Klein tunneling at the surface of a topological insulator Can a surface electron be confined by applying a potential barrier? Teuntje Tijssen (5887623) Bachelorthesis Physics and Astronomy, 12 EC Carried out between 01-05-2011 and 25-08-2011 Supervisor: Prof.dr. C.J.M. Schoutens Institute for Theoretical Physics FNWI, University of Amsterdam Abstract Topological insulators are recently discovered materials with special physical features. One of the current hot topics in theoretical physics is describing the electronic properties on the two dimensional boundary of these materials. Topological insulators have the special property that their surface states allow only helical electrons. These electrons have due to the properties of the bulk material a linear dispersion relation. This makes topological insulators good conducting materials without possible backscattering of the electrons from impurities. The behaviour of an electron at the surface of a topological insulator is investigated by deriving its wavefunction. The tunneling behaviour of the electron is investigated, to find out if it shows Klein tunneling or can be confined by a potential barrier.
Transcript
Klein tunneling at the surface of atopological insulator
Can a surface electron be confined by applying a potential barrier?
Teuntje Tijssen (5887623)Bachelorthesis Physics and Astronomy, 12 ECCarried out between 01-05-2011 and 25-08-2011
Supervisor: Prof.dr. C.J.M. Schoutens
Institute for Theoretical PhysicsFNWI, University of Amsterdam
Abstract
Topological insulators are recently discovered materials with special physical features.One of the current hot topics in theoretical physics is describing the electronic propertieson the two dimensional boundary of these materials. Topological insulators have thespecial property that their surface states allow only helical electrons. These electronshave due to the properties of the bulk material a linear dispersion relation. This makestopological insulators good conducting materials without possible backscattering of theelectrons from impurities. The behaviour of an electron at the surface of a topologicalinsulator is investigated by deriving its wavefunction. The tunneling behaviour of theelectron is investigated, to find out if it shows Klein tunneling or can be confined by apotential barrier.
6 Discussion 296.1 Conditions at the origin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296.2 Combination of m and the width of the potential to capture the electron . . . 29
7 Conclusion 30
8 Acknowledgments 30
A Derivation of wavefunction 32
B Calculating the energy by the Dirac equation 33
C Calculation of tunneling coefficient 34
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1 Dutch summary
Stel je voor dat je door een muur kon lopen, alsof die er helemaalniet is..... De wereld zou er totaal anders uitzien. ”Dat kan niet!”zeg je nu waarschijnlijk, maar dit is niet helemaal waar. Een menskan het inderdaad niet, maar elektronen, de kleinste bouwstenenwaaruit we bestaan, kunnen dit wel! Dit komt doordat elektronenworden beschreven door de quantummechanica, waarin er geenzekerheden meer zijn, maar alleen kansen.
De kans dat een elektron door een barriere heen ”tunnelt” is meestal erg klein, maar wel groter dannul. In sommige gevallen, echter, is deze kans gelijk aan 1! Dan is het dus zeker dat het elektron aande andere kant van de barriere komt. Dit verschijnsel wordt Klein tunneling genoemd en is voorspelddoor Oskar Klein in 1929, maar pas sinds 2009 zijn er experimenten mogelijk die dit kunnen aantonen.Nu heb ik een artikel1 gelezen waarin voor elektronen wordt berekend of zij door een barriere kunnentunnelen. Deze kans kan varieren tussen 0 en 1, afhankelijk van de bewegingsrichting van de elektronenten opzichte van de barriere. Als ze er recht op af komen gaan ze er zonder problemen doorheen, maarals ze er bijna langs scheren wordt de kans dat ze er doorheen tunnelen gelijk aan nul. Dit verschijnselis dus niet hetzelfde als Klein tunneling, maar lijkt er wel heel erg op.
Deze elektronen hebben wel een speciale eigenschap, namelijk dat hun spin in dezelfde richtingwijst als waarin ze bewegen. Elektronen kunnen om hun as draaien, en de richting van deze as wordtde richting van hun spin genoemd. Verder bewegen de elektronen zich alleen aan het oppervlak vaneen materiaal, dus in slechts twee dimensies in plaats van drie.
Nu heb ik zelf voor mijn project berekend wat de kans is dat een elektron kan tunnelen aanhet oppervlak een topologische isolator. Topologische isolatoren zijn nog vrij onbekende materialenen hebben de nuttige eigenschap dat elektronen op hun oppervlak vrijwel zonder weerstand kunnenbewegen. Ook is de spin van deze elektronen altijd hetzelfde georienteerd ten opzichte van hun be-wegingsrichting. Omdat deze elektronen zich dus bijna hetzelfde gedragen als in het artikel dat ik hadgelezen, en er nog niet onderzocht was of deze elektronen ook Klein tunneling vertoonden, was dit eenerg interessant onderzoek om te gaan doen.
Dit onderzoek heb ik gedaan door eerst een beschrijving (een golffunctie) te geven van een elektronmet de genoemde eigenschappen, zowel buiten als binnen de barriere. Vervolgens moesten deze tweebeschrijvingen mooi in elkaar overlopen op de randen van de barriere. Uit deze eis heb ik kunnenafleiden wat de kans was het elektron aan de andere kant van de barriere te vinden. Er bleek dat alshet elektron weinig draai-impuls heeft, dus vrij recht op de barriere afgaat, de kans erg groot is dathet elektron door de barriere heen tunnelt. Ook moet de barriere niet te breed zijn. Dit komt dusovereen met het artikel dat ik had gelezen.
1M. I. Katsnelson, K.S. Novoselov, A.K. Geim, Chiral tunneling and the Klein paradox in graphene, NaturePhysics, 2, 620-625 (2006) ArXiv:cond-mat/0604323v2
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2 Summary
Topological insulators are recently discovered materials with special physical features. One ofthe current hot topics in theoretical physics is describing the electronic properties on the twodimensional boundary of these materials. Topological insulators have the special property thattheir surface states allow only helical electrons. These electrons have due to the propertiesof the bulk material a linear dispersion relation. This makes topological insulators goodconducting materials without possible backscattering of the electrons from impurities. Firstsome background information will be given, followed by the calculation of the wavefunction foran electron at the surface of a topological insulator. It is checked whether this wavefunctioncorrectly describes a spin-momentum coupled electron with a linear dispersion relation.The main question of this thesis is whether an electron at the surface of a topological insulatorwill be able to tunnel through a potential with a transmission equal to unity, as predicted byKlein in 1929. This is a very special phenomenon, as ordinary quantum mechanics predictsthe possibility of transmission to be only smaller than one. To answer this question a circularpotential is applied and by demanding continuity of the wavefunction on the boundaries thetransmission is calculated numerically. For this electron the relationship between the angle ofincidence and the tunneling probability. The results show at perpendicular incidence perfecttransmission (Klein tunneling), that decreases with the angle the electron makes with thepotential boundary.
