjm(d†lσdiσ + d†iσdlσ))d†iσdiσ

= − UV 2

∆(U −∆)

∑<lm>σ

∑<ij>

(δil(p†jσpmσ + p†mσpjσ)− δjm(d†lσdiσ + d†iσdlσ))d†iσdiσ

= − UV 2

∆(U −∆)

∑<lm>σ

∑<ij>

δil(p†jσpmσ + p†mσpjσ)d†iσdiσ +

UV 2

∆(U −∆)

∑<lm>σ

∑<ij>

δjm(d†lσdiσ + d†iσdlσ)d†iσdiσ

= − UV 2

∆(U −∆)

∑<im>σ

∑<ij>

(p†jσpmσ + p†mσpjσ)d†iσdiσ +UV 2

∆(U −∆)

∑<lj>σ

∑<ij>

(d†lσdiσ + d†iσdlσ)d†iσdiσ

(55)

So putting (ii1) and (ii2) together, we get the final form for (ii) as the following,

(ii) =UV 2

∆(U −∆)

∑<im>σ

∑<ij>

(d†iσpjσ − p†jσdiσ)(d†iσpmσ − p

†mσdiσ)

− UV 2

∆(U −∆)

∑<im>σ

∑<ij>

(p†jσpmσ + p†mσpjσ)d†iσdiσ +UV 2

∆(U −∆)

∑<lj>σ

∑<ij>

(d†lσdiσ + d†iσdlσ)d†iσdiσ

(56)

21

6.3 Solving H2

We are now in a position where we can combine our (i) and (ii) together to get the commutator denotedby [S,H1] .

[S,H1] =V 2

∆

∑<im>σ

∑<ij>

(p†jσpmσ + p†mσpjσ)− V 2

∆

∑<lj>σ

∑<ij>

(d†lσdiσ + d†iσdlσ)

+UV 2

∆(U −∆)

∑<im>σ

∑<ij>

(d†iσpjσ − p†jσdiσ)(d†iσpmσ − p

†mσdiσ)

− UV 2

∆(U −∆)

∑<im>σ

∑<ij>

(p†jσpmσ + p†mσpjσ)d†iσdiσ +UV 2

∆(U −∆)

∑<lj>σ

∑<ij>

(d†lσdiσ + d†iσdlσ)d†iσdiσ

=∑

<im>σ

∑<ij>

(p†jσpmσ + p†mσpjσ)(V 2

∆− UV 2

∆(U −∆)d†iσdiσ)

+∑<lj>σ

∑<ij>

(d†lσdiσ + d†iσdlσ)(−V2

∆+

UV 2

∆(U −∆)d†iσdiσ)

+UV 2

∆(U −∆)

∑<im>σ

∑<ij>

(d†iσpjσ − p†jσdiσ)(d†iσpmσ − p

†mσdiσ)

(57)

By introducing the factor of 12 we can transform [H1, S] into H2, so we get the subsequent answer,

H2 =∑

<im>σ

∑<ij>

(p†jσpmσ + p†mσpjσ)(+V 2

2∆− UV 2

2∆(U −∆)d†iσdiσ) (58)

+∑<lj>σ

∑<ij>

(d†lσdiσ + d†iσdlσ)(−V2

2∆+

UV 2

2∆(U −∆)d†iσdiσ) (59)

+UV 2

2∆(U −∆)

∑<im>σ

∑<ij>

(d†iσpjσ − p†jσdiσ)(d†iσpmσ − p

†mσdiσ) (60)

Another way of writing this is,

H2 =V 2

2

∑<im>σ

∑<ij>

(p†jσpmσ + p†mσpjσ)(1

∆− U

∆(U −∆)d†iσdiσ) (61)

− V 2

2

∑<lj>σ

∑<ij>

(d†lσdiσ + d†iσdlσ)(1

∆− U

∆(U −∆)d†iσdiσ) (62)

+UV 2

2∆(U −∆)

∑<im>σ

∑<ij>

(d†iσpjσ − p†jσdiσ)(d†iσpmσ − p

†mσdiσ) (63)

Finally a more physically intuitive form is,

H2 =V 2

2

∑<im>σ

∑<ij>

(p†jσpmσ + p†mσpjσ)(d†iσdiσU −∆

− 1− d†iσdiσ∆

) (64)

− V 2

2

∑<lj>σ

∑<ij>

(d†lσdiσ + d†iσdlσ)(d†iσdiσU −∆

− 1− d†iσdiσ∆

) (65)

+UV 2

2∆(U −∆)

∑<im>σ

∑<ij>

(d†iσpjσ − p†jσdiσ)(d†iσpmσ − p

†mσdiσ) (66)

22

So we have completed our mathematical derivation for the form of the second order hopping correctionto our Hamiltonian H2, and rewritten the final form in several ways. The reason for this is that the finalform seen just above allows us to clearly see the energy costs and gains resulting from our model. Forexample if d†iσdiσ = 1, then the coefficients simplify down to 1/(U − ∆). The earlier forms are moreuseful for the further mathematical manipulations to the Hamiltonian we will do in later chapters.

23

7 H2 for Hole-Doping

Whilst we have now calculated a form for H2 we need to recognise and address the fact that some of theterms in the Hamiltonian will take us out of the ground state configuration. We shall address each ofthe terms individually and explain what they correspond to and if they take us out of the ground state,to produce a Hamiltonian that exactly descibes our model.

The ground state before we start hole doping is where we have every copper site with a single holeon it and there are no holes on any of the oxygen sites. As we then start doping, we introduce someadditional holes to the system which reside on the oxygen atoms in the lattice.

Consider the term in line (58).

∑<im>σ

∑<ij>

(p†jσpmσ + p†mσpjσ)(+V 2

2∆− UV 2

2∆(U −∆)d†iσdiσ)

We can keep both the (p†jσpmσ + p†mσpjσ) and (p†jσpmσ + p†mσpjσ)d†iσdiσ terms as we want to allow thepossibility of oxygen hole hopping. We clearly have terms that are valid when m=j and m 6=j whichcorrespond to the cases where we end with singly occupied oxygen sites and doubly occupied oxygensites respectively. Both of these cases are allowed as we have not forbidden these configurations sincedoping only introduces relatively few holes into the system.

Consider the term in line (59).

∑<lj>σ

∑<ij>

(d†lσdiσ + d†iσdlσ)(−V2

2∆+

UV 2

2∆(U −∆)d†iσdiσ)

We see that this represents the possibility of copper to copper hole hopping. However as we have ex-plained earlier the ground state has every single copper ion having a single hole located on it. This meansthat we can only have the case l=i, as if this was not the case we would end up with doubly occupiedcopper sites which need to be forbidden.

Consider the term in line (60).

+UV 2

2∆(U −∆)

∑<im>σ

∑<ij>

(d†iσpjσ − p†jσdiσ)(d†iσpmσ − p

†mσdiσ)

We clearly need to further expand out this term to better understand its meaning, so leaving aside thesum and coefficients for the moment,

(d†iσpjσ − p†jσdiσ)(d†iσpmσ − p

†mσdiσ) = d†iσpjσd

†iσpmσ − d

†iσpjσp

†mσdiσ − p

†jσdiσd

†iσpmσ + p†jσdiσp

†mσdiσ

= −d†iσd†iσpjσpmσ + d†iσp

†mσpjσdiσ + p†jσd

†iσdiσpmσ − p

†jσp†mσdiσdiσ

(67)

We can see that we now have four terms that are within our Hamiltonian, but again some of these termstake us out of the ground state. We can observe that the first term vacates two holes from oxygen sitesand moves them to the same copper site, causing double occupancy. The reverse process happens inthe fourth term, where we lose holes from copper sites, which mean that the resulting copper sites haveno holes at all. As a result we need to remove these terms from the Hamiltonian. The remaining twoterms correspond to the situation where a copper hole is hopped to a neighbouring oxygen, to make thatoxygen site doubly occupied. Then another oxygen hole, that is adjacent to the copper, is hopped ontothe copper site leaving a unoccupied oxygen.

(d†iσpjσ − p†jσdiσ)(d†iσpmσ − p

†mσdiσ) = +d†iσp

†mσpjσdiσ + p†jσd

†iσdiσpmσ

= +d†iσp†mσpjσdiσ + d†iσp

†jσpmσdiσ

(68)

24

Thus the result we obtain for H2 is,

H2 =∑

<im>σ

∑<ij>

(p†jσpmσ + p†mσpjσ)(V 2

2∆− UV 2

2∆(U −∆)d†iσdiσ) (69)

+∑<ij>σ

(d†lσdiσ + d†iσdlσ)(−V2

2∆+

UV 2

2∆(U −∆)d†iσdiσ) (70)

+UV 2

2∆(U −∆)

∑<im>σ

∑<ij>

d†iσp†mσpjσdiσ + d†iσp

†jσpmσdiσ (71)

We can notice that (70) is of the same form as H0 and represents just a small change to it as a result ofour analysis. We could in fact incorporate this term into H0 and remove it from our form for H2. If wedo this then H2 becomes,

H2 = +V 2

2∆

∑<im>σ

∑<ij>

(p†jσpmσ + p†mσpjσ) (72)

− UV 2

2∆(U −∆)

∑<im>σ

∑<ij>

(p†jσpmσ + p†mσpjσ)d†iσdiσ (73)

+UV 2

2∆(U −∆)

∑<im>σ

∑<ij>

d†iσp†mσpjσdiσ + d†iσp

†jσpmσdiσ (74)

Now if we combine some more terms together we get the following expression.

H2 =V 2

2∆

∑<im>σ

∑<ij>

(p†jσpmσ + p†mσpjσ) (75)

− UV 2

2∆(U −∆)

∑<im>σ

∑<ij>

p†jσpmσd†iσdiσ + p†mσpjσd

†iσdiσ − d

†iσp†mσpjσdiσ − d

†iσp†jσpmσdiσ (76)

Yet more rearrangement of the mixed operator terms produces the next form for H2.

H2 = +V 2

2∆

∑<im>σ

∑<ij>

(p†jσpmσ + p†mσpjσ) (77)

− UV 2

2∆(U −∆)

∑<im>σ

∑<ij>

d†iσp†jσpmσdiσ + d†iσp

†mσpjσdiσ − d

†iσp†mσpjσdiσ − d

†iσp†jσpmσdiσ (78)

As we are summing over σ we are allowed to interchange the σ and σ in individual terms, which will bebeen done for the first term of the line above. We shall also switch around the third and fourth terms.

− UV 2

2∆(U −∆)

∑<im>σ

∑<ij>

d†iσp†jσpmσdiσ + d†iσp

†mσpjσdiσ − d

†iσp†jσpmσdiσ − d

†iσp†mσpjσdiσ (79)

As we are summing over both j and m, we shall switch their labels around in both the second and fourthterms above as we are entitled to do so.

− UV 2

2∆(U −∆)

∑<im>σ

∑<ij>

d†iσp†jσpmσdiσ + d†iσp

†jσpmσdiσ − d

†iσp†jσpmσdiσ − d

†iσp†jσpmσdiσ (80)

Finally we can now factor these terms into the following form which represents the formation of a singletbetween an oxygen hole and a hole on a copper site.

− UV 2

2∆(U −∆)

∑<im>σ

∑<ij>

(d†iσp†jσ − d

†iσp†jσ)(pmσdiσ − pmσdiσ) (81)

25

We should also recognise the fact that the two terms in the oxygen hopping term will give the same valuewhen summed over.

H2 = +V 2

∆

∑<im>σ

∑<ij>

p†jσpmσ −UV 2

∆(U −∆)

∑<im>σ

∑<ij>

(d†iσp†jσ − d

†iσp†jσ)(pmσdiσ − pmσdiσ) (82)

This is our interpretation of the correct form of H2 to use for our hole doping model. It contains both aterm to do with oxygen hole hopping and a second term to do with singlet hopping. This form will beshown to be fully equivalent to other representations of H2, such as the Zhang-Rice singlet form in thenext section, but it is this form that will be the basis of our future work.

26

7.1 Derivation of Zhang-Rice Singlet

The aim of this section is to rigorously show that the previous hole doping Hamiltonian we found in 7.0is able to be re-described in the form of a singlet and triplet state.

To do this we will turn the oxygen hole hopping term into the same form as the other terms by adding ina factor d†iτdiτ = 1, at the end. This is valid as it is merely the fact that every copper site has one holeon it in the ground state, so it does not change the problem. Using this idea turns the oxygen hoppingterm from (77) into the following.∑

<im>

∑<ij>

∑σ

∑τ

p†jσpmσd†iτdiτ + p†mσpjσd

†iτdiτ (83)

∑<im>

∑<ij>

∑σ

∑τ

d†iτp†jσpmσdiτ + d†iτp

†mσpjσdiτ (84)

Now we can now merely rewrite this as we have a sum over j and m as,

2∑<im>

∑<ij>

∑σ

∑τ

d†iτp†jσpmσdiτ (85)

We now need to carefully evaluate out this sum. We note that we have the cases where σ = τ and σ 6= τ .When sigma and tau are parallel we have σ =↑, τ =↑ and the case σ =↓, τ =↓. When sigma and tauare anti-parallel we have σ =↑, τ =↓ or σ =↓, τ =↑.

The first case is where sigma and tau are both parallel to each other.

2∑<im>

∑<ij>

(d†i↑p†j↑pm↑di↑ + d†i↓p

†j↓pm↓di↓) (86)

The second case is where sigma and tau are both anti-parallel to each other.

2∑<im>

∑<ij>

(d†i↓p†j↑pm↑di↓ + d†i↑p

†j↓pm↓di↑) (87)

There exists an elegant factorisation of the second case, seen below, where we produce a singlet and partof a triplet state.∑

<im>

∑<ij>

[(d†i↑p†j↓ + d†i↓p

†j↑)(pm↓di↑ + pm↑di↓) + (d†i↑p

†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓)] (88)

We can now recombine both the parallel and anti-parallel cases back together noting they constitute asinglet and a triplet state.

