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Exploring the Dirac equation Mackovic, Brie 2018

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Exploring the Dirac EquationbyBrie MackovicA THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THEREQUIREMENTS FOR THE DEGREE OFMaster of ScienceinTHE FACULTY OF GRADUATE AND POSTDOCTORAL STUDIES(Physics)The University of British Columbia(Vancouver)July 2018c©Brie Mackovic, 2018The following individuals certify that they have read, and recommend to the Faculty ofGraduate and Postdoctoral Studies for acceptance, a thesis/dissertation entitled:Exploring the Dirac Equationsubmitted by Brie Mackovic in partial fulfillment of the requirements for the degree of Mas-ter of Science in Physics.Examining Committee:Dr. Gordon Semenoff, Physics (Supervisor)Dr. Joanna Karczmarek, Physics (Supervisory Committee Member)iiAbstractIn this thesis low energy excitations of perfectly dimerized trans-polyacetylene are mod-elled using the one-dimensional Dirac equation. The system is solved on both the half-lineand segment, and the solutions are used to explore quantum phenomena. It is discoveredthat the zero mode of the half-line is a Majorana fermion quasiparticle. It is also foundthat dominate zero mode coupling to an electron on a scanning tunnelling microscope isachieved with a sufficiently large mass gap of the quantum wire. This allows scattering stateexcitations to be ignored in calculations in this thesis. It is also shown that the zero modecan facilitate entanglement of two electrons, each in proximity to opposite ends of a long seg-ment of trans-polyacetylene. An algorithm is also developed which teleports the spin stateof an electron on a segment of trans-polyacetylene. The quantum measurement used in thisalgorithm conserves fermion parity symmetry, however charge superselection is violated forthree-fourths of the measurement operators. In the thermally isolated system teleportationis successful for all of the measurement operators on the ground state. However, decoher-ence occurs in the non-thermally isolated system due to thermal mixing of nearly degeneratestates, leading to teleportation being successful for only half of the measurement operationson the thermal state.iiiLay SummaryIn this thesis we model how electrons behave on a special material called polyacetylene.This material is considered to be quasi one-dimensional since it is a chain of single atoms.This reduced dimensionally causes electrons on the material to behave in unusual ways. Weexplore this unusual behaviour and determine how it can be harnessed for the creation ofdevices made from polyacetylene.ivPrefaceThis thesis is an original intellectual product of the author, B. Mackovic. Work inspiredby this material led to the co-authorship of a paper which has been published on the arXivand submitted to a journal as M. Ghrear, B. Mackovic, and G. W. Semenoff (2018).vTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiLay Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viList of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viiAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viiiDedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ix1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12. The Dirac Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33. Polyacetylene . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74. The Dirac Equation on the Half-Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.1. Bound and Scattering States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.2. Charge Conjugation and the Majorana Fermion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184.3. Zero Mode Coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234.4. Electronic Entanglement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315. The Dirac Equation on the Segment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365.1. Bound and Scattering States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365.2. Quantum Teleportation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416. Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51Appendix A. Normalization: Half-Line Scattering States . . . . . . . . . . . . . . . . . . . . . . . . . . . 52Appendix B. Completeness: Half-Line States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55viList of Figures1. The configuration of trans-polyacetylene . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82. Polyacetylene modeled along the quasi-one-dimensional symmetry axis of the chain.Phases A and B represent the two degenerate ground states of the system. . . . . . . . . 83. Scattering state coupling, c, as a function of changing mass gap, m, using thefixed parameters z0 = 5 × 10−10m, M = .5 × 106eV, k1 =√2M , k2 =√20M andC = D = 1/2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304. Bound state coupling, c0, as a function of changing mass gap, m, using the fixedparameters z0 = 5 × 10−10m, M = .5 × 106eV, k1 =√2M , k2 =√20M andC = D = 1/2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30viiAcknowledgementsI am deeply grateful for my supervisor, Dr. Gordon Semenoff, for his guidance and en-couragement throughout the course of my graduate studies. I will always hold fondly in mymemory the time we shared discussing the quantum world together.I would also like to thank NSERC for kindly funding my research.viiiDedicationI would like to dedicate this work to my family, for always believing in my dreams andlovingly supporting my journey towards them. I cannot imagine a better team to have bymy side.ix1. IntroductionThe Dirac equation is a relativistic and quantum mechanical equation originally developedby Dirac to describe spin 1/2 free electrons [1]. The development of this beautiful equationled to the ground-breaking discovery of anti-matter, as Dirac described the negative energysolutions of the equation as oppositely charged anti-electrons. This postulation was laterconfirmed experimentally through the discovery of positrons.To this day Dirac’s equation is still leading to new discoveries through various extensionsand deformations of the equation. In particular, it has been confirmed that Dirac-like equa-tions can be used to describe low-energy excitations of condensed matter systems. This hasrevealed novel excitations in condensed matter systems when physically interpreting spectralpeculiarities of the Dirac equation. In this thesis it is shown that the Dirac equation canrepresent low energy excitations in the quasi-one-dimensional material trans-polyacetylene.Interesting physical properties and applications of the solutions of the system are then ex-plored.In Sec.2 a brief derivation of the Dirac equation is given. Then in Sec.3 it is shown thatthe Dirac equation can be used to represent the Hamiltonian of perfectly dimerized trans-polyacetylene at low energies. In this thesis trans-polyacetylene is referred to simply aspolyacetylene. In Secs.4 and 5 the bound and scattering states of the Dirac equation aresolved on the half-line and segment, respectively. In each of these sections analysis is doneto explore the interesting physical properties of the solutions and how they can be used toachieve quantum phenomena. In Subsec.4.2 it is seen that the Dirac equation has a chargeconjugation symmetry and it is discovered that the zero mode of the half-line is a Majoranafermion quasiparticle. In Subsec.4.3 it is found that if polyacetylene has a sufficiently largemass gap then zero mode coupling to a scanning tunnelling microscope (STM) electronwill dominate over scattering state coupling. This allows scattering state excitations to beignored in calculations in this thesis. In Subsec.4.4 it is found that two electrons, one inproximity to each end of a long chain of polyacetylene, can entangle via the zero mode.Finally, in Subsec.5.2 a teleportation algorithm is developed which teleports the spin of anelectron on a segment of polyacetylene. The quantum measurement used in this algorithmconserves fermion parity symmetry, but three-fourths of the measurement operators violatecharge superselection. In the thermally isolated case teleportation is successful for all of the1measurement operations on the ground state. However, in the non-thermally isolated casedecoherence occurs as teleportation fails half of the time on the thermal state resulting fromthermal mixing of nearly degenerate states.22. The Dirac EquationThe energy-momentum relationship of a relativistic particle isgµνpµpν = pνpν = m2where pµ = (E, ~p) is the contravariant four-momentum of the particle in 3 + 1-dimensions,gµν is the metric tensor with the signature (1,−1,−1,−1), and m is the particle’s propermass. Therefore,E2 − |~p|2 = m2and substituting in the differential operators,Eˆ → i ∂∂t, pˆ→ −i~∇,leads to the relativistic Klein-Gordon equation(− ∂2∂t2+ ~∇2)ψ(x) = m2ψ(x)(1)where we have used the notation (x) = (t, ~x).There are two fundamental issues with the Klein-Gordon equation. For one thing theequation requires negative energy solutions, since E = ±√m2 + |~p|2, which leads to thequestion of what the physical interpretation of the negative energies would be. In addition,the equation requires the possibility of negative probability density, ρ(x). This can be seenby considering the continuity equation∂µjµ(x) = 0where jµ(x) = (ρ(x),~j(x)), with ρ(x) the probability density and~j(x) the probability current.In the relativistic casejµ(x) =i2m(ψ∗(x)∂µ(ψ(x))− ∂µ(ψ∗(x))ψ(x))where ∂µ = (∂t,−∂x,−∂y,−∂z) is the four-gradient. Looking at the first componentρ(x) = j0(x) =i2m(ψ∗(x)∂∂t(ψ(x))− ∂∂t(ψ∗(x))ψ(x))we can see the second problem with the Klein-Gordon equation: ρ(x) can be positive ornegative depending on ψ(x) and ∂ψ(x)∂t. Since the Klein-Gordon equation is a second order3partial differential equation the functionsψ(t = 0, ~x),∂∂tψ(t = 0, ~x)can arbitrarily be chosen at t = 0, and as such there could be negative values for ρ(t = 0, ~x):negative probability density. In light of this issue Dirac stepped in to develop a relativisticwave equation with first order time and space derivatives.The Dirac equation is described as a “square root” of the Klein-Gordon equation, itsprobability density is always positive but it still includes negative energy solutions. Bydefinition the Dirac equation represents a relativistic wave equation for a free electron andcan be written as(iγ0∂∂t+ i~γ · ~∇−m)ψ(x) = 0,(2)or equivalently as(iγµ∂µ −m)ψ(x) = 0where γµ = (γ0, ~γ) = (γ0, γ1, γ2, γ3). By multiplying the Dirac equation, Eqn.2, by itsconjugate and requiring it to be equal to the Klein-Gordon equation, Eqn.1, we find therestrictions on the γµ coefficients. Looking atψ†(x)(−iγ0 ∂∂t− i~γ · ~∇−m)(iγ0 ∂∂t+ i~γ · ~∇−m)ψ(x) = 0(3)and comparing Eqn.3 to Eqn.1 gives(γ0)2 = I, (γn)2 = −I, {γµ, γν} = 0 for µ 6= ν(4)where n = 1, 2, 3 and µ, ν = 0, 1, 2, 3. We can write the Dirac equation in a slightly differentform by multiplying Eqn.2 by γ0 and using the properties of the gamma matrices, Eqn.4, toget(−i~α · ~∇+ γ0m)ψ(x) = i ∂∂tψ(x)(5)where we have defined αn = γ0γn which gives(γ0)2 = I, (αn)2 = I, {αn, αm} = 0 for n 6= m, {αn, γ0} = 0(6)with n,m = 1, 2, 3. In the (3 + 1)-dimensional case the minimum dimension of the matricesmust be 4 × 4 in order to satisfy the conditions in Eqn.6. Using separation of variables4the solution to Eqn.5 has the form Ψ(~x, t) = ψ(~x)e−iEt, where ψ(~x) satisfies the eigenvalueequation(−i~α · ~∇+ γ0m)ψ(~x) = Eψ(~x).