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3 Introduction
This bachelor thesis is about topological insulators and will go into the theoretical aspectsof these materials. It is a quantum mechanical approach to understand them. The physicsinvolved is new, challenging and important. I will give a short explanation why these topo-logical insulators are a hot topic in current physics. Further explanation will follow later.
Atoms, discovered centuries ago, can give rise to different states of matter, for examplecrystalline solids, magnets or superconductors. Condensed matter physics is interested inthese states and tries to classify them. This can be done by spontaneous symmetry break-ing. This classification method is based on the fact that a material is symmetric in a certainproperty. This means that different outcomes of a measurement will have the same proba-bility. But when this property is measured only one outcome will be possible, and hence thesymmetry is broken.For example, ferromagnetic materials are invariant under spatial rotations. The parameterhere is the magnetization, measuring the magnetic dipole density. Above the Curie tem-perature the magnetization is zero, and there is no symmetry breaking. Below the Curietemperature, however, the magnetization acquires a constant non vanishing value, pointingin a certain direction. There are now rotations possible that do not leave the orientation ofthe magnetization invariant, and thus these rotations are spontaneously broken.
This was a straightforward way of classification, until the discovery of the Quantum Halleffect. This effect causes the electrical conductance of a material under special conditions tobe quantized. It can only take integer values of e2/ ~, and this value is independent of thematerial details. Smooth changes in parameters do not have any effect on this quantization.Another result of the Quantum Hall effect is that the current is only carried along the edgeof the material sample, not through the bulk. This means that in the bulk of the material,since it is insulating, there is a gap between the conduction and valence band. But at thesurface there is no gap; it is conducting and there are states that cross the gap.This discovery was the start of a topological classification of phases of matter.
Figure 1: All objects contain just one hole and are hence not topologically different
This new classification method can be understood by considering geometrical objects, asthese can also be classified topologically. For example, two dimensional surfaces are classifiedby the number of holes in them. Two objects belong to the same topological class if theycan be deformed into each other without creating or destroying any holes. So, the surface ofa perfect sphere is topologically equivalent to the surface of an ellipsoid, and a coffee cup istopologically equivalent to a donut, since they both contain one hole. Here small details canbe forgotten, only fundamental differences between the shapes are important. Translatingthis to the Hall conductance, small changes in parameters change the Hamiltonian. This willresult in a change in the band structure. But the states that cross the gap cannot be removed.You can think of these states as holes in an object.
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Now this thesis is about topological insulators. These materials show behaviour justlike the Quantum Hall effect; the boundary is conducting while the inside of the materialis insulating. Further, the conducting surface allows no backscattering of the electrons, andthey behave as if they are massless. These properties are protected by symmetries and canonly change if these symmetries are broken. Topological insulators are a very recent subjectof investigation. The first theoretical predictions of their existence were in 2006 and 2007,2
quickly followed by the first observations in 2007 and 2008.3 As they are so recently discovered,little is known about them, and we would like to understand more about these materials.
The subject ”Topological insulators” had been proposed to me for a short presentation.While I was looking for information about these strange types of matter I quickly realized thatthis was a very difficult subject with many concepts involved. I have always liked quantummechanics and wanted to know more about the physics involved. But a project of just gath-ering information and translating this to a level understandable for undergraduate studentswas not very attractive. To get more feeling for what is going on calculating something myselfwould help me more.
The goal of my project was to understand, calculate and describe a part of the physicsof a topological insulator. I did this by looking at an electron at its surface. This electronwill have its spin and momentum coupled. It is expected from quantum mechanics that thedispersion relation is linear. Now what will happen if there is some defect at the surface, ahole maybe? At this point the wavefunction must go to zero. But what consequences doesthis have for the behaviour of the particle? Will it be possible to catch it, to capture it in apotential?
This thesis will begin with background information and theory for the general idea of thephysical aspects of these systems. Then the calculations will be presented. As an introductiona free particle will be described in cylindrical coordinates in the plane. This derivation showsthe appearance of Bessel functions, which will come back in all other calculations. It is followedby the derivation of the wavefunction for the electron with its spin coupled to its momentum.As a check the energy will be calculated by the Dirac Hamiltonian. This paper will end withcalculations concerning the possibility of tunneling through a potential. These details are ofphysical interest for anyone who wants to understand two dimensional topological non-trivialphysics.
2Bernevig, Hughes and Zhang, Science 314, 1757 (2006) & Fu and Kane, Phys. Rev. B 76, 045302 (2007)3Konig et al., Science 318, 766 (2007) & Hsieh et al., Nature 452, 970 (2008)
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4 Theory
In this section a brief and physical introduction to quantum physics and topology is presented.This will give the reader a good basis of knowledge about the topics of two dimensionaltopological insulators.
4.1 Electrons in materials
This thesis will be about electrons in two dimensional materials. Now these systems arevery difficult, if not impossible, to realize in nature. The only real two dimensional materialknown is graphene. (See section 4.1.1) All other two dimensional systems are actually theboundary of a three dimensional system. Here only spatial dimensions are considered, thetime evolution is not taken into account.In materials there are available states for electrons, with corresponding energies. The stateswith the lowest energies will be filled with electrons first. There can be pointed out a ”bound-ary” energy that is the energy of the highest occupied level. Below this energy all states arefilled and above this energy all states are empty. This energy (at T = 0 K) is called theFermi energy εF . When one particle is added to a system in its ground state the energy ofthe system will increase, and this increase will be equal to εF . Now properties of materialsare largely determined by the states that have energies around the Fermi energy. Electronsin these states have the most possibilities of changing states, hereby changing the electronconfiguration. It also plays a role in the difference between conducting and insulating systems.This is explained in the next section.(4.2)
4.1.1 Graphene
Graphene is a recently discovered material (by A.K. Geim and K.S. Novoselov, rewarded forthis discovery in 2011 by the Nobel Prize). It is a single layer of carbon-atoms linked uptogether, creating a perfect two dimensional system. Hence it is the thinnest layer possiblein nature. It shows many resemblances with the physics of topological insulators. It is a verygood conductor, due to a linear dispersion at the point where the conduction- and valenceband touch, showing Dirac points. (Explained in section 4.5.) It also shows the QuantumHall Effect, which is explained in section 4.3.
4.2 Band structure
To describe the electronic structure of insulating states the band theory of solids is needed.In this theory a solid is considered to be a lattice of atoms, separated by the lattice constanta. This gives a periodic potential V (x) = V (x + a) in which the electrons move. Now Bloch’stheorem says that in a periodic lattice the electron wavefunctions are plane wave functionsmultiplied with a periodic function:
Ψn(x) = eik·xunk(x) (1)
with unk(x) = unk(x + a). The electrons are called ”nearly free” as the potential is a smallperturbation on the free electron potential in which the wave functions would be plane waves.The periodicity of the lattice changes the dispersion relation (the relation between the energyand momentum) of a free particle, which is parabolic, to a broken dispersion relation, as can
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be seen in Figure 2. Instead of the unbroken dashed line, energy bands and energy gaps areformed in the energy spectrum of the electrons. The gaps divide the range in momentum ofthe electron in Brillouin zones. As a result, the energy becomes a function of the (crystal)momentum. Crystal momentum is the momentum of the electron with respect to the latticeand the momentum of the lattice combined. Every solution (1) can be associated with a wavevector k.