=∑<im>

∑<ij>

(d†i↑p†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓) (89)

+ 2∑<im>

∑<ij>

d†i↑p†j↑pm↑di↑ (90)

+∑<im>

∑<ij>

(d†i↑p†j↓ + d†i↓p

†j↑)(pm↓di↑ + pm↑di↓) (91)

+ 2∑<im>

∑<ij>

d†i↓p†j↓pm↓di↓ (92)

The singlet and triplet states are more intuitively seen by grouping them together as (93) and (94) below.

=∑<im>

∑<ij>

(d†i↑p†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓) (93)

+∑<im>

∑<ij>

2d†i↑p†j↑pm↑di↑ + (d†i↑p

†j↓ + d†i↓p

†j↑)(pm↓di↑ + pm↑di↓) + 2d†i↓p

†j↓pm↓di↓ (94)

27

We now substitute this expression back into our original H2 for the oxygen hopping term.

H2 = +V 2

2∆

∑<im>

∑<ij>

(d†i↑p†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓) (95)

+V 2

2∆

∑<im>

∑<ij>

2d†i↑p†j↑pm↑di↑ + (d†i↑p

†j↓ + d†i↓p

†j↑)(pm↓di↑ + pm↑di↓) + 2d†i↓p

†j↓pm↓di↓ (96)

− UV 2

2∆(U −∆)

∑<im>σ

∑<ij>

(d†iσp†jσ − d

†iσp†jσ)(pmσdiσ − pmσdiσ) (97)

It is clear to see that we can evaluate out the sum over all possible spins for the third term, which willcancel the factor of a half.

H2 = +V 2

∆

∑<im>

∑<ij>

1

2(d†i↑p

†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓) (98)

+V 2

∆

∑<im>

∑<ij>

d†i↑p†j↑pm↑di↑ +

1

2(d†i↑p

†j↓ + d†i↓p

†j↑)(pm↓di↑ + pm↑di↓) + d†i↓p

†j↓pm↓di↓ (99)

− UV 2

∆(U −∆)

∑<im>

∑<ij>

(d†i↑p†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓) (100)

Finally adding together the two singlet terms and being careful with the coefficients we produce the formof a singlet and a triplet state as found by Zhang and Rice [22].

H2 = +V 2

∆

∑<im>

∑<ij>

d†i↑p†j↑pm↑di↑ +

1

2(d†i↑p

†j↓ + d†i↓p

†j↑)(pm↓di↑ + pm↑di↓) + d†i↓p

†j↓pm↓di↓ (101)

− V 2(U + ∆)

∆(U −∆)

∑<im>

∑<ij>

1

2(d†i↑p

†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓) (102)

It is clear from this Hamiltonian that the most stable term is the second term, which is known as theZhang-Rice singlet in the literature [22].

(a) Zhang-Rice Singlet Orbitals [22]. (b) Doped hole localisation [15].

Figure 12: Illustrations of the Zhang-Rice Singlet.

This can be best visualised in Figure 12b which illustrates how a doped hole localises into a singlet acrossfour oxygen sites adjacent to a copper. As we have allowed second order hopping, if the hole is initially

28

in a singlet involving the oxygen directly below the copper site, the hole can hop to any of the otheradjacent oxygen sites to that copper. This is seen through the < im >< ij > notation and in essencemeans the doped hole is in a singlet across all the oxygen sites.

These are the key to the work of Zhang-Rice and their t-J model, with the assumption being thatthey move around on the lattice and are never broken. Indeed when the observation of these Zhang-Ricesinglets was experimental verified, they became a major reason as to why the t-J model was some widelyaccepted. It is hence important that we have also been able to predict their existence.

29

8 H2 for Electron-Doping

As we were careful when we were defining our model and the manipulation of it, our H2 is valid forelectron doping and hole doping. This means that in a similar way to that what we did for hole dop-ing, we need to merely remove the terms from H2 that take us out of the ground state or are simply invalid.

Consider the term in line (58).

∑<im>σ

∑<ij>

(p†jσpmσ + p†mσpjσ)(+V 2

2∆− UV 2

2∆(U −∆)d†iσdiσ)

As we have earlier discussed we simply do not have any oxygen holes introduced in the case of electrondoping, so we should remove all the terms which involve oxygen p operators. Hence the term abovecompletely disappears from the model.

Consider the term in line (59).

∑<lj>σ

∑<ij>

(d†lσdiσ + d†iσdlσ)(−V2

2∆+

UV 2

2∆(U −∆)d†iσdiσ)

We see that this represents the possibility of copper to copper hole hopping, without explicitly mention-ing any terms in p. There are two cases to consider, when l=i and when l 6=i. We should keep only thecase of l 6=i in our form for H2, as l=i represents just a small change to H0 and can be incorporated intoit. The first term involving just (d†lσdiσ + d†iσdlσ) can be kept as it is purely copper to copper hopping.

The second term, (d†lσdiσ + d†iσdlσ)d†iσdiσ, however has a different meaning. Here we either move a holefrom a doubly occupied site to a neighbouring site or move a hole to another site already occupied witha hole to cause double occupancy. We need to remove this double occupancy from the model as we havealready accounted for it by extracting out the intermediate states needed for hopping earlier on.

Consider the term in line (60).

+UV 2

2∆(U −∆)

∑<im>σ

∑<ij>

(d†iσpjσ − p†jσdiσ)(d†iσpmσ − p

†mσdiσ)

Once again we observe that every term in the line above involves p, so we should discard the entire termin generality.

Thus the result we obtain for H2 is merely,

H2 = −V2

2∆

∑<lj>σ

∑<ij>

(d†lσdiσ + d†iσdlσ) (103)

We can now remove the now unnecessary index j and recognise the fact that the two terms are hermitianconjugates of each other and will be equivalent when summed over to obtain.

H2 = −V2

∆

∑<il>σ

d†lσdiσ (104)

We have hence obtained the appropriate electron doping Hamiltonian which will serve as the basis ofour model of electron-doped cuprate superconductors.

30

9 Superexchange

9.1 The Principle of Superexchange

It should be noted that we have neglected to mention the existence of an additional interaction betweenthe particles which is often considered in the literature. Whilst we have considered physical hoppingthere is also an additional interaction term to virtual hopping of the spins. This was first proposed byAnderson in 1959 [1] and is in fact the reason why the undoped compounds are antiferromagnetic asstated in Chapter 2. It turns out that this is an order V4 interaction which promotes an antiferromagneticalignment of copper spins as a way of allowing for virtual double occupations of sites.

Figure 13: Possible virtual double occupations.

We shall derive this antiferromagnetic interaction from the Hubbard model in the strong-coupling limitfor simplicity.

9.2 Derivation from the Hubbard Model in Strong-Coupling Limit

We shall start with the classic Hubbard model which has the parameter t representing the kinetic energyof electrons hopping between atoms and U the energy required to doubly occupy a site.

H = Hhop +Honsite = −t∑<il>σ

d†lσdiσ + U∑i

d†iσdiσd†iσdiσ (105)

As U is now the dominant term in this strong-coupling limit, U >> t ,we consider the hopping term asa small perturbation to the Hamiltonian, i.e. Honsite +Hhop = H0 +H1. Then it is necessary to utilisesecond order degenerate perturbation theory to calculate out the H2 for this Hubbard model.

Using degenerate perturbation theory we find that the second order correction is,

ε2n =∑j

〈0|H1 |j〉 〈j|H1 |0〉ε0 − εj

= −〈0|H1H1 |0〉U

(106)

as the energy required to reach any of the intermediate states is always U (εj = U, ε0 = 0). This meansthat we essentially want to calculate −H1H1/U and project it onto the ground state. Calculating thisout we find that,

H2 = − t2

U

∑<il>σ

d†lσdiσ∑

<i′l′>τ

d†l′τdi′τ (107)

It is clear to see that as we wish to remain in the ground state, then we need to recreate any particledestroyed in the same state, so l = i′ and i = l′ are required in general.

= − t2

U

∑<il>στ

d†lσdiσd†iτdlτ (108)

31

This gives us four combinations due to the two independent spin subscripts,

= − t2

U

∑<il>

(d†l↑di↑d

†i↑dl↑ + d†l↑di↑d

†i↓dl↓ + d†l↓di↓d

†i↑dl↑ + d†l↓di↓d

†i↓dl↓

)(109)

Through standard algebra we can switch the order of the operators to get us into a more intuitive form,

= +t2

U

∑<il>

(−d†l↑dl↑(1− d

†i↑di↑) + d†l↑dl↓d

†i↓di↑ + d†l↓dl↑d

†i↑di↓ − d

†l↓dl↓(1− d

†i↓di↓)

)(110)

We can see that we have terms that are essentially components of the spin-operators,

S+ = d†↑d↓ (111)

S− = d†↓d↑ (112)

Sx =S+ + S−

2=

1

2(d†↑d↓ + d†↓d↑) (113)

Sy =S+ − S−

2i=

1

2i(d†↑d↓ − d

†↓d↑) (114)

Sz =1

2(d†↑d↑ − d

†↓d↓) (115)

So terms 2 and 3 become,

d†l↑dl↓d†i↓di↑ + d†l↓dl↑d

†i↑di↓ = S+

l S−i + S−l S

+i (116)

We know that every copper site is singly occupied,

d†i↑di↑ + d†i↓di↓ = 1 (117)

The definition of the z-component of the spin operator is,

2Szi = d†i↑di↑ − d†i↓di↓ (118)

So combining these two facts together we recieve the following statement,

d†i↑di↑ =1

2+ Szi (119)

d†i↓di↓ =1

2− Szi (120)

(121)

So terms 1 and 4 become,

−d†l↑dl↑(1− d†i↑di↑)− d

†l↓dl↓(1− d

†i↓di↓) = −(

1

2+ Szl )(

1

2− Szi )− (

1

2− Szl )(

1

2+ Szi ) (122)

= −(1

2− 2Szl S

zi ) (123)

So our overall equation becomes,

=t2

U

∑<il>

(−1

2+ 2Szl S

zi + S+

l S−i + S+

i S−l

)(124)

=2t2

U

∑<il>

−1

4+

(Szl S

zi +

1

2(S+l S−i + S+

i S−l )

)(125)

=2t2

U

∑<il>

Si.Sl −1

4(126)

=2t2

U

∑<il>

(Si.Sl −1

4) (127)

32

As we have to consider both spins virtually hopping to site i and to site l, we actually have the total

energy that can be gained by having an antiferromagnetic alignment is J = 4t2

U .

HAF = J∑<il>

(Si.Sl −1

4) (128)

We should note that in our earlier work we have calculated the hopping parameter t=V2

∆ , so we have

J = 4V4

∆2U

as the coefficient.

As we will see later in the paper, if the adjacent spins i and l are parallel, then Si.Sl = 14 and if

they are anti-parallel Si.Sl = − 34 . Using these values in the above expression it is clear to see that we

gain an energy of 4t2

U by having an anti-parallel alignement of spins, whereas we gain nothing if they areparallel.

It is important to recognise that this antiferromagnetic interaction exists, indeed it is the J in thet-J model we mentioned earlier. However for the rest of our work we have neglected to directly study itfurther on the grounds that it is dominated by physical hopping.

33

10 Solving our Hamiltonians

10.1 Models and Geometries

To summarise the previous work we have developed two models of the cuprate square lattices for bothelectron doping and hole doping.

The electron doping model is a single parameter Hamiltonian which turns out to be exactly the form as

that of the Hubbard model if we define the hopping term t=V2

∆ ,

H2 = −V2

∆

∑<il>σ

d†lσdiσ (129)

The hole doping model is more complex as it is two parameter Hamiltonian, one of which is a freeparameter,

H2 = +V 2

∆

∑<jm>σ

p†jσpmσ −UV 2

∆(U −∆)

∑<im>

∑<ij>

(d†i↑p†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓) (130)

which we have achieved by evaluating out both spins in order to drop the factor of a half.

It turns out there are two limits that we want to investigate for the case of hole doping. As seenback in Figure 8, we need to consider the case where ∆ ≈ U and the case where ∆ � U . These corre-spond to different ions being produced physically. In the limit ∆ ≈ U we should see the emergence ofCu3+ ions. These is best thought of as when it is very easy to doubly occupy copper sites with holes.In the case of ∆ � U , we have that the energy required to doubly occupy is extremely large, so it isespecially hard to doubly occupy copper sites. This suggests the presence of Cu+ ions. If we defineκ = ∆

U , then these cases correspond to the case when κ = 1 and κ = 0 respectively.

By rewriting the hole-doping Hamiltonian in terms of κ and then taking the limit κ → 1, we producethe Cu3+ model.

H2 = +V 2

∆

∑<jm>σ

p†jσpmσ −V 2

κ(U −∆)

∑<im>

∑<ij>

(d†i↑p†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓) (131)

H2 = +V 2

∆

∑<jm>σ

p†jσpmσ −V 2

(U −∆)

∑<im>

∑<ij>

(d†i↑p†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓) (132)

It is clear to see that as ∆ ≈ U the singlet term is divergent and hence dominates the oxygen hoppingterm, so we can drop that term from the Hamiltonian.

H2 = − V 2

(U −∆)

∑<im>

∑<ij>

(d†i↑p†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓) (133)

If we now take the other limit that of κ→ 0, we produce the Cu+ model.

H2 = +V 2

∆

∑<jm>σ

p†jσpmσ −V 2

∆(1− k)

∑<im>

∑<ij>

(d†i↑p†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓) (134)

H2 = +V 2

∆

∑<jm>σ

p†jσpmσ −V 2

∆

∑<im>

∑<ij>

(d†i↑p†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓) (135)

In this form both the singlet and the oxygen terms are of the same magnitude but are of opposite phase.

