[TOC]
参考
- 《Topological Insulators and Topological Superconductors》 - B. Andrei Bernevig
- Chapter-7 Graphene
Chapter-7 Graphene
In the previous chapters, we have primarily investigated phenomena on the square lattice. In this chapter, we switch gears and look at another two-dimensional lattice, the hexagonal, or honeycomb, lattice. This is not just a simple variation of our work so far. The hexagonal lattice is important for of both experimental and theoretical reasons. Experimentally, the hexagonal lattice is realized in graphene, an amazing material of fundamental importance and interest. The synthesis of graphene sheets via the "scotch tape" method was awarded with the Nobel Prize in Physics for 2010. Graphene is currently a popular subject of research and is likely to remain so for many years. From a theoretical point of view, graphene is also interesting: it has two sites per unit cell (A-sites and \(B\) -sites), so the minimal model for graphene has to be a two-band model. It turns out that the hexagonal lattice with nearest-neighbor hopping is a semimetal with gapless Dirac fermions at the Fermi level. Gaps for these Dirac fermions can be opened in different ways, and the insulator obtained can exhibit fascinating properties, such as a zero-field quantized Hall effect (i.e., the Chern insulator or quantum anomalous Halleffect state).
We focus on graphene in this long chapter, We show that the nearest-neighbor-hopping model of graphene has Dirac nodes and then ask what symmetries protect the nodes from opening a gap. We learn that in two spatial dimensions, both inversion and TR symmetry are needed to keep Dirac fermions gapless. These two conditions endow Dirac fermions with a "reality" constraint, and we relate their gaplessness to the Wigner-vonNeumann classification. We then keep inversion and time reversal intact but break the \(C_{3}\) symmetry of the graphene nearest-neighbor model and show that, although Dirac nodes are locally stable, they are not globally stable: two Dirac nodes can annihilate and open a gap. This can happen, however,only if the two nodes have opposite "vorticity," which is the same effect the Berry phase accumulated around a constant-energy contour encircling the Dirac point. We hence learn that the Berry phase is a vorticity that keeps Dirac nodes locally stable. We then move on to show that gapless graphene has edge modes that link the two Dirac nodes. We obtain analytic solutions for these modes and then argue that the opening of different types of gaps at the Dirac nodes gives rise to different types of insulators. This provides a simple and heuristic way of understanding the Chern insulator presented in a later chapter of this book.
Hexagonal Lattices
The graphene lattice, shown in figure \(7.1\), has the following translation vectors: \[ a _{1}=\frac{a}{2}(3, \sqrt{3}), \quad a _{2}=\frac{a}{2}(3,-\sqrt{3}) \] where \(a\) is the bond length. The vectors give rise to the reciprocal lattice vectors \[ b _{1}=\frac{2 \pi}{3 a}(1, \sqrt{3}), \quad b _{2}=\frac{2 \pi}{3 a}(1,-\sqrt{3}), \] which satisfy \(b _{i} \cdot a _{j}=2 \pi \delta_{i j} .\) Notice that the lattice has a two-site unit cell because \(A\) -sites (see fig. 7.1) can be moved into \(A\) -sites only by the primitive wavevectors. In practice, the \(BZ\) can be taken to be the parallelogram formed by the wavevectors \(b _{1}, b _{2}\), although in most work on the subject, the \(BZ\) is taken to be the hexagon with the two vertices at points \(\left(K, K^{\prime}\right)\), where \[ K =\frac{2 \pi}{3 a}\left(1, \frac{1}{\sqrt{3}}\right), \quad K ^{\prime}=\frac{2 \pi}{3 a}\left(1,-\frac{1}{\sqrt{3}}\right) . \] These two points are TR partners of each other, which is clear in the parallelogram \(BZ\) but which can easily be shown to be the case in the hexagonal BZas well because \(- K = K ^{\prime}- b _{1}- b _{2}\). The simplest tight-binding Hamiltonian for graphene contains hoppings to nearest- neighbor sites. It is a \(2 \times 2\) Hamiltonian, which is easy to diagonalize in the \(\left(c_{A}, c_{B}\right)\) basis of second quantized operators for sites \(A, B\). For hopping \(t_{1}, t_{2}, t_{3}\) along the \(\delta _{1}, \delta _{2}, \delta _{3}\) bonds (see fig. 7.1) the Hamiltonian Fourier- transformed in momentum space is \[ H=\sum_{k}\left(\begin{array}{ll} c_{A k}^{\dagger} & c_{B k}^{\dagger} \end{array}\right)\left(\begin{array}{cc} 0 & \sum_{i=1}^{3} t_{i} e^{i k \cdot \delta_{i}} \\ \sum_{i=1}^{3} t_{i} e^{-i k \cdot \delta_{i}} & 0 \end{array}\right)\left(\begin{array}{l} c_{A k} \\ c_{B k} \end{array}\right) \] The distances between the three nearest neighbors of one of the sites is \[ \delta _{1}=\frac{a}{2}(1, \sqrt{3}), \quad \delta _{2}=\frac{a}{2}(1,-\sqrt{3}), \quad \delta _{3}=a(-1,0) . \] Unfortunately, in this basis, the Hamiltonian is not in Bloch form, i.e., \(h( k + G ) \neq h( k )\), where \(G=b_{1}, b_{2}\), because \(b _{i} \cdot \delta _{j} \neq 2 \pi \delta_{i j} .\) We could work in this basis and forego Bloch form, but this would not be consistent with results in the past chapters and would also preempt the automatic identification of time reversal invariant points \(G / 2\) and \(- G / 2\), up to reciprocal lattice wavevectors. We decide to restore Bloch form by making a gauge transformation of the preceding Hamiltonian on \(B\) -sites, \[ c_{B k} \rightarrow c_{B k} e^{i k \cdot \delta_{3}} \] to get the Bloch Hamiltonian \(H=\sum_{ k } c_{ k }^{\dagger} h(k) c_{ k }\), where \[ h(k)=\left(\begin{array}{cc} 0 & -t_{a} e^{i k \cdot a _{1}}-t_{b} e^{i k \cdot a _{2}}-t_{c} \\ -t_{a} e^{-i k \cdot a _{1}}-t_{b} e^{-i k \cdot a _{2}}-t_{c} & 0 \end{array}\right) \] The gauge transformation that makes the Hamiltonian of Bloch form corresponds to a different gauge choice of orbitals in the graphene lattice.