Figure 2: The Brillouin zones in the nearly free electron model [10]
Now what makes the difference between an insulator and a conductor? In a conductorthe highest occupied band (the valence band) is not completely filled. There are empty statesdirectly besides occupied states, making it possible to shift the whole electron distributionin k-space and creating an electron-carrying state. But in an insulator the valence band isfilled and the conduction band is empty, and to change states an electron needs to cross thelarge energy gap in between. There are different kinds of insulating states. For example asemiconductor that has a small energy gap. If the electrons have enough energy to crossthe gap (i.e. if the temperature is high enough) an electron carrying state is possible. Butinsulating states like an insulator and semiconductor are topologically equivalent; there is anenergy gap and although the system can be changed smoothly it will never close the gap.Hence it can be said that these systems are topologically equivalent to the vacuum, which isan insulator after all.But this is not the case for all electronic configurations with gaps, for example the QuantumHall state.
4.3 Quantum Hall Effect
If a sufficiently large magnetic field (B) is applied perpendicular to a two dimensional layerof metal, the electrons will exhibit a cyclotron motion in the layer due to the Lorentz force.The temperature needs to be low enough to prevent thermal excitations from occurring. Theelectrons have quantized energies En = ~ωc(n + 1/2) in which ωc = eB/me, the cyclotronfrequency. Every value of n corresponds to an orbital called a Landau level. These Landau
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Figure 3: The cyclotron motion of the electrons at the surface (a) results in a net currentalong the edge (b)
levels can be seen as a band structure. When a magnetic field is applied N Landau levelswill be filled while the others will stay empty. If this N in an integer this will result in anenergy gap of ~ωc. In general, N will not be an integer, and hence there will be no energygap. The number of states for every Landau level per unit area of the material depends onthe magnitude of the magnetic field and is given by:
Nl =eB
h
If the magnetic field strength is increased, the number of states per Landau level grows. Asa result the number of lowest states needed to accommodate all electrons decreases and theLandau levels move with respect to the Fermi energy. Hence the energy gap between thestates can be closed, which is not possible in the case of a normal insulator. This is the mainreason the Quantum Hall state is not topologically equivalent to the vacuum. The gap doesnot really matter any more, as the magnetic field strength changes the electrons are ”pushed”to other states on the other side of the gap. When the Fermi energy falls in between twoLandau levels, the electric conductivity will be zero as this will resemble a insulating state.Only at the moment the Fermi energy is in the Landau level the resistance will be unequalto zero and finite. Now the classical Hall resistance is given by ρ = B
Me , with M the totalnumber of electrons. When there are N Landau levels below the Fermi energy, all completelyfilled with Nl electrons, the Hall resistance will be given by:
ρ =B
e
1NNl
=B
e
h
NeB=
h
Ne2
And the Hall conductivity:
σ =Ne2
h
The conductivity is quantized, and this quantization is very robust. Under these conditionsthe electrons, move along the edge of the material with the counter flows spatially separated
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(See Figure 4). As a result the top of the material only contains half the degrees of freedomcompared to a normal surface. Electrons on the top surface cannot turn around, as thecounter flow is on the bottom surface. Hence the conductivity cannot change continuouslyand is dependent on the number of Landau levels below the Fermi energy.
Figure 4: The Quantum Hall Effect [1]
This behaviour does not depend on the geometry of the material. It can be cut in halfwithout any effect. But if it is bent into a closed loop, or if a hole is made in it, (in other wordsthe material is changed topologically) the behaviour will be affected. To enable the QuantumHall Effect a magnetic field is applied, breaking time (T ) symmetry. This can be understoodby the following reasoning: assume the magnetic field bends the current of electrons to theright. When time is reversed, the electrons move backwards and their trajectory is bent tothe left. But this requires a magnetic field pointing in the opposite direction. Hence if amagnetic field is applied there can be made a distinction between the two cases.
4.4 Quantum Spin Hall Effect
Now, as electrons carry spin, instead of two channels actually four can be distinguished. Anelectron moving forward could have its spin up or down, and this is also the case when it ismoving backwards. These four channels can be split into two pairs, based on the phenomenoncalled spin-orbit coupling (SOC). When electrons are moving their angular momentum (L)and spin magnetic moment (S) can couple to the total magnetic moment (J).4 If L and Spoint in the same direction J will be of a larger magnitude than when they point in oppositedirections.5 With this angular moment J a quantum numbers j is associated, and a differentJ will result in another quantum number j. The channels will pair up with the same j. Onepair will consist of forward moving electrons with spin up and backward moving electronswith spin down and the other pair with the other two combinations. As a result, there willbe a net spin transport. These electrons are called helical, as the spin-momentum orientationdoes not change.
4J is the vector sum of L and S.5These states are degenerate, and this degeneracy can be lifted by applying a magnetic field. Then the
state in which L and S are aligned parallel will have a lower energy than the state in which they are alignedantiparallel.
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Figure 5: The Quantum Spin Hall Effect [1]
In contrary to the Quantum Hall effect (QH) the Quantum Spin Hall effect (QSH) doesnot require a magnetic field to be applied, leaving T -symmetry unbroken, but can be realizedby SOC. In the QSH-case there will be a perfect transmission through the material andabsolutely no reflection on impurities. This is because of two processes.First if there is scattering from an impurity the reflected beams will interfere destructively witheach other. An electron can turn clockwise or anti-clockwise around an impurity, resulting ina turn of the spin by +π or −π respectively. Hence the two possibilities will differ 2π. Nowthe wavefunction of a spin-1
2 particle picks up a -1 if the spin rotates over 2π, so the twodifferent paths will cancel each other and there will be no reflection.
Figure 6: The spin can rotate in two different ways [1]
This can be compared with anti-reflective coating on sunglasses. The coating has a thick-ness of exactly 1
4 -th wavelength. As a result the reflected waves from the coating and the glasswill have a path difference of 1
2 -th wavelength and cancel out each other. But, in contrary tothe electron case, this phenomenon requires many stable variables, such as the thickness ofthe coating, and is very sensitive to the optical wavelength. In the case of the electrons thesituation is not changed easily. The only way to do this is to add a magnetic impurity. Thisbreaks the T -symmetry, and as a result there is no destructive interference. It could be saidthat the robustness of the QSH-state is protected by T -symmetry.