(7)There are two different ways in which Dirac physically interpreted the negative energysolutions. In one case Dirac developed the Dirac sea model, in which the vacuum is redefinedso that all the negative energy states are filled with electrons, and all the positive energystates are empty. Therefore, the vacuum state has infinite negative energy, infinite negativecharge, and zero momentum since for every electron with momentum ~p there is an electronwith −~p. Dirac predicted the existence of antiparticles through his postulation of a hole statewhich accompanies the lifting of an electron out of the Dirac sea into a positive energy state.A hole state is a state where all the negative energy states are filled except for one, and aparticle state is a state where all the negative energy states are filled, plus one positive energystate is filled. If the original electron in the sea has momentum ~p, energy E = −√~p2 +m2and charge −|e|, then the hole left in its place has momentum −~p, energy E = √~p2 +m2 andcharge |e|. Therefore, the hole state of energy E and charge −|e| represents the antiparticle,positron, to the particle state, electron, of energy E and charge |e|. Experimental discoveryconfirmed the existence of positrons shortly after Dirac’s postulation.Both the hole and particle states are multi-fermion states, therefore in order to determinetheir wave functions we must construct the Slater determinant consisting of the infinite setof occupied negative energy states minus one negative energy state for the hole, and plusone positive energy state for the particle. The Slater determinant for an N -electron systemhas the formψ( ~x1, ~x2, ..., ~xN) =∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣ψ1(~x1) ψ2(~x1) · · · ψN(~x1)ψ1(~x2) ψ2(~x2) · · · ψN(~x2)· · ·· · ·· · ·ψ1(~xN) ψ2(~xN) · · · ψN(~xN)∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣where ψj(~xi) denotes the wave function of electron j with spatial coordinates ~xi. Con-structing the wave function from an infinite Slater determinant is clearly an impossible task.5Therefore, the Dirac sea model predicts the positron wave function but does not give itswave function, or the wave function of the electron for that matter.An alternative way in which Dirac deals with the negative energy electron wave functionsis to redefine them as positron wave functions. That is, the electron wave function withnegative energy −E and momentum −~p, is used to describe a positron with positive energyE and momentum ~p: Ψe+(E, ~p) = Ψe−(−E,−~p). This seems logical since the Dirac equationdescribing an electron is in fact a definition in itself, since particle charge does not appear inthe equation. If positrons had been known when Dirac first developed the equation he couldjust as easily defined it to represent positrons. Furthermore, since we associate free electronswith plane waves it seems perfectly natural that positrons should also be associated withplane waves. Either method of dealing with the negative energy states is appropriate.It turns out that Dirac-like equations can be used to approximate low-energy excitations incondensed matter systems, as will be seen in Sec.3. The positive and negative energy statesrepresent the free conduction and bound valence electrons, respectively. This makes sensewhen considering that the ground state of a material would have no conduction electrons,but all valence electron states would be filled. The mass term, m, in the energy eigenvalueexpression, E = ±√p2 +m2, produces a gap between the conduction and valence bands.Therefore, the material is an insulator for a non-zero mass term and a conductor if otherwise.63. PolyacetyleneIn this section we will show how the Dirac equation can represent the equation of motionfor conduction and valence electrons on trans-polyacetylene, at low energies. Polyacetyleneis classified as a semiconducting quantum wire, or nanowire. Quantum wires are consideredto be quasi one-dimensional because they are essentially a long chain of atoms. This reduceddimensionality allows for quantum effects to influence the transport of charge. Semiconductornanotechnology is a rapidly growing field, leading to the possibility of semiconductor devicesmade from polyacetylene.Polyacetylene is a quasi one-dimensional polymer of Carbon atoms, where two of the fourvalence electrons of Carbon are used to create strong covalent bonds to nearest neighbouringCarbon atoms, one is used to bond covalently to a Hydrogen atom, and the remaining electronacts as a conduction electron. We consider the tight-binding model in which the conductionelectron of Carbon is sitting in an orbital localized near the atom. There are two differentconfigurations of polyacetylene, known as trans-polyacetylene and cis-polyacetylene, both ofwhich have an optimal bond angle of 120o between bonds. In this article we will focus onlyon trans-polyacetylene, configuration in Fig.1, and henceforth it will be referred to simply aspolyacetylene. Equilibrium of the polymer is achieved only after the Carbon atoms shift byabout 0.04 angstroms to the left or right. This shifting is called Peierls instability and leadsto alternating double and single Carbon-Carbon bonds on the chain. Letting un representthis shift from equilibrium the degenerate ground state has either un < 0 and un−1, un+1 > 0to have a double bond to the left and single bond to the right of the Carbon atom at atomicsite n, or the reverse case, as seen in Fig.2. Peierls instability leads to a gap in the electronicspectrum of polyacetylene at the Fermi surface, making the material a semiconductor. Inabsence of Peierls instability polyacetylene would otherwise behave as a conductor. For someof the origional literature on polyacetylene see references [2]-[9].Defining “a” to be the lattice spacing of unshifted polyacetylene, the position of the nthatom on the chain of one of the degenerate ground states is xn = na+ un. The Lagrangianfor the chain of CH groups in polyacetylene with N atomic sites is thenL =12∑nMu˙2n +∑n 6=mu(xn − xm)7Figure 1. The configuration of trans-polyacetyleneFigure 2. Polyacetylene modeled along the quasi-one-dimensional symmetryaxis of the chain. Phases A and B represent the two degenerate ground statesof the system.where the first term represents the KE of the CH group, with M the mass of the CHgroup, and the second term is the potential energy between atomic sites. Assuming thereis sufficient screening of Coulomb interactions we make the approximation that only nearestatomic neighbour interactions need be considered, which givesL =12∑nMu˙2n +∑nu(xn+1 − xn).Furthermore, we make a harmonic approximation by assuming the un are small:u(x) = u(a) +k2(x− a)2 + ....So for nearest neighbour interactions we haveu(xn+1 − xn) = u(a) + k2(xn+1 − xn − a)2 + ....Given that,xn+1 = (n+ 1)a+ un+18andxn = na+ un,thenu(xn+1 − xn) = u(a) + k2(un+1 − un)2 + ...which gives the approximationL = Nu(a) +12∑nMu˙2n +k2∑n(un+1 − un)2where the first term will be ignored since it is included in the cohesion energy.Including hopping energy between nearest atomic neighbours the Hamiltonian isH = −∑nσ(t1 − α(un+1 − un))(c†(n+1),σcn,σ + c†n,σc(n+1),σ) + Lwhere c†nσ, cn,σ are the creation and annihilation operators, respectively, for electrons on then-th site with spin σ = ±1/2, and t1 and α are some constants. This Hamiltonian is knownas the Su-Schrieffer-Heeger (SSH) model. The Born-Oppenheimer approximation has beenused in this model, which assumes the ions to be “frozen” in their equilibrium positions. Forperfectly dimerized polyacetylene, un = (−1)nu, the Lagrangian isL =∑nk2(−1)n2(−u− u)2 =∑nk42u2 = 2Nku2.In order to get our Hamiltonian into a form that resembles the Dirac equation we will ignorethis constant term: we will ignore the PE between nearest neighbour atoms. This leaves,H = −∑nσ(t1 + α(−1)n2u)(c†(n+1),σcn,σ + c†n,σc(n+1),σ).Let t2 = α2u, thenH = −∑nσ(t1 + t2(−1)n)(c†(n+1),σcn,σ + c†n,σc(n+1),σ).Also, let the operators be off by a phase: cn = inan and c†n = (−i)na†n, giving,H = i∑nσ(t1 + t2(−1)n)(a†(n+1),σan,σ − a†n,σa(n+1),σ).Notice that the hopping constants alternate, between t1 + t2 and t1 − t2, so the chain nowhas an overall lattice constant of 2a and we can model the N -site chain as an N/2-site even9system plus an N/2-site odd system. Hopping between shorter bonds has a larger probability,|t1 + t2|2, than hopping between longer bonds, |t1 − t2|2, as expectedH = i∑nσ((t1 + t2)(a†(2n+1),σa2n,σ − a†2n,σa(2n+1),σ) + (t1 − t2)(a†2n,σa(2n−1),σ − a†(2n−1),σa2n,σ)).Since the even and odd atomic sites are different the even and odd creation and annihilationoperators must be different. Transforming to k-space these operators area2n,σ =1√N∑keik2naak,σanda(2n+1),σ =1√N∑keik(2n+1)abk,σ.For our model we ignore the spin of the electron, thereforeH =iN∑n,k,k′((t1 + t2)(e−ik(2n+1)ab†keik′2naak′ − e−ik′2naa†k′eik(2n+1)abk)+ (t1 − t2)(e−ik′2naa†k′eik(2n−1)abk − e−ik(2n−1)ab†keik′2naak′)).Using the property 1N∑n ei(k−k′)n = δk,k′ , thenH = i∑k((t1 + t2)(e−ikab†kak − eikaa†kbk) + (t1 − t2)(e−ikaa†kbk − eikab†kak)).(8)If we had Periodic Boundary Conditions (PBCs) on our system, so that site 1 ≡ N + 1, thenaN+1 =1√N∑keik(N+1)aak → a1 = 1√N∑keikaakso we needeikNa = 1kNa = 2pink =2pinNawhere n → −N2...N2for unique solutions, therefore k ∈ (−pia, pia]. In Sec.5 we find that theenergy momentum relationship for the scattering states of the Dirac equation solved on thesegment are: E = mcos kLand E2 = m2 + k2. These energy momentum relationships can becombined to give a restriction on k,k = m tan (kL),10which can have an infinite number of unique solutions as L → ∞. So we do not have aBrillion zone in our case. However, as we will see below, the Dirac equation approximatesthe Hamiltonian of polyacetylene for only small values of k, so not having a Brillion zoneshould not be a problem.To transition from a discrete to continuous spectrum we write∑kf(k)→ L2pi∑kf(k)δk =Na2pi∑kf(k)δk → Na2pi∫f(k)dktherefore our Hamiltonian in Eqn.8 can be written asH =L2pi∫i((t1 + t2)(e−ikab†kak − eikaa†kbk) + (t1 − t2)(e−ikaa†kbk − eikab†kak))dk.Writing the Hamiltonian in matrix formH =L2pi∫  0 i(−eika(t1 + t2) + e−ika(t1 − t2))i(e−ika(t1 + t2)− eika(t1 − t2)) 0 dkwhere the top component of a vector in this basis represents an electron on an even atomicsite and the bottom component represents an electron on an odd atomic site. Simplifyingthe HamiltonianH =L2pi∫  0 i(−t12i sin (ka)− t22 cos (ka))i(−t12i sin (ka) + t22 cos (ka)) 0 dk.Finding the energy eigenvalues of our Hamiltonian from det(H − EI) = 0 givesE2 = t214 sin2 (ka) + t224 cos2 (ka).Since t2 = α2u we can see that with Peierls instability, u 6= 0, polyacetylene is a semicon-ductor, however in its undimerized state, u = 0, it is a conductor. For small k: sin (ka) ' kaand cos (ka) ' 1 thereforeH =L2pi∫  0 i(−t12i(ka)− t22)i(−t12i(ka) + t22) 0 dk.Choose t22 = m and at12 = 1, thenH =L2pi∫  0 k − imk + im 0 dk.(9)11Notice in this form the energy eigenvalues of the Hamiltonian areE2 − (k + im)(k − im) = 0→ E2 = k2 +m2which is what we expect for the energy eigenvalues of the relativistic Dirac equation. Herewe can see that dimerized polyacetylene is an insulator with a mass gap of size “m” in theelectronic spectrum. Converting from position to momentum space, using the fact that k = pin natural units, we replacekˆ = −i ddxin Eqn.9H =L2pi∫  0 −i ddx − im−i ddx+ im 0 dx.So we see that the Dirac equation allows for an electron travelling on polyacetylene tobe viewed as a relativistic particle travelling in free space, if we give it the effective massm rather than the true electron mass me. This viewpoint is equivalent to determining theequation of motion of the electron with its true electron mass, me, existing on polyacetylenewith all of the forces the polymer is actually exerting on it.124. The Dirac Equation on the Half-Line4.1. Bound and Scattering States. Using the results of Sec.2 we can write the 1-dimensionaltime independent Dirac equation as(−iα1 ddx+ γ0m)ψ(x) = Eψ(x)where ψ(x) is the Dirac spinor, and α1 and γ0 must satisfy(γ0)2 = (α1)2 = I, {α1, γ0} = 0.The Pauli matrices, σn, and the identity, I, satisfy these requirements. Since all 2 × 2Hermitian matrices can be written as linear combinations of σn and I, they are a naturalchoice for α1 and γ0. Choosing α1 = σx and γ0 = σz gives the HamiltonianH = −i ddxσx +mσz = m −i ddx−i ddx−m(10)which is related to the Hamiltonian we found in Sec.3,H ′ = −i ddxσx +mσy = 0 −i ddx − im−i ddx+ im 0 ,by a simple rotation. The Pauli matrices can be rotated into each other by the followingoperationH = e−i(nˆ·~σ)φ/2H ′ei(nˆ·~σ)φ/2.Using the identityei(nˆ·~σ)φ/2 = I cosφ/2 + i(nˆ · ~σ) sinφ/2wherenˆ = Pauli axis of rotationφ = angle of rotation.In our case nˆ = xˆ and φ = pi/2 so we can write the rotation asei(nˆ·~σ)φ/2 = I1√2+ iσx1√2.13The rotation can be seen by considering the property of Pauli matrix multiplication: σaσb =δab + iεabcσc. Therefore, the spinors of the two Hamiltonians are related bye−i(nˆ·~σ)φ/2ψ′(x) =1√2 1 −i−i 1u′(x)v′(x) = 1√2 u′(x)− iv′(x)−iu′(x) + v′(x) = ψ(x)where in Sec.3 we saw that u′(x) and v′(x) represent the even and odd atomic wave functions,respectively, of dimerized polyacetylene at location x.The eigenvalue equation of H is m −i ddx−i ddx−mu(x)v(x) = Eu(x)v(x)which is equivalent to the two equations(m− E)u(x)− i ddxv(x) = 0(11)and−i ddxu(x)− (m+ E)v(x) = 0.