Dirac Fermions
The simple tight-binding graphene Hamiltonian displays the interesting physics of massless Dirac fermions. For the case of isotropic-hopping matrix elements \(t_{a}=t_{b}=t_{c}=1\), if we expanded around the points \(K , K ^{\prime}\), at \(k = K + \kappa\), with \(\kappa \ll K\), we find that the Hamiltonian has the Dirac form \[ h( K + \kappa )=\kappa_{x} \sigma_{x}+\kappa_{y} \sigma_{y} \] Expanding around \(K ^{\prime}\) to first order, we obtain another Dirac Hamiltonian: \[ h\left( K ^{\prime}\right)=H(- K + \kappa )=-\kappa_{x} \sigma_{x}+\kappa_{y} \sigma_{y} \] There are no other independent Dirac nodes in the problem. The presence of these nodes renders graphene to be a semimetal, with fundamentally different properties from an insulator, because low-energy excitations are always present in such a system.
The question one should now ask is whether these Dirac points are stable to perturbations. Our expansion of the Hamiltonian has showned us the existence of two Dirac fermions. However, the Hamiltonian we used was by no means generic. For example, it contained only nearest-neighbor coupling with \(C_{3}\) symmetry, and it did not allow for different on-site energies of the \(A\) and \(B\) -sites in the unit cell, etc. Would adding small perturbations to the graphene lattice result in the gapping of these Dirac fermions? What are the perturbations we are allowed to add while keeping the system a semimetal? What kind of perturbations open a gap? For all these questions, we need to look at the symmetries of graphene. ## Symmetries of a Graphene Sheet
Two main symmetries characterize the hexagonal lattice with identical atoms on \(A\) and \(B\) -sites. The first symmetry is time reversal, which is present regardless of the spatial distribution of hopping matrix elements, as long as they are real. The second symmetry, inversion, is present if hoppings in the lattice are symmetric upon inversion with the inversion center either in the middle of the unit cell bond or in the middle of the hexagonal lattice: it is clearly present
1 Time Reversal
As previously discussed, time reversal for spinless fermion Bloch Hamiltonians takes the form \[ \operatorname{Th}( k ) T^{-1}=h(- k ) \longrightarrow h( k )^{*}=h(- k ) \] If we impose TR invariance on the system, this symmetry fixes the form of the Hamiltonian at \(K^{\prime}\) once the Hamiltonian at \(K\) is known \[ \begin{array}{c} h( K + \kappa )^{*}=h(- K - \kappa )=\left(\kappa_{x} \sigma_{x}+\kappa_{y} \sigma_{y}\right)^{*}=\kappa_{x} \sigma_{x}-\kappa_{y} \sigma_{y} \\ \longrightarrow h(- K + \kappa )=-\kappa_{x} \sigma_{x}+\kappa_{y} \sigma_{y} \end{array} \] Notice that \(T\) by itself does not protect the Dirac fermions from opening a gap as long as the gap at points \(K, K^{\prime}\) has special properties: if at \(K\) we have a Bloch Hamiltonian \(h( K + \kappa )=\) \(\kappa_{x} \sigma_{x}+\kappa_{y} \sigma_{y}+m \sigma_{z}\), whereas at \(K^{\prime}\) we have the Bloch Hamiltonian \[ h(- K - \kappa )=h( K + \kappa )^{*}=\kappa_{x} \sigma_{x}-\kappa_{y} \sigma_{y}+m \sigma_{z} \] the system preserves time reversal. This type of mass, which preserves \(T\) but opens a gap in the Dirac spectrum, physically represents a different on-site energy for the atoms on the \(A\) -sites and \(B\) -sites. It corresponds to adding a momentum-independent term \(m \sigma_{z}\) in the full Bloch Hamiltonian. For example, the Hamiltonian of boron nitride, (BN) a system that also has a hexagonal lattice structure but, as the name suggests, as different atoms on the \(A\) - and \(B\) -sites, has this form and is a gapped insulator.