It has been determined experimentally as we mentioned in Chapter 2 that the physical limit is ac-tually the Cu+ limit in our work, since V = 0.75eV � U = 10ev. However in order to compare between

34

the limits we shall continue to try to solve both hole-doped models as well as the electron-doped model.

There are several geometries of interest for us as we seek to build up towards the full solution of asquare lattice. We shall first consider what happens on a linear chain, and then on a closed ring of sites.The reasoning for this is that the mathematics for each of the arrangements gets progressively moredifficult, so tackling them in this order will allow us to build up our competence in forming solutions. Amore tangible argument for this approach can be deduced by considering the square lattice itself. Motionin the full CuO2 plane can either linear or periodic, so by studying the cases for linear chains and ringswe are able to isolate each of these possibilities for motion on the full square lattice.

35

10.2 Electron Doping - 1D Linear Chain

The electron doping Hamiltonian does not contain any terms involved with oxygen orbitals explicitly.This means that we are effectively simplifying the problem down to only the copper sites, so we onlyneed to consider a linear chain of copper sites.

H2 = −V2

∆

∑<il>σ

d†lσdiσ (136)

Initially every copper site would have a hole on it representing the undoped parent compound, then if wewere to introduce additional electrons to the system they would remove holes on the chain to produceempty sites. The key thing to notice in this problem is that we have the idea of blocking becomingapparent in the linear chain. Our Hamiltonian only allows the holes to move to adjacent sites, but ourtheory prevents us from doubly occupying a copper site with holes. This means that the overall motionalong the chain is a kind of concertina effect, as seen below.

Figure 14: The holes are only able to hop onto empty sites as double occupation is forbidden.

It is clear to see that this preserves the overall alignment of spins along the linear chain, so if the statewas initially in an antiferromagnetic alignment it would remain so. In order for the alignment to bealtered we would have to introduce the weaker superexchange interaction to allow for the possibility ofchanging the ordering of copper spins.

10.3 Electron Doping - Ring

We shall now consider the case where we have the ability to produce motion periodic motion by turningour linear chain into a ring. This can clearly be done by applying periodic boundary conditions, namelythe constraint that the first and last copper sites should be connected. This connection will allow us tocalculate the effects of hopping between these coppers sites on a ring, indeed it also turns out that thecase where we have an infinite number of sites in our ring is exactly the same as the infinite linear chainproblem.

Figure 15: A ring composed of N total copper sites.

36

The key to the solution of this problem lies within the geometry of the ring itself. It is clear to see thatif we introduce a single hole onto the ring, then there are only two possible states it can move to with asingle hop: forwards one site or backwards one site. For example a hole on site 1 can move to 2 or backto 0. This pattern of hopping to nearest neighbours means that we can represent the hopping in termsof a translation matrix, T, as seen below.

T = −V2

∆

0 1 0 0 · · · · · · 11 0 1 0 · · · · · · 00 1 0 1 0

0 0 1 0. . . 0

......

. . .. . .

. . ....

......

. . .. . . 1

1 0 0 · · · · · · 1 0

(137)

Here each row represents the possible movements available to a hole currently on that site. For examplea hole on site 2 can move to site 1 or site 2. This matrix can be further simplified into two matricesrepresenting the forwards and the backwards movements.

T = −V2

∆(Tforward + Tbackward) (138)

where we have

Tforward =

0 1 0 0 · · · 00 0 1 0 · · · 00 0 0 1 · · · 0...

......

.... . .

...0 0 0 0 · · · 11 0 0 0 · · · 0

(139)

(140)

Tbackward =

0 0 0 · · · 0 11 0 0 · · · 0 00 1 0 · · · 0 00 0 1 · · · 0 0...

......

. . ....

...0 0 0 · · · 1 0

(141)

It is plain to see both from the matrices and the ring itself that Tforward and Tbackward are inverses toeach other. This means that all we need to do is find the eigenvalues of the forwards hopping and theeigenvalues of the backwards hopping will be the reciprocal of them.

Av = λv =⇒ A−1Av = λA−1v =⇒ A−1v =1

λv (142)

The eigenvaluesλ, of the forwards hopping are contained in the following eigenvalue equation equationthat we need to solve,

−λ 1 0 0 · · · 00 −λ 1 0 · · · 00 0 −λ 1 · · · 0...

......

.... . .

...0 0 0 0 · · · 11 0 0 0 · · · −λ

a0

a1

a2...

an−2

an−1

=

000...00

(143)

37

Expanding out this yields,

−λa0 + a1 = 0⇒ a1 = λa0

−λa1 + a2 = 0⇒ a2 = λ2a0

−λa2 + a3 = 0⇒ a3 = λ3a0

−λan−1 + a0 = 0⇒ an−1 = λn−1a0

We then substitute these results back into the general eigenvector,

a0

a1

a2...

an−2

an−1

=

a0

λa0

λ2a0...

λn−2a0

λn−1a0

= a0

1λ

λ2

...

λn−2

λn−1

(144)

Hence we obtain our unnormalised eigenvectors as,

1λ

λ2

...

λn−2

λn−1

(145)

Now we can use this eigenvector in the eigenvalue equation seen earlier,

−λ 1 0 0 · · · 00 −λ 1 0 · · · 00 0 −λ 1 · · · 0...

......

.... . .

...0 0 0 0 · · · 11 0 0 0 · · · −λ

1λ

λ2

...

λn−2

λn−1

=

000...00

(146)

We can see that all lines apart from the bottom line are already self-consistent with the right hand side.In order to complete this self-consistency, we can see from the last row that we require, 1 − λn = 0 i.e.λn = 1.

This equation tells us that the eigenvalues of the problem are simply the nth roots of unity. Theresults are obtained through an elementary substitution of λ = eik, where i is the imaginary number andk is the variable. Then 1 = eikN , where N is the fixed number of sites. Clearly this just describes theunit circle in the complex plane, where k = 2π

N n, where n is a real number between 0 and 2π.

Now we can put this result for the eigenvalues back into the original equation for our ground state|0〉, which contains a single hole.

H |0〉 = −V2

∆(λ+

1

λ) |0〉 = −V

2

∆(eik + e−ik) |0〉 = −2V 2

∆cos k |0〉 (147)

E = −2V 2

∆cos k (148)

We only have a discrete number of values of k, one for each site in the ring. In the case of an infinitenumber of sites we get more and more values of k squeezed even tighter together, eventually forming the

38

continuous cosine function.

This means we can plot out a graph of these points which will be in the shape of a cosine function.However as we increase the number of sites towards infinity we get more k values squeezed even tightertogether to eventually form a continuous line.

Clearly the ground state is the minimum of this function, so it is −2V 2/∆ regardless of the numberof sites in the ring and preserves the initial alignment of spins. If the initial alignment is antiferromag-netic it will remain antiferromagnetic around the ring.

39

10.4 Cu3+ Limit - 1D Linear Chain

In this section we will try to solve the singlet hopping Hamiltonian on an infinite linear chain to un-derstand the motion of the singlet. This will require the use of some complicated mathematics, whichdirectly proceeds on from here.

The Hamiltonian that we have derived and have to consider is,

H2 = − V 2

(U −∆)

∑<im>

∑<ij>

(d†i↑p†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓) (149)

This problem is more difficult than the Hubbard model, as we know have to account for oxygen orbitals.We start by describing the linear chain in a mathematical form. We shall assume a chain which hasa singlet located in it; the case where the oxygen hole is to the right of the nth copper shall be called|n+ > and when the oxygen hole is the to the left of the nth copper hole it shall be called |n− >.

|n+ >=∑{~σ}

d†1σ1d†2σ2

. . . d†n−1σn−1[d†n↑p

†n+↓ − d

†n↓p†n+↑]d

†n+1σn+1

. . . |0 > (150)

|n− >=∑{~σ}

d†1σ1d†2σ2

. . . d†n−1σn−1[d†n↑p

†n−↓ − d

†n↓p†n−↑]d

†n+1σn+1

. . . |0 > (151)

The sum over {~σ} purposefully excludes the spin of the doped oxygen hole and nth copper hole as seenabove. We now need to apply the Hamiltonian to these states separately and then recombine the resultsto give us the solution to this system, e.g. we want to find H2|n+ > and H2|n− >.

We begin by applying the Hamiltonian to the ground state.

− V 2

(U −∆)

∑<im>

∑<ij>

∑{~σ}

(d†i↑p†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓)d

†1σ1

. . . d†n−1σn−1[d†n↑p

†n+↓ − d

†n↓p†n+↑]d

†n+1σn+1

. . . |0 >

(152)

Ideally we would like to move the Hamiltonian terms through to just before the singlet terms which canbe done via operator identities. We will illustrate the first of these shifts, as it can clearly be generalisedfor the rest of the required copper holes. Considering the first copper hole,

(d†i↑p†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓)d

†1σ1

(153)

We can see that if i 6= 1 the di↑ and di↓ terms will anti-commute with the d1↑ term. Then standard

relations between the d, p, p† terms mean that d1↑ will anti-commute with the rest of the terms to produce,

d†1σ1(d†i↑p

†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓) (154)

This idea can clearly be generalised so that we can move all the copper hole terms up to and in-cluding d†n−1σn−1

past the Hamiltonian terms. The only restriction we need mathematically is then

i /∈ {1, 2, . . . n− 1}, which corresponds to the doped oxygen hole not being adjacent to that copper site.

So we end up with,

− V 2

(U −∆)

∑<im>

∑<ij>

∑{~σ}

d†1σ1d†2σ2

. . . d†n−1σn−1(d†i↑p

†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓)[d

†n↑p†n+↓ − d

†n↓p†n+↑]d

†n+1σn+1

. . . |0 >

(155)

10.4.1 Term 1

We shall first consider the case when the copper site is on the left hand side of the doped hole, i = n.We now need to calculate the effect of Hamiltonian on the singlet term itself. The relevant part of theequation in line (155) is,

(d†i↑p†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓)[d

†n↑p†n+↓ − d

†n↓p†n+↑] (156)

40

We need to ensure that we only keep the cross terms that do not anticommute. If they did we couldmove the Hamiltonian all the way to the end and eventually have the d terms acting on the |0 >, whichis defined to be zero (d|0 >= 0|0 >= 0) as you are destroying a non-existent hole, invalidating our work.This means that the only terms that should remain are,

(d†i↑p†j↓ − d

†i↓p†j↑)(pm↓di↑d

†n↑p†n+↓ + pm↑di↓d

†n↓p†n+↑) (157)

which require us to select that i=n and keep the terms that have parallel spin. Further simplificationthrough the anti-commutation relations means we can drop those terms to get,

(d†i↑p†j↓ − d

†i↓p†j↑)(pm↓p

†n+↓ + pm↑p

†n+↑) (158)

Once again we have that the p and p† terms will only anti-commute if m=n+, leaving the following.

2(d†i↑p†j↓ − d

†i↓p†j↑) (159)

We can now put the result in the line above back into line (155), along with the fact that i=n and m=n+.

− V 2

(U −∆)

∑<nn+>

∑<nj>

∑{~σ}

d†1σ1d†2σ2

. . . d†n−1σn−12(d†n↑p

†j↓ − d

†n↓p†j↑)d

†n+1σn+1

. . . |0 > (160)

Now we have to evaluate out the summation terms. As < jn > means evaluate for both oxygen nearestneighbours to copper site n, we have that j = n+ and j = n−. The other summation essentially meansthe same thing as well. This means that the result we get is,

H2 |n+〉 = − V 2

(U −∆)(2 |n+〉+ 2 |n−〉) (161)

10.4.2 Term 2

Now we have to do the other case when applying the Hamiltonian which is when i = n + 1, when thecopper site on the right hand side of the doped hole.

(d†i↑p†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓)[d

†n↑p†n+↓ − d

†n↓p†n+↑]d

†n+1σn+1

(162)

In a similar manner to before, we get the following expression by ensuring we keep the terms with parallelspins.

(d†i↑p†j↓ − d

†i↓p†j↑)(pm↓di↑d

†n↑p†n+↓ + pm↑di↓d

†n↓p†n+↑)d

†n+1σn+1

(163)

Now in this case we have to ensure the fact that i = n+ 1 to make further progress.

(d†n+1↑p†j↓ − d

†n+1↓p

†j↑)(pm↓dn+1↑d

†n↑p†n+↓ + pm↑dn+1↓d

†n↓p†n+↑)d

†n+1σn+1

(164)

Now if we temporarily ignore the first bracket and focus on the latter half of (164),

(pm↓dn+1↑d†n↑p†n+↓d

†n+1σn+1

+ pm↑dn+1↓d†n↓p†n+↑d

†n+1σn+1

) (165)

We now want to swap the order of the operators in these terms around so that we can eliminate pairsthat anti-commute.

(pm↓dn+1↑d†n↑p†n+↓d

†n+1σn+1

+ pm↑dn+1↓d†n↓p†n+↑d

†n+1σn+1

) (166)

=(dn+1↑d†n↑pm↓p

†n+↓d

†n+1σn+1

+ dn+1↓d†n↓pm↑p

†n+↑d

†n+1σn+1

) (167)

=− (d†n↑pm↓p†n+↓dn+1↑d

†n+1σn+1

+ d†n↓pm↑p†n+↑dn+1↓d

†n+1σn+1

) (168)

=− (d†n↑dn+1↑d†n+1σn+1

+ d†n↓dn+1↓d†n+1σn+1

) (169)

41

Here we had to pick the constraint that m = n+. It is possible to further simplify the line above bynoticing that if σn+1 =↑, we get −d†n↑ and if σn+1 =↓, we get −d†n↓. Using this fact we then can deduce,

− V 2

(U −∆)

∑<im>

∑<ij>

∑{~σ}

∑σn+1

d†1σ1d†2σ2

. . . d†n−1σn−1(d†n+1↑p

†j↓ − d

†n+1↓p

†j↑)− d

†nσn+1

d†n+2σn+2. . . |0 >

(170)

− V 2

(U −∆)

∑<im>

∑<ij>

∑{~σ}

∑σn+1

d†1σ1d†2σ2

. . . d†n−1σn−1− d†nσn+1

(d†n+1↑p†j↓ − d

†n+1↓p

†j↑)d

†n+2σn+2

. . . |0 >

(171)

Once more we need to remember the sum over < jm > now meaning j = (n+ 1)+ and j = (n+ 1)−, sowe will end up with those states as part of the answer.