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Secondly an electron in a channel moving forwards cannot be scattered to the other channelit is paired up with without reversing its spin. It could move backwards in the channel on theother side of the material, but it cannot get there. This is the case if there is just one pairof channels on each side of the material. But if there are two pairs on each side an electroncan be scattered to another channel moving backwards without reversing its spin. This is notjust the case for two pairs, but for every even number. As a consequence, an odd number ofchannels moving forward and an odd number of channels moving backwards is needed for theQSH-state to occur.A topological insulator is actually a QSH-insulator. This is relevant, as the electrons at thesurface behave exactly as described here.This counterpropagation of spins can be realized by SOC and the effect will be larger forheavy elements, as the SOC in these elements is stronger than in lighter elements. The mech-anism behind this is band inversion. Here the ordering of the conduction band (p-orbitals)and valence band (s-orbitals) is inverted. Mercury, for example, would be a good element.It is predicted in 20066 that nanoscopic mercury-telluride layers (quantum wells) sandwichedbetween other materials would show this behaviour if the thickness exceeded the critical thick-ness dc. The coupling strength increases with the thickness, so if the layer is too thin therewill be no band inversion. In the inverted regime a pair of edge states will appear, carryingopposite spin. These edge states connect the valence band with the conduction band andcross each other one time. This crossing is required by T -symmetry and cannot be removed.In these states no reflection can occur as explained earlier, and they both contribute onequantum of conductance e2
~ . This is very different from a ”normal” insulator that does nothave states crossing the band gap and does have a vanishing conductance.
Figure 7: The behaviour of a mercury telluride quantum well depends on the thickness of thelayer. Top: For quantum wells thinner than a critical thickness dc = 6.5 nm, the energy ofthe conduction band, labeled E1, is higher than that of the valence band, labeled H1. Butfor d > dc, those electron and hole bands are inverted. Bottom: The energy spectra of thequantum wells. The thin quantum well has an insulating energy gap, but inside the gap inthe thick quantum well are edge states, shown by red and blue lines.[1]
6Bernevig, Hughes and Zhang, Science 314, 1757 (2006)
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4.5 Dirac points and cones
Near the crossing of the two edge states in the gap the dispersion is linear. This looks like thedispersion relation that follows from the Dirac equation for a massless relativistic particle.7 Acrossing like this is called a Dirac point.8 Now the effective mass of a particle is proportionalto the second derivative of its dispersion. Since the dispersion is linear the second derivativewill be zero and the particle will behave like it is massless. This Dirac point can be extendedfrom one to two dimensions in which the dispersion relation will form a Dirac cone. This coneis located at a T -invariant point, and just as in the one dimensional case the degeneracy atthe crossing in two dimensions is protected by T -symmetry.A claim of quantum physics is that Dirac cones always appear in even numbers, so it isimpossible to have just one Dirac cone in a two dimensional system. It is not yet understoodwhy this is exactly the case. If the two dimensional system is actually the boundary of a threedimensional system, the two members of a pair will be spatially separated by the interior ofthe system.The states inside the gap are twofold degenerate, labeled in Figure 8 with Γa and Γb. Awayfrom these points SOC will split the degeneracy. Now by changing the Hamiltonian nearthe surface the electronic dispersion can be changed. All different shapes and kinks can beformed, resulting in a different number of times crossing the Fermi energy. But the ”net effect”will be the same; the difference in channels with a positive (right moving) and negative (leftmoving) group velocity will remain the same. This number ∆n is an integer characterizing theinterface. If the states connect pairwise, like in Figure 8(a), they will cross the Fermi energyan even number of times, so ∆n will be zero. These states can be eliminated, pushed out ofthe gap. But if the states do not connect pairwise, like in Figure 8(b), they will intersect theFermi energy an odd number of times. Now ∆n = 1 and these states cannot be completelyeliminated; there will always be a state crossing the Fermi energy.[2] This state crossing theFermi energy results in a Dirac cone.
Figure 8: Electronic dispersion between two Dirac points [2]
7See Section 4.6.8Remember that this is all in k-space, so it is actually a Dirac-momentum.
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4.6 Connection with the project
Knowing about these relevant aspects of the behaviour of an electron at the surface of atopological insulator, it is time to move on to the calculations. As the surface electron isexpected to have a linear dispersion relation, it will behave relativistically. Hence the energyof the particle will not be given by the Schrodinger equation, but by the Dirac equation:(
αipi + βm
)Ψ = EΨ (2)
in which αi = σi the Pauli spin matrices (this is for spin-1/2 particles like electrons). Theseare given by:
σx =(
0 11 0
)σy =
(0 −ii 0
)σz =
(1 00 −1
)As the mass of the particle can be considered to be zero and it can only move in the x,y-plane,equation (2) simplifies to:
(σxpx + σypy) Ψ = EΨ (3)
This is the equation the wavefunction that is constructed has to satisfy. This is checked inSection 5.3.
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5 Calculations
Now it is time to start describing a relativistic particle at the surface of a topological insulator.First a derivation of the wavefunction for a massive electron in 2D-cylindrical coordinates isgiven, without taking the spin into account. This is followed by the derivation of a spin-orbitcoupled wavefunction for a relativistic (regarded as massless) electron, with the check thatthe spin momentum and angular (orbit) momentum are really coupled. It is checked whetherthis wavefunction satisfies the massless Dirac equation. Finally the problem of tunneling willbe encountered in subsection 5.4.