2 Inversion Symmetry
The graphene lattice has another symmetry which is spatial inversion (see fig. \(7.2\) for the \(t_{1}, t_{2}, t_{3}\) -hopping model and inversion center). If the hopping and on-site matrix elements do not change upon the action of the inversion operator \[ I:(x, y) \rightarrow(-x,-y) \] we have that the Hamiltonian also has inversion symmetry. Inversion is a unitary operator. It does not contain the complex conjugation operator, and it leaves the \([x, p]\) commutator unchanged, due to the fact that both \(x, p\) change under inversion, unlike in the \(T\) case, in which only \(p\) transformed under inversion. Under inversion with respect to the middle of the unit cell, the second quantized operators transform as \[ I c_{i, A} I^{-1}=c_{-i, B}, \quad I c_{i, B} I^{-1}=c_{-i, A} \] or, in compact form, \[ I c_{i, a} I^{-1}=\sigma_{a, a^{\prime}}^{\chi} c_{-i, a^{\prime}} \] where \(a=A, B\). The Fourier component of the second quantized operators transforms as \[ I c_{ k , a} I^{-1}=\sigma_{a, a^{\prime}}^{x} \frac{1}{\sqrt{N}} \sum_{j} e^{-i k r_{j}} c_{j a^{\prime}}=\sigma_{a a^{\prime}}^{x} c_{- k , a^{\prime}} \] For the inversion-symmetric second quantized graphene Hamiltonian with \([H, I]=0\), we find the transformation on the Bloch Hamiltonian: \[ \begin{aligned} I H I^{-1} &=\sum_{ k } \sigma_{\theta a}^{x} c_{- k \theta}^{\dagger} I h_{a \beta}( k ) I^{-1} \sigma_{\beta, \delta}^{x} c_{- k \delta}=\sum_{ k } \sigma_{\theta a}^{x} c_{- k \theta}^{\dagger} h_{a \beta}( k ) \sigma_{\beta, \delta}^{\chi} c_{- k \delta} \\ &=\sum_{ k } \sigma_{\theta a}^{x} c_{ k \theta}^{\dagger} h_{a \beta}(- k ) \sigma_{\beta, \delta}^{\chi} c_{ k \delta} . \end{aligned} \] If inversion is a good symmetry, then \(I H I^{-1}=H\) and, hence, \[ h( k )=\sigma_{x} h(- k ) \sigma_{x} \] In this equation, the matrix representation of the inversion operator is \(\sigma_{x}\) in the graphene lattice. We will later see other representations, such as \(\sigma_{z} .\)
By itself, inversion also does not guarantee the stability of the Dirac points. If at \(K\) we open a gap \(m, h( K + \kappa )=\kappa_{x} \sigma_{x}+\kappa_{y} \sigma_{y}+m \sigma_{z}\), then we can have a perfectly inversion-symmetric Hamiltonian if the Hamiltonian at the \(K^{\prime}\) point is \[ h(- K - \kappa )=\sigma_{x}\left(\kappa_{x} \sigma_{x}+\kappa_{y} \sigma_{y}+m \sigma_{z}\right) \sigma_{x}=\kappa_{x} \sigma_{x}-\kappa_{y} \sigma_{y}-m \sigma_{z} \] This type of mass, however, cannot come from a momentum-independent term (in the full Bloch Hamiltonian). Indeed, a full lattice realization of such a mass term was first found by Haldane in the first example of a topological insulator, the Chern insulator. We analyze it in chapter 8 .