H2 |n+〉 = − V 2

(U −∆)(− |(n+ 1)+〉 − |(n+ 1)−〉) (172)

We can now combine both parts of the Hamiltonian back together to produce the total answer for the|n+〉 state.

H2 |n+〉 = − V 2

(U −∆)(2 |n+〉+ 2 |n−〉 − |(n+ 1)+〉 − |(n+ 1)−〉) (173)

10.4.3 Solving the Remainder

It can be worked out by similar methods or an argument based in the symmetry of the problem, thatthe correct answer for the |n−〉 state is as follows,

H2 |n−〉 = − V 2

(U −∆)(2 |n+〉+ 2 |n−〉 − |(n− 1)+〉 − |(n− 1)−〉) (174)

It would now make sense to define a new state that is a composite of the |n+〉 and |n−〉 states, |n〉 =|n+〉 + |n−〉, which essentially is a state that contains singlets on both sides of the doped oxygen hole.Then we can add both of our previous expressions together to produce,

H2(|n+〉+ |n−〉) = − V 2

(U −∆)(4 |n+〉+ 4 |n−〉 − |(n− 1)+〉 − |(n− 1)−〉 − |(n+ 1)+〉 − |(n+ 1)−〉)

(175)

This can be written in our new notation as below.

H2 |n〉 = − V 2

(U −∆)(4 |n〉 − |n− 1〉 − |n+ 1〉) (176)

Here we essentially have an energy associated with the singlet staying where it is |n〉 which can be ignored

as it would in fact come from a H0 like contribution. Then we have a term |n+ 1〉 corresponding to the

forward hopping of the singlet and |n− 1〉 which is a backward hopping of the singlet. We can again usethe idea of translation matrices like we employed to solve the electron-doped limit to rewrite the hoppingstates in the following way.

H2 |n〉 = − V 2

(U −∆)(−T−1 |n〉 − T |n〉) (177)

H2 |n〉 = − V 2

(U −∆)(−2 cos k) |n〉 (178)

42

This means that we can write down the energy of the singlet hopping on the linear chain as,

E =2V 2

(U −∆)cos k (179)

We can clearly see that this is very similar to what we derived for the case of electron doping. Theonly difference is that we have a slightly different constant and we have an additional minus sign due to

changing from electrons to holes. Here the minimum energy is − 2V2

(U−∆) .

10.5 Cu3+ Limit - Triangle

We shall try and solve the model on a triangular lattice, the simplest case where we can start to examinethe possibility of periodic motion. In this case the geometry of the problem presents an extra complica-tion to the mathematics.

The first thing to notice is if we introduce a single hole to one of the three sites, we have four pos-sibilities for the spins of the remaining two sites. These can be though of as the parallel and anti-parallelcases. We shall focus on the case where we have anti-parallel spins because they turn out to be the moreinteresting case mathematically and are more relevant as the system is initially in an antiferromagneticalignment when we start doping.

We can define the 6 possible states of the system which are all connected by the hopping of the hole asseen below. Here we choose to enforce a strict ordering of spins, so that the only thing that changes isthe number of the site. This allows us to keep a hold of the mathematics more firmly.

|1〉 = d†1↑d†2↓ |0〉 (180)

|2〉 = d†3↑d†2↓ |0〉 (181)

|3〉 = d†3↑d†1↓ |0〉 (182)

|4〉 = d†2↑d†1↓ |0〉 (183)

|5〉 = d†2↑d†3↓ |0〉 (184)

|6〉 = d†1↑d†3↓ |0〉 (185)

The application of the Hamiltonian to each of these states allows us to see that we can either move

forward or backwards by a state (Y = − V2

(U−∆) ).

H2 |1〉 = Y (|2〉+ |6〉) (186)

H2 |2〉 = Y (|3〉+ |1〉) (187)

H2 |3〉 = Y (|4〉+ |2〉) (188)

H2 |4〉 = Y (|5〉+ |3〉) (189)

H2 |5〉 = Y (|6〉+ |4〉) (190)

H2 |6〉 = Y (|1〉+ |5〉) (191)

43

Figure 16: The triangular superlattice is composed of six possible states since there is a single hole inthe system. Site 1 is bottom left, 2 is bottom right, 3 is top.

We can hop the hole forward or hop the hole backwards to get into the next or previous state on thetriangular superlattice [12]. Then we can again write down the general hopping matrix that we usedearlier in the electron doping model, which in this case is,

T =

0 1 0 0 0 11 0 1 0 0 00 1 0 1 0 00 0 1 0 1 00 0 0 1 0 11 0 0 0 1 0

(192)

We now need to apply this to an general amount of each state, e.g. a linear combination (which spansthe whole of the subspace) i.e. a0 times state 1 + a1 times state 2 etc. Hence as we know this will be aneigenvalue problem, we can assume the existence of eigenvalues and find the unormalised eigenvectors aswell. Recalling the previous solution and as illustrated in Figure 16, we will find that the eigenvectorsmade up of the eigenvalues z are,

1z

z2

z3

z4

z5

(193)

This clearly corresponds physically to the fact that if you start in a state it requires one hop (eigenvaluez) to get to the next state, so two hops (z2) will be needed to get to get to the state after and so on andso forth. The equation for the eigenvalue z is then defined in the equation z6 = 1.

The solution is obviously once again the sixth roots of unity, but there is also something else worthnoticing in this example. The equation z6 = 1, is the same as (z3 − 1)(z3 + 1) = 0, which means thatz3 = 1 and z3 = −1 are also solutions contained within it. If we look at Figure 16 we can immediately seethat we can now combine |1〉 and |4〉 together, |2〉 and |5〉 as well as |3〉 and |6〉, by using these values of z3.This idea was not possible in the electron doping case as there were no anti-periodic boundary conditions.

44

We can rewrite these three new spinwave functions as,

|ψ1〉 = (d†1↑d†2↓ + z3d†1↓d

†2↑) |0〉 = |1〉+ z3 |4〉 (194)

|ψ2〉 = (d†2↑d†3↓ + z3d†2↓d

†3↑) |0〉 = |2〉+ z3 |5〉 (195)

|ψ3〉 = (d†3↑d†1↓ + z3d†3↓d

†1↑) |0〉 = |3〉+ z3 |6〉 (196)

We can determine which of these states are ferromagnetic and which of these states are antiferromagneticin a trivial manner for this simple triangular case. As we only have two spin- 1

2 particles, it is clear to see

that if we use z3 = 1 this will correspond to the cases of triplets, which have total spin 1 and are henceferromagnetic. The case of z3 = −1 will correspond to the existence of singlet states, which have total

spin 0 and are antiferromagnetic. Mathematically the cases where z = e−2iπ

3 , 1, e2iπ3 are ferromagnetic

and when z = e−iπ3 ,−1, e

iπ3 they exhibit antiferromagnetism.

Now we can put this result for the eigenvalues back into the original operator equation,

H2 |n〉 = Y (z +1

z) |n〉 = Y (eik − e−ik) |n〉 = 2Y cos k |n〉 (197)

So the energy of the triangular lattice is given by the formula,

E = − 2V 2

(U −∆)cos k (198)

We can then illustrate the information we have derived visually in the graph below as Figure 17 with

the ground state having energy − 2V2

(U−∆) .

Figure 17: Energy of the states composing the triangular superlattice.

Here we can see that the lowest energy state is clearly an antiferromagnetic one, however if we hadan additional minus sign the graph would be reflected in the horizontal axis and we would have aferromagnetic ground state.

45

10.6 Cu3+ Limit - Square

This is clearly a more complicated case than the triangle due to the existence of an additional site in theloop. However this example is of more importance to us as it is physically relevant to the structure ofCuO2 layers in the cuprate superconductors.

We shall examine the case where we have two spin up electrons and one spin down electron. Clearly viathe same logic as we did on the triangle we will end up with a superlattice containing 12 possible states.

|1〉 = d†1↑d†2↓d†3↑ |0〉 (199)

|2〉 = d†4↑d†2↓d†3↑ |0〉 (200)

|3〉 = d†4↑d†1↓d†3↑ |0〉 (201)

|4〉 = d†4↑d†1↓d†2↑ |0〉 (202)

|5〉 = d†3↑d†1↓d†2↑ |0〉 (203)

|6〉 = d†3↑d†4↓d†2↑ |0〉 (204)

|7〉 = d†3↑d†4↓d†1↑ |0〉 (205)

|8〉 = d†2↑d†4↓d†1↑ |0〉 (206)

|9〉 = d†2↑d†3↓d†1↑ |0〉 (207)

|10〉 = d†2↑d†3↓d†4↑ |0〉 (208)

|11〉 = d†1↑d†3↓d†4↑ |0〉 (209)

|12〉 = d†1↑d†2↓d†4↑ |0〉 (210)

If we apply the Hamiltonian to each of these states then we merely hop forwards or backwards a state

in the superlattice (Y = − V2

(U−∆) ).

H2 |1〉 = Y (|2〉+ |12〉) (211)

H2 |2〉 = Y (|3〉+ |1〉) (212)

H2 |3〉 = Y (|4〉+ |2〉) (213)

H2 |4〉 = Y (|5〉+ |3〉) (214)

H2 |5〉 = Y (|6〉+ |4〉) (215)

H2 |6〉 = Y (|7〉+ |5〉) (216)

H2 |7〉 = Y (|8〉+ |6〉) (217)

H2 |8〉 = Y (|9〉+ |7〉) (218)

H2 |9〉 = Y (|10〉+ |8〉) (219)

H2 |10〉 = Y (|11〉+ |9〉) (220)

H2 |11〉 = Y (|12〉+ |10〉) (221)

H2 |12〉 = Y (|1〉+ |11〉) (222)

46

Figure 18: The twelve possible states of the superlattice for the square. The hole hops in an anticlockwisedirection. The yellow circle is the single doped hole.

It is sufficient to see that this problem is defined by the equation for the twelfth roots of unity, z12 = 1.

On this occasion we shall denote the three cube roots as 1, w = ei2π3 and w∗ = e−i

2π3 , such that w = z4

and (z4)3 = 1. This means that we can combine states together by using the values of the z4.

This just results in every spin wave function being founded on a set of cyclic permutations for thepositions of each spin,

|ψ1〉 = d†1↑d†2↓d†3↑ |0〉+ z4d†3↑d

†1↓d†2↑ |0〉+ z8d†2↑d

†3↓d†1↑ |0〉 = |1〉+ z4 |5〉+ z8 |9〉 (223)

|ψ2〉 = d†4↑d†2↓d†3↑ |0〉+ z4d†3↑d

†4↓d†2↑ |0〉+ z8d†2↑d

†3↓d†4↑ |0〉 = |2〉+ z4 |6〉+ z8 |10〉 (224)

|ψ3〉 = d†4↑d†1↓d†3↑ |0〉+ z4d†3↑d

†4↓d†1↑ |0〉+ z8d†1↑d

†3↓d†4↑ |0〉 = |3〉+ z4 |7〉+ z8 |11〉 (225)

|ψ4〉 = d†4↑d†1↓d†2↑ |0〉+ z4d†2↑d

†4↓d†1↑ |0〉+ z8d†1↑d

†2↓d†4↑ |0〉 = |3〉+ z4 |7〉+ z8 |11〉 (226)

It is still just about possible to avoid having to do a difficult calculation to determine which of thesestates are ferromagnetic and which are antiferromagnetic. It can be observed from a Pascal triangle likeresult that three spin- 1

2 particles couple together to produce two spin- 12 states and one spin- 3

2 state. The

case z4 = 1 is where we have |↑↑↓〉+ |↑↓↑〉+ |↓↑↑〉, which results from applying S−Total = S−1 + S−2 + S−3to the state |↑↑↑〉. This means that the case with w=1 has total spin 3/2 and is therefore ferromag-netic. This means that the two other combinations must be the spin 1/2 states and are antiferromagnetic.