5.1 Wavefunction for a free particle in cylindrical coordinates in 2 dimen-sions
From this derivation it is shown that the wavefunction of a particle in a two dimensionalsystem described in cylindrical coordinates (r,α) contains Bessel functions and is dependenton two quantum numbers. The particle is still treated as non-relativistic, so instead of theDirac equation the Schrodinger equation has to be considered to calculate the energy of theparticle:
− ~2
2me∇2Ψ = EΨ (4)
In cylindrical coordinates (r,α) equation (4) becomes:
− ~2me
(∂2
∂r2+
1r
∂
∂r+
1r2
∂2
∂α2
)Ψ = EΨ
Now by using separation of variables, i.e. Ψ(r, α) = R(r)A(α) the equation can be spit in twoparts depending on different variables. These two differential equations can only be equal toeach other if they are constant. Thus, equating both independent parts of the equation to m2
(the m here should not be confused with the mass of the particle me) gives for the angularpart:
A(α) ∝ eimα (5)
As A(α+ 2π) = A(α) then it is required that e2πim = 1. From this it follows that m must bean integer. For the radial part the differential equation becomes:
r2R′′ + rR′ +(
2meE
~2r2 −m2
)R = 0 (6)
This differential equation is a Bessel equation, with corresponding Bessel functions as solution:
R(r) = c1Jm(kr) + c2J−m(kr)
In which k =√
2meE/~2, the classical momentum. Since for integer m the two solutions arelinearly dependent, as Jm(kr) = (−1)mJ−l(kr), another set of solutions is needed. These arethe Bessel functions of the second kind;9
Ym(kr) =cos (mπ)Jm(kr)− J−m(kr)
sin (mπ)9See p.366 of [7]
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The general solution of equation (6) is:
R(r) = CmJm(kr) +BmYm(kr) (7)
There are two boundary conditions this solution has to satisfy:The first condition is that the wavefunction vanishes at the origin; if there is a defect placedat the origin Ψ(0, α) = 0. This yields Bm = 0 for all m, as limr→0 Ym(kr) = −∞.10 Forthe remaining Bessel functions of the first kind, Jm(0) = 0 for all m 6= 0. J0(0) = 1 so thenA0 = 0. Now, if there is no defect at the origin, this does not apply and it is possible thatΨ(0) 6= 0.The second condition is the wavefunction vanishing at the boundary, at some radius a. Callingthe n-th zero of the m-th Bessel function βnm it is clear that ka = βnm, thus k = βnm
a . Anapproximation to the Bessel function, for large arguments, is the following:
Jm(kr) ∼√
2πkr
cos(kr − mπ
2− π
4
)(8)
For the n-th zero the argument of the cosine should be equal to π2 (2n−1), from this it follows
that βnm is given by:βnm = (n+m)
π
2+π
4Putting everything together, the wavefunction becomes, for m 6= 0:
Ψnm(r, α) = AmJm(βnmr/a)eimα (9)
The energy is given by
Enm =~2k2
2me(10)
=~2β2
nm
2mea2(11)
Which is, explicitly showing the dependence on the quantumnumbers n and m:
Enm =~2π2
8mea2
(n+m+
12
)2
5.2 Derivation of the spin-orbit coupled wavefunction
To construct the wavefunction, it is necessary to consider what spinor correctly describesthe spin-momentum orientation of an electron at the surface of a topological insulator: thespin vector always points in the direction that is phase-shifted by +π/2 with respect to thedirection of the momentum vector. Let the momentum of particle be defined by the sphericalcoordinates (θ,φ). Then the spinor will be given by:11
|χ〉 = cos(θ/2)| ↑〉+ ei(φ+π/2) sin(θ/2)| ↓〉 (12)10Ym(kr) ∝ (kr)−m11Remember that | ↑〉 is an eigenstate of the Sz operator. Hence when the particle is in the state | ↑〉 it is
said to have its spin oriented in the +z-direction. This can also be represented by
(10
)
15
This can be checked for all directions as follows:For the momentum pointing in the positive x-direction, θ = π/2, φ = 0. Now (12) reduces to1√2
(1i
), which indeed is the eigenvector of Sy corresponding to the positive eigenvalue.
And for the momentum pointing in the positive y-direction, θ = π/2, φ = π/2 (12) reduces to1√2
(1−1
)which indeed is the eigenvector of Sx corresponding to the negative eigenvalue.
If the momentum is pointing in the positive or negative z-direction, the spin vector would doas well, because when θ = 0 φ can take any value. This is not really a problem, since theelectron can only move in the x,y-plane. Now in the x,y-plane (12) reduces to:
|χ〉 =1√2
(1ieiφ
)The most general wavefunction for a free particle with this spinor is given by:
Ψ(r) =∑k
akeik·r(
1ieiφ
)(13)
Here ak is a constant. Using separation of variables for ak (i.e. ak = a(k)eimφ) and as k iscontinuous replacing the sum by an integral results in:
Ψ(r) =∫dk k a(k)
∫ 2π
0eimφeik·r
(1ieiφ
)dφ (14)
To calculate the inner product of k and r it is necessary to consider what coordinate systemis worked in, see Figure 9. The particle is confined to the x,y-plane. From this it follows thatk · r can be rewritten as kr cos(φ− α):12
Ψ(r) =∫dk k a(k)
∫ 2π
0eimφeikr cos(φ−α)
(1ieiφ
)dφ
Figure 9: The coordinate system
12Here kr is |k||r|
16
Now a couple of steps need to be taken. First (φ−α) can be replaced by θ, the boundarieschanged and after using eimπ/2 = im the wavefunction reads:13
)The sine part in both integrals is equal to zero.14 To simplify the answer the following identityis needed:
Jm(x) =1
2π
∫ π
−πcos(mθ − x sin(θ))dθ [7] (15)
This leads to the elegant expression for the wavefunction:
Ψ(r) = 2πimeimα∫dk k a(k)
(Jm(kr)
−eiαJm+1(kr)
)(16)
5.2.1 Probability Distribution
The density probability of the particle is given by ρ =∫V |Ψ(r)|2dτ . To simplify this calcula-
tion, (16) is considered to be an integral over k of wavefunctions for every value of k:
Ψ(r) =∫dk Ψk(r)
Here
Ψk(r) = 2πka(k)(
imeimαJm(kr)−imei(m+1)αJm+1(kr)
)The density probability is now calculated for every Ψk(r):
ρk = 4π2k2a(k)2(Jm(kr)2 + Jm+1(kr)2) (17)
The probability for just one value of k, for three different values of m are shown below:13This derivation, including these steps, is shown in Appendix A.14This is because near x = 2πk, sin(x) ∼ x and hence sin(nx − sin(x)) ∼ sin(x mod 2π) which is an odd
function, while cos(nx − sin(x)) ∼ cos(x mod 2π) and this is an even function, giving an overall positivecontribution to the integral.
17
Figure 10: The probability distribution for m = 1, 2, 3 for k = 1.
As Bessel functions decrease proportional to 1√r, the density probability will decrease
proportional to 1r . So the density probability does not go to zero fast enough to be normalized.
Plane wave described by sines and cosines show this behaviour as well.
5.2.2 Check on the Spin-Orbit Coupling
An interesting calculation now would be the expectation values of the spin operators,15 to seehow the spin behaves. But the spin alone turns out to be not well defined:
〈Sx〉 = 4π2~∑k
k2a(k)2Jm(kr)Jm+1(kr)(− sin(α))
〈Sy〉 = 4π2~∑k
k2a(k)2Jm(kr)Jm+1(kr) cos(α)
〈Sz〉 =4π2~
2
∑k
k2a(k)2(Jm(kr)2 − Jm+1(kr)2
)The expectation values of Sx and Sy depend on α, which cannot be exactly right as thisangle defines the position vector, while the spin only depends on the momentum vector. Itis not strange, though, as these expectation values are calculated for a fixed combination ofmomentum k and angular momentum m. This combination represents one single state thatcan also be described by a fixed radius r and angle α, it is just another coordinate-system.This problem can be solved by noting that, as a result of the spin-orbit coupling, (16) is not
15These are given by: Sx = ~2σx, Sy = ~
2σy, Sz = ~
2σz
18
an eigenfunction of Lz16 nor Sz any more but of Jz. This can be seen from the following:
LzΨ(r) = −2πi~∫dk k a(k)
∂
∂α
(imeimαJm(kr)
−imei(m+1)αJm+1(kr)
)=(
−~m−~(m+ 1)
)Ψ(r)
= ~(−m 0
0 −(m+ 1)
)Ψ(r) (18)
SzΨ(r) =~2
(1 00 −1
)Ψ(r) (19)
= ~(
12−1
2
)Ψ(r)
Adding (18) to (19) leads to:
JzΨ(r) = (Lz + Sz)Ψ(r)
= ~(−m+ 1
2 00 −(m+ 1)− 1
2
)Ψ(r)
= ~(−m+
12
)Ψ(r) (20)
This shows that Ψ(r) is an eigenfunction of Jz with eigenvalue j = 12−m. There is something
special about the state with m = 0 or m = −1. In these cases the wavefunction cannot satisfythe condition Ψ(0) = 0, as in both cases one of the components of the wavefunction will beJ0(kr). This component is equal to one at r = 0.It is expected that (16) is now an eigenfunction of J2 instead of L2 and S2.