3 Local Stability
Local Stability of Dirac Points with Inversion and Time Reversal
The reason that inversion and time reversal separately do not protect the Dirac fermions is because they both link a generic \(k\) to \(- k\), thereby not really imposing any constraints on the Hamiltonian at a generic \(k\). However, when both these symmetries are present and when they are used together, they relate \(k\) to \(k\) and impose constraints on the form of the Bloch Hamiltonian at each \(k\) separately: \[ h( k )=\sigma_{x} h(- k ) \sigma_{x}=h^{*}(- k ) \longrightarrow h( k )=\sigma_{x} h^{*}( k ) \sigma_{x} \] The first equality uses \(I\), and the second uses \(T\). The equation (29) is the matrix representation of the operator product of inversion and time reversal, \(\left.(T I) h( k )(T I)^{-1}=h( k )\right)\). For a generic, two-level Hamiltonian of Bloch form, \[ H=d_{i}( k ) \sigma_{i}+\epsilon( k ) I_{2 \times 2} \] \(T\) and inversion impose the conditions (we drop the explicit \(k\) -dependence) \[ d_{i} \sigma_{i}+\epsilon=\sigma_{x}\left(d_{i} \sigma_{i}+\epsilon\right)^{*} \sigma_{x}=d_{x} \sigma_{x}+d_{y} \sigma_{y}-d_{z} \sigma_{z}+\epsilon \] We see that as a result of the combined inversion and time reversal, we obtain \[ d_{z}( k )=-d_{z}( k )=0 \] As such, if both inversion and time reversal are respected, no \(\sigma_{z}\) term can arise, and the Dirac points are locally stable. What does this mean for a Dirac Hamiltonian? If, around the point \(K\) in the BZ, the Hamiltonian is \(k_{x} \sigma_{x}+k_{y} \sigma_{y}\) (where \(k\) is the deviation from \(K\) ), we are not allowed to add a \(\sigma_{z}\) -term to the Hamiltonian by any perturbation. Although \(\sigma_{x}\) -and \(\sigma_{y}\) -terms are not forbidden by time reversal and inversion, their only result, if small, is to shift the Dirac points in momentum space. Explicitly, \[ H^{\prime}=k_{x} \sigma_{x}+k_{y} \sigma_{y}+a_{1} \sigma_{x}+a_{2} \sigma_{y} \] and as long as \(a_{1}\) and \(a_{2}\) are small, \(H^{\prime}\) has nodes at \(k_{x}=-a_{1}\) and \(k_{y}=-a_{2}\). Hence, with inversion and time reversal, a single Dirac node is locally stable-no small perturbation can open a gap. Large perturbations can and do open a gap, but the gap-opening mechanism is very different.
As long as the Dirac fermions at \(K\) and \(- K\) are forbidden to open a gap by \(T\) and inversion, they carry a vortex in the Bloch wavefunction, which is the same as saying that their Berry phase \(\int A_{i}(\kappa) d \kappa_{i}\) (where \(A_{i}(\kappa)\) is the Berry potential and the integral is performed over any closed contour enclosing the Dirac point) equals \(\pm \pi\). Because the matrix \(\sigma_{z}\) is forbidden (whether multiplied by a constant term or even by an even or odd power of \(k\) ), we thus have that the Hamiltonian at both \(K\) and \(K ^{\prime}\) can be written as a matrix: \(H(k)=\kappa_{i} A_{i j} \sigma_{j} .\) The Berry phase of such a Dirac fermion is given by \(\pi \operatorname{sign}(\operatorname{Det}(A))\), which we be prove later. Note that the lack of \(\sigma_{z}\) is important because it allows us to write the Hamiltonian with both \(i, j=1,2\) rather than \(H(\kappa)=\kappa_{i} A_{i a} \sigma_{a}\), with \(a=1,2,3-\) in this case \(A_{i a}\) would not be a square matrix and defining a determinant (vorticity) would be impossible. The fact that vorticities can be defined when the dimension of the BZ (in this case, 2 ) is equal to the codimension of the crossing (in this case, also 2 because we have two Pauli matrices) is no accident. The Wigner-vonNeumann classification says that a generic crossing has codimension 3 - we need to tune three parameters to obtain a degeneracy because there exist three anticommuting Pauli matrices whose coefficients need to be tuned to zero. However, the BZ provides for two parameters \(\left(k_{x}, k_{y}\right)\), which are tuned automatically, leaving us with only one tunable parameter needed to obtain a degeneracy. This would be the Dirac fermion mass, which would need to be fine-tuned to vanish. But, if time reversal and inversion are present, then the matrix that would couple to this remaining tunable parameter cannot exist, and degeneracies can happen without tuning-i.e., they are locally stable. The presence of inversion and time reversal is said to impose a reality condition on the Hamiltonian (the Hamiltonian can be made real by a gauge transformation).
For the Dirac fermion at \(K\) we have \(A_{11}=A_{22}=1\), whereas for the one at \(K ^{\prime}\), we have \(A_{11}=-A_{22}=-k .\) Hence, the Berry phase, which is \(\pi \operatorname{sign}(\operatorname{Det}(A))\), is opposite for the two fermions. In graphene, the Dirac fermion at \(K\) carries Berry phase \(\pi\), whereas the Dirac fermion at \(K ^{\prime}=- K\) carries Berry phase \(-\pi\). Because the wavefunctions have vorticity, the nodes cannot be removed by themselves. For example, for the Dirac fermion at \(K\), the wavefunction has a vortex (which gives it a Berry phase \(\pi\) ) and the inclusion of extra terms that break the \(C_{3}\) symmetry (for example,the simplest way is to make \(t_{1} \neq t_{2} \neq t_{3}\) ) but respect \(T\) and \(I\) can only move the Dirac point away from \(K\) but cannot open a gap.