(12)We can solve Eq.11 for u(x)u(x) =−iE −mddxv(x)and substitute u(x) into Eq.12 to get(p2 +d2dx2)v(x) = 0where p2 = E2 −m2. For the bound state solutions, |E| < m, we make the ansatzv(x) = Aeipx +Be−ipx = Ae−kx +Bekx,u(x) =−ikE −m(−Ae−kx +Bekx),E = ±ω(k), ω(k) ≡√−k2 +m2, p = ik.In order to have a normalizable wave function we must set B = 0, thereforeψBk = A ikE−m1 e−kx.(13)14We must ensure that the Hamiltonian remains Hermitian on the half line, a requirementthat places a restriction on the possible boundary conditions. An operator is Hermitian if< ψ|Hψ >=< Hψ|ψ >,therefore using our time independent 1-D Dirac equation, equate< ψ|Hψ >=∫ ∞0[u(x)∗ v(x)∗] m −i ddx−i ddx−mu(x)v(x) dxand< Hψ|ψ >=∫ ∞0( m −i ddx−i ddx−mu(x)v(x))†u(x)v(x) dxwhich leads to the following boundary condition0 = (u(x)∗v(x) + v(x)∗u(x))∣∣∣∞0.From here we see that the Dirac current, u(x)∗v(x)+v(x)∗u(x), must vanish at the boundariesin order for the Dirac equation to be Hermitian. At infinity the current vanishes sinceu(x) and v(x) are assumed to be square integrable, therefore the only remaining boundarycondition isu∗(0)v(0) + v∗(0)u(0) = 0.A boundary condition which would satisfy this isu(0) = −iv(0).(14)Applying Eqn.14 to the wave function solution in Eqn.13 leads to the following restrictionon the possible momentum valuesk = m−√−k2 +m2.Considering all the positive and negative momentum states the boundary condition is sat-isfied by our bound state only when k = 0 and k = m. However, k = 0 gives the E = |m|constant function throughout space, so this cannot be a solution. The k = m solution givesthe normalized wave function for E = 0ψB = ψ0 =√m−i1 e−mx.(15)15The larger the mass gap of the wire, m, the steeper the wave function’s incline to its peakat the edge of the wire and the closer it hugs the axis away from the edge. Having azero energy solution is a surprising result, one that Dirac did not predict when originallydeveloping the Dirac equation. Therefore, he did not include a zero energy state in his Diracsea interpretation, so there is a question of whether this state should be empty or filled.To transfer from bound to scattering state equations, |E| > m, all we need to do is replaceik → k which is equivalent to replacing k → −ik.v(x) = Aeikx +Be−ikx,u(x) =kE −m(Aeikx −Be−ikx),E = ±ω(k) , ω(k) =√k2 +m2, k = p.As with the bound state wave functions, take the boundary condition of u(0) = −iv(0). Thisleads to the relationship between the constantsA =(q − i)(q + i)Bwhere we have defined q = kE−m , giving the scattering state wave functionsψS = Bq( (q−i)(q+i)eikx − e−ikx)(q−i)(q+i)eikx + e−ikx(16)which oscillate throughout the wire.In Appx.A we find B = ((q2 +1)2pi)−1/2 when the wave function is Dirac normalized alongthe half line ∫ ∞0ψ†SkψSk′dx = δ(k − k′)and in Appx.B the bound and scattering states, Eqns.15 and 16, are used to show complete-ness of the solutions of the system∫ ∞−∞ψxψ†x′dk = Iδ(x− x′).As we wish to analyze many particle systems of electrons we employ the above singleparticle wave functions in second quantization formulation. The second quantization field16operator for a complex fermionic state, relativistic or non-relativistic, isΨ(x, t) =∑E>0ψE(x)e−iEtaE +∑E<0ψE(x)e−iEtb†−E(17)where aE is the annihilation operator for a particle with energy E and b†−E is the creationoperator for a hole with energy −E. The operators obey the anticommutator relationships,{aE, a†E′} = δEE′ , {bE, b†E′} = δEE′ ,and ψE(x) are the eigenstates of the Hermitian Hamiltonian, H0, for a single non-interactingparticleH0ψE(x) = EψE(x).The full wave functions satisfy the Schrodinger equationi∂∂tΨ(~x, t) = H0Ψ(~x, t)and for Ψ(~x, t) to be a second quantized field operator it must also obey the equal-timeanticommutation relation{Ψ(~x, t),Ψ(~y, t)} = δ(~x− ~y).Considering the Dirac sea interpretation the ground state of the system has all negativeenergy states filled and all positive energy states emptyaE|0 >= 0 = bE|0 > .The excited states of the system result from creating particles and holes in the ground state,giving them the forma†E1 ...a†Emb†E1 ...b†En|0 > .Together with the ground state these excited states form a basis for the Fock space of secondquantization theory.174.2. Charge Conjugation and the Majorana Fermion. Charge conjugation is an op-eration where all particles are replaced by their antiparticles, and visa versa,ψE(x)→ Cψ∗−E(x)where C is a t-independent matrix that must satisfy the relationships −CH∗ = HC andC∗C = CC∗ = I. This can be seen by considering the Schro¨dinger equationi∂∂tΨE(x, t) = HΨE(x, t) = EΨE(x, t).(18)Taking the complex conjugate of Eqn.18 and multiplying it from the left by matrix C givesi∂∂tCΨ∗E(x, t) = −CH∗Ψ∗E(x, t) = −ECΨ∗E(x, t).(19)Therefore CΨ∗E(x, t) satisfies the Dirac equation if ΨE(x, t) satisfies the Dirac equation andC has the property−CH∗ = HC.Assuming the Hamiltonian is t-independent then from Eqn.19 we haveHCψ∗E(x) = −ECψ∗E(x).Therefore we need ψE(x) → Cψ∗−E(x) for charge to change sign but energy to remain thesame: for the particles and antiparticles to swap. So the energy spectrum of the particlestates is the same as the antiparticle states, but charge is opposite. When this occursthe Hamiltonian is said to have a charge conjugation symmetry, or particle-antiparticlesymmetry. A particle is said to be self-conjugate if it satisfiesψE(x) = ±Cψ∗−E(x)(20)that is the particle is its own antiparticle. When considering the E = 0 case this conditionimplies that C must also have the propertyC∗C = CC∗ = I.Self-conjugate fermions are called Majorana fermions. Italian physicist Ettore Majoranafirst postulated the Majorana fermion in 1937, as a means to avoid the negative energy statesin the Dirac sea, by identifying the positive and negative energy states as manifestations ofthe same excited state [10]. He originally developed the Majorana theory in the contextof particle physics, however most modern day Majorana research takes place in condensed18matter physics where Majorana quasiparticle states are sought out. Unlike charged particlesand antiparticles which have complex quantum fields, neutral particles that are their ownantiparticles have real fields. The neutral spin-zero pion and the spin-one photon are exam-ples of bosonic neutral self-conjugate particles. The complex electron field that has a chargeconjugation symmetry can be split into two emergent Majorana fermions by treating theparticle and antiparticle with the same energy as a single excitation: the real and imaginarycomponents, Re(ψE(x)) and Im(ψE(x)) respectively, of the complex fermion:Re(ψE(x)) =12(ψE(x) + Cψ∗−E(x))andIm(ψE(x)) =12i(ψE(x)− Cψ∗−E(x)).Since these linear combinations are self-conjugate(ψE(x)± Cψ∗−E(x)) = ±C(ψ−E(x)± Cψ∗E(x))∗both solutions are considered to be emergent Majorana fermions. When a Hamiltonian hasa charge conjugation symmetry it is always possible to impose this mathematical constraintthat decomposes the electron into two Majorana fermions. However, this constraint is noteasily held in reality as these states are not eigenstates of the full Hamiltonian of quantumelectrodynamics. Once decomposing the wave function into its real and imaginary partsquantum fluctuations re-mix them very quickly: there is no accessible time frame in whichthey stay unmixed. This mixing is due to electromagnetic interactions of the system causingvery quick emission and absorption of low energy photons. Therefore, in order to maintaina Majorana mode this mixing must be damped. This is why superconductors are presentlyused to find Majorana fermions, as they stop photons, or at least low energy photons, frominteracting with the electron [11].An emergent Majorana state can, however, be sustainable if the system is self conjugate:ψE(x) = ±Cψ∗−E(x). The wave function of a complex fermion can always be written asψE(x) = Re(ψE(x)) + iIm(ψE(x)) = 1/2(ψE(x) +Cψ∗−E(x)) + 1/2(ψE(x)−Cψ∗−E(x)), there-fore if the state is self conjugate the wave function is either purely real, ψ(x) = Re(ψE(x)),or purely imaginary, ψ(x) = iIm(ψE(x)). So the degrees of freedom is reduced in half for19a Majorana fermion. Quantum computing has already been making use of single-particlestates which obey a Majorana condition [12]-[14].Since particles and antiparticles with the same energy are equivalent for a Majoranafermion, they are treated as a single excitation: the fermion does not have both particlesand antiparticles. Therefore the second quantization field operator isΨ(x, t) =∑E>0(ψE(x)e−iEtaE + Cψ∗E(x)eiEta†E)(21)where the creation and annihilation operators satisfy the algebra{aE, a†E′} = δEE′ .The ground state is annihilated by all annihilation operators,aE|0 >= 0 ∀aE,and the excited states come from the creation operators, a†E, acting on the ground state,a†E1a†E2...a†Ek |0 > .The field operator obeys the anticommutation relation{Ψ(x, t), Ψ†(y, t)} = δ(x− y)and is considered to be pseudo-real since it obeysΨ(x, t) = CΨ∗(x, t).If the Majorana fermion had a single zero mode then the second quantization field operator,Eqn.21, would have an additional zero mode termΨ(x, t) = ψ0(x)α +∑E>0(ψE(x)e−iEtaE + Cψ∗E(x)eiEta†E)where the zero mode operator α is real, α = α†, and satisfies the algebraα2 = 1/2, {α, aE} = 0 = {α, a†E}.For our Dirac Hamiltonian on the half line, given in Eqn.10, the charge conjugation matrixisC =0 ii 020as it satisfies the requirement −CH∗ = HC. Therefore, in our system, charge conjugationmaps the wave function asψ(x) =u(x)v(x) →iv∗(x)iu∗(x) .In Sec.4.1 we solved the system with the boundary condition u(0) = −iv(0). This boundarycondition allows the Dirac equation to remain Hermitian on the half line. Charge conju-gation, ψ(x) → Cψ∗(x), doesn’t change the relationship between u(x) and v(x), so theantiparticle also satisfies the boundary condition if it is satisfied by the particle. Looking atcharge conjugation of the bound state found in Sec.4.1ψ0 =√m−i1 e−mxwe see that it is an eigenstate of charge conjugationCψ∗0 =√m i−1 e−mx = −ψ0.Therefore, the zero mode is an emergent Majorana fermion! Next look at the scatteringstates found in Sec.4.1ψS = B1(q + i)q((q − i)eikx − (q + i)e−ikx)(q − i)eikx + (q + i)e−ikxwhere q = kE−m and B = ((q2 + 1)2pi)−1/2. They are not eigenstates of charge conjugationCψ∗S = Bi(q − i) (q + i)e−ikx + (q − i)eikxq((q + i)e−ikx − (q − i)eikx) 6= ±ψS.Since the only self-conjugate state is the zero mode the second quantization field operatorof our system is complex (not Majorana), with the formΨ(x, t) = ψ0(x)α +∑E>mψE(x)e−iEtaE +∑E<−mψE(x)e−iEtb†−Ewhere the zero mode operator obeys the algebra{α, α†} = 1(22)21and anticommutes with all other creation and annihilation operators of the system. Havinga zero mode with this algebra leads to a degeneracy of the fermion spectrum. The groundstate must still be annihilated by all of the annihilation operators, aE and bE ∀E > m, butnow the algebra of the zero mode, Eqn.22, must also be satisfied. This leads to a degeneracyof the spectrum since the minimal representation of the algebra is two dimensional: thereare two ground states, |+ > and |− >, which satisfy the propertiesaE|+ >= 0 = aE|− >, bE|+ >= 0 = bE|− >andα†|− >= |+ >, α†|+ >= 0,α|− >= 0, α|+ >= |− > .Therefore, the degenerate excited states of the system have the forma†E1 ...a†Emb†E1 ...b†En|− >anda†E1 ...a†Emb†E1 ...b†En|+ > .Due to the Dirac Hamiltonian satisfying charge conjugation the energy spectrum of theparticles and holes are the same, but with opposite