We shall now illustrate this more clearly for the case of |ψ1〉.

|ψ1:1〉 = d†1↑d†2↓d†3↑ |0〉 .1 + d†3↑d

†1↓d†2↑ |0〉 .1 + d†2↑d

†3↓d†1↑ |0〉 .1 (227)

|ψ1:w〉 = d†1↑d†2↓d†3↑ |0〉 .1 + d†3↑d

†1↓d†2↑ |0〉 .w + d†2↑d

†3↓d†1↑ |0〉 .w

2 (228)

|ψ1:w∗〉 = d†1↑d

†2↓d†3↑ |0〉 .1 + d†3↑d

†1↓d†2↑ |0〉 .w

∗ + d†2↑d†3↓d†1↑ |0〉 .(w

∗)2 (229)

47

Using the fact that w∗ = w2 and (w∗)2 = w4 = w3.w = w, we obtain the following three states,

|ψ1:1〉 = d†1↑d†2↓d†3↑ |0〉 .1 + d†3↑d

†1↓d†2↑ |0〉 .1 + d†2↑d

†3↓d†1↑ |0〉 .1 (230)

|ψ1:w〉 = d†1↑d†2↓d†3↑ |0〉 .1 + d†3↑d

†1↓d†2↑ |0〉 .w + d†2↑d

†3↓d†1↑ |0〉 .w

2 (231)

|ψ1:w∗〉 = d†1↑d

†2↓d†3↑ |0〉 .1 + d†3↑d

†1↓d†2↑ |0〉 .w

2 + d†2↑d†3↓d†1↑ |0〉 .w (232)

Clearly the first state |ψ1:1〉 is a ferromagnetic state, as every spin is symmetric under exchange withanother. We shall take linear combinations of |ψ1:w〉 and |ψ1:w

∗〉 to better understand what these statescorrespond to.

|ψ1:w〉 − |ψ1:w∗〉 = d†1↑d

†2↓d†3↑ |0〉 .(1− 1) + d†3↑d

†1↓d†2↑ |0〉 .(w − w

2) + d†2↑d†3↓d†1↑ |0〉 .(w

2 − w) (233)

= (w − w2)(d†3↑d†1↓d†2↑ − d

†2↑d†3↓d†1↑) |0〉 (234)

= (w − w2)(d†2↑d†3↑d†1↓ − d

†2↑d†3↓d†1↑) |0〉 (235)

= (w − w2)d†2↑(d†3↑d†1↓ − d

†3↓d†1↑) |0〉 (236)

|ψ1:w〉 − w |ψ1:w∗〉 = d†1↑d

†2↓d†3↑ |0〉 .(1− w) + d†3↑d

†1↓d†2↑ |0〉 .(w − 1) + d†2↑d

†3↓d†1↑ |0〉 .(w

2 − w2) (237)

= (1− w)(d†1↑d†2↓d†3↑ |0〉 − d

†3↑d†1↓d†2↑) |0〉 (238)

= (1− w)(d†3↑d†1↑d†2↓ |0〉 − d

†3↑d†1↓d†2↑) |0〉 (239)

= (1− w)d†3↑(d†1↑d†2↓ |0〉 − d

†1↓d†2↑) |0〉 (240)

|ψ1:w〉 − w2 |ψ1:w

∗〉 = d†1↑d†2↓d†3↑ |0〉 .(1− w

2) + d†3↑d†1↓d†2↑ |0〉 .(w − w) + d†2↑d

†3↓d†1↑ |0〉 .(w

2 − 1) (241)

= (1− w2)(d†1↑d†2↓d†3↑ − d

†2↑d†3↓d†1↑) |0〉 (242)

= (1− w2)(d†1↑d†2↓d†3↑ − d

†1↑d†2↑d†3↓) |0〉 (243)

= (1− w2)d†1↑(d†2↓d†3↑ − d

†2↑d†3↓) |0〉 (244)

It is clear to see that these three solutions are exactly those of a singlet state coupled to another spin- 12

state as we earlier mentioned. Hence the total spin of each of these states will be one half and result inan antiferroagnetic alignment.

We can clearly repeat this process for the states |ψ2〉, |ψ3〉 and |ψ4〉 to produce similar results. Mathe-matically we can see that the cases which are ferromagnetic are given by,

z ∈ {e−iπ, e−iπ2 , 1, e

iπ2 }

The cases which produce antiferroagnetic states are,

z ∈ {e−5iπ

6 , e−4iπ

6 , e−2iπ

6 , e−iπ6 , e

iπ6 , e

2iπ6 , e

4iπ6 , e

5iπ6 }

Now we can put this result for the eigenvalues back into the original operator equation,

H2 |n〉 = Y (z +1

z) |n〉 = (eik − e−ik) |n〉 = 2Y cos k |n〉 (245)

So the ground state energy of the square lattice is given by,

E = − 2V 2

(U −∆)cos k (246)

This can be represented graphically as seen below in Figure 19,

48

Figure 19: Energy of the states composing the superlattice for the square.

This result differs from the case of the triangular lattice as we find that we have a ferromagnetic groundstate with energy -2|Y|. It is also clear to see that regardless of the plus or minus signs involved, we willalways have a ferromagnetic ground state for periodic motion around a square. It is worth noting thatthis interaction will directly compete with the smaller superexchange interaction as we dope holes intothe system.

It is possible to determine the approximate energy that the system gains by driving the spins into alow spin state over a high spin state from the graph above in Figure 19. The lowest state is ferromag-netic and has k=0, which produces an energy of -2|Y|. The lowest antiferromagnetic state has k= −π6and has an energy of −

√3|Y|. Thus the energy difference between the two spin states is (2 −

√3)|Y|

≈ 0.27|Y|.

10.7 Connection with Nagaoka’s Theorem

The results we have derived for periodic motion around a triangular and square lattice are actuallyconsequences of a more general proof developed by Nagaoka [13]. In his paper he managed to exactlysolve the case of periodic motion for any number of sites in a ring provided that they form a closed loop.The main conclusion we can draw from his work that it is useful for us, is that if the closed loop is madeup of an even number of sites the ground state is always ferromagnetic. If the closed loop is made upof an odd number of sites, then the ground state is either the state with maximum or minimum totalspin. Both of these results exactly match what we found for the cases of the triangle and square, so weeffectively have Nagaoka ferromagnetism introduced into the square lattice problem.

We can also attempt to estimate the relative strength of the superexchange interaction and the Na-gaoka ferromagnetism. The Nagaoka ferromagnetism should be directly proportional to the number ofholes in the system nh, the number of hops made by each hole z, and the energy gained by each ofthe hops t. So ε ∝ nhzt. The superexchange interaction we found has a strength J across each pair of

49

sites, so ε ∝ 0.5zJ for the low doping limit. Therefore these energy scales are equal when nh ∝ J/t.Using experimental values for J≈ 0.1eV and t= V 2/∆ ≈ 0.28eV for the square lattice, we find thatthe concentration of holes required to produce similarly sized interactions is nh = 0.36. This impliesthat superexchange is the main interaction below nh = 0.36 and produces an antiferromagnetic squarelattice, whereas above nh = 0.36 Nagaoka ferromagnetism dominates and produces a ferromagnetic ar-rangement. Referring back to the experimental phase diagram in Figure 10, we can indeed see that thereis indeed no apparent transition to ferromagnetism. This example demonstrates the fact that the Cu3+

limit cannot be the physical limit.

50

11 Spin Physics

11.1 Singlet and Triplet Operators

As we have seen earlier there are several different ways of making states that all have the same totalspin. This causes us some problems, so we introduce the spin projection operators to help us gain morecontrol over the situation.

P0 =1

4− S1.S2 (247)

P1 =3

4+ S1.S2 (248)

(249)

These project the wavefunction onto the component of it that is either spin-0 (a singlet) or spin-1 (atriplet). In other words P0 give us the probability that the two spins S1 and S2 are in a singlet and P1

the probability that they are in a triplet. We can clearly see that P0 +P1 = 1, so that the two spins canonly be in either a singlet or a triplet state.

The basic components of the spin operators are the following.

S+ |↑〉 = 0 (250)

S+ |↓〉 = |↑〉 (251)

S− |↑〉 = |↓〉 (252)

S− |↓〉 = 0 (253)

Sz |↑〉 =1

2|↑〉 (254)

Sz |↓〉 = −1

2|↓〉 (255)

These can be used to build up the complete form of a two-spin problem as seen below.

S1.S2 = Sz1 Sz2 +

1

2(S+

1 S−2 + S+

2 S−1 ) (256)

The next section of this chapter will be to apply the spin operator for two particles to each of the fourbasis states of the two particle spin states. The reasoning for this is we will be able to work out the rele-vant probabilities of the particles being in either a singlet or triplet state which will be useful for our work.

We now want to calculate the eigenvalues of the S1.S2 operator for each of basis spin states of boththe singlet and the triplet combinations.

S1.S2

1√2

(|↑↓〉 − |↓↑〉) (257)

S1.S2 |↑↑〉 (258)

S1.S2

1√2

(|↑↓〉+ |↓↑〉) (259)

S1.S2 |↓↓〉 (260)

It is relatively trivial to calculate the eigenvalues of (330) and (332) as both 1/4 respectively. We shallexplicitly calculate the eigenvalues of the more challenging mixed states.

51

Calculating the eigenvalues of the mixed triplet state,

S1.S2

1√2

(|↑↓〉+ |↓↑〉) =1√2S1.S2(|↑↓〉+ |↓↑〉) (261)

=1√2{Sz1 S

z2 +

1

2(S+

1 S−2 + S+

2 S−1 )(|↑↓〉+ |↓↑〉)} (262)

(263)

The x and y components of this problem yield the following,

(S+1 S−2 + S+

2 S−1 )(|↑↓〉+ |↓↑〉) = S+

1 S−2 |↑↓〉+ S+

1 S−2 |↓↑〉+ S+

2 S−1 |↑↓〉+ S+

2 S−1 |↓↑〉 (264)

= 0.0. |↑↓〉+ |↑↓〉+ 0.0. |↑↓〉+ |↑↓〉 (265)

= |↑↓〉+ |↑↓〉 (266)

Whilst the z component produces an additional term,

Sz1 Sz2 (|↑↓〉+ |↓↑〉) = Sz1 S

z2 |↑↓〉+ Sz1 S

z2 |↓↑〉 (267)

=1

2.− 1

2|↑↓〉+−1

2.1

2|↓↑〉 (268)

= −1

4(|↑↓〉+ |↓↑〉) (269)

Combining together all of the components is the final step.

S1.S2

1√2

(|↑↓〉+ |↓↑〉) =1√2S1.S2(|↑↓〉+ |↓↑〉) (270)

=1√2{Sz1 S

z2 +

1

2(S+

1 S−2 + S+

2 S−1 )(|↑↓〉+ |↓↑〉)} (271)

=1√2

(−1

4(|↑↓〉+ |↓↑〉) +

1

2(|↑↓〉+ |↑↓〉) (272)

=1√2

(1

4(|↑↓〉+ |↓↑〉) (273)

=1

4

1√2

(|↑↓〉+ |↓↑〉) (274)

So we can see that the eigenvalue of the mixed triplet state is indeed also 14 , the same as the two other

triplet states.

We shall now work out the x and y components of the singlet state,

(S+1 S−2 + S+

2 S−1 )(|↑↓〉 − |↓↑〉) = S+

1 S−2 |↑↓〉 − S

+1 S−2 |↓↑〉+ S+

2 S−1 |↑↓〉 − S

+2 S−1 |↓↑〉 (275)

= 0.0. |↑↓〉 − |↑↓〉+ |↓↑〉 − 0.0. |↑↓〉 (276)

= −(|↑↓〉 − |↓↑〉) (277)

Whilst the z components produce an additional term,

Sz1 Sz2 (|↑↓〉 − |↓↑〉) = Sz1 S

z2 |↑↓〉 − S

z1 S

z2 |↓↑〉 (278)

=1

2.− 1

2|↑↓〉 − −1

2.1

2|↓↑〉 (279)

= −1

4(|↑↓〉 − |↓↑〉) (280)

52

This means that overall we receive the final form for the singlet as the following,

S1.S2

1√2

(|↑↓〉 − |↓↑〉) =1√2S1.S2(|↑↓〉 − |↓↑〉) (281)

=1√2{Sz1 S

z2 +

1

2(S+

1 S−2 + S+

2 S−1 )(|↑↓〉 − |↓↑〉)} (282)

=1√2

(−1

4(|↑↓〉 − |↓↑〉)− 1

2(|↑↓〉 − |↑↓〉) (283)

=1√2

(−3

4(|↑↓〉 − |↓↑〉) (284)

= −3

4

1√2

(|↑↓〉 − |↓↑〉) (285)

So we find that the eigenvalue of the singlet state is actually − 34 .

Now we have completed the necessary calculations we have found that the singlet state is indeed themore stable configuration as S1.S2 |S〉 = − 3

4 |S〉 and S1.S2 |T 〉 = 14 |T 〉. There is no distinction that can

be made in terms of the energy associated with each of the triplet states, degeneracy is present in thiscase.

11.2 Exchange Operator

Finally we will introduce a quantity known as the exchange operator, described mathematically asP = P1 − P0 = 1

2 + 2S1.S2. It’s name suggests that it will effectively interchange the spin of the twostates it is applied to.

We shall demonstrate this by applying it to all of the four basis states once again.

P |↑↓〉 =1

2|↑↓〉+ 2S1.S2 |↑↓〉 (286)

=1

2|↑↓〉+ 2Sz1 S

z2 |↑↓〉+ S+

1 S−2 |↑↓〉+ S+

2 S−1 |↑↓〉 (287)

=1

2|↑↓〉+ 2

1

2.− 1

2|↑↓〉+ |↓↑〉 (288)

=1

2|↑↓〉 − 1

2|↑↓〉+ |↓↑〉 = |↓↑〉 (289)

P |↓↑〉 =1

2|↓↑〉+ 2S1.S2 |↓↑〉 (290)

=1

2|↓↑〉+ 2Sz1 S

z2 |↓↑〉+ S+

1 S−2 |↓↑〉+ S+

2 S−1 |↓↑〉 (291)

=1

2|↓↑〉+ 2.− 1

2.1

2|↓↑〉+ |↑↓〉 (292)

=1

2|↓↑〉 − 1

2|↓↑〉+ |↑↓〉 = |↑↓〉 (293)

P |↑↑〉 =1

2|↑↑〉+ 2S1.S2 |↑↑〉 (294)

=1

2|↑↑〉+ 2Sz1 S

z2 |↑↑〉+ S+

1 S−2 |↑↑〉+ S+

2 S−1 |↑↑〉 (295)

=1

2|↑↑〉+ 2.