Global Stability of Dirac Points
The stability of the Dirac nodes proved in the previous section is valid for any perturbation that respects \(T\) and \(I\), as long as it is small. For example, we can break \(C_{3}\) and make the hopping on the bonds different, or we can add second and third nearest-neighbor hoppings. The second nearest-neighbor hopping does nothing because it is diagonal in the sublattice space (couples identical sublattices). What happens if the perturbation is large? What happens if we add large, arbitrary-range hopping terms in our Bloch Hamiltomian? It is clear that the Hamiltonian can be gapped. For example, pick \(t_{a}=t_{b}=0\) in our graphene model, i.e., make a model of very anisotropic graphene with nonzero hopping in only one direction. The energy levels would then be \(\pm t_{c}\), thereby giving a fully gapped Hamiltonian. How did this happen? It turns out that Dirac modes can and will open a gap by coming together and annihilating at a TR-symmetric point; at this point, the dispersion has to be quadratic in one direction. This section analyzes this aspect of the problem. We show that if the Hamiltonian has \(C_{3}\) symmetry (on top of \(I\) and \(T\) ), then the Dirac nodes are globally stable, and their position is fixed at the \(K\) and \(K ^{\prime}\) points in the BZ. If, however, \(C_{3}\) symmetry is broken, then the Dirac nodes can move off the K, K' points-and, upon large-enough perturbations, the two Dirac nodes can meet up and annihilate at a TR-invariant point in the BZ.
1 \(C_{3}\) Symmetry and the Position of the Dirac Nodes
We now prove that, in the case where the Hamiltonian has the three symmetries \(T, I\), and \(C_{3}\), the position of the Dirac nodes does not change and stays at \[ K =\frac{2 \pi}{3 a}\left(1, \frac{1}{\sqrt{3}}\right), \quad K ^{\prime}=\frac{2 \pi}{3 a}\left(1,-\frac{1}{\sqrt{3}}\right) . \] As stated previously, \(T\) and \(I\) guarantee the absence of \(\sigma_{z}\) terms in the Hamiltonian. However, with just \(T\) and \(I\), the hopping can be anisotropic, the simplest example being a model with different nearest-neighbor hoppings. An extra \(C_{3}\) symmetry, i.e., rotation by \(\frac{2 \pi}{3}\) around a hexagon center in the lattice or around the points \(A\) or \(B\), adds extra constraints to the system. The \(C_{3}\) symmetry is manifest by the fact that the hopping matrix elements must be invariant upon the cyclic change of \[ \delta _{1} \rightarrow \delta _{2} \rightarrow \delta _{3} \rightarrow \delta _{1} \] where, as previously defined, \(\delta _{1}=\frac{a}{2}(1, \sqrt{3}), \delta _{2}=\frac{a}{2}(1,-\sqrt{3}), \delta _{3}=a(-1,0)\) are the \(A \rightarrow B\) bond vectors. This \(C_{3}\) transformation is obvious if we take the \(\frac{2 \pi}{3}\) rotation center to be either the center of the hexagon, one of the \(A\) -sites, or one of the \(B\) -sites. Notethat this symmetry implies equal hopping parameters for the nearest-neighbor hopping \(A \rightarrow B\), but, in general, for \(n\) th next nearest-neighbor hopping, it does not imply equal hopping parameters. For example, for the hopping matrix element from site \(A\) to the third-nearest-neighbor site \(B\) (there are two third-nearest neighbor sites \(B\), as given in fig. 7.3), we have perfect \(C_{3}\) symmetry even though the six matrix elements can break up into two different matrix elements: \[ \begin{array}{l} t_{1 n n n}\left(e^{i k \cdot\left(\delta_{1}-\delta_{3}+\delta_{1}\right)}+e^{i k \cdot\left(\delta_{3}-\delta_{2}+\delta_{3}\right)}+e^{i k \cdot\left(\delta_{2}-\delta_{1}+\delta_{2}\right)}\right) \\ +t_{2 m n n}\left(e^{i k \cdot\left(\delta_{1}-\delta_{2}+\delta_{1}\right)}+e^{i k \cdot\left(\delta_{3}-\delta_{1}+\delta_{3}\right)}+e^{i k \cdot\left(\delta_{2}-\delta_{3}+\delta_{2}\right)}\right), \end{array} \] where \(t_{1 n m n} \neq t_{2 m n n} .\) In other words, \(C_{3}\) symmetry does not require symmetry over all permutations of \(\delta_{1,2,3}\), but over only the cyclic permutations. It is clear from figure \(7.3\) that the preceding term respects \(C_{3}\) even though \(t_{1 m n n} \neq t_{2 n m n}\)