charge. The particle and hole state ofthe zero mode differ by a charge of one, so for the two states to have opposite charge the holestate must have charge −1/2 and the particle state must have charge 1/2: fractional charge[15]-[17]. There has been experimental confirmation of this unexpected result of chargefractionalization by observing that it explains some of the unique features of polyacetylene.Charge conjugation demands that only particle/hole pairs with the same energy can be liftedinto each other, so the Dirac sea remains at the same energy but with opposite charge. Thismakes the zero mode of the spectrum of great interest to quantum computation since it istopologically protected: perturbing the system respects charge conjugation symmetry, so thezero mode state cannot be lifted because it has no pair to be lifted to.Since having an electron in an isolated zero mode state holds so much appeal the nextsubsection focuses on how the parameters of our quantum wire system on the half-line canbe modified so that an electron interacting with the system would preferably occupy theempty zero mode state rather than an empty scattering state.224.3. Zero Mode Coupling. For the electron source of our system we will use a scanningtunnelling microscope (STM). We are only interested in how strongly the zero mode, ψ0,is coupled to the STM wave function, ψT , verses the scattering states, ψk, so to model thecoupling we can just consider the overlap of the wave functions. For example, the couplingof the bound state will bec0 =∫ψ†Tψ0d3x · h.c.where h.c. is the Hermitian conjugate of the integral.As an approximation the tip can be modelled locally as an asymptotic spherical potentialwellV (r) = 0 if r < a,= V1 if r > aand solved for the bound, Et < V1, asmyptotic, r →∞, solution of the Schro¨dinger equationto getψT = A1k1|~r − ~r0|e−k1|~r−~r0|(23)where k1 =√2M(V1 − Et) is the wave number of the tip electron, M is the mass of theelectron, A is a normalization constant, and ~r0 is the center of the spherical well potential.Normalizing the wave function givesA =√k312pi.The tunnelling current, I, measured from an STM tip depends on 4 variables: x, y, z,and V . Where V is a bias voltage applied to the STM tip, or sample, causing the Fermilevel of the tip to be a higher energy than the Fermi level of the sample so that tunnellingoccurs from tip to sample, or visa versa. The most common use of STM measurements isto reveal the topography of a surface of a material by holding I and V constant, V being abias voltage to cause tunnelling from sample to STM tip, and measuring the z at each x, ylocation that is needed to keep I constant. In other words “z” is not exactly the “heightof the surface”, but rather the tip-sample separation needed for the tunnelling current toremain constant at a fixed bias voltage. Another common use for STM measurements is tounderstand the density of states of the surface of a material. In this case the I and z are23held fixed, and the bias V changes at each x, y location as needed so that the tunnellingcurrent from sample to tip remains constant.In our case the role we want the STM tip to play is simply an electron source at the edgeof the quantum wire. In Sec.4.1 we found that the bound state solution on the half lineis peaked at the end of the quantum wire, and the scattering states oscillate throughoutthe quantum wire. Since we wish to couple an electron on the STM tip to the bound statemode, we make the approximation that the STM tip can be centered on the quantum wireaxis and brought in proximity to the end of the quantum wire. Assuming the quantum wirelies along the z axis we take x = y = 0 and apply a bias voltage V0 to the tip, thereforeEt → Et + V0, then for a fixed I the system would measure some value z = z0. Sincewe are considering a theoretical toy model of the system, and not actually performing anyexperimental measurements, we make a best guess at what z0 would be knowing that z0typically ranges from 4-7 A˚.We assume the electron wave functions on the quantum wire are separable: ψ0(z)ψ⊥(ρ, φ)and ψS(z)ψ⊥(ρ, φ). When the electron is on the quantum wire it acts as a relativistic particlemoving through a vacuum, in the z direction, with an emergent mass, m, equal to the massgap of the quantum wire. However, when removed from the quantum wire the electron actsas a non-relativistic particle with its true electron mass M . The electrons are bound to thequantum wire in the perpendicular ρ, φ direction, so in analog to the bound STM wavefunction we estimate the perpendicular bound states by solving the Schro¨dinger equationwith a circular potential well. The Schro¨dinger equation with a circular potential isV (ρ) = 0 if ρ < a,= V2 if ρ > a.Since quantum wires are quasi 1-dimensional let a→ 0, so we are only interested in the wavefunction in the region ρ > a. We wish to find the bound state wave functions, because wewant the electrons to be bound to the wire, so we only consider the case Ew < V2 where Ewis the energy of an electron on the quantum wire.The Scho¨dinger equation we must solve is− 12M(1ρ∂∂ρ(ρ∂∂ρ) +1ρ2∂2∂φ2)ψ(ρ, φ) = (Ew − V2)ψ(ρ, φ)24where we use the true mass of the electron, M , because we are solving for the wave functionoutside of the quantum wire ρ > a. Using separation of variables ψ⊥(ρ, φ) = R(ρ)Φ(φ), gives− 12M(1ρR(ρ)∂∂ρ(ρ∂R(ρ)∂ρ) +1ρ21Φ(φ)∂2Φ(φ)∂φ2) = (Ew − V2).(24)Let −α2 be the separation constant1Φ(φ)∂2Φ(φ)∂2φ= −α2,which has the solutionΦ(φ) = Aeiαφ(25)where A is the normalization constant and α = ... − 1, 0, 1... so that the solution is single-valued under φ-rotations of 2pi. Substituting Eqn.25 into Eqn.24 gives− 12M(1ρR(ρ)∂∂ρ(ρ∂R(ρ)∂ρ) +1ρ2B−1e−iαφ(iα)2Beiαφ) = (Ew − V2)− 12M(ρ∂∂ρ(ρ∂R(ρ)∂ρ) +R(ρ)(iα)2) = ρ2R(ρ)(Ew − V2)ρ2∂2R(ρ)∂ρ2+ ρ∂R(ρ)∂ρ+ (k22ρ2 − α2)R(ρ) = 0, ρ > a where k22 = 2M(Ew − V2).(26)Since we are in a bound state, Ew < V2, k2 is imaginary. Eqn.26 is a Bessel equation,therefore the general solution isR(ρ) = BJα(k2ρ), ρ > a(27)where B is a normalization constant. Using Eqns.25 and 27ψ⊥(ρ, φ) = R(ρ)Φ(φ) = ABJα(k2ρ)eiαφ, ρ > a.Since we are considering a spherically symmetric potential we want the solution to haveno angular dependence, so we choose α = 0. For every fixed α the infinite sequence ofeigenfunctions Jα(k2ρ) are normalizable on the infinite interval as∫ ∞0Jα(k′2ρ)Jα(k2ρ)ρdρ =δ(k2 − k′2)k2,therefore our normalized solution isψ⊥(ρ, φ) = Jα(k2ρ), ρ > a(28)where k2 = ±√2M(Ew − V2).25In Sec.4.1 the scattering state (E > m) quantum wire wave functions on the half line werefound to beψS(z) = Bq( (q−i)(q+i)eikz − e−ikz)(q−i)(q+i)eikz + e−ikz(29)where q = kE−m and B = ((q2 + 1)2pi)−1/2. The overlap of the STM tip wave function andthe scattering state wave functions is therefore∫ψ†TψSd3x =∫ψ†T (~r)ψS(z)ψ⊥(ρ, φ)d3x.Since ψS(z) is only defined where the wire exists cylindrical coordinates are the natural choicefor the integration region. The spherical parameter r can be written in terms of cylindricalcoordinates asr =√ρ2 + z2.In order for an electron to tunnel from the STM tip to the quantum wire the two mustbe separated by a distance on the order of angstroms. We make the assumption that thisseparation is along the z axis, represented by z0. Let the spherical waves of the STM tip beshifted by a distance −z0 from the origin and integrate the coupling over the region whereψS(z) exists|~r − ~r0| =√ρ2 + (z + z0)2.Since all the negative E scattering states are assumed to be filled in the Dirac sea interpre-tation we only include the positive E states in our coupling integral. Also, in Appx.A we seethat the positive and negative momentum scattering states, ψS(z), are linearly dependentso we only consider k > 0 in our integral.Using Eqns.23, 28 and 29 the coupling can be written as∫ψ†TψSd3x = 2pi∫ ∞0∫ ∞0A1k1√ρ2 + (z + z0)2e−k1√ρ2+(z+z0)2B(Cq((q − i)(q + i)eikz − e−ikz) +D((q − i)(q + i)eikz + e−ikz))J0(k2ρ)ρdzdρwhere we have taken a linear combination of the top and bottom components of the spinorψS(z), C and D being arbitrary constants, since the components are linear combinations of26the even and odd atomic wave functions of polyacetylene at location z. Scaling the variables:ρ→ ρk2and z → zk1does not change the limits but dρ→ 1k2dρ and dz → 1k1dz∫ψ†TψSd3x = 2piAk1k22∫ ∞0∫ ∞01k1√( ρk2)2 + ( z+z0k1)2e−k1√(ρ/k2)2+((z+z0)/k1)2B(Cq((q − i)(q + i)eikz/k1 − e−ikz/k1) +D((q − i)(q + i)eikz/k1 + e−ikz/k1))J0(k2ρk2)ρdzdρ∫ψ†TψSd3x = 2piAk1k22∫ ∞0∫ ∞01√(k1ρk2)2 + (z + z0)2e−√(k1ρ/k2)2+(z+z0)2B(Cq((q − i)(q + i)eikz/k1 − e−ikz/k1) +D((q − i)(q + i)eikz/k1 + e−ikz/k1))J0(ρ)ρdzdρ.There should be a lot less tunnelling from the perpendicular components of the wire, there-fore from the STM tip we need k1 =√2M(V1 − Et) << |k2| =√2M(V2 − Ew), so k1|k2| << 1,and we can make the approximation:√(k1ρk2)2 + (z + z0)2 =√−( k1ρ|k2|)2 + (z + z0)2 ≈ (z+z0).Then,∫ψ†TψSd3x = 2piAk1k22∫ ∞0∫ ∞01z + z0e−(z+z0)B(Cq((q − i)(q + i)eikz/k1 − e−ikz/k1) +D((q − i)(q + i)eikz/k1 + e−ikz/k1))J0(ρ)ρdzdρ∫ψ†TψSd3x = 2piAk1k22∫ ∞0∫ ∞01z + z0e−(z+z0)B((q − i)(q + i)(Cq+D)eikz/k1+(−Cq+D)e−ikz/k1)J0(ρ)ρdzdρ∫ψ†TψSd3x = 2piAk1k22B((q − i)(q + i)(Cq +D)I1 + (−Cq +D)I2)whereI1 =∫ ∞0∫ ∞01z + z0e−(z+z0)eikz/k1J0(ρ)ρdzdρ,I2 =∫ ∞0∫ ∞01z + z0e−(z+z0)e−ikz/k1J0(ρ)ρdzdρ.In order for the above integrals to converge we must take the upper limit of ρ to befinite. Since there is minimal tunnelling of the electrons from the quantum wire to thesurrounding vacuum taking a finite upper limit should have minimal effect on the overallintegral. Typically z0 ranges from 4-7 A˚ for tunnelling from STM tip to sample, therefore27taking an upper limit of ρ = 10m should be more than sufficient. These integrals are solvedusing Mathematica:I1 =∫ 100e−ikz0/k1J0(ρ)(Γ(0, z0−ikz0/k1)−Log(1−ik/k1)+Log(1/z0)+Log(z0−ikz0/k1))ρdρwhich is a conditional expression of the z integral that is satisfied since Re[z0] > 0 andIm[k/k1] = 0 > −1,I1 = 10e−ikz0/k1J1(10)(Γ(0, z0 − ikz0/k1)− Log(1− ik/k1) + Log(1/z0) + Log(z0 − ikz0/k1)).Similarly,I2 =∫ 100eikz0/k1J0(ρ)(Γ(0, z0 + ikz0/k1)−Log[1 + ik/k1] + Log[1/z0] + Log[z0 + ikz0/k1])ρdρwhich is a conditional expression of the z integral that is satisfied since Im[k/k1] = 0 <1 and Re[z0] > 0,I2 = 10eikz0/k1J1(10)(Γ(0, z0 + ikz0/k1)− Log(1 + ik/k1) + Log(1/z0) + Log(z0 + ikz0/k1)).Depending on the state of the system an electron is either annihilated on the STM tip andcreated in a scattering state, or the reverse occurs. Therefore, the interaction Hamiltonianfor the scattering states isHint =∫(ckα†EtaE + c∗ka†EαEt)dk,where the coefficent can be approximated asck = c∗k =∫ψ†SψTd3x∫ψ†TψSd3x.To compare the size of this coefficient relative to the bound state coefficient we must integrateover all possible k states in which the Hamiltonian of the system can be described by theDirac equation. In Sec.3 we found this to be the domain of k values where sin(ka) ≈ kaand cos(ka) ≈ 1, with a being the lattice spacing of the chain. For simplicity take a to haveunit value, then we can approximate the Dirac equation to hold in the domain where k hasvalues less then 1/10th the sinusoidal period:c =∫ pi/50ckdk =∫ pi/50∫ψ†SψTd3x∫ψ†TψSd3xdk.(30)28As we see in Appx.B the positive and negative k states are linearly dependent on the half line,therefore we only integrate over the positive momentum states. This integral is too compli-cated to evaluate exactly, but since it is being integrated over a small domain Mathematicacan provide a numerical approximation of the integral.In Sec. 4.1 the bound state (E < m) wave function for Dirac fermions on the half line wasfound to beψB(z) =√m−i1 e−mz.Following the same steps as we did for the scattering state coupling, the bound state couplingis ∫ψ†TψBd3x = 2piAk1k22∫ 100∫ ∞01z + z0e−(z+z0)√m(−Ci+D)e−mz/k1J0(ρ)ρdzdρ∫ψ†TψBd3x = 2piAk1k22√m(−Ci+D)I3where,I3 =∫ 100∫ ∞01z + z0e−(z+z0)e−mz/k1J0(ρ)ρdzdρ.Solving the integral in Mathematica givesI3 =∫ 100emz0/k1J0(ρ)(Γ(0, (k1+m)z0/k1)−Log((k1+m)/k1)+Log(1/z0)+Log((k1+m)z0/k1))ρdρwhich is a conditional expression that is satisfied since Re[(k1+m)/k1] > 0 and 1+Re[m/k1] ≥0 and Re[z0] > 0 and z0 6= 0 are all satisfied,I3 = 10emz0/k1J1(10)(Γ(0, (k1+m)z0/k1)−Log((k1+m)/k1)+Log(1/z0)+Log((k1+m)z0/k1)).The interaction Hamiltonian for the bound state isHint = c0α†Eta0 + c∗0a†0αEt ,where the coefficent isc0 = c∗0 =∫ψ†BψTd3x∫ψ†TψBd3x.