In spherical coordinates, the operators are given by:
L± = ~e±iα(i cot(θ)
∂
∂α± ∂
∂θ
)L2 = ~2
[1
sin2(θ)∂2
∂α2+
1sin(θ)
(sin(θ)
∂
∂θ
)]S2 =
3~2
4
Now a problem occurs: L± involves a derivative with respect to θ. But since the electroncan only move in the x,y-plane, θ cannot change. So this operator can not be applied to thiswavefunction. Hence the expression (21) does not have any meaning. The only componentof J that is well defined is Jz, as in (20). However, if the electron would move on a sphericalsurface instead of on the x,y-plane, the calculation of J would make sense.
16Lz = −i~ ∂∂α
19
5.3 Calculating the energy by the Dirac equation
To find out whether this wavefunction (16) satisfies the Dirac equation (3) the energy needsto be calculated.17 For this purpose the Dirac equation is written the following way:
(pxσx + pyσy)Ψ = −i~(
0 ∂x − i∂y∂x + i∂y 0
)Ψ (22)
To apply (22) to (16) it is necessary to find out what the derivatives with respect to x and yare. For this the chain rule18 is needed, giving:
∂Jm(kr)eimα
∂x= k cos(α)J ′me
imα − im sin(α)r
Jm(kr)eimα (23)
∂Jm(kr)eimα
∂y= k sin(α)J ′me
imα +im cos(α)
rJm(kr)eimα (24)
Using the expression for the derivative of a Bessel function 2J ′m(kr) = Jm−1(kr)− Jm+1(kr)and combining (23) and (24) leads to:
(p · σ)Ψk =~k2
Ψk − π~k2a(k)(
imeimαJm+2(kr)−imei(m+1)αJm−1(kr)
)(25)
+2~πka(k)
r
(imeimα(m+ 1)Jm+1(kr)−imei(m+1)αmJm(kr)
)This can be simplified further by using Jm−1(kr) + Jm+1(kr) = 2m
kr Jm(kr):
(p · σ)Ψk =~k2
Ψk + π~k2a(k)(
imeimαJm(kr)−imei(m+1)αJm+1(kr)
)Giving the final answer:
(p · σ)Ψk =~k2
Ψk +~k2
Ψk
= ~kΨk (26)
This shows that indeed the energy of the particle is proportional to k, and hence the dispersionrelation is linear!
17The full calculation is shown in Appendix B.18 ∂Jm(kr)eimα
∂x= ∂Jm(kr)eimα
∂kr∂kr∂r
∂r∂x
+ ∂Jm(kr)eimα
∂α∂α∂x
20
5.4 Tunneling through a potential
5.4.1 Klein tunneling
In non relativistic quantum mechanics tunneling is observed with exponential damping. Thisis a consequence of the Schrodinger equation for the electron:
− ~2m
∂2ψ
∂x2+ V (x)ψ = Eψ
This can be rewritten as follows:∂2ψ
∂x2= −p
2
~2ψ
Withp(x) =
√2m[E − V (x)]
Outside the potential, p(x) is real. Inside the potential, if E < V (x), p(x) is imaginary. Sincethe general wavefunction of the electron is given by:
ψ(x) = Ae±ipx/~
this solution represents running waves outside the potential, while inside the potential theamplitude of the wavefunction is decreasing exponentially. Now the transmission of theincoming wave in this process is given by:
T =|t|2
|j|2(27)
In which j is the incident amplitude, and t is the transmitted amplitude. As the amplitudedecreases inside the potential the transmission will always be smaller than one. Hence therewill always be some reflection.Now in 1929, Oskar Klein applied the Dirac equation to the problem of electron scatteringfrom a potential. He found that there is no exponential decrease in amplitude. As a matterof fact, he found that the transmission increases with the potential height, and that if thepotential approaches infinity, there is no reflection and the electron is always transmitted.This is called Klein tunneling. Imagine the following situation.
A massless relativistic particle, moving in the positive x-direction, is approaching a po-tential barrier that is given by:
V (x) ={
0 x < 0V0 x > 0
Now the wavefunction is a solution of (σxp+V (x))ψ = Eψ and given by the following, in thetwo different regions:
ψ1(x) = Aeipx(
11
)+A′e−ipx
(−11
)x < 0 (28)
ψ2(x) = Beip′x
(11
)x > 0
Here p = E and p′ = E − V0. Equating the wavefunction in both regions at the boundaryresults in A′ = 0 and B = A. From this it follows that the transmission is equal to one and
21
there is no reflection. There have not been many experiments that confirm this Klein tun-neling. One of them, 19 investigated the oscillations of the conductance in narrow structuresof graphene. Here a resonant cavity was formed in between two bipolar junctions. At lowmagnetic fields, the conductance showed a phase shift. This shift is a signature of perfecttransmission of the electrons through the junctions, at normal incidence. So this is an ex-perimental observation of Klein tunneling. Moreover, there have been, calculations on helicalelectrons in graphene.[4] These calculations show a close approximation to Klein tunneling.Depending on the angle of incidence of the electrons on the potential, and the width andheight of the potential, the transmission varies between 0 and 1. From these calculations, thetransmission is given by:
T =cos2(φ)
1− cos2(qxD) sin2(φ)(29)
Here qx is the momentum in the x-direction inside the potential (and hence a measure ofthe height of the potential), D the width of the potential and φ the angle of incidence ofthe electrons on the potential. An angle φ = 0 corresponds to normal incidence. Here it isexpected that the energy of the electron is much smaller than the potential barrier. (E � V0)The relationship (29) is shown in Figure 11.
Figure 11: The transmission T as function of the angle of incidence for a potential barrier of200 (red) and 285 (blue) MeV for electrons with an energy of 80 MeV. [4]
As the wavefunction (16) represents a helical electron as well, the question that arises is:will this electron also show some analogy to Klein tunneling? And if so, is there a dependenceon the potential width and height, and the amount of angular momentum the electron carries?The angular momentum is expressed in the quantum number m, and can be compared to the
19A.F. Young, and P. Kim, Quantum interference and carrier collimation in graphene heterojunctions, NaturePhysics, 5, 222-226 (2009).
22
angle of incidence mentioned earlier. The more angular momentum, the larger the angle ofincidence will be. Normal incidence corresponds to no angular momentum, so m = 0.