To see the effect of the \(C_{3}\) symmetry on the graphene band structure, we first analyze the nearest-neighbor Hamiltonian-where hopping occurs only through the nearest-neighbor bond. It is convenient to work in the basis in which the hopping occurs through \(\delta _{1}, \delta _{2}, \delta _{3}\). It is in this basis that the \(C_{3}\) symmetry is easily imposed, and we can go to the basis \(1, a _{1}, a _{2}\) through a gauge transformation. The off-diagonal matrix element of this Hamiltonian is \[ -t\left(e^{i k \cdot \delta_{1}}+e^{i k \cdot \delta_{2}}+e^{i k \cdot \delta_{3}}\right)=-t e^{i k \cdot \delta_{3}}\left(1+e^{i k a _{2}}+e^{i k \cdot a _{1}}\right) . \] The gapless point happens at the \(k\) for which \[ 1+e^{i k \cdot a _{2}}+e^{i k \cdot a _{1}}=0 \] Because these are complex numbers of unit absolute value, they can be represented by a vector on the unit circle. The vector 1 is on the positive real axis, whereas the other two vectors, \(e^{i k \cdot a _{2}}, e^{i k \cdot a _{1}}\), are above and below the real negative axis at the same angle so that their imaginary part cancels; however, their real parts add up to cancel \(1 .\) The only two solutions are \(K \cdot a _{2}=\frac{2 \pi}{3}, K \cdot a _{1}=\frac{4 \pi}{3}\) and \(K ^{\prime} \cdot a _{1}=\frac{2 \pi}{3}, K ^{\prime} \cdot a _{2}=\frac{4 \pi}{3}\), which are, of course, time reversals of each other (up to reciprocal lattice vectors). The gapless equality can then be written as \[ 1+z+z^{2}=0 ; \quad z=e^{i K \cdot a _{2}}=e^{i \frac{2 \pi}{3}}, \quad z^{2}=e^{i K \cdot a _{1}}=e^{i \frac{4 \pi}{3}} \quad z^{3}=1, \] where \(z\) is the third primitive root of unity. We now show that longer- range hopping not only cannot open a gap in perturbation theory (which was known from the local stability of Dirac points studied in section 7.3.3) but also cannot move the Dirac points \(K , K ^{\prime}\), no matter how large the perturbation. However, we cannot guarantee the absence of other gapless points in the BZ upon the introduction of longer-range hopping. To prove that the gapless points are stable, we have to prove that any off-diagonal matrix element vanishes at \(k = K\) when \(C_{3}\) symmetry is present (if so, T guarantees that the matrix element will also vanish at \(\left.K ^{\prime}\right)\). Consider one \(A\) -site and the hoppings from it to the other sites. The hoppings are of two kinds: 1. There are hoppings that go through an even number of bonds (i.e., the vector that links a certain \(A\) -site with the site we want to hop onto can be expressed as a linear combination of a total even number of \(\delta _{1,2,3}-\) the number of \(\delta _{1}, \delta _{2}, \delta _{3}\), when added, is even). These hoppings couple sites in the same sublattice, i.e., \(A\) to \(A\) and \(B\) to \(B\). Because we cannot have any \(\sigma_{z}\) matrix by \(T\) and \(I\) combined, the term induced by these hoppings must be diagonal and proportional to the identity matrix (this is true even if \(C_{3}\) symmetry is absent, just due to \(T\) and \(I)\). Hence, these terms are just an energy shift, which breaks the perfect particle-hole symmetry of graphene but cannot open a gap at the Dirac points. 