(31)As we wish to obtain an isolated zero mode we are interested in how the parameters ofthe system can be modified so that the bound state coupling, Eqn.31, dominates over thescattering state coupling, Eqn.30. A promising variable to consider is the mass gap, m, ofthe quantum wire, since the larger the mass gap the sharper the bound state is peaked at theboundary. When deriving a Dirac-like equation from the Hamiltonian of polyacetylene in29Sec.3 we saw that the greater the difference between alternating bond lengths, t2, the largerthe mass gap m. Research conducted by Grant and Batra [18] agree with this finding as theycalculated a mass gap of m = 0.8eV for polyacetylene with weak bond alternation, doublebonds of length 1.36A˚ and single bonds of length 1.43A˚, and a mass gap of m = 2.3eV forpolyacetylene with strong bond alternation, double bonds of length 1.34A˚ and single bondsof length 1.54A˚.Figure 3. Scattering state coupling, c, as a function of changing mass gap,m, using the fixed parameters z0 = 5× 10−10m, M = .5× 106eV, k1 =√2M ,k2 =√20M and C = D = 1/2Figure 4. Bound state coupling, c0, as a function of changing mass gap, m,using the fixed parameters z0 = 5 × 10−10m, M = .5 × 106eV, k1 =√2M ,k2 =√20M and C = D = 1/2In Fig.3 we see that the scattering state coupling, c, decreases sharply towards zero as mincreases. Conversely, in Fig.4 we see that the bound state coupling, c0, increases linearly as30m increases. This is a wonderful result as it tells us that there is indeed a way to modify thesystem for favourable coupling to the isolated zero mode. For example, polyacetylene witha mass gap m = 2eV has bound state coupling that is two orders of magnitude larger thanthe scattering state coupling.4.4. Electronic Entanglement. Consider the Hamiltonian of a segment of polyacetylenewith an STM tip at each end. We assume the length of the quantum wire is long enoughso that each end can be treated as a half line solution. We also assume the mass gap, m, islarge enough so that coupling from the STM tips to the scattering states on the quantumwire can be ignored, as found in Sec.4.3. The STM tips are treated as two-level systemsthat have energy 1, 2 when occupied and zero energy otherwise, and the probability oftunnelling from tip to bound state is given by amplitudes λ1, λ2. Therefore the HamiltonianisH = H1 +H2(32)whereH1 = 1α†α + iλ1(α + α†)(a0 + a†0),H2 = 2β†β + λ2(β + β†)(a0 − a†0),and α†, α, β†, β, and a†0, a0 are the creation and annihilation operators of the STM tip onsides 1 and 2, and the bound state, respectively. The only non-vanishing anticommutatorsare{α, α†} = 1, {β, β†} = 1, {a0, a†0} = 1,they act on three two-level systems with states defined byα†|0 >1= |1 >1, α†|1 >1= 0, α|0 >1= 0, α|1 >1= |0 >1,a†0|0 >0= |1 >0, a†0|1 >0= 0, a0|0 >0= 0, a0|1 >0= |0 >0,β†|0 >2= |1 >2, β†|1 >2= 0, β|0 >2= 0, β|1 >2= |0 >2,We can define Majorana zero mode operators,γ1 =1√2(a0 + a†0),γ2 =i√2(a0 − a†0),31where γ†1 = γ1 and γ†2 = γ2, and the only non-vanishing anticommutator is {γi, γ†j} = δij.These properties can be seen by considering the fact that the bound state is a two-levelsystem, therefore a20 and a†20 give zero when acting on the basis states. The bound statewave function created and annihilated by a†0, a0 is peaked at each end of the quantum wire,as we have combined two half line solutions. However, γ1 and γ2 create/annihilate particlesthat exist at only one end of the quantum wire, ends 1 and 2 respectively.Solving for the eigenvectors of H, Eqn.32, givesψσ,τ,η =(−iλ1η|0 >1 +E1,σ|1 >1)|η >0 (λ2η|0 >2 +E2,τ |1 >2)√E21,σ + λ21√E22,τ + λ22where,|η >0= 1√2(|0 >0 +η|1 >0), η = ±1and the eigenvalues are,E = E1,σ + E2,τ ,Ei,σ =i2+ σ√(i2)2+ λ2i , σ = ±1.Since Eiσ does not depend on η, the eigenstates ψσ,τ,1 and ψσ,τ,−1 have the same energy:H(ψσ,τ,1 + ψσ,τ,−1) = Hψσ,τ,1 +Hψσ,τ,−1 = Eσ,τ (ψσ,τ,1 + ψσ,τ,−1).Therefore instead of 8 eigenvalues we have 4, that is we have 4 two-fold degenerate levels.Consider the general linear combination of degenerate states:(Aψσ,τ,1 +Bψσ,τ,−1) = (−iλ1λ2(A+B)|0 >1 |0 >0 |0 >2 +E1σE2τ (A−B)|1 >1 |1 >0 |1 >2−iλ1E2τ (A−B)|0 >1 |0 >0 |1 >2 −iλ1λ2(A−B)|0 >1 |1 >0 |0 >2 −iλ1E2τ (A+B)|0 >1 |1 >0 |1 >2+E1σλ2(A−B)|1 >1 |0 >0 |0 >2 +E1σE2τ (A+B)|1 >1 |0 >0 |1 >2 +E1σλ2(A+B)|1 >1 |1 >0 |0 >2)2−1/2(E21,σ + λ21)−1/2(E22,τ + λ22)−1/2.At the quantum level fermion parity symmetry states that there is no physical process thatcan create or destroy an isolated fermion: the number of fermions in the system is conservedmod 2 [19]. Therefore, we can either set A = B for eigenstates with fermion number 0 mod322 or A = −B for eigenstates with fermion number 1 mod 2:(33)ψσ,τ |0mod2 = A(ψσ,τ,−1 + ψσ,τ,1) = A(−iλ1λ2|0 >1 |0 >0 |0 >2 −iλ1E2τ |0 >1 |1 >0 |1 >2+ E1σE2τ |1 >1 |0 >0 |1 >2 +E1σλ2|1 >1 |1 >0 |0 >2)(E21,σ + λ21)−1/2(E22,τ + λ22)−1/2,(34)ψσ,τ |1mod2 = A(ψσ,τ,−1 − ψσ,τ,1) = A(E1σE2τ |1 >1 |1 >0 |1 >2 −iλ1E2τ |0 >1 |0 >0 |1 >2− iλ1λ2|0 >1 |1 >0 |0 >2 +E1σλ2|1 >1 |0 >0 |0 >2)(E21,σ + λ21)−1/2(E22,τ + λ22)−1/2.For each linear combination, Eqns.33 and 34, we can measure the entanglement betweenthe bipartitions of the whole system: 1|02, 0|21, and 2|10. To measure the entanglement ofour system we consider the following. First of all, the definition of entanglement is:a bipartite system, A|B, is entangled if it cannot be written as a direct product of two eigen-vectors, one from each subsystem.For example, systems A and B are not entangled ifψ = ψA ⊗ ψB.Entropy of entanglement, E(ρ), can be used as a measure of entanglement. Entropy ofentanglement is defined using von Neumann entropy,S(ρ) = −Tr{ρln(ρ)} = −∑ipilnpiwhere pi are the eigenvalues of ρ and ρ is the density matrix of the system. If the system isin a single state, |ψ >, then the system is given the label pure state and ρ is defined asρ = |ψ >< ψ|.However, if the system has probability pj of being in state |ψj >, not necessarily orthogonal,then the system is given the label mixed state and ρ is defined asρ =N∑j=1pj|ψj >< ψj|33where N could be anything, it is not limited to the dimension of the Hilbert space, and theweights pj satisfy0 < pj ≤ 1;N∑j=1pj = 1.To check if the density matrix is pure or not one can test if ρ2 = ρ: pure state, or ρ2 6= ρ:mixed state.The von Neumann entropy vanishes for a pure state, S(ρ) = S(|ψ >< ψ|) = 0, and is amaximum for the completely mixed state, S(ρ) = S( 1NI) = ln(N), where N is the dimensionof the Hilbert space. The density matrix of a subsystem, called the reduced density matrix,can be found by taking a partial trace of the density matrix of the system:ρA = TrB(ρ) =k∑j=1< ψBj |ρ|ψBj >where |ψB1 >, ....|ψBk > are eigenvectors for the basis of system B. The entropy of entangle-ment is defined as the von Neumann entropy of one of the reduced density matrices,E(ρ) = S(ρA) = S(ρB).If the system can be written as a product state, |ψ >= |ψ >A |ψ >B then its density matrixcan be written as ρ = |ψ >A |ψ >B< ψ|A < ψ|B and therefore E(ρ) = S(ρA) = S(ρB) = 0.Considering this together with our earlier definition of entanglement, we can conclude:subsystems A and B of a bipartitioned system, A|B, are entangled if E(ρ) 6= 0.Taking the partial trace of the density matrix of our system with the bipartition 2|10 givesρ2 = Tr10ρ = Tr10|ψ >< ψ|ρ2 =< 0|0 < 0|1ψ >< ψ|0 >0 |0 >1 + < 0|0 < 1|1ψ >< ψ|0 >0 |1 >1+ < 1|0 < 0|1ψ >< ψ|1 >0 |0 >1 + < 1|0 < 1|1ψ >< ψ|1 >0 |1 >1 .34For the 0 mod 2 case the normalized density matrix isρ2|0mod2 = (λ1λ2)2 + (E1σλ2)2(λ1λ2)2 + (λ1E2τ )2 + (E1σE2τ )2 + (E1σλ2)2|0 >2< 0|2+(E1σE2τ )2 + (λ1E2τ )2(λ1λ2)2 + (λ1E2τ )2 + (E1σE2τ )2 + (E1σλ2)2|1 >2< 1|2.For the 1 mod 2 case the normalized density matrix isρ2|1mod2 = (E1σλ2)2 + (λ1λ2)2(E1σE2τ )2 + (λ1E2τ )2 + (λ1λ2)2 + (E1σλ2)2|0 >2< 0|2+(λ1E2τ )2 + (E1σE2τ )2(E1σE2τ )2 + (λ1E2τ )2 + (λ1λ2)2 + (E1σλ2)2|1 >2< 1|2.For both reduced density matrices the eigenvalues are 0 < pi < 1, thus there is entanglementbetween the bipartition 2|10. This tells us that bringing an STM tip electron in proximityto one side of the quantum wire will entangle it with an STM tip electron at the other!As a check we set λ1 = λ2 = 0, which givesρ2|0mod2 = 0|0 >2< 0|2 + |1 >2< 1|2andρ2|1mod2 = 0|0 >2< 0|2 + |1 >2< 1|2.Since both reduced density matrices are pure states they have zero entropy, therefore thereis no entanglement between the bipartition 2|10. This is the expected result since settingλ1 = λ2 = 0 means the electron never leaves either STM tip, so there cannot be entanglementof the tips via the quantum wire.355. The Dirac Equation on the Segment5.1. Bound and Scattering States. In this section we will solve the bound (|E| < m)and scattering (|E| > m) state solutions of the Dirac equation on a segment 0 ≤ x ≤ L. InSec.4.1 we found that in order for the Dirac HamiltonianH = −i ddxσx +mσz = m −i ddx−i ddx−mto remain Hermitian on the segment the Dirac current must satisfy0 = (u(x)∗v(x) + v(x)∗u(x))∣∣∣L0.(35)It turns out that not all boundary conditions that satisfy Eqn.35 will produce solutions ofthe Hamiltonian. For example, by applying the boundary condition of u(x) or v(x) to bezero at x = 0 and x = L on H it can be seen that the only solution for the bound states isk = 0: the E = |m| non-normalizable solution. Since in Sec.4.1 we found a zero mode for theboundary condition u(0) = −iv(0) a possibility we might choose is an additional boundarycondition of u(L) = iv(L) to satisfy Eqn.35. Below we see that this boundary condition doesindeed have solutions.In Sec.4.1 we found that to rotate H intoH ′ = −i ddxσx +mσy = 0 −i ddx − im−i ddx+ im 0 ,the form of the Dirac equation that was derived from the Hamiltonian of polyacetylene inSec.3, we must perform the rotationH ′ = ei(nˆ·~σ)φ/2He−i(nˆ·~σ)φ/2whereei(nˆ·~σ)φ/2 = I1√2+ iσx1√2.We must also perform this rotation on the boundary conditions in order to be solving thesame systemψ′(0) = ei(nˆ·~σ)φ/2ψ(0) =1√21 ii 1−i1 v(0)→ ψ′(0) = 1√202 v(0)36andψ′(L) = ei(nˆ·~σ)φ/2ψ(L) =1√21 ii 1i1 v(L)→ ψ′(L) = 1√22i0 v(L)which is the boundary condition u′(0) = 0 and v′(L) = 0. This is a realistic boundarycondition as u′(x) and v′(x) represent even and odd atomic sites, respectively, then it justrequires the chain to begin on an even site and end on an odd one.To find the bound states of the system consider eigenvalue equation 0 −im− i ddxim− i ddx0u(x)v(x) = εu(x)v(x)which is equivalent to the two equations−imv(x)− i ddxv(x) = εu(x)(36)andimu(x)− i ddxu(x) = εv(x).(37)We can solve Eqn.36 for u(x)u(x) =1ε(−imv(x)− i ddxv(x))(38)and substitute u(x) into Eqn.37 to get(p2 +d2dx2)v(x) = 0where p2 = E2 −m2. For the bound state solutions, |E| < m, we make the ansatzv(x) = Aeipx +Be−ipx = Ae−kx +Bekx,u(x) =iε(A(−m+ k)e−kx −B(m+ k)ekx),ε = ±ω(k), ω(k) =√−k2 +m2, p = ik.We want to impose the boundary conditionsu(0)v(0) =u(0)0 ,u(L)v(L) = 0v(L) .37Imposing the first boundary condition gives the relationship between the components,ψε = A iε((−m+ k)e−kx + (m+ k)ekx)(e−kx − ekx) .Imposing the second boundary condition gives the energy momentum relationship,(m− k)(m+ k)= e2kL.(39)Using ε2 = m2 − k2 we can write this as,εm+ k= ekL.(40)Using Eqns.39 and 40 to simplify the spinor and then normalizing it givesψε =1√λi sinh (k(L− x))sinh (kx)whereλ = L(−1 + e2kL4Lk+e−2kL−4Lk ).Notice that the top component of the spinor is peaked at x = 0 and the bottom componentis peaked at x = L.For the negative energy solutions find a matrix that anticommutes with the Hamiltonian:CH ′ = −H ′C. For H ′ this matrix isC =1 00 −1and therefore, taking ε > 0, the negative energy solution isψ−ε = Cψε =1√λi sinh (k(L− x))− sinh (kx) .Using Eqn.40, ε2 = m2 − k2, and hyperbolic trig identities, the energy-momentum rela-tionship for the bound states isε =mcosh (kL)(41)which can be written asm tanh (kL) = k.