5.4.2 Introducing the potential barrier
Since the system is described in cylindrical coordinates, the potential needs to be cylindricalas well. The potential will be assumed to be as shown in Figure 12:
Figure 12: The cylindrical potential applied to the system of height V0 and width R2 −R1.
The transmission is again given by (27) now with the incident amplitude scaled to 1. Tocalculate this, the ingoing and outgoing wavefunctions drawn in Figure 12 need to be defined.As a single Bessel function of the first or second kind represents a standing wave, and not arunning wave, so a combination of these two has to be made. What combination this will becan be deduced from the approximations of both for large arguments. For Bessel functionsof the first kind this is given by (8), for Bessel functions of the second kind this is the sameapproximation with the cosine replaced by sine. From this it follows that the moving wavescan be constructed as follows:
φ+m(kr) = Jm(kr) + iYm(kr) ∼
√2πkr
ei[kr−mπ2−π
4] (30)
φ−m(kr) = Jm(kr)− iYm(kr) ∼√
2πkr
ei[−kr+mπ2
+π4] (31)
This combination of Bessel functions of the first and second kind is called a Hankel function;H
(1)m = Φ+
m and H(2)m = Φ−m. The part of the two components of the wavefunction containing
Bessel functions can be defined in the three different regions as follows:
Ψ1(kr) =
φ+m(kr) + wφ−m(kr) 0 < r < R1
aφ+m(qr) + bφ−m(qr) R1 < r < R2
tφ+m(kr) R2 < r
(32)
Ψ2(kr) =
s [φ+
m(kr) + wφ−m(kr)] 0 < r < R1
s′ [aφ+m(qr) + bφ−m(qr)] R1 < r < R2
stφ+m(kr) R2 < r
(33)
Here s is the sign of E and s′ the sign of E − V0. Important is to note that k = E~ and
q = |E−V0|~ . The main difference between the two components is the quantum number m. For
23
the second component this is m+ 1.In [4] there was a difference in + and − signs between the two components. Following thearticle, the second component would be described by:
Ψ2(kr) =
s [φ+
m(kr)− wφ−m(kr)] 0 < r < R1
s′ [aφ+m(qr)− bφ−m(qr)] R1 < r < R2
stφ+m(kr) R2 < r
(34)
This is needed to keep the right orientation of the spin, even if the wave is running in theother direction. It can be compared with the orientation of the magnetic and electric field ina polarized electromagnetic wave. If the wave is reflected, the electric field will point in thesame direction, but the magnetic field will be turned by 180◦. This is needed to maintainthe Poynting vector be pointing in the direction of propagation. Here a similar conditionhas to be satisfied. But in the spherical case these minus signs are not needed, because theapproximations of the Bessel functions contain a eimπ/2 term. For the second component, theonly difference compared to the first is that m is changed into m+1. This results in a e3imπ/2
factor for the right-moving wave and a e−3imπ/2 factor for the left-moving wave, while thefirst component only has the factors e±imπ/2. Multiplying the second component by e−iπ toreduce these terms to e±imπ/2 yields the minus sign in front of the left moving waves.
5.4.3 Calculating the transmission
Demanding the continuity of both components results in the next four equations:
φ+m(kR1) + wφ−m(kR1) = aφ+
m(qR1) + bφ−m(qR1) (35)aφ+
m(qR2) + bφ−m(qR2) = tφ+m(kR2) (36)
ss′(φ+m+1(kR1) + wφ−m+1(kR1)
)= aφ+
m+1(qR1) + bφ−m+1(qR1) (37)ss′(aφ+
m+1(qR2) + bφ−m+1(qR2))
= tφ+m+1(kR2) (38)
Solving this exact is not an easy job. In the paper [4] it is done in Cartesian coordinates.This involves working with exponentials, which are easy to multiply with each other, andcan be combined to give sines and cosines. Unfortunately, Bessel functions do not have theseproperties, and the calculation can not (yet) be done by hand. With the help of Mathematicaa numerical solution has been found.
5.4.4 Effect of the parameters on the transmission
As the solution is found numerically, it is not possible to show a formula expressing the trans-mission as a function of the momentum k, the angular momentum m, the radii R1 and R2 orthe height of the potential V0. The effect of the parameters separately will be discussed below.For the values of k and V0 one unit step corresponds to 80 meV (so V0 = 2.5 corresponds to200 meV), for units of length (R1 and R2) a step of 2π corresponds to 50 nm.
24
The width of the potential R2-R1.The next figure shows the effect of the width of the potential barrier. The inner radius iskept the same, while the outer radius is increased. It shows that a narrower potential resultsin less oscillations, of larger amplitude.
The radius R1.By changing R1 without changing the width of the potential barrier, it can be seen that theoscillations become larger for a barrier that is closer to the origin. This effect can be comparedto increasing m. It can be explained by the reasoning that a larger angular momentum willresult in a (by comparison) smaller momentum in the radial direction. As a result, a potentialbarrier would seem to be further away from the origin. To maintain the same apparentsituation, the barrier has to be closer to the origin.
Figure 14: The transmission for R1 = 8 (blue), R1 = 24 (purple), R1 = 40 (brown), m = 8,V0 = 2.5, R2 −R1 = 16
The momentum k and the potential height V0.As can be seen from all graphs, at the point k = V0 the transmission goes down to zero. This isthe same in all different cases, the potential height just makes the whole graph shift in k-space.
26
The quantum number m.
Figure 15: The transmission for m = 0 (blue), m = 4 (purple), m = 20 (brown) for V0 = 2.5,R1 = 20π, R2 = 24π
This figure shows the transmission for m = 0, 4, 20 for a barrier that is relatively far awayfrom the origin. As can be seen, for small m there is almost no effect of the barrier on thetransmission. For larger m, however, the transmission goes to zero when the momentum ofthe electron approaches the potential height. It can also be seen that the momentum at whichthe transmission becomes larger than zero shifts to the right as m becomes larger. It can beconcluded that the transmission is equal to zero, if m is large, for the next values of k:
0 < k < m/R1 (39)|V0 − k| < m/R1 (40)
It can also be seen that for a larger value of m the oscillations of the transmission becomelarger in the region where k < V0 , while the maximum height decreases. For very large m,the peaks have almost vanished. In the region where k > V0 the oscillations also increase withm, but the transmission always ends up being equal to 1. For larger m this will be reachedat a larger value of k.
27
Figure 16: The transmission for m = 80, V0 = 2.5, R1 = 20π, R2 = 24π
Combination of m and the width of the potential to capture the electronFrom the section about the effect of m on the transmission it could be concluded that thetransmission was zero when 0 < k < m/R1 and |V0− k| < m/R1. So, a larger m would resultin a larger range of zero transmission. But when m is smaller than a certain critical value, thetransmission decreases but does not reach zero. The width of the potential has a comparableeffect. The wider the potential, the closer the transmission will get to zero. But, in contraryto the effect of m, a wider potential does not increase the range in which the transmission iszero.