2. More importantly, there are hoppings that go through an odd number of bonds and that couple \(A\) -sites with \(B\) -sites. These are important and must be treated with care. Its obvious that on the graphene lattice, \(\delta _{1,2,3}\) span the space, so any vector can be written in terms of them, but more importantly, any vector coupling an \(A\) -site and a \(B\) -site can be written in the form \[ v _{A B}=n_{1} \delta _{1}+n_{2} \delta _{2}+n_{3} \delta _{3} \] where \(n_{1,2,3}\) are integers that crucially satisfy \[ n_{1}+n_{2}+n_{3}=1 \bmod 3 \] The preceding is true because of the chosen vector orientation of \(\delta _{1,2,3}\). We write each \(n\) as a multiple of 3 plus a remainder: \[ n_{i}=3 p_{i}+a_{i}, \quad i=1,2,3 \] where, of course, \(a_{i} \in[0,1,2]\) and, crucially, because of equation (7.30), \[ a_{1}+a_{2}+a_{3}=-3\left(p_{1}+p_{2}+p_{3}\right)+1=1 \bmod 3 \] Per \(C_{3}\) symmetry, the off-diagonal matrix element coupling sites \(A\) and \(B\) is a cyclic permutation of \(e^{i k \cdot v _{A B}}\) over the cyclic permutation of \(\delta\) 's: \[ e^{i k \cdot\left(n_{1} \delta _{1}+n_{2} \delta _{2}+n_{3} \delta _{3}\right)}+e^{i k \cdot\left(n_{1} \delta _{2}+n_{2} \delta _{3}+n_{3} \delta _{1}\right)}+e^{i k \cdot\left(n_{1} \delta _{3}+n_{2} \delta _{1}+n_{3} \delta _{2}\right)} \] We also have \(a _{1}= \delta _{1}- \delta _{3}, \quad a _{2}= \delta _{2}- \delta _{3}\) and so the above matrix element becomes: \[ \begin{array}{l} \exp \left(i k \cdot\left(n_{1} \delta _{1}+n_{2} \delta _{2}+n_{3} \delta _{3}\right)\right)+\exp \left(i k \cdot\left(n_{1} \delta _{2}+n_{2} \delta _{3}+n_{3} \delta _{1}\right)\right)+\exp \left(i k \cdot\left(n_{1} \delta _{3}+n_{2} \delta _{1}+n_{3} \delta _{2}\right)\right) \\ =\exp \left(i k \cdot \delta _{3}\right)\left[\exp \left(i k \cdot\left(3 p_{1} a _{1}+3 p_{2} a _{2}\right)\right) \exp \left(i k \cdot\left(a_{1} a _{1}+a_{2} a _{2}\right)\right)\right. \\ \quad+\exp \left(i k \cdot\left(3 p_{1} a _{2}+3 p_{3} a _{1}\right)\right) \exp \left(i k \cdot\left(a_{1} a _{2}+a_{3} a _{1}\right)\right) \\ \left.+\exp \left(i k \cdot\left(3 p_{2} a _{1}+3 p_{3} a _{2}\right)\right) \exp \left(i k \cdot\left(a_{2} a _{1}+a_{3} a _{2}\right)\right)\right] \end{array} \] We now particularize at \(k = K -\) the point for which the nearest-neighbor matrix element vanishes. Since \[ e^{i 3 K \cdot a _{i}}=\left(e^{i K \cdot a _{i}}\right)^{3}=1 \] for \(i=1,2\), we see that the off-diagonal matrix element at \(K\) is \[ \begin{array}{c} e^{i k \cdot \delta_{3}}\left(e^{i K \cdot\left(a_{1} a _{1}+a_{2} a _{2}\right)}+e^{i K \cdot\left(a_{1} a _{2}+a_{3} a _{1}\right)}+e^{i K \cdot\left(a_{2} a _{1}+a_{3} a _{2}\right)}\right) \\ =e^{i k \cdot \delta_{3}}\left(z^{2 a_{1}+a_{2}}+z^{2 a_{3}+a_{1}}+z^{2 a_{2}+a_{3}}\right) \end{array} \] Because we know \(a_{i} \in 0,1,2\) and \(a_{1}+a_{2}+a_{3}=1 \quad\) mod \(\quad 3\), we can have only three distinct combinations of the numbers \(a_{i} .\) First, \(a_{1}=0, a_{2}=0, a_{3}=1:\) the matrix element is \(z^{0}+z^{2}+z=0 .\) Second, \(a_{1}=1, a_{2}=1, a_{3}=2\) : the matrix element is \(z^{3}+z^{5}+z^{4}=1+z^{2}+z=0\) Third, \(a_{1}=0, a_{2}=2, a_{3}=2\) : the matrix element is \(z^{2}+z^{4}+z^{6}=z^{2}+z+1=0\). Permutations of these three values of \(a\) obviously do not make a difference.
We hence proved that the matrix elements for any \(A\) to \(B\) hopping vanish at \(k = K\) when \(C_{3}, T\), and \(I\) are respected. We can say that the Dirac nodes are locally protected by \(T\) and \(I\) (we will see that two of them can annihilate and gap out, so they are not globally protected), whereas they are globally protected by \(C_{3}\) symmetry.
2 Breaking of \(C_{3}\) Symmetry
When \(C_{3}\) symmetry is broken (we assume \(T\) and \(I\) are unbroken), there is nothing that stops the Dirac points from moving away from \(K , K ^{\prime}\). It is easy to see that in the limit of high anisotropy, the graphene Hamiltonian is fully gapped. For example, for \(t_{3} \gg t_{1}, t_{2}\), we have \(h(k)=-t_{3} \sigma_{x}\), with eigenvalues \(E=\pm\left|t_{3}\right|\) and a gap equal to \(2\left|t_{3}\right| .\) It is hence clear that the absence of \(C_{3}\) symmetry spoils the protection of Dirac modes. However, we know that for small anisotropy \(t_{c} \approx t_{a} \approx t_{b}\), the Dirac nodes are stable because of the proof in section 7.4.1. How do the Dirac nodes gap?