(42)38From Eqn.41 we clearly see that |ε| < m, since cosh (k) ≥ 1, as expected for the boundstate solutions. Eqn.42 will be satisfied at most twice, once at k = 0 and once more whenf(k) = k outgrows g(k) = m tanh kL, assuming that f ′(0) < g′(0). The zero vector resultsfrom k = 0, so this solution is ignored, and the second solution depends on m and L of thesystem. Since cosh (kL) increases exponentially with k, ε will be an infinitesimal value. AsL → ∞, cosh kL → ∞ and tanh kL → 1, therefore the solution of the system has k → mand ε→ 0: the half line solution.For the scattering states k = −iκ, using the relationships sinh (x) = −i sin (ix), cosh (x) =cos (ix) and tanh (x) = −i tan (ix) then,E = m2 + κ2,E =mcos (κL),m tan (κL) = κ,ψE =1√λ,sin (κ(L− x))−i sin (κx) ,ψ−E = CψE =1√λsin (κ(L− x))i sin (κx) .The scattering states oscillate throughout the length of the wire.The second quantization field operator for the complex fermionic state of the system isΨσ(x, t) = ψε(x)e−iεtaεσ + ψ−ε(x)eiεtbεσ +∑E>m[ψE(x)e−iEtaEσ + ψ−E(x)eiEtb†Eσ]where σ =↑, ↓ labels the electron spin state, the creation operator for a hole with energy εhas been unconventionally labelled as bεσ, and the anticommutators of the operators are{aεσ, a†ετ} = δστ , {bεσ, b†ετ} = δστ ,{aEσ, a†E′τ} = δστδEE′ , {bEσ, b†E′τ} = δστδEE′ .In Sec.4.3 we found that if polyacetylene has a sufficiently large mass gap, m, then the boundstate coupling dominates such that the scattering state coupling can be ignored. In this casethe field operator is approximated asΨσ(x, t) ≈ ψε(x)e−iεtaεσ + ψ−ε(x)eiεtbεσ39Ψσ(x, t) ≈ 1√λi sinh (k(L− x))sinh (kx) e−iεtaεσ + 1√λi sinh (k(L− x))− sinh (kx) eiεtbεσ.Since the bound state wave function is peaked at the ends of the quantum wire we can definenew operators that are localized at the ends of the wireΨσ(x, t) ≈√2λi sinh (k(L− x))βεσsinh (kx)αεσwhere αεσ(t)βεσ(t) = 1√2e−iεt −eiεte−iεt eiεtaεσbεσ(43)and α†εσ(t), αεσ(t) and β†εσ(t), βεσ(t) are operators that create/annihilate electrons on thequantum wire at x = L and x = 0, respectively. These operators have a t-dependencebecause the wave functions they create are generally not eigenstates of the Hamiltonian.The only nonvanishing anti-commutators are{αεσ(t), α†ετ (t)} = δστ , {βεσ(t), β†ετ (t)} = δστ .The α− system is a two qubit system with the four dimensional basis|0 >αε↑ |0 >αε↓,α†ε↑(t)|0 >αε↑ |0 >αε↓= |1 >αε↑ |0 >αε↓,α†ε↓(t)|0 >αε↑ |0 >αε↓= |0 >αε↑ |1 >αε↓,α†ε↑(t)α†ε↓(t)|0 >αε↑ |0 >αε↓= |1 >αε↑ |1 >αε↓,where the direct product symbol has been omitted: we are using notation such that |0 >αε↑⊗|0 >αε↓= |0 >αε↑ |0 >αε↓. The other systems are defined in a similar manner and we canwrite the vacuum state of the whole system as |0 >= |0 >αε↑ |0 >αε↓ |0 >βε↑ |0 >βε↓ |0 >S or|0 >= |0 >aε↑ |0 >aε↓ |0 >bε↑ |0 >bε↓ |0 >S where |0 >S represents the filled negative energyand empty positive energy scattering states. For convenience we have defined the vacuumstate, |0 >, to have both the positive and negative energy bound states unoccupied: the twohole states with spin ↑, ↓ and energy ε are excited. Therefore, this state is annihilated bythe following operatorsaεσ|0 >= 0 = bεσ|0 >, aEσ|0 >= 0 = bEσ|0 > where |E| > m40using Eqn.43 we see the state is also annihilated by the new operatorsαεσ(t)|0 >= 0 = βεσ(t)|0 > .The neutral ground state, |gs >, of the system has all the negative energy states filled andthe positive energy states empty, therefore it is related to the vacuum state, |0 >, by theannihilation of the two hole states with energy ε: |gs >= b†ε↑b†ε↓|0 >, or |0 >= bε↓bε↑|gs >.Inverting Eqn.43 aεσbεσ = 1√2 eiεt eiεt−e−iεt e−iεtαεσ(t)βεσ(t)(44)allows us to write the ground state in terms of the new operators|gs >= b†ε↑b†ε↓|0 >=e2iεt2(α†ε↑(t)− β†ε↑(t))(α†ε↓(t)− β†ε↓(t))|0 > .(45)5.2. Quantum Teleportation. Quantum information is transmitted by microparticles fromsender to receiver. All modern technologies process data by microparticles, photons andelectrons, so quantum information is already in effect in these devices. However, since thedevices are macroscopic in size quantum mechanical effects can be ignored and only classicalinformation theory used. If, however, these devices are shrunk down to the nanoscale, thenquantum mechanical effects can no longer be ignored. Polyacetylene is an excellent candidatefor the possible creation of quantum information communication devices.Many exciting new technologies are expected to emerge from quantum information, oneof particular intrigue is the quantum computer which promises an exponential speed upof computational power, enabling currently unsolvable problems to be solved. Quantumcryptography is another emerging quantum technology with great hype, making use of theproperty that quantum mechanical measurements always destroy the state of a system tosecure transmission of a cryptographic key.Since the state of a quantum system cannot be detected or duplicated then classicalteleportation of these systems cannot occur. State transfer between quantum systems occursinstead by quantum teleportation, which makes use of quantum entanglement. Unfortunatelyentanglement is a rather elusive phenomenon. One obstacle is the highly ordered state ofthe system required for its establishment. This is an unfavourable state for the systemto be in since the laws of thermodynamics always favour the system to be in the state of41highest disorder: entangled pairs are susceptible to decoherence by interactions with theenvironment and each other. Experimentalists usually tackle this thermodynamic barrierby bringing the system to ultra-low temperatures and applying very large magnetic fieldsor chemical reactions. Another major obstacle is the heavy reliance on optical componentsthat most experimental setups require. Photon detectors are extremely sensitive devices.Major scientific advancements are needed to improve their detection efficiency, spectral range,signal-to-noise ratio, and ability to resolve photon number [20]. Although there have beenhuge efforts to improve photon detectors worldwide, much work still remains to be done. Weaim to develop electron teleportation that could be performed at ambient temperatures andwithout the use of photons. To achieve this we look towards the wave function solutions wefound for a segment of polyacetylene.Consider the ground state of the system, given in Eqn.45, where we will drop the t-argument of the α(t), β(t) operators from here on out|gs >= b†ε↑b†ε↓|0 >|gs >= e2iεt2(α†ε↑ − β†ε↑)(α†ε↓ − β†ε↓)|0 >|gs >= e2iεt2(α†ε↑α†ε↓ − α†ε↑β†ε↓ − β†ε↑α†ε↓ + β†ε↑β†ε↓)|0 > .The ground state is a pure state, however taking a partial trace of the density matrix,ρ = |gs >< gs|, with respect to the α and β subsystems gives non-zero entanglement.Considering the eigenstates of the α-system, |0 >,α†ε↑|0 >,α†ε↓|0 >,α†ε↑α†ε↓|0 >, and the β-system, |0 >, β†ε↑|0 >, β†ε↓|0 >, β†ε↑β†ε↓|0 >, the partial traces of the system are ρα = Trβρ =14Iα and ρβ = Trαρ =14Iβ. Looking at the entanglement entropy of the partial tracesS(ρα) = S(ρβ) = −Tr{ρβln(ρβ)} = −∑i pβilnpβi = ln(4) for all basis states: the maximumpossible entropy of a system. Therefore the α and β systems are maximally entangled. Thisentanglement can be used for quantum teleportation.If another electron is brought up to one side of the quantum wire the entangled α − βpair can be used to teleport its spin from one side to the other. This additional electron onthe STM tip with energy E has creation/annihilation operators γ†Eσ, γEσ and is in the spinstate |γ >= (g1γ†E↑ + g2γ†E↓)|0 > where the state is normalized, g1g∗1 + g2g∗2 = 1. We choose42the γ-system to be on the α-system side of the quantum wire. The state of the system is|γ > |gs >=(g1γ†E↑ + g2γ†E↓) e2iεt2(α†ε↑α†ε↓ − α†ε↑β†ε↓ − β†ε↑α†ε↓ + β†ε↑β†ε↓)|0 >where the γ-system has the basis |0 >, γ†E↑|0 >, γ†E↓|0 >, γ†E↑γ†E↓|0 > and the only non-vanishing anticommutator is{γEσ, γ†Eτ} = δστ .To simplify the notation write |0 >= |00 >1 |00 >2 |00 >β |0 >S, where we have defined|00 >1= |0 >γE↑ |0 >αε↑, |00 >2= |0 >γE↓ |0 >αε↓, and |00 >β= |0 >βε↑ |0 >βε↓. We wish totransfer the α − β entanglement into γ − α entanglement, such that the β-system electronwill be left in the original spin state of the γ-system electron: thus teleporting the γ spinstate. We can write the state of the system as|γ > |gs >= e2iεt2g1γ†E↑(α†ε↑α†ε↓ − α†ε↑β†ε↓ + α†ε↓β†ε↑ + β†ε↑β†ε↓)|0 >+e2iεt2g2(−α†ε↑γ†E↓α†ε↓ + α†ε↑γ†E↓β†ε↓ + γ†E↓α†ε↓β†ε↑ + γ†E↓β†ε↑β†ε↓)|0 >(46)|γ > |gs >= e2iεt2(g1(|11 >1 |01 >2 |00 >β −|11 >1 |00 >2 |01 >β +|10 >1 |01 >2 |10 >β+|10 >1 |00 >2 |11 >β)+g2(−|01 >1 |11 >2 |00 >β +|01 >1 |10 >2 |01 >β +|00 >1 |11 >2 |10 >β+ |00 >1 |10 >2 |11 >β))|0 >S .We will use a classic “Alice/Bob” teleportation protocol in which Alice is on one sideof the quantum wire with access to the α, γ-systems, and Bob is on the other side withaccess to the β-system. Alice takes Bell measurements of the γE↑, αε↑-systems to convert theentanglement between the αε↑, βε↑ qubits into entanglement between the αε↑, γE↑ qubits,leaving the βε↑-system in the original spin state of the γE↑-system, and similarly for the spindown qubits. The Bell states of a two qubit system are|φ± >= 1√2(|00 > ±|11 >),|ψ± >= 1√2(|01 > ±|10 >).Therefore, there are 16 Bell projective measurements that can be taken of the γ − α systemof |gs >. Since Alice cannot act on Bob’s state, the β-system electron, the gate Iβ must be43acted on Bob’s state. In the Bell basis for systems 1 and 2, where |00 >1= |0 >γE↑ |0 >αε↑and |00 >2= |0 >γE↓ |0 >αε↓, four of these Bell measurements areM1 = |φ+ >1 |φ+ >2 ·h.c.⊗ Iβ ⊗ IS = [12(|00 > +|11 >)1(|00 > +|11 >)2] · h.c.⊗ Iβ ⊗ IS= [12(|00 >1 |00 >2 +|00 >1 |11 >2 +|11 >1 |00 >2 +|11 >1 |11 >2)] · h.c.⊗ Iβ ⊗ IS,M5 = |φ+ >1 |ψ+ >2 ·h.c.⊗ Iβ ⊗ IS = [12(|00 > +|11 >)1(|01 > +|10 >)2] · h.c.⊗ Iβ ⊗ IS= [12(|00 >1 |01 >2 +|00 >1 |10 >2 +|11 >1 |01 >2 +|11 >1 |10 >2)] · h.c.⊗ Iβ ⊗ IS,M9 = |ψ+ >1 |φ+ >2 ·h.c.⊗ Iβ ⊗ IS = [12(|01 > +|10 >)1(|00 > +|11 >)2] · h.c.⊗ Iβ ⊗ IS= [12(|01 >1 |00 >2 +|01 >1 |11 >2 +|10 >1 |00 >2 +|10 >1 |11 >2)] · h.c.⊗ Iβ ⊗ IS,M13 = |ψ+ >1 |ψ+ >2 ·h.c.⊗ Iβ ⊗ IS = [12(|01 > +|10 >)1(|01 > +|10 >)2] · h.c.⊗ Iβ ⊗ IS= [12(|01 >1 |01 >2 +|01 >1 |10 >2 +|10 >1 |01 >2 +|10 >1 |10 >2] · h.c.⊗ Iβ ⊗ IS.The remaining 12 measurements have the form of one of the above measurements, but withdifferent relative signs between terms. They areM2 = |φ− >1 |φ+ >2 ·h.c.⊗ Iβ ⊗ IS, M3 = |φ− >1 |φ− >2 ·h.c.⊗ Iβ ⊗ IS,M4 = |φ+ >1 |φ− >2 ·h.c.⊗ Iβ ⊗ IS, M6 = |φ− >1 |ψ+ >2 ·h.c.⊗ Iβ ⊗ IS,M7 = |φ− >1 |ψ− >2 ·h.c.⊗ Iβ ⊗ IS, M8 = |φ+ >1 |ψ− >2 ·h.c.⊗ Iβ ⊗ IS,M10 = |ψ− >1 |φ+ >2 ·h.c.⊗ Iβ ⊗ IS, M11 = |ψ− >1 |φ− >2 ·h.c.⊗ Iβ ⊗ IS,M12 = |ψ+ >1 |φ− >2 ·h.c.⊗ Iβ ⊗ IS, M14 = |ψ− >1 |ψ+ >2 ·h.c.⊗ Iβ ⊗ IS,M15 = |ψ− >1 |ψ− >2 ·h.c.⊗ Iβ ⊗ IS, M16 = |ψ+ >1 |ψ− >2 ·h.c.