1

2.1

2|↑↑〉 (296)

=1

2|↑↑〉+

1

2|↑↑〉 = |↑↑〉 (297)

53

P |↓↓〉 =1

2|↓↓〉+ 2S1.S2 |↓↓〉 (298)

=1

2|↓↓〉+ 2Sz1 S

z2 |↓↓〉+ S+

1 S−2 |↓↓〉+ S+

2 S−1 |↓↓〉 (299)

=1

2|↓↓〉+ 2.− 1

2.− 1

2|↓↓〉 (300)

=1

2|↓↓〉+

1

2|↓↓〉 = |↓↓〉 (301)

The explicit calculation of each of the states above has now shown us that indeed the exchange operatordoes do what its name suggests, meaning that the anti-parallel spin states are flipped whereas the parallelspin states are effectively unchanged. This operator will be made use of later on in Chapter 12.

54

12 1D Linear Chain - Cu+ Model

In this section we shall focus our efforts on trying to solve the more complicated Cu+ model. This has twodirectly competing interactions, oxygen hole hopping and the singlet hopping, which are of equal strength.

The Hamiltonian we shall use is,

H2 =V 2

∆

∑<jm>σ

p†jσpmσ −V 2

∆

∑<im>

∑<ij>

(d†i↑p†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓) (302)

We need to try and apply this to a particular state that is representative of the 1D linear chain whichwe shall construct below.

|n〉 =∑{~σ}

d†1σ1d†2σ2

. . . d†n−1σn−1d†nσnp

†jσ0d†n+1σn+1

. . . |0 > (303)

Here we have an infinite chain of copper holes, where we have introduced a single doped hole on theoxygen site j. In order to keep as big a generality as possible, we are again allowing for every spin to betotally independent, as indicated by the sum over {~σ}. An example of one of the configurations describedby this state is given beneath in Figure 20.

Figure 20: One possible scenario for the single doped hole, where we have spin up next to doped spindown.

Now lets apply the Hamiltionian to this linear chain,

V 2

∆

∑<jm>σ

p†jσpmσ∑{~σ}

d†1σ1. . . d†n−1σn−1

d†nσnp†jσ0d†n+1σn+1

. . . |0 > (304)

−V2

∆

∑<im>

∑<ij>

(d†i↑p†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓)

∑{~σ}

d†1σ1. . . d†n−1σn−1

d†nσnp†jσ0d†n+1σn+1

. . . |0 > (305)

We shall again solve both components separately and then recombine them to get our complete answer.

12.1 Oxygen Hopping Part

Looking at the oxygen hole hopping part, it simplifies down to the following,

V 2

∆

∑{~σ}

∑<im>

∑<ij>σ

p†jσpmσd†1σ1

. . . d†n−1σn−1d†nσnp

†jσ0d†n+1σn+1

. . . |0 > (306)

Clearly the p†jσpmσ can be moved through the copper hole terms via the properties of the anti-commutationrelations in general, so we reach,

V 2

∆

∑{~σ}

∑<im>

∑<ij>σ

d†1σ1. . . d†n−1σn−1

d†nσnp†jσpmσp

†jσ0d†n+1σn+1

. . . |0 > (307)

55

In order to remove the pmσ term, we require that σ = σ0 and j = m. The definition of j and m is thatthey are nearest neighbours to copper site n, so in this case we have them being constrained to be thesame nearest neighbour.

V 2

∆

∑{~σ}

∑<im>

∑<ij>

d†1σ1. . . d†n−1σn−1

d†nσnp†jσ0d†n+1σn+1

. . . |0 > (308)

As we mentioned before j and m are the same oxygen nearest neighbours to the copper site n, soj = m = n + 1/2 or j = m = n − 1/2. This means we can drop the summation over j and m byevaluating both of these contributions. Thus the final result is,

V 2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

d†nσnp†(n+1/2)σ0

d†n+1σn+1. . . |0 > (309)

−V2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

p†(n−1/2)σ0d†nσnd

†n+1σn+1

. . . |0 > (310)

12.2 Singlet Part

Now lets look at the singlet making part of the Hamiltonian,

−V2

∆

∑{~σ}

∑<im>

∑<ij>

(d†i↑p†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓)d

†1σ1

. . . d†n−1σn−1d†nσnp

†jσ0d†n+1σn+1

. . . |0 > (311)

Once again we can move through the terms from the Hamiltonian to in front of the d†nσnp†jσ0

by thecommutator identities to produce,

−V2

∆

∑{~σ}

∑<im>

∑<ij>

d†1σ1. . . d†n−1σn−1

(d†i↑p†j↓ − d

†i↓p†j↑)(pm↓di↑ − pm↑di↓)d

†nσn

p†jσ0d†n+1σn+1

. . . |0 > (312)

We must assume that i=n, which means that j and m can both be either n+1/2 or n-1/2 independent

of each other. Let us also consider the case where σn =↑ so that d†nσn can anti-commute with d†i↑,

−V2

∆

6=σn∑{~σ}

∑<im>

∑<ij>

d†1σ1. . . d†n−1σn−1

(d†n↑p†j↓ − d

†n↓p†j↑)(pm↓dn↑d

†n↑p†n+ 1

2σ0)d†n+1σn+1

. . . |0 > (313)

Then this reduces to,

−V2

∆

6=σn∑{~σ}

∑<im>

∑<ij>

d†1σ1. . . d†n−1σn−1

(d†n↑p†j↓ − d

†n↓p†j↑)(pm↓p

†n+ 1

2σ0)d†n+1σn+1

. . . |0 > (314)

To avoid this term evaluating to zero, we also need to make m=n+1/2 and σ0 =↓. (Note that we stillhave j being either j=n+1/2 and j=n-1/2). Hence we can expand out the sum to obtain,

−V2

∆

6=σ0,σn∑{~σ}

d†1σ1. . . d†n−1σn−1

(d†n↑p†n+ 1

2↓− d†n↓p

†n+ 1

2↑)d†n+1σn+1

. . . |0 > (315)

−V2

∆

6=σ0,σn∑{~σ}

d†1σ1. . . d†n−1σn−1

(d†n↑p†n− 1

2↓− d†n↓p

†n− 1

2↑)d†n+1σn+1

. . . |0 > (316)

It is important to notice that in this case the sum over all sigma terms should exclude σn and σ0, asthey have been determined in the derivation. In this instance we have σn =↑ and σ0 =↓.

56

If we had calculated the other case where σn =↓ we would have produced,

−V2

∆

6=σn∑{~σ}

∑<im>

∑<ij>

d†1σ1. . . d†n−1σn−1

(d†n↑p†j↓ − d

†n↓p†j↑)(−pm↑p

†n+ 1

2σ0)d†n+1σn+1

. . . |0 > (317)

Then we utilise the fact that i=n, m=n+1/2 and σ0 =↑, and again evaluating out the two cases for thepossible values of j to obtain,

+V 2

∆

6=σ0,σn∑{~σ}

d†1σ1. . . d†n−1σn−1

(d†n↑p†n+ 1

2↓− d†n↓p

†n+ 1

2↑)d†n+1σn+1

. . . |0 > (318)

+V 2

∆

6=σ0,σn∑{~σ}

d†1σ1. . . d†n−1σn−1

(d†n↑p†n− 1

2↓− d†n↓p

†n− 1

2↑)d†n+1σn+1

. . . |0 > (319)

Again we have already selected the values of σn and σ0, so they are excluded from the summation. Thisis the complementary solution for the singlet, where σn =↓ and σ0 =↑.

12.3 Combining the combinations

We now need to recombine the results from the oxygen hopping and singlet hopping back together tounderstand what happens in the Cu+ limit. As we had to specify the orientation of the spins of σ0 andσn we shall produce solutions for each of the combinations of these spins.

The first combination is for σn =↑ &σ0 =↑. Here we only have a contribution from the oxygen holehopping term.

V 2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

d†n↑p†(n+1/2)↑d

†n+1σn+1

. . . |0 > (320)

+V 2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

d†n↑p†(n−1/2)↑d

†n+1σn+1

. . . |0 > (321)

The second combination is for σn =↑ &σ0 =↓. In this instance we have a terms resulting from bothoxygen hole hopping and the singlet hopping interactions.

V 2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

d†n↑p†(n+1/2)↓d

†n+1σn+1

. . . |0 > (322)

+V 2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

d†n↑p†(n−1/2)↓d

†n+1σn+1

. . . |0 > (323)

−V2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

(d†n↑p†n+ 1

2↓− d†n↓p

†n+ 1

2↑)d†n+1σn+1

. . . |0 > (324)

−V2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

(d†n↑p†n− 1

2↓− d†n↓p

†n− 1

2↑)d†n+1σn+1

. . . |0 > (325)

We note that the hole hopping terms (322) & (323), exactly cancel the first terms from the singletcontribution, (324) & (325), which leaves us with the following result,

+V 2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

(d†n↓p†n+ 1

2↑)d†n+1σn+1

. . . |0 > (326)

+V 2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

(d†n↓p†n− 1

2↑)d†n+1σn+1

. . . |0 > (327)

57

The third combination is for σn =↓ &σ0 =↑. Here we again get a contribution from both oxygenand singlet hopping.

V 2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

d†n↓p†(n+1/2)↑d

†n+1σn+1

. . . |0 > (328)

+V 2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

d†n↓p†(n−1/2)↑d

†n+1σn+1

. . . |0 > (329)

+V 2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

(d†n↑p†n+ 1

2↓− d†n↓p

†n+ 1

2↑)d†n+1σn+1

. . . |0 > (330)

+V 2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

(d†n↑p†n− 1

2↓− d†n↓p

†n− 1

2↑)d†n+1σn+1

. . . |0 > (331)

We note that the hole hopping terms (328) & (329), exactly cancel the second terms from the singletcontribution, (330) & (331), which leaves us with the following result,

+V 2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

(d†n↑p†n+ 1

2↓)d†n+1σn+1

. . . |0 > (332)

+V 2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

(d†n↑p†n− 1

2↓)d†n+1σn+1

. . . |0 > (333)

The fourth and final combination is when σn =↓ &σ0 =↓. Similar to the first case we only have acontribution from oxygen hole hopping.

V 2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

d†n↓p†(n+1/2)↓d

†n+1σn+1

. . . |0 > (334)

+V 2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

d†n↓p†(n−1/2)↓d

†n+1σn+1

. . . |0 > (335)

58

12.4 Illustration of States

We shall now illustrate the case when σn =↑ & σ0 =↑.

V 2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

d†n↑p†n+ 1

2↑d†n+1σn+1

. . . |0 > (336)

−V2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

p†n− 1

2↑d†n↑d

†n+1σn+1

. . . |0 > (337)

Figure 21: Possible hops of spins

We shall now illustrate the case when σn =↑ & σ0 =↓.

+V 2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

(d†n↓p†n+ 1

2↑)d†n+1σn+1

. . . |0 > (338)

−V2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

(p†n− 1

2↑d†n↓)d

†n+1σn+1

. . . |0 > (339)

Figure 22: Possible hops of spins

59

We shall now illustrate the case when σn =↓ & σ0 =↑.

+V 2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

(d†n↑p†n+ 1

2↓)d†n+1σn+1

. . . |0 > (340)

−V2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

(p†n− 1

2↓d†n↑)d

†n+1σn+1

. . . |0 > (341)

Figure 23: Possible hops of spins

We shall illustrate the case when σn =↓ & σ0 =↓.

V 2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

d†n↓p†n+ 1

2↓d†n+1σn+1

. . . |0 > (342)

−V2

∆

∑{~σ}

d†1σ1. . . d†n−1σn−1

p†n− 1

2↓d†n↓d

†n+1σn+1

. . . |0 > (343)

Figure 24: Possible hops of spins

These resulting states can be produced by thinking about the physics that is occurring. The first stateoccurs when the copper spin hops onto the oxygen, then the oxygen spin hops back to where the copperspin was originally. This is essentially the result of a spin exchange operator that we discussed in theprevious chapter. The second state is where each of the spins shuffles along the line, first the copperhops to the oxygen and the oxygen to the copper site.

60

13 Redefined Hamiltonian - Cu+ Model

We have seen that new mechanisms emerge from applying the Cu+ Hamiltonian to the linear chain statein the previous section. The fact that the results are indifferent to the nature of the spins of the systemsuggests that we can think about the spins and holes separating from each other. We can then rewriteour Hamiltonian to describe the exchange and shuffle purely in terms of the doped hole in the system,where every site has its terms own individual spin wave function and the hole is able to hop to each ofthese sites. In essence this approach now labels the relative order of spins along the chain, not the actuallocations on it.

H = −V2

∆(∑j

h†jhj+1 + h†j+1hj) +V 2

∆

∑j

h†jhj(Pj−1j + Pjj+1) (344)

The first term causes the holes to hop to the left or to the right on the linear chain, while the secondterm exchanges the spins between sites j and j + 1 or j and j − 1.

We shall now define our state which contains the location of the hole and the spin wave function ofthe site itself.

|ψ〉 =∑j

h†j |Sj〉 (345)

〈ψ| =∑j

〈Sj |hj (346)

Here h†j tells us where the hole is and |Sj〉 is the spin wave function of site j. Note that the coefficient

X = V2

∆ is used onwards from here in this section.