Figure 17: The transmission for m = 1, V0 = 2.5, R1 = 8, R2 = 12 (left), R2 = 40 (right)
28
6 Discussion
6.1 Conditions at the origin
Now one initial condition that is not satisfied yet is the condition for the wavefunction at theorigin: the wavefunction should vanish for r = 0. Clearly, a Hankel function is not zero forr = 0, as Bessel functions of the second kind blow up at zero. The only case in which there isno problem is when w is exactly equal to 1. This may sound not to difficult, as it can be seenfrom the figures that the transmission goes to zero for a range of values of k, which meansthat the reflection will go to 1. But r = |w|2. So the imaginary part of the reflection shouldbe equal to zero, i.e. the argument of w should be equal to zero as well. The next graphshows the argument of the reflection:
Figure 18: The argument for the reflection for m = 4, V0 = 2.5, R1 = 8, R2 = 24
From this graph it can be concluded that in the range 2 < k < 3 it happens three timesthat the reflection is real. For other values of the variables there will be another density ofthese real points, but they will always be distinct.
6.2 Combination of m and the width of the potential to capture the elec-tron
It is unknown the relationship between the width of the potential and m is. There are a coupleof problems involved, in determining the width necessary for the transmission to be zero fora given m. First, setting the value of k equal to V0 gives some errors in the calculation. So khas to be taken approximately equal to V0. For this a minimum difference has to be chosen.Also, the transmission becomes very small but remains larger than 1. So what value of T issmall enough?
29
7 Conclusion
The derivation of the wavefunction has been successful, considering that it indeed describesa spin-orbit coupled electron and that it satisfies the Dirac equation. This is the expectedbehaviour of an electron at the surface of a topological insulator. To answer the questionwhether this electron shows an analogy to Klein tunneling, it is needed to find out for whatproperties of the potential the transmission is equal to 1. As mentioned earlier, if m is largeenough there is a range in k for which the transmission is equal to zero. This range is given by(39) and (40). On the other hand, for smaller m a wider potential will make the transmissiondrop as well, only does not increase the range in which the transmission is zero. So yes,perfect transmission is possible, for values of k that are not in the range given by (39) and(40). And for m/R1 < k < V0−m/R1 the transmission can only get to 1 if m is small enough,comparable with almost perpendicular incidence. For large k the transmission will always goto one.
8 Acknowledgments
First of all I would like to say that doing this project was a whole new experience for me. Ithas been a period in which I got stuck on calculations so many times that I wondered howthis would ever yield results. Especially the last section about the tunneling problem hastaken me more than a month to solve. It has been a series of different attempts, trying overand over again. Just a week before I wanted to be finished, with the help of my supervisor,I understood the calculation in the paper.[4] This gave me the motivation to work out myproblem the same way, but during this calculation I had to make assumptions that I couldnot justify, see appendix C. Plugging in the equations in Mathematica finally gave an answer.And even with the help of Mathematica there were problems. I would like to thank mysupervisor for giving me support and helping with the calculations, especially during the lastfew days. I would also like to thank the ones who have given me the mental support neededwhen calculations failed, and the ones who helped me correcting my thesis.
30
References
[1] X.-L. Qi, S.-C. Zhang, The quantum spin Hall effect and topological insulators, PhysicsToday, 63, 33-38 (2010) ArXiv:1001.1602v1
[3] X.-L. Qi, S.-C. Zhang, Topological insulators and superconductors, ArXiv:1008.2026v1
[4] M. I. Katsnelson, K.S. Novoselov, A.K. Geim, Chiral tunneling and the Klein paradox ingraphene, Nature Physics, 2, 620-625 (2006) ArXiv:cond-mat/0604323v2
[5] J.C.Y. Teo, C.L. Kane, Topological Defects and Gapless Modes in Insulators and Super-conductors, Phys.Rev.B, 82, 115-120 (2010) ArXiv:1006.0690v2
[6] J. Moore, Topological insulators, the next generation, Nature Physics, 5, 378-380 (2009)
[7] Z. X. Wang D. R. Guo, Special Functions, World Scientific (1989) ISBN 9971-50-659-9
[8] D.J. Griffiths, Introduction to Quantum Mechanics, Pearson Education (2005) ISBN 0-13-191175-9
[9] E. Kreyszig, Advanced Engineering Mathematics, Wiley (2006) ISBN 978-0-471-72897-9
[10] J.R. Hook & H.E. Hall, Solid State Physics, Wiley (1991) ISBN 978-0471-92805-8
[11] http://www.warwick.ac.uk/˜phsbm/qhe.htm
[12] http://en.wikipedia.org/wiki/Quantum Hall effect
[13] http://en.wikipedia.org/wiki/Bessel function
31
A Derivation of wavefunction
Starting with the spinor:
|χ〉 = cos(θ/2)| ↑〉+ ei(φ+π/2) sin(θ/2)| ↓〉
In the x,y-plane this reduces to:
|χ〉 =1√2
(1ieiφ
)The most general wave function for a free particle with this spinor reads:
Ψ(r) =∑k
akeik·r(
1ieiφ
)Using separation of variables for ak as a function of k and φ (i.e. ak = a(k)eimφ) and replacingthe sum by an integral:
Ψ(r) =∫dk k a(k)
∫ 2π
0eimφeik·r
(1ieiφ
)dφ
Since k · r can be rewritten as kr cos(φ− α):
Ψ(r) =∫dk k a(k)
∫ 2π
0eimφeikr cos(φ−α)
(1ieiφ
)dφ
Replacing φ− α by θ:
Ψ =∫dk k a(k)
∫ 2π+α
0+αeimθeimαeikr cos(θ)
(1
ieiθeiα
)dθ
Now cos(θ + π/2− π/2) = − sin(θ − π/2) = − sin(θ′). Replacing θ − π/2 by θ′:
Ψ =∫dk k a(k)
∫ 3π/2+α
−π/2+αeimπ/2eimθ
′eimαe−ikr sin(θ′)
(1
ieiπ/2eiθ′eiα
)dθ′
As the integral ranges over a whole cycle, the boundaries can be changed into −π and +π
The next assumption to be made is that Hankel functions can be multiplied the same wayexponentials can;
Φ+m(x)Φ+
m(y) = Φ+m(x+ y)
This turns out not to be the case. First, in using the approximation makes clear that thefactor in front of the exponential should be taken into account as well. It is well known that√x√y 6=√x+ y. But compensating for this factor makes the calculation impossible to do.
To exclude the coefficients a and b it is needed to write the combination of Hankel functionsas:
Φ+m(qW )− Φ−m(qW ) = 2i sin(qW )
However, as Φ+m(qW ) is actually the product of Φ+
m(qR2) and Φ−m(qR1), and Φ−m(qW ) is nota product of two other Hankel functions, this can not be done.