⊗ Iβ ⊗ IS.A measurement in quantum mechanics consists of a set of Hermitian operators, M †n = Mn,which satisfy ∑nM †nMn = I(47)44where I is the identity for the Hilbert space that the measurement is acting on [21]. Ourmeasurement operators satisfy M †nMn = M2n = Mn, where the first equality holds since ourmeasurements are Hermitian and the last equality holds since we have projective measure-ment operators. Then, since the Bell states are a basis for the two qubit γ − α space thatour measurement operators are acting on, Eqn.47 is satisfied.The measurement operators obey fermion parity symmetry which, at the quantum level,states that there is no physical process that can create or destroy an isolated fermion: thenumber of fermions in the system is conserved 0 mod 2. However, unfortunately three-fourths of these measurements violate charge superselection. Charge superselection statesthat no local action can create or destroy a charge, since destroying a charge would requiredestroying the electric field lines extending to infinity and no local procedure can achievethis [21]. Applying the measurement operators to our system in Eqn.46M1,2|γ > |gs >= |φ+,− >1 |φ+,+ >2 e2iεt4(∓g1|01 >β +g2|10 >β)|0 >S,M3,4|γ > |gs >= −|φ−,+ >1 |φ−,− >2 e2iεt4(∓g1|01 >β +g2|10 >β)|0 >S,M5,6|γ > |gs >= |φ+,− >1 |ψ+,+ >2 e2iεt4(±g1|00 >β +g2|11 >β)|0 >S,M7,8|γ > |gs >= −|φ−,+ >1 |ψ−,− >2 e2iεt4(±g1|00 >β +g2|11 >β)|0 >S,M9,10|γ > |gs >= |ψ+,− >1 |φ+,+ >2 e2iεt4(±g1|11 >β −g2|00 >β)|0 >S,M11,12|γ > |gs >= −|ψ−,+ >1 |φ−,− >2 e2iεt4(±g1|11 >β −g2|00 >β)|0 >S,M13,14|γ > |gs >= |φ+,− >1 |φ+,+ >2 e2iεt4(±g1|10 >β +g2|01 >β)|0 >S,M15,16|γ > |gs >= −|φ−,+ >1 |φ−,− >2 e2iεt4(±g1|10 >β +g2|01 >β)|0 >S .The normalized state of the system after these measurements is therefore 4M |γ > |gs >.Alice then classically communicates the outcome of these Bell projective measurementsto Bob, which lets him know which single qubit gates to apply to his β-system in order toconvert it to the original spin of the γ-system. Expressing the state of a single qubit invector forma|0 > +b|1 >→ab45then the quantum gates we are interested in areX =0 11 0 ,Z =1 00 −1 .For M1,3 Bob applies the X-gate to both of his qubits, and then the Z-gate to his first qubitXβ1Xβ24M1,3|γ > |gs >= ±|φ+,− >1 |φ+,− >2 (−g1|10 >β +g2|01 >β)|0 >SZβ1Xβ1Xβ24M1,3|γ > |gs >= ±|φ+,− >1 |φ+,− >2 (g1|10 >β +g2|01 >β)|0 >S .For M2,4 Bob needs to apply the X-gate to both of his qubits, for M5,7 the X-gate to hisfirst qubit, for M6,8 the X-gate then the Z-gate to his first qubit, for M9,11 the X-gate thenthe Z-gate to his second qubit, for M10,12 the X-gate to his second qubit, for M13,15 nothing,and finally for M14,16 the Z-gate to his first qubit.Since the bound state energy levels are separated by an infinitesimal energy gap thenunless we have a completely thermally isolated system we must consider the possibility ofthe bound state electrons in the lower energy state jumping to the higher energy state andvisa versa: decoherence of the state. That is, we must consider the other states that arespinless and chargless like the ground state but vary infinitesimally in energy. The basis forthis system is {b†ε↑b†ε↓|0 >, a†ε↑a†ε↓|0 >,1√2(a†ε↑b†ε↓ + b†ε↑a†ε↓)|0 >}.Notice that the states are all zero under the operation S+ = S1+ + S2+ telling us that theyare all spinless, and fermion number has not changed so the system remains chargeless.The state of a quantum system at thermal equilibrium is described by the thermal statedensity matrixρ =d∑i=1pi|i >< i| =∑di=1 e−βEi|i >< i|Zwhere pi = e−βEi/Z is the Boltzmann distribution at temperature T , Z =∑di=1 e−βEi isthe partition function, Ei is the energy of the quantum state |i >, β = 1/kBT , and kB isBoltzmann’s constant. We consider the basis |1 >= (g1γ†E↑ + g2γ†E↓)b†ε↑b†ε↓|0 > to be theground state so it has E1 = 0, and so |2 >= (g1γ†E↑ + g2γ†E↓)a†ε↑a†ε↓|0 > has E2 = 4ε, and46|3 >= (g1γ†E↑ + g2γ†E↓) 1√2(a†ε↑b†ε↓ + b†ε↑a†ε↓)|0 > has E3 = 2ε. Therefore, Z = 1 + e−4βε + e−2βεandρ =1Z[(g1γ†E↑ + g2γ†E↓)b†ε↑b†ε↓|0 > · h.c.]+e−4βεZ[(g1γ†E↑ + g2γ†E↓)a†ε↑a†ε↓|0 > · h.c.]+e−2βε2Z[(g1γ†E↑ + g2γ†E↓)(a†ε↑b†ε↓ − a†ε↓b†ε↑)|0 > · h.c.].We translate from the positive and negative energy bound state operators that existthroughout the wire, aεσ and bεσ, to the operators located at the edges, αεσ and βεσ, usingthe relationship given in Eqn.44:aεσbεσ = 1√2 eiεt eiεt−e−iεt e−iεtαεσβεσ .The vacuum state is defined as before: |0 >= |00 >1 |00 >2 |00 >β |0 >S, where |00 >1=|0 >γE↑ |0 >αε↑, |00 >2= |0 >γE↓ |0 >αε↓, and |00 >β= |0 >βε↑ |0 >βε↓. This givesρ = (14Z[(g1(|11 >1 |01 >2 |00 >β −|11 >1 |00 >2 |01 >β +|10 >1 |01 >2 |10 >β +|10 >1 |00 >2 |11 >β)+ g2(−|01 >1 |11 >2 |00 >β +|01 >1 |10 >2 |01 >β +|00 >1 |11 >2 |10 >β +|00 >1 |10 >2 |11 >β)) · h.c.]+e−2βε4Z[(g1(|11 >1 |01 >2 |00 >β +|11 >1 |00 >2 |01 >β −|10 >1 |01 >2 |10 >β +|10 >1 |00 >2 |11 >β)+ g2(−|01 >1 |11 >2 |00 >β −|01 >1 |10 >2 |01 >β −|00 >1 |11 >2 |10 >β +|00 >1 |10 >2 |11 >β)) · h.c.]+e−βε2Z[(g1(|11 >1 |01 >2 |00 >β −|10 >1 |00 >2 |11 >β)+ g2(−|01 >1 |11 >2 |00 >β −|00 >1 |10 >2 |11 >β)) · h.c.])|0 >S< 0|S.Applying the Bell state projective measurements to the density matrix givesM †1,2ρM1,2 =(14Z− e−2βε4Z)14[|φ+,− >1 |φ+,+ >2 (∓g1|01 >β +g2|10 >β) · h.c.]|0 >S< 0|S,M †3,4ρM3,4 =(14Z− e−2βε4Z)14[|φ−,+ >1 |φ−,− >2 (∓g1|01 >β +g2|10 >β) · h.c.]|0 >S< 0|S,M †5,6ρM5,6 =(14Z+e−2βε4Z)14[|φ+,− >1 |ψ+,+ >2 (±g1|00 >β +g2|11 >β) · h.c.]|0 >S< 0|S+e−βε8Z[|φ+,− >1 |ψ+,+ >2 (±g1|00 >β −g2|11 >β) · h.c.]|0 >S< 0|S,47M †7,8ρM7,8 =(14Z+e−2βε4Z)14[|φ−,+ >1 |ψ−,− >2 (±g1|00 >β +g2|11 >β) · h.c.]|0 >S< 0|S+e−βε8Z[|φ−,+ >1 |ψ−,− >2 (±g1|00 >β −g2|11 >β) · h.c.]|0 >S< 0|S,M †9,10ρM9,10 =(14Z+e−βε4Z)14[|ψ+,− >1 |φ+,+ >2 (±g1|11 >β −g2|00 >β) · h.c.]|0 >S< 0|S+e−βε8Z[|ψ+,− >1 |φ+,+ >2 (∓g1|11 >β −g2|00 >β) · h.c.]|0 >S< 0|S,M †11,12ρM11,12 =(14Z+e−βε4Z)14[|ψ−,+ >1 |φ−,− >2 (±g1|11 >β −g2|00 >β) · h.c.]|0 >S< 0|S+e−βε8Z[|ψ−,+ >1 |φ−,− >2 (∓g1|11 >β −g2|00 >β) · h.c.]|0 >S< 0|S,M †13,14ρM13,14 =(14Z− e−2βε4Z)14[|φ+,− >1 |φ+,+ >2 (±g1|10 >β +g2|01 >β) · h.c.]|0 >S< 0|S,M †15,16ρM15,16 =(14Z− e−2βε4Z)14[|φ−,+ >1 |φ−,− >2 (±g1|10 >β +g2|01 >β) · h.c.]|0 >S< 0|S.Notice that the basis states b†ε↑b†ε↓|0 > and a†ε↑a†ε↓|0 > give the same state when projectedon by the measurement operators, turning the mixed state thermal density matrix into apure state density matrix. However, the basis state 1√2(a†ε↑b†ε↓ + b†ε↑a†ε↓)|0 > gives a differentstate when acted on by the measurements. Therefore, projective measurements that projectonto b†ε↑b†ε↓|0 > and/or a†ε↑a†ε↓|0 >, as well as 1√2(a†ε↑b†ε↓ + b†ε↑a†ε↓)|0 > are not purified. Thistells us that for measurements M5−M12 the state cannot be teleported as the thermal statedoes not collapse into a pure state density matrix. However, measurements M1 −M4 andM13 −M16 collapse the thermal state into a pure state density matrix, and so teleportationcan occur by Bob applying the same single qubit gates to the β-system as he did for theground state case in thermal isolation.486. SummaryWe found that the hopping energy of the Su-Schrieffer-Heeger (SSH) Hamiltonian of per-fectly dimerized polyacetylene can be written as a Dirac-like equation at low energies. ThatisH = −∑nσ(t1 + t2(−1)n)(c†(n+1),σcn,σ + c†n,σc(n+1),σ)→ H =L2pi∫  0 k − imk + im 0 dkwhere the top component of a vector in this basis represents an electron on an even atomicsite and the bottom component represents an electron on an odd atomic site. This allowedus to give our solutions to the Dirac equation a physical interpretation.When analyzing the half-line solutions of the Dirac equation we discovered that the zeromode on the half-line is an eigenstate of charge conjugationCψ∗0 = −ψ0.Therefore, the zero mode has half the degrees of freedom of an electron: it is a Majoranaquasiparticle. This Majorana quasiparticle is topologically protected.We then found that if the mass gap of polyacetylene is sufficiently large an electron onan STM will dominatly couple to the zero mode over the scattering states. For example, ifpolyacetylene has a mass gap m = 2eV then zero mode coupling is two orders of magnitudelarger than scattering state coupling. This allowed us to approximate the second quantizationfield operator asΨ(x, t) = ψ0(x)α +∑E>mψE(x)e−iEtaE +∑E<−mψE(x)e−iEtb†−E ≈ ψ0(x)αand for the remainder of the thesis we were able to ignore scattering state excitations in ourcalculations.We also found that entanglement of two electrons could be achieved via the zero mode ona long segment of polyacetylene. We assumed the segment to be long enough so that eachend could be treated as a half-line solution, and found that the bipartition of the zero modeand an electron on one side, systems 0 and 1 respectively, was entangled with the electronon the other side, system 2. Specifically, the entropy of entanglement of this bipartition,conserving fermion parity symmetry, was measured to be non-zero: E(ρ) = S(ρ2|0mod2) 6= 0and E(ρ) = S(ρ2|1mod2) 6= 0.49When analyzing the bound state solutions of the Dirac equation on the segment we noticedthat the wave function was peaked at both ends of the quantum wireψε =1√λi sinh (k(L− x))sinh (kx) .We used this phenomenon to teleport the spin state of an electron interacting with the boundstate peak on one end to the peak on the other end. Bell projective measurements were usedfor this teleportation, providing a quantum measurement which conserved fermion paritysymmetry but violated charge superselection for three-fourths of the measurements. Spinwas successfully teleported for every Bell measurement on the ground state in the thermallyisolated case. However, since there are two bound states with infinitesimal energy levels,−ε and ε, possible hopping between the two levels needed to be considered. Therefore, Bellprojective measurements were performed on the thermal density matrix that included allspinless and chargeless basis states, varying infinitesimally in energy to the ground state.In this case spin was only successfully teleported half of the time, as only half of the Bellprojective measurements purified the mixed state thermal density matrix.50References[1] P.A.M. Dirac, Proc. Roy. Soc. A117, 610 (1928); P. A. M. Dirac, Proc. Roy. Soc. A118, 351 (1928).[2] W.P. Su, J.R. Schrieffer and A.J. Heeger, Phys. Rev. Lett. 42 1698 (1979).[3] M.J. Rice, Phys. Lett. A71, 152 (1979).[4] S.A. Brazovskii, JETP Lett. 28, 656 (1978).[5] B. Horovitz and J.A. Krumhansl, Solid State Commun. 26, 81 (1978).[6] H. Shirakawa, T. Ito and S. Ikeda, Die Macromol. Chem. 197, 1565 (1978).[7] H. Yakayama, Y. Lin-Liu and K. Maki, Phys. Rev. B21, 2388 (1980).[8] R. Jackiw and J.R. Schrieffer, Nucl. Phys. B 190, 253 (1981).[9] A. Heeger, Comments Solid State Phys. 10, 53 (1981).[10] E. Majorana, Nuovo Cimento 14, 171 (1937).[11] G. W. Semenoff and P. Sodano, Electron. J. Theor. Phys. 10, 157 (2006).[12] A. Kitaev, [arXiv:cond-mat/0010440].[13] L.S. Levitov, T.P. Orlando, J.B. Majer and J.E. Mooij, [arXiv:cond-mat/0108266].[14] U. Dorner, P. Fedichev, D. Jaksch, K. Lewenstein and P. Zoller, [arXiv:quant-ph/0212039].[15] R. Jackiw and C. Rebbi, Phys. Rev. D 13, 3398 (1976).[16] A.J. Niemi and G.W. Semenoff, Phys. Rept. 135, 99 (1986).[17] A.J. Niemi and G.W. Semenoff, Phys. Rev. D 30, 809 (1984).[18] P.M. Grant and I.P. Batra, Solid State Commun. 29, 225 (1979).[19] R.F. Streater and A.S. Wightman, PCT, Spin and Statistics, and All That (Advanced Book Classics),(Redwood City, USA, 1989).[20] R. H. Hadfield, Nat. Photonics 3, 696 (2009).[21] H.-K. Lo, S. Popescu, and T. P. Spiller, Introduction to Quantum Computation, (Singapore, 1998), pp.255-256.51Appendix A. Normalization: Half-Line Scattering StatesWe wish to Dirac normalize the scattering states:∫ ∞0ψ†SkψSk′dx = δ(k − k′).For |E| > m the wave functions for the 1-dimensional Dirac equation solved on the halfline were found in Sec.4.1 to beψSk(E) = Bkq( (q−i)(q+i)eikx − e−ikx)(q−i)(q+i)eikx + e−ikxwhere Bk is the normalization constant, q =kE−m , and sinceE = ±√k2 +m2the scattering state wave function is not defined for k = 0. Writing the scattering states asψSk = Bk(q + i)−1q((q − i)eikx − (q + i)e−ikx)(q − i)eikx + (q + i)e−ikxand taking k → −k, which gives q → −q, leads toψS−k = −B−k(−q + i)−1q((q − i)eikx − (q + i)e−ikx)(q − i)eikx + (q + i)e−ikxtherefore the negative and positive momentum scattering states of ψSk are linearly dependent.Normalizing the wave functions:∫ ∞0ψ†SkψSk′dx =∫ ∞0BkB∗k′[q( (q+i)(q−i)e−ikx − eikx) (q+i)(q−i)e−ikx + eikx]q′( (q′−i)(q′+i)eik′x − e−ik′x)(q′−i)(q′+i)eik′x + e−ik′x dx(48)∫ ∞0ψ†SkψSk′dx = BkB∗k′(qq′+1)(q + i)(q − i)(q′ − i)(q′ + i)∫ ∞0eix(k′−k)dx+BkB∗k′(−qq′+1)(q + i)(q − i)∫ ∞0e−ix(k+k′)dx+BkB∗k′(−qq′ + 1)(q′ − i)(q′ + i)∫ ∞0eix(k+k′)dx+BkB∗k′(qq′ + 1)∫ ∞0eix(k−k′)dx.To solve the integral below we will use the Cauchy principal value, which is a method usedto evaluate improper integrals. Consider,∫ ∞0eikxdx = lim→0∫ ∞0ei(k+i)xdx52then for some small  ∫ ∞0ei(k+i)xdx =1i(k + i)eikx−x∣∣∞0.At negative infinity the negative exponential will dominate, making this term zero, so weare left with ∫ ∞0ei(k+i)xdx =ikk2 + 2+k2 + 2∫ ∞0ei(k+i)xdx = iPV (1k) + piδ(k)and similarly, ∫ ∞0ei(−k+i)xdx =−ikk2 + 2+k2 + 2∫ ∞0ei(−k+i)xdx = −iPV (1k) + piδ(k)where PV represents the principle value of the function: a term that is not to be evaluatedat its singularity. Using the above result in Eqn.48 gives,∫ ∞0ψ†SkψSk′dx = BkB∗k′(qq′ + 1)(q + i)(q − i)(q′ − i)(q′ + i)(iPV (1k′ − k ) + piδ(k′ − k))+BkB∗k′(−qq′ + 1)(q + i)(q − i)(−iPV (1k + k′) + piδ(k + k′))+BkB∗k′(−qq′+1)(q′ − i)(q′ + i)(iPV (1k + k′)+piδ(k + k′))+BkB∗k′(qq′+1)(iPV (1k − k′ )+piδ(k − k′)).Since the wave equations of this system are