Now we shall try and calculate out H|ψ〉 as it is necessary to solve the problem,

H |ψ〉 = −X(∑j

h†jhj+1 + h†j+1hj)∑j′

h†j′ |Sj′〉+X(

∑j

h†jhj(Pj−1j + Pjj+1))∑j′

h†j′ |Sj′〉 (347)

= −X∑j

∑j′

(h†jhj+1h

†j′ |Sj′〉+ h†j+1hjh

†j′ |Sj′〉

)+X

∑j

∑j′

(h†jhj(Pj−1j + Pjj+1)h†

j′ |Sj′〉

)(348)

Now in the first term we are required to set j′ = j + 1 and in the second term j′ = j in order to ensurethat we get a non-zero contribution. In the third term j′ = j, as the h†

j′ can be moved through the spin

exchange operators without the need for a minus sign, so we can obtain the following result,

H |ψ〉 = −X∑j

(h†j |Sj+1〉+ h†j+1 |Sj〉

)+X

∑j

(h†j(Pj−1j + Pjj+1) |Sj〉

)(349)

This is easier to understand if we let j = j − 1 in the second term, which is fine as we are summing overj anyway,

H |ψ〉 = −X∑j

(h†j |Sj+1〉+ h†j |Sj−1〉

)+X

∑j

(h†j(Pj−1j + Pjj+1) |Sj〉

)(350)

As we know that H|ψ〉 =E|ψ〉, we get the following equation,

H∑j

h†j |Sj〉 = −X∑j

(h†j |Sj+1〉+ h†j |Sj−1〉

)+X

∑j

(h†j(Pj−1j + Pjj+1) |Sj〉

)(351)

E∑j

h†j |Sj〉 = −X∑j

(h†j |Sj+1〉+ h†j |Sj−1〉

)+X

∑j

(h†j(Pj−1j + Pjj+1) |Sj〉

)(352)

61

Thus the equation for the ground state energy that we have derived is the following,

E |Sj〉 = −X |Sj+1〉 −X |Sj−1〉+X(Pj−1j + Pjj+1) |Sj〉 (353)

This is the exact equation that we would like to solve in general, however it is beyond our ability to solveit analytically. As a result we shall try to obtain some approximate solutions to it which we can use todraw conclusions about the model.

The first assumption that we shall make is that the spin wave function is the same for every site,i.e. |Sj〉 = |S〉. Then our equation is greatly simplified down,

E |S〉 = −X |S〉 −X |S〉+X(Pj−1j + Pjj+1) |S〉 (354)

〈S|E |S〉 = 〈S| − 2X |S〉+ 〈S|X(Pj−1j + Pjj+1) |S〉 (355)

So the equation for the approximate ground state energy becomes,

E = −2X +X 〈S| (Pj−1j + Pjj+1) |S〉 (356)

We again want to minimise the ground state energy. This implies that the expectation of the spinexchange operators must also be minimised.

13.1 First Approximation

We shall now consider what happens in the vicinity of the doped hole, by performing a simple calculationwhere the hole is in a tightly bound singlet. Then we only have to consider three sites one of which isthe doped oxygen site. We shall have σ1, the doped oxygen hole anti-parallel to σ0 & σ2, the copperholes. Then the system wants to minimise the energy, which means minimising the expectation of theexchange operators. In this approximation we have,

Pj−1j + Pjj+1 = (1

2+ 2S0.S1) + (

1

2+ 2S1.S2) (357)

= 1 + 2S1.(S0 + S2) (358)

Via some mathematical trickery we can rewrite the second term in the following form which allows us tobetter understand the problem,

2S1.(S0 + S2) = (S1 + S0 + S2)2 − S1.S1 − (S0 + S2)2 (359)

To minimise this expression we essentially need to minimise the first term and maximise the second andthird terms. For a single spin− 1

2 particle it is s= 12 . So S2 = 1

2 .32 = 3

4 . For two spin− 12 particles it is

either s=1 or s=0, and we want to have the maximum. So we have that S2 = 1.2 = 2. For three spin− 12

particles it is either s=3/2 or s=1/2, and we want to have the minimum. So we have that S2 = 12 .

32 = 3

4 .Substituting these max and min values into the equation we obtain the following result,

Min 1 + 2S1.(S0 + S2) = 1 + (3

4− 2− 3

4) = −1 (360)

Hence the expectation of the two spin operators must be greater than or equal to minus 1. Thus,

E = −2X +X 〈S| (Pj−1j + Pjj+1) |S〉 (361)

E0 = −2X +X 〈S| (−1) |S〉 (362)

E0 ≥ −3X (363)

So we have found that the lower bound for ground state is −3X which is the case where the hole is in atightly bound Zhang-Rice singlet.

As we have found 2S1.(S0 + S2) = −2, this means that S1.(S0 + S2) = −1. As the interaction on

62

either side are the same, S1.S0 = S1.S2 = − 12 . This means we can work out the probability that we form

singlets and triplets,

P0 =1

4− S1.S2 =

1

4+

1

2=

3

4(364)

P1 =3

4+ S1.S2 =

3

4− 1

2=

1

4(365)

So we have found that if the hole is fixed then the probability that it forms a singlet is 0.75, which is thebest it could do.

13.2 Second Approximation

In this approximation we shall consider the limit where the hole is equally likely to be found anywherealong the linear chain. This is the case where it is completely delocalised, the opposite limit to what wedid in the first approximation.

It turns out that there is a connection with the Heisenberg spin- 12 chain cunningly disguised within

the spin exchange operators.

Pj−1j + Pjj+1 = (1

2+ 2Sj−1.Sj) + (

1

2+ 2Sj .Sj+1) = 1 + 2Sj .(Sj−1 + Sj+1) (366)

Essentially as we want to minimise this function, the second term is that of the Heisenberg model, fromwhich have we can utilise the well known ground state solution [8],

Min Sj .(Sj−1 + Sj+1) = Min

2∑j=1

Sj .Sj+1 = −2 ln 2 +1

2(367)

Hence we get that,

E = −2X +X 〈S| (Pj−1j + Pjj+1) |S〉 (368)

E0 = −2X +X 〈S| (1 + 2(−2 ln 2 +1

2)) |S〉 (369)

E0 = −2X +X 〈S| (1 +−4 ln 2 + 1) |S〉 (370)

E0 = −2X +X 〈S| (−4 ln 2 + 2) |S〉 (371)

E0 = X 〈S| (−4 ln 2) |S〉 (372)

E0 = −4X ln 2 ≈ −2.77X (373)

Here we have calculated the energy of the ground state if the hole was completely delocalised along thechain which we know is closer to the actual solution. This yield an improved ground state energy ofaround −2.77X.

Now we want to work out what the probability of the hole being in a singlet with one of the sitesand see if it differs from the case where the hole was fixed in a tightly bound singlet. As we haveS1.(S0 + S2) = −2 ln 2 + 1

2 , we can clearly see that S1.S0 = S1.S2 = − ln 2 + 14 . Using these values we

find that,

P0 =1

4− S1.S2 =

1

4− (− ln 2 +

1

4) = ln 2 ≈ 0.69 (374)

P1 =3

4+ S1.S2 =

3

4+ (− ln 2 +

1

4) = 1− ln 2 ≈ 0.31 (375)

It transpires that in this second case where the holes are free to move and are equally likely to beanywhere along the chain the probability of the hole being in a singlet is still only ln 2 ≈ 0.69, which isnot much less than if the hole was fixed in a tightly bound singlet, 0.75. So we can see that the presenceof holes increases the likelihood of forming singlets in its presence, something that we will use in the nextapproximation.

63

13.3 Third Approximation

Our final improvement will be to add in an extra term to the second approximation that causes thereto be an increased likelihood of the hole forming a singlet where it is, in addition to the ability for thehole to delocalise along the chain. This should correspond to a more realistic description of the physicalpictures. We shall introduce this with a variational parameter α and then perform a variational approx-imation to give us the energy for this ground state in terms of α. Our final task will then be to minimisethis ground state energy with respect to α, which should produce a solution even closer to experimentalresults than our first two approximation.

The equation defining the variational approximation is the following,

E ≈ 〈ψ| H |ψ〉〈ψ|ψ〉

(376)

The form of the |ψ〉 is as given below where the term with the parameter α is the new addition to thewavefunction.

|ψ〉 =∑j

h†j |S〉+ α∑j

h†j(Pjj−1 + Pjj+1) |S〉 (377)

|ψ〉 =∑j

h†j(1 + α(Pjj−1 + Pjj+1)

)|S〉 (378)

13.3.1 Denominator of Variational Approximation

Then we shall calculate out the denominator of the approximation, 〈ψ|ψ〉 as follows,

〈ψ|ψ〉 =∑jj′

〈S|(

1 + α(Pj′j′−1 + Pj′j′+1))hj′h

†j

(1 + α(Pjj−1 + Pjj+1)

)|S〉 (379)

Clearly we need to set j′ = j and then drop the j,

〈ψ|ψ〉 =∑j

〈S|(1 + α(Pjj−1 + Pjj+1)

) (1 + α(Pjj−1 + Pjj+1)

)|S〉 (380)

=∑j

〈S|(

1 + 2α(Pjj−1 + Pjj+1) + α2(Pjj−1 + Pjj+1)2)|S〉 (381)

=∑j

1 + 2α 〈S| (Pjj−1 + Pjj+1) |S〉+ α2 〈S| (Pjj−1 + Pjj+1)2 |S〉 (382)

Now we can set j=2 and multiply by N as we are making use of the fact that this expression is the sameover every site j to yield,

〈ψ|ψ〉 = N(

1 + 2α 〈S| (P21 + P23) |S〉+ α2 〈S| (P21 + P23)2 |S〉)

(383)

= N(

1 + 2α(< P21 > + < P23 >) + α2(< P 221 > + < P 2

23 > + < P21P23 > + < P23P21 >))

(384)

We can further simplify this result by using the fact that exchanging spins twice results in the spinsreturning to their original order, so that < P 2

21 >=< P 223 >= 1. It is also fine to utilise the fact that the

spin interaction between nearest neighbours is always the same meaning terms like Pjj+1 are equal forany j. Thus we can simplify our equation down to reach,

〈ψ|ψ〉 = N(

1 + 4α < P21 > +α2(2+ < 2P21P23 >))

(385)

Note that also by the ideas described above we can deduce that, Pjj+1 = Pj+1j , so we finally can obtaina form for 〈ψ|ψ〉.

〈ψ|ψ〉 = N(

1 + 4α < P12 > +α2(2+ < 2P12P23 >))

(386)

64

13.3.2 Numerator of Variational Approximation

We nnow wish to work out the more complicated numerator term in the variational approximation,〈ψ| H |ψ〉. First of all we will need to apply the Hamiltonian to our state.

H |ψ〉 = −X∑j

(h†jhj+1 + h†j+1hj

)+X

∑j

(h†jhj(Pj−1j + Pjj+1)

)∑j′

h†j′

(1 + α(Pj′j′−1 + Pj′j′+1)

)|S〉

(387)

= −X(∑j

h†jhj+1 + h†j+1hj)∑j′

h†j′

(1 + α(Pj′j′−1 + Pj′j′+1)

)|S〉 (388)

+X∑j

(h†jhj(Pj−1j + Pjj+1)

)∑j′

h†j′

(1 + α(Pj′j′−1 + Pj′j′+1)

)|S〉 (389)

= −X∑jj′

(h†jhj+1 + h†j+1hj)h†j′

(1 + α(Pj′j′−1 + Pj′j′+1)

)|S〉 (390)

+X∑jj′

(h†jhj(Pj−1j + Pjj+1)

)h†j′

(1 + α(Pj′j′−1 + Pj′j′+1)

)|S〉 (391)

Then as we can move the hole operator past the spin exchange operators we get,

H |ψ〉 = −X∑jj′

h†jhj+1h†j′

(1 + α(Pj′j′−1 + Pj′j′+1)

)|S〉 (392)

−X∑jj′

h†j+1hjh†j′

(1 + α(Pj′j′−1 + Pj′j′+1)

)|S〉 (393)

+X∑jj′

(h†jhjh

†j′(Pj−1j + Pjj+1)

)(1 + α(Pj′j′−1 + Pj′j′+1)

)|S〉 (394)

In the first term we set j′ = j+1, and in the second and third terms j′ = j. This means we can eliminatej from the equation.

H |ψ〉 = −X∑j′

h†j′−1

(1 + α(Pj′j′−1 + Pj′j′+1)

)|S〉 (395)

−X∑j′

h†j′+1

(1 + α(Pj′j′−1 + Pj′j′+1)

)|S〉 (396)

+X∑j′

(h†j′(Pj′−1j

′ + Pj′j′+1))(

1 + α(Pj′j′−1 + Pj′j′+1))|S〉 (397)

Now we need to combine this with the other term 〈ψ| in order to complete the calculation.

〈ψ| H |ψ〉 =∑m

〈S| (1 + α(Pmm−1 + Pmm+1))hm multiplied by the following lines (398)

−X∑j′

h†j′−1

(1 + α(Pj′j′−1 + Pj′j′+1)

)|S〉 (399)

−X∑j′

h†j′+1

(1 + α(Pj′j′−1 + Pj′j′+1)

)|S〉 (400)

+X∑j′

(h†j′(Pj′−1j

′ + Pj′j′+1))(

1 + α(Pj′j′−1 + Pj′j′+1))|S〉 (401)

65

We expand this out to get,

〈ψ| H |ψ〉 = −X∑mj′

〈S| (1 + α(Pmm−1 + Pmm+1))hmh†j′−1

(1 + α(Pj′j′−1 + Pj′j′+1)

)|S〉 (402)

−X∑mj′

〈S| (1 + α(Pmm−1 + Pmm+1))hmh†j′+1

(1 + α(Pj′j′−1 + Pj′j′+1)

)|S〉 (403)

+X∑mj′

〈S| (1 + α(Pmm−1 + Pmm+1))hm

(h†j′(Pj′−1j

′ + Pj′j′+1))(

1 + α(Pj′j′−1 + Pj′j′+1))|S〉

(404)

In the first equation we set m = j′ − 1, the second m = j′ + 1 and the third m = j′. Employing thisreduction yields,

〈ψ| H |ψ〉 = −X∑j′

〈S|(

1 + α(Pj′−1j′−2 + Pj′−1j

′))(

1 + α(Pj′j′−1 + Pj′j′+1))|S〉 (405)

−X∑j′

〈S|(

1 + α(Pj′+1j′ + Pj′+1j

′+2))(

1 + α(Pj′j′−1 + Pj′j′+1))|S〉 (406)

+X∑j′

〈S|(

1 + α(Pj′j′−1 + Pj′j′+1))(

(Pj′−1j′ + Pj′j′+1)

)(1 + α(Pj′j′−1 + Pj′j′+1)

)|S〉

(407)

Continuing the algebra,

= −X∑j′

〈S|(

1 + α(Pj′−1j′−2 + Pj′−1j

′ + Pj′j′−1 + Pj′j′+1) + α2(Pj′−1j′−2 + Pj′−1j

′)(Pj′j′−1 + Pj′j′+1))|S〉

(408)

−X∑j′

〈S|(

1 + α(Pj′+1j′ + Pj′+1j

′+2 + Pj′j′−1 + Pj′j′+1) + α2(Pj′+1j

′ + Pj′+1j′+2)(Pj′j′−1 + Pj′j′+1)

)|S〉

(409)

+X∑j′

〈S|(

1 + α(Pj′j′−1 + Pj′j′+1))(

(Pj′−1j′ + Pj′j′+1) + α(Pj′−1j

′ + Pj′j′+1)(Pj′j′−1 + Pj′j′+1))|S〉

(410)

Now we shall simplify things by dividing by N and setting j′ = 2 to get us down to the next line,

〈ψ| H |ψ〉N

= −X 〈S|(

1 + α(P10 + P12 + P21 + P23) + α2(P10 + P12)(P21 + P23))|S〉 (411)

−X 〈S|(

1 + α(P32 + P34 + P21 + P23) + α2(P32 + P34)(P21 + P23))|S〉 (412)

+X 〈S| (1 + α(P21 + P23)) ((P12 + P23) + α(P12 + P23)(P21 + P23)) |S〉 (413)

Now collecting like terms for the coefficients of α,

= −X 〈S| (2− (P12 + P23)) |S〉 (414)

− αX 〈S| (P10 + P12 + P21 + P23) + (P32 + P34 + P21 + P23)− (P21 + P23)(P12 + P23)− (P12 + P23)(P21 + P23) |S〉(415)

− α2X 〈S| (P10 + P12)(P21 + P23) + (P32 + P34)(P21 + P23)− (P21 + P23)(P12 + P23)(P21 + P23) |S〉(416)

66

Finally we shall utilise the fact that < P10 >=< P01 > and < P01 >=< P12 > for example to makefurther simplifications.

= −2X + 2X < P12 > (417)

− 8αX < P12 > +αX(< P21P21 > + < P23P23 > + < P21P23 > + < P23P12 >) (418)

+ αX(< P12P21 > + < P23P21 > + < P12P23 > + < P23P23 >) (419)

− α2X(< P10P21 > + < P10P23 > + < P12P21 > + < P12P23 >) (420)

− α2X(< P32P21 > + < P32P23 > + < P34P21 > + < P34P23 >) (421)

+ α2X 〈S| (P21P12P21 + P21P23P21 + P23P12P21 + P23P23P21) |S〉 (422)

+ α2X 〈S| (P21P12P23 + P21P23P23 + P23P12P23 + P23P23P23) |S〉 (423)

When we further group terms together we get that,

= −2X + 2X < P12 > (424)

− 8αX < P12 > +αX(1 + 1+ < P21P23 > + < P23P12 >) (425)

+ αX(1 + 1+ < P23P21 > + < P12P23 >) (426)

− α2X(< P01P12 > + < P01P23 > +1+ < P12P23 >) (427)

− α2X(< P32P21 > +1+ < P43P21 > + < P43P32 >) (428)

+ α2X 〈S| (P21P12P21 + P21P23P21 + P23P12P21 + P23P23P21) |S〉 (429)

+ α2X 〈S| (P21P12P23 + P21P23P23 + P23P12P23 + P23P23P23) |S〉 (430)

So finally we get,

〈ψ| H |ψ〉N

= X(−2 + 2 < P12 >) (431)

+ αX(4− 8 < P12 > +4 < P12P23 >) (432)

+ α2X(−2 + 6 < P12 > −4 < P12P23 > −2 < P12P34 > +2 < P12P23P12 >) (433)

We have still to calculate out the values of the terms we just derived in this section. In particular weneed to derive forms for < P12 >and the more complicated < P12P23 >, < P12P34 > and < P01P12P01 >.

13.3.3 Spin Exchange Operator Simplifications

It turns out that either through long and tedious spin physics manipulations or an elementary knowledgeof permutation theory, < P12P23P12 >=< P13 >. We have been unable to find any solutions for what< P12P34 > is so we will use a numerical approximation of it in our work, found by Dr Long to be0.39252. We now need to work out what < P12P23 > is.

< P12P23 > =< (1

2+ 2 < S1.S2 >)(

1

2+ 2 < S2.S3 >) > (434)

=<1

4+ S1.S2 + S2.S3 + 4(S1.S2)(S2.S3) > (435)

=1

4+ < S1.S2 > + < S2.S3 > +4 < (S1.S2)(S2.S3) > (436)

As we know that < S1.S2 >=< S2.S3 >= 14 − ln 2, we can clearly work out the values of the simpler

terms in the line above. The more challenging part is the final term which has to be calculated in terms

67

of spin physics identities.

(S1.S2)(S2.S3) + (S2.S3)(S1.S2) = Sα1 Sα2 S

β2 S

β3 + Sβ2 S

β3 S

α1 S

α2 (437)

= Sα1 Sβ3 (Sα2 S

β2 + Sβ2 S

α2 ) (438)

= Sα1 Sβ3 (

1

2δαβ) (439)

=1

2Sα1 S

α3 (440)

=1

2S1.S3 (441)

So it is clear to see that < (S1.S2)(S2.S3) >= 14 < S1.S3 >. This means that we get terms for nearest

and next nearest neighbours below.

< P12P23 > =1

4+ 2 < S1.S2 > + < S1.S3 > (442)

13.3.4 Results from Takahashi

After reviewing the literature surrounding the area of the ground state of the spin- 12 Heisenberg linear

chain, we found that Takahashi [20] has calculated out the exact form of the more complicated nextnearest neighbour spin correlations.

< σiσi+1 >0= 1− 4 ln 2 (443)

< σiσi+2 >0= 1− 16 ln 2 + 9ξ(3) (444)

Here ξ(3) is defined to be the Riemann Zeta function of 3. This means that the expectation of theexchange operators becomes,

< P12 > =1

2+ 2 < S1.S2 > (445)

=1

2+ 2(

1

4− ln 2) (446)

= 1− 2 ln 2 (447)

< P13 > =1

2+ 2 < S1.S3 > (448)

=1

2+ 2(

1

4− 4 ln 2 +

9

4ξ(3)) (449)

= 1− 8 ln 2 +9

2ξ(3) (450)

< P12P23 > =1

4+ 2 < S1.S2 > + < S1.S3 > (451)

=1

4+ 2(

1

4− ln 2) +

1

4− 4 ln 2 +

9

4ξ(3) (452)

= 1− 6 ln 2 +9

4ξ(3) (453)

13.3.5 Minimising the Variational Approximation

We can now minimise the energy that we have calculated over alpha by doing a standard differentation.

E ≈ 〈ψ| H |ψ〉〈ψ|ψ〉

(454)

68

This differentiation will be best achieved by introducing constants for each power of α as seen below.

E ≈ A2α2 +A1α+A0

B2α2 +B1α+B0

(455)

These constants are written in terms of what we calculated earlier for the numerator and denominatorof the variational approximation.

• A2 = X(−2 + 6 < P12 > −4 < P12P23 > −2 < P12P34 > +2 < P12P23P12 >)

• A1 = X(4− 8 < P12 > +4 < P12P23 >)

• A0 = X(−2 + 2 < P12 >)

• B2 = 2 + 2 < P12P23 >

• B1 = 4 < P12 >

• B0 = 1

Doing this differentiation gives us the following quadratic equation that we need to solve.

α2(A1B2 −A2B1) + 2α(A0B2 −A2B0) + (A0B1 −A1B0) = 0 (456)

Making use of the quadratic formula we can then substitute our values into the line below to obtain theanswer for α.

α =−(A0B2 −A2B0)±

√(A0B2 −A2B0)2 − (A1B2 −A2B1)(A0B1 −A1B0)

(A1B2 −A2B1)(457)

We can now use the values we found to calculate out each of the constants we have and use them tosolve for the minimum value of alpha, which we find to be αmin ≈ −0.2168 numerically. This value ofalpha is indeed a small correction to the general Heisenberg state, which we can substitute back into thevariational approximation to get an estimate of the ground state energy.

Now we can put the negative solution of α into the following variational approximation to give usthe estimate of the ground state energy E0.

E0 ≈−1.55763Xα2 + 5.27333Xα− 2.77259X

1.09149α2 − 1.54518α+ 1≈ −2.87749X (458)

So have found our improved ground state energy for the linear chain to be E0 ≈ −2.88X to 2 decimalplaces. This shows that we gain a little energy from the hole being able to form singlets where it iscompared to the second approximation where it was unable to, about 0.11X.

The actual solution for this Cu+ linear chain has been calculated numerically in other works. Theyfound an answer of −2.89535X to 5 decimal places. This suggests that our approximation is a very goodone as it is about 99 % of the actual value.

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14 Conclusions

In this project we have been studying and trying to model the two types of high temperature cupratesuperconductors, that of hole-doped cuprates such as La2−xSrxCuO4 and electron-doped cuprates likeNd2−xCexCuO4. By considering the similarities between all the cuprates in Chapter 1 we were able toshow that the key physics for both types is contained within the 2D CuO2 layers and the relevant orbitalsin that layer are copper 3d

x2−y2 and oxygen 2px and 2py. We further showed in Chapter 2 that the

cases of electron and hole doping differ, because the doped holes would rather occupy the oxygen sitesand the doped electrons want to create full 3d copper orbitals. We illustrated the fact that experimentspredict an asymmetry in the phase diagram between the cases of hole doping and electron doping, incontrast to the prediction of many conventionally used models such as the t-J and Hubbard models.

Our work in Chapter 4 was dedicated at constructing an appropriate ground state for the copper oxidelayer based on the three band model proposed by Anderson. Then we went on to try and determine ahigher order correction to this ground state by considering the effects of second order hopping, eventuallyproducing three models. The first was the electron doping model which was identical to the Hubbardmodel, the second was for the case of hole doping in the Cu3+ limit. We showed that this solution wasessentially the same as the t-J model with J=0 and predicted a ferromagnetic ground state not seenexperimentally. This lead us to the idea that the t-J model is an incomplete description for the cupratesuperconductors. Finally we developed a hole doping model for the Cu+ limit which we found to be inagreement with the physical system itself.

When we studied the results of the electron doping model on the linear chain we found that therethe system was degenerate. As we were only hopping copper holes, the overall alignment of the systemwas preserved due to the blocking effect and the system has no preference for antiferromagnetism overferromagnetism. When we studied the case for periodic motion around a ring we found the system againhas no preference for antiferromagnetism or ferromagnetsim. The ground state energy was found to beabout −0.56eV in both cases.

In Chapter 10 we started to analyse what happens for our hole doping model in the Cu3+ limit. As thismodel corresponds to hopping of singlets only, we found that there is again massive degeneracy in thelinear chain geometry as no spin state is preferred over another. When we considered the model on asquare ring we found that the spins did have an important role to play. For the cases where we had amix of up and down spins, the hole motion was able to permute the order of the spins. This resultedin the appearance of Nagaoka ferromagnetism, which lowered the energy of the high spin ground stateby about 0.3|Y| compared to the low spin state. We also showed that even the addition superexchangecould not prevent the prediction of this ferromagnetic ground state at high doping, which is not observedexperimentally.

We finally delved into the correct Cu+ hole doping limit in Chapters 12 and 13. The bulk of the workwas dedicated to understanding the problem of the linear chain, for which we found a different result tothe previous results. The combined motion of holes and singlets allowed us to recast the problem intoa form where the holes and spins have separated from each other. This meant that the solution on thelinear chain is very similar to that of the nearest neighbour Heisenberg spin- 1

2 chain, so the ground stateis an overall antiferromagnetic low spin state. We solved the problem exactly for the case of a single holewith the ground state of the low spin state being -2.88X ≈ −0.81eV as our best approximation, whereaspeople have used impurity theory to find the energy of the high spin state as -2.47X ≈ −0.69eV.

The next phase of our work would be to go on to try and solve the Cu+ limit for a ring of sitesand then compare the results with what we found for electron doping and Cu3+ limit. This should allowus to gain an understanding into what is actually happening for the complete CuO2 layer as we wouldthen have studied both cases of linear and periodic motion.

As we have not had time to solve our model completely for the square lattice it is rather difficultto make any connections with experiments, since the cuprate layers in the superconductors all have thisgeometry. However the compound YBa2Cu3O6+δ seen back in Figure 1, which is also known as the 1-2-3

70

compound actually contains layers of CuO in addition to the standard CuO2 planes. This means that intheory we could determine an experiment to isolate the effects of these linear chains from the 2D planesand we could measure experimentally the linear chain system in practice.

Finally although we have studied the problem for only a single doped hole we should conclude bytouching upon what we suspect happens for the cases of doping multiple holes. As we have explainedthe ground state is initially in a pure Heisenberg spin ground state, but the motion of the doped holesdistort this Heisenberg ground state in a local region around them. The doped holes form tightly boundZhang-Rice singlets which leave a trail of spin correlations behind them when they hop, which eventuallyfade away back into the global antiferromagnetic background. The more holes that are doped into thesystem, the closer they become together on average and the more the overall antiferromagnetic back-ground will decays. This overall process destroys the antiferromagnetic background much quicker thanin the case of electron doping, as it is of order V2 rather V4 and is the mechanism that we propose toexplain the asymmetry in the phase diagram.

71

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