linearly dependent for the positive and negativemomentum solutions only positive momentum is integrated over. Therefore, we can neverhave k = −k′, so δ(k + k′) will always be zero and these terms can be eliminated,∫ ∞0ψ†SkψSk′dx = BkB∗k′(qq′ + 1)iPV (1k′ − k )((q + i)(q − i)(q′ − i)(q′ + i)− 1)+BkB∗k′(qq′+1)pi((q + i)(q − i)(q′ − i)(q′ + i)+1)δ(k′ − k)+BkB∗k′(−qq′+1)iPV (1k′ + k)(−(q + i)(q − i)+(q′ − i)(q′ + i)).As only the terms where k = k′ will survive in front of δ(k′ − k) we can simplify the middleterm,∫ ∞0ψ†SkψSk′dx = BkB∗k′(qq′+1)iPV (1k′ − k )((q + i)(q − i)(q′ − i)(q′ + i)−1)+|Bk|2(q2 +1)2piδ(k′ − k)+BkB∗k′(−qq′ + 1)iPV (1k′ + k)(−(q + i)(q − i) +(q′ − i)(q′ + i)).53Since we wish to Dirac normalize the scattering state wave functions we must choose Bksuch that|Bk|2(q2 + 1)2pi = 1.For simplicity assume that Bk is a positive, real constant, thereforeBk = ((q2 + 1)2pi)−1/2.We hope that the remaining terms in the equation are zero:BkB∗k′(qq′+1)iPV (1k′ − k )((q + i)(q − i)(q′ − i)(q′ + i)−1)+BkB∗k′(−qq′+1)iPV (1k′ + k)(−(q + i)(q − i)+(q′ − i)(q′ + i)).We can drop the PV in the first term by assuming |k| 6= |k′|, as well as the PV in the secondterm since k > 0 and k′ > 0. Therefore,BkB∗k′if(k, k′) = BkB∗k′i((qq′+1)(1k′ − k )((q + i)(q − i)(q′ − i)(q′ + i)−1)+(−qq′+1)( 1k′ + k)(−(q + i)(q − i)+(q′ − i)(q′ + i))).Since Bk 6= 0 we must have f(k, k′) = 0 ∀k, k′. Writing f(k, k′) asf(k, k′) =1(q′ + i)(q − i)((qq′ + 1)(1k′ − k )((q + i)(q′ − i)− (q − i)(q′ + i))+ (−qq′ + 1)( 1k′ + k)(−(q + i)(q′ + i) + (q′ − i)(q − i)))f(k, k′) =1(q′ + i)(q − i)g(k, k′).Ignoring the overall factor we now want g(k, k′) = 0 ∀k, k′. Using q = kE−m and E2 = k2+m2,k can be found in terms of qk =2qm(q − 1)(q + 1) .Sub this into g(k, k′) and simplify to getg(k, k′) =(q′ − 1)(q′ + 1)(q − 1)(q + 1)2m((qq′ + 1)((q + i)(q′ − i)− (q − i)(q′ + i))(q′(q2 − 1)− q(q′2 − 1))+ (−qq′ + 1)(−(q + i)(q′ + i) + (q′ − i)(q − i))(q′(q2 − 1) + q(q′2 − 1)) )g(k, k′) =(q′ − 1)(q′ + 1)(q − 1)(q + 1)2mj(k, k′).54Ignoring the overall factor we now want j(k, k′) = 0, ∀k, k′,j(k, k′) = ((q′(q2 − 1) + q(q′2 − 1))(q′(q2 − 1)− q(q′2 − 1)))−1((qq′ + 1)((q + i)(q′ − i)− (q − i)(q′ + i))(q′(q2 − 1) + q(q′2 − 1))+ (−qq′ + 1)(−(q + i)(q′ + i) + (q′ − i)(q − i))(q′(q2 − 1)− q(q′2 − 1)))j(k, k′) = ((q′(q2 − 1) + q(q′2 − 1))(q′(q2 − 1)− q(q′2 − 1)))−1m(k, k′).Ignoring the overall factor we now want m(k, k′) = 0, ∀k, k′. Consider the factors((q + i)(q′ − i)− (q − i)(q′ + i)) = 2i(−q + q′),(49)(−(q + i)(q′ + i) + (q′ − i)(q − i)) = −2i(q′ + q),(50)(qq′ + 1)(−q + q′) = −q′(q2 − 1) + q(q′2 − 1),(51)(−qq′ + 1)(q + q′) = −q′(q2 − 1)− q(q′2 − 1).(52)Using Eqn.49 - 52 in m(k, k′) givesm(k, k′) = 2i((−q′(q2−1)+q(q′2−1))(q′(q2−1)+q(q′2−1))−(q′(q2−1)+q(q′2−1))(−q′(q2−1)+q(q′2−1)) = 0.Therefore, altogether we have ∫ ∞0ψ†SkψSk′dx = δ(k − k′)when Bk = ((q2 + 1)pi2)−1/2.Appendix B. Completeness: Half-Line StatesTo show completeness of the eigenstates of the 1-dimensional Dirac Hamiltonian solvedon the half line in Sec.4.1 we wish to prove∫ ∞−∞ψxψ†x′dk = Iδ(x− x′)where ψx represents the eigenstates of the system.For |E| > m the scattering state wave functions were found to beψSx(E) = Bk(q + i)−1q((q − i)eikx − (q + i)e−ikx)(q − i)eikx + (q + i)e−ikx(53)55where q = kE−m , Bk is found in Appx.A to be Bk = ((q2 + 1)2pi)−1/2 and sinceE = ±√k2 +m2the scattering state wave functions are not defined for k = 0. In Appx.A we saw thatthe positive and negative momentum states of ψSx are linearly dependent, therefore in thecompleteness integral we only consider the positive momentum scattering states. q = kE−m =√E2−m2E−m =√E+mE−m , therefore for the negative energy states E → −E, assuming E > 0, wesimply replace q →√E−mE+m= q−1ψSx(−E) = ((1 + q2)2pi)−1/2(i− q)−1 (i+ q)eikx − (i− q)e−ikxq((i+ q)eikx + (i− q)e−ikx) .(54)As with the positive energy states it can be seen that the negative energy states of positiveand negative momentum are linearly dependent.Using Eqns.53 and 54 completeness of the scattering states is therefore∫ ∞0ψSxψ†Sx′dk =12∫ ∞−∞B2kq( (q−i)(q+i)eikx − e−ikx)(q−i)(q+i)eikx + e−ikx[q( (q+i)(q−i)e−ikx′ − eikx′) (q+i)(q−i)e−ikx′ + eikx′]dk+12∫ ∞−∞B2k (i+q)(i−q)eikx − e−ikxq( (i+q)(i−q)eikx + e−ikx)[ (i−q)(i+q)e−ikx′ − eikx′ q( (i−q)(i+q)e−ikx′+ eikx′)]dkwhere we have extended the integral to the whole real line since ψSxψ†S′xis an even function.Taking the outer product,(55)∫ ∞0ψSxψ†Sx′dk =12∫ ∞−∞B2kq2( (q−i)(q+i)eikx − e−ikx)( (q+i)(q−i)e−ikx′ − eikx′) q( (q−i)(q+i)eikx − e−ikx)( (q+i)(q−i)e−ikx′ + eikx′)( (q−i)(q+i)eikx + e−ikx)q( (q+i)(q−i)e−ikx′ − eikx′) ( (q−i)(q+i)eikx + e−ikx)( (q+i)(q−i)e−ikx′ + eikx′) dk+12∫ ∞−∞B2k ( (i+q)(i−q)eikx − e−ikx)( (i−q)(i+q)e−ikx′ − eikx′) ( (i+q)(i−q)eikx − e−ikx)q( (i−q)(i+q)e−ikx′ + eikx′)q( (i+q)(i−q)eikx + e−ikx)( (i−q)(i+q)e−ikx′ − eikx′) q( (i+q)(i−q)eikx + e−ikx)q( (i−q)(i+q)e−ikx′+ eikx′) dk.For |E| ≤ m the only bound state wave function found was the zero modeψBx =√m−i1 e−mx.56Taking k → −k leads to an identical solution so the positive and negative momentum boundstates are also linearly dependent. Therefore,∫ ∞0ψBxψ†Bx′dk =12∫ ∞−∞ψBxψ†Bx′dk =12me−m(x+x′)−i1[i 1] ,where we have extended the integral to the whole real line since ψBxψ†Bx′is an even function,∫ ∞0ψBxψ†Bx′dk =12me−m(x+x′)1 −ii 1 .(56)Adding the scattering and bound state completeness integrals, Eqns.55 and 56, gives thefull completeness integral of the system with the following matrix components:[M ]1,1 =12∫ ∞−∞B2kq2(eik(x−x′) − (q − i)(q + i)eik(x+x′) − (q + i)(q − i)e−ik(x+x′) + e−ik(x−x′))dk+12∫ ∞−∞B2k(eik(x−x′) − (i+ q)(i− q)eik(x+x′) − (i− q)(i+ q)e−ik(x+x′) + e−ik(x−x′))dk +12me−m(x+x′),[M ]1,2 =12∫ ∞−∞B2kq(eik(x−x′) +(q − i)(q + i)eik(x+x′) − (q + i)(q − i)e−ik(x+x′) − e−ik(x−x′))dk+12∫ ∞−∞B2kq(eik(x−x′) +(i+ q)(i− q)eik(x+x′) − (i− q)(i+ q)e−ik(x+x′) − e−ik(x−x′))dk − 12ime−m(x+x′),[M ]2,1 =12∫ ∞−∞B2kq(eik(x−x′) − (q − i)(q + i)eik(x+x′) +(q + i)(q − i)e−ik(x+x′) − e−ik(x−x′))dk+12∫ ∞−∞B2kq(eik(x−x′) − (i+ q)(i− q)eik(x+x′) +(i− q)(i+ q)e−ik(x+x′) − e−ik(x−x′))dk + 12ime−m(x+x′),[M ]2,2 =12∫ ∞−∞B2k(eik(x−x′) +(q − i)(q + i)eik(x+x′) +(q + i)(q − i)e−ik(x+x′) + e−ik(x−x′))dk+12∫ ∞−∞B2kq2(eik(x−x′) +(i+ q)(i− q)eik(x+x′) +(i− q)(i+ q)e−ik(x+x′) + e−ik(x−x′))dk +12me−m(x+x′).Combining terms,[M ]1,1 =∫ ∞−∞B2kq2(eik(x−x′)−(q − i)(q + i)eik(x+x′))dk+∫ ∞−∞B2k(eik(x−x′)−(i+ q)(i− q)eik(x+x′))dk+12me−m(x+x′),[M ]1,2 =∫ ∞−∞B2kq(eik(x−x′)+(q − i)(q + i)eik(x+x′))dk+∫ ∞−∞B2kq(eik(x−x′)+(i+ q)(i− q)eik(x+x′))dk−12ime−m(x+x′),57[M ]2,1 =∫ ∞−∞B2kq(eik(x−x′)−(q − i)(q + i)eik(x+x′))dk+∫ ∞−∞B2kq(eik(x−x′)−(i+ q)(i− q)eik(x+x′))dk+12ime−m(x+x′),[M ]2,2 =∫ ∞−∞B2k(eik(x−x′)+(q − i)(q + i)eik(x+x′))+∫ ∞−∞B2kq2(eik(x−x′)+(i+ q)(i− q)eik(x+x′))dk+12me−m(x+x′).Defining the integrals,I1 =∫ ∞−∞B2kq2 (q − i)(q + i)eik(x+x′)dk,I2 =∫ ∞−∞B2kq(q − i)(q + i)eik(x+x′)dk,I3 =∫ ∞−∞B2k(q − i)(q + i)eik(x+x′)dk,I4 =∫ ∞−∞B2kq2 (i+ q)(i− q)eik(x+x′)dk,I5 =∫ ∞−∞B2kq(i+ q)(i− q)eik(x+x′)dk,I6 =∫ ∞−∞B2k(i+ q)(i− q)eik(x+x′)dk,I7 =∫ ∞−∞B2kq2eik(x−x′)dk,I8 =∫ ∞−∞B2kqeik(x−x′)dk,I9 =∫ ∞−∞B2keik(x−x′)dk.Then,[M ]1,1 = (I7 − I1 + I9 − I6) + 12me−m(x+x′),[M ]1,2 = (2I8 + I2 + I5)− 12ime−m(x+x′),[M ]2,1 = (2I8 − I2 − I5) + 12ime−m(x+x′),[M ]2,2 = (I9 + I3 + I7 + I4) +12me−m(x+x′).For the first 6 integrals, I1 − I6, since x and x′ are positive we can extend the contourintegral by adding an additional arc at infinity on the upper half of the complex k plane,because that section gives an exponentially damped contribution. Then we can use Cauchy’sintegral formula: if f(z) is analytic inside and on the simple closed contour Γ oriented inthe counterclockwise direction, and if z0 is any point inside Γ, then,f(z0) =12pii∫Γf(z)z − z0dz.58First look at I1, where Bk = ((q2 + 1)2pi)−1/2I1 =∫ ∞−∞B2kq2 (q − i)(q + i)eik(x+x′)dkI1 = (2pi)−1∫Γq2(q + i)2eik(x+x′)dkwhere q = kE−m and E =√k2 +m2, therefore,I1 = (2pi)−1∫Γk2(k + i(E −m))2 eik(x+x′)dk.The singularities of the integrand are k = 0 and k = im. Since the singularity k = 0 is alongthe boundary, and leads to a singular integral that converges when the integrand is evaluatedalong the real line, we can ignore it. So the only singularity of the integrand on the simplyconnected domain contained by Γ occurs at k = im. Therefore re-write the integrand asI1 = (2pi)−1∫Γk2(√(k − im) + i√(k + im))−2(k − im) eik(x+x′)dk.Notice that f(k) = k2(√(k − im)+ i√(k + im))−2 is analytic for every value in the domain.Therefore, we can use Cauchy’s integral formula to solve the above functionI1 = (2pi)−1∫Γk2(√(k − im) + i√(k + im))−2(k − im) eik(x+x′)dk = (2pi)−1(2pii)f(im)I1 =12me−m(x+x′).Evaluating I2 and I3 in a similar manner leads to,I2 =12ime−m(x+x′), I3 = −12me−m(x+x′).Next look at I4, where Bk = ((q2 + 1)2pi)−1/2,I4 =∫ ∞−∞B2kq2 (i+ q)(i− q)eik(x+x′)dkI4 = −(2pi)−1∫Γq2(q − i)2 eik(x+x′)dkwhere q = kE−m and E =√k2 +m2, therefore,I4 = −(2pi)−1∫Γk2(k + im)(√(k + im)− i√(k − im))2 eik(x+x′)dk.The singularities of the integrand are k = 0 and k = −im. Since the singularity k = 0 isalong the boundary, and leads to a singular integral that converges when the integrand is59evaluated along the real line, we can ignore it. So the only singularity of the integrand onthe simply connected domain is k = −im, which occurs outside of our simply connecteddomain, therefore by Cauchy’s integral theorem the integral is zeroI4 = 0.I5 and I6 can be solved in a similar manner to findI5 = I6 = 0.Now consider integrals I7, I8, and I9. For each integral there are three cases for x− x′:• Case 1: (x− x′) > 0: add an arc at infinity along the top half of the complex plane,where the integral is exponentially damped.• Case 2: (x − x′) < 0: add an arc at infinity along the bottom half of the complexplane, where the integral is exponentially damped.• Case 3: (x− x′) = 0First consider I7, where Bk = ((q2 + 1)2pi)−1/2,I7 =∫ ∞−∞B2kq2eik(x−x′)dkI7 = (2pi)−1∫Γk2k2 + (E −m)2 eik(x−x′)dkI7 = (4pi)−1∫Γk2((k2 +m2)−√k2 +m2m)eik(x−x′)dkI7 = (4pi)−1∫Γk2((k − im)(k + im)−√(k − im)(k + im)m)eik(x−x′)dk.The singularities of the above expression occur at k = 0, k = im, and k = −im. Since thesingularity k = 0 is along the boundary, and leads to a singular integral that converges whenthe integrand is evaluated along the real line, we can ignore it.For Case 1 the only singularity is k = im because we added an arc at infinity along thetop half of the complex plane. Therefore,I7 = (4pi)−1∫Γk2√(k − im)(k − im)((k + im)√(k − im)−√(k + im)m)eik(x−x′)dkI7 = (4pi)−1∫Γ1(k − im)f(k)dk60using Cauchy’s integral formulaI7 = (4pi)−1(2pii)f(im) = 0.For Case 2 the only singularity is k = −im because we added an arc at infinity along thebottom half of the complex plane. Therefore,I7 = (4pi)−1∫Γk2√(k + im)(k + im)((k − im)√(k + im)−√(k − im)m)eik(x−x′)dkusing Cauchy’s integral formulaI7 = (4pi)−1∫Γ1(k + im)f(k)dkI7 = (4pi)−1(2pii)f(−im) = 0.Now consider Case 3I7 = (4pi)−1∫ ∞−∞k2((k2 +m2)−√k2 +m2m)dkit can be seen on Mathematica that this integral diverges, so we assume the integral isinfinite. Therefore since I7 = 0 for x > x′ and x < x′ but is infinite for x = x′ we canwrite it as a delta function, and without loss of generality multiply the delta function by anarbitrary constantI7 =12δ(x− x′).By a similar argument it can be found that I8 = I9 = 0 for Cases 1 and 2. For Case 3I8 = (4pi)−1∫Γk(√k2 +m2 −m)((k2 +m2)−√k2 +m2m)dkwhich is an odd function integrated over an even interval so it is zero, therefore altogetherI8 = 0.Also for Case 3I9 = (4pi)−1∫Γ(√k2 +m2 −m)2((k2 +m2)−√k2 +m2m)dkit can be seen on Mathematica that this integral diverges, so we assume the integral is infiniteand thereforeI9 =12δ(x− x′).61Using the above results together with our earlier resultsI1 =12me−m(x+x′), I2 =12ime−m(x+x′), I3 = −12me−m(x+x′),I4 = I5 = I6 = 0,the components of the completeness integral are[M ]1,1 = (I7 − I1 + I9 − I6) + 12me−m(x+x′) = δ(x− x′),(57)[M ]1,2 = (2I8 + I2 + I5)− 12ime−m(x+x′) = 0,(58)[M ]2,1 = (2I8 − I2 − I5) + 12ime−m(x+x′) = 0,(59)[M ]2,2 = (I9 + I3 + I7 + I4) +12me−m(x+x′) = δ(x− x′).(60)Considering Eqns.57-60 altogether the completeness integral is proven:∫ ∞−∞ψxψ†x′dk = Iδ(x− x′).62

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