In the anisotropic case, Dirac nodes come at points \(K _{0}(\neq K )\) and \(- K _{0}\) related by TR invariance. As such, as long as \(K _{0} \neq- K _{0}\left(\bmod b _{1,2}\right)\), the Dirac nodes are stable and cannot gap because of the vorticity that they carry. However, when: \[ K _{0}=- K _{0}\left(\bmod b _{1,2}\right) \] the two Dirac points of vorticity \(\pm \pi\) meet up and can annihilate, as in figure \(7.4\), at a TRinvariant point. The Hamiltonian at the point at which the Dirac nodes annihilate is quadratic (or flat) in the momentum multiplying one Pauli matrix and linear in the other. This has to be so for several reasons. First, the dispersion cannot be linear in both Pauli matrices because the wavefunction would then exhibit vorticity. We know this is not the case because the Dirac nodes annihilate, and the end result is a gapped Hamiltonian with zero vorticity. The second way of seeing this is the following: with \(T\) and \(I\), the Hamiltonian for graphene takes the form \[ h( k )=d_{x}( k ) \sigma_{x}+d_{y}( k ) \sigma_{y}+\epsilon( k ) \] with no \(\sigma_{z}\) term. The \(\epsilon( k )\) term is diagonal and, hence, does not influence the spectral gap, so we drop it in the further discussion. The time reversal \(\left(h(- k )=h^{*}( k )\right)\) implies \(d_{x}(- k )=d_{x}( k )\), \(d_{y}(- k )=-d_{y}( k )\), and \(\epsilon( k )=\epsilon(- k ) .\) At a \(T\) -invariant point \(K_{0}\), where the two Dirac nodes finally touch as anisotropy is increased, we have \(K _{0}=- K _{0}\left(\bmod b _{1,2}\right)\), which means \[ d_{x}\left( K _{0}\right)=d_{x}\left(- K _{0}\right) ; \quad d_{y}\left( K _{0}\right)=d_{y}\left(- K _{0}\right)=-d_{y}\left( K _{0}\right)=0 . \] We now see that only \(d_{x}\) is present in the Hamiltonian at \(k = K _{0}\). Infinite-simally away from a \(T\) -invariant point \(k = K _{0}+\delta k\), we have \[ d_{x}\left( K _{0}+\delta k \right)=d_{x}\left(- K _{0}-\delta k \right)=d_{x}\left( K _{0}-\delta k \right) . \] gives Fourier-transforming to first order gives \[ d_{x}\left( K _{0}\right)+\left.\delta k _{i} \frac{\partial d_{x}}{\partial k_{i}}\right|_{k=K_{0}}=d_{x}\left( K _{0}\right)-\left.\delta k _{i} \frac{\partial d_{x}}{\partial k}\right|_{k=K_{0}}, \] so the first order is missing: \(\left.\frac{\partial d_{x}}{\partial k_{i}}\right|_{k=K_{0}}=0\) and \[ \left.d_{y}\left( K _{0}+\delta k \right) \approx \delta k _{i} \frac{\partial d_{y}}{\partial k_{i}}\right|_{k=K_{0}} \] The Hamiltonian is then \[ \left.h\left( K _{0}+\delta k \right) \approx \delta k _{i} \frac{\partial d_{y}}{\partial k_{i}}\right|_{k=K_{0}} \sigma_{y}+\left.\delta k _{i} \delta k _{j} \frac{\partial^{2} d_{x}}{\partial k_{i} \partial k_{j}}\right|_{k=K_{0}} \sigma_{x}+d_{x}\left( K _{0}\right), \] with gap \(2\left|d_{x}\left( K _{0}\right)\right| .\) When the Dirac nodes are touching, the gap is \(2\left|d_{x}\left( K _{0}\right)\right|=0 .\) At the touching point, the Hamiltonian is linear in one direction but quadratic in the other. We want to stress that the preceding scenario for the gapping of Dirac nodes is absolutely necessary: for high anisotropy, the band structure is gapped,and for low anisotropy it is gapless, so there must be a place in between where the gap opens. Because the gapless Dirac nodes are locally stable (as proved earlier), to open a gap we need to bring them together first and annihilate them. This happens only at a \(T\) -invariant point.
Edge Modes of the Graphene Layer
So far in this chapter we have analyzed graphene with periodic boundary conditions and have focused on the symmetries of the problem and the stability of Dirac points. It turns out that graphene exhibits spectacular physics not only for the bulk (periodic boundary conditions), but also for the edge. We now focus on the new physics that arises in graphene once edges are placed on the sample. We will see that a careful analysis of this situation strongly hints to the existence of the first topological insulator, the Chern insulator.
We analyze one graphene layer with edges on it. Instead of the transfer-matrix method, we will use the direct diagonalization of the Hamiltonian. As before, the isotropic Hamiltonian is \[ H=-t \sum_{i \neq j} c_{i}^{\dagger} c_{j}+\text { h.c. } \] with \(i, j\) on the hexagonal lattice and where we assume spinless fermions. For fermions with spin, we add an extra spin quantum number. We consider open-boundary conditions in the \(y\) -direction and periodic-boundary conditions in the \(x\) -direction as in figure \(7.5\), with "zigzag" chains (shown in the figure) oriented at an angle of \(\pi / 6\) angle with respect to the horizontal, represented by the color red in figure \(7.5\). Because we have distinct \(A\) - and \(B\) sites in the unit cell, we can count them along one red chain (fig. 7.5) as even and odd\(2 j-1,2 j, 2 j+1,2 j+2, \ldots-\) sites. If we have periodic boundary conditions in the \(x\) -direction, we make the Fourier transform \[ c_{x, y}^{\dagger}=\frac{1}{N_{x}} \sum_{k_{x}} e^{i k_{x} x} c_{k_{x}, y} \] where \(N_{x}\) is the number of sites in the \(x\) -direction. The diagonalization of the Hamiltonian proceeds easily: \[ H=-t \sum_{j}\left(c_{2 j, k_{x}}^{\dagger} c_{2 j+1, k_{x}}+\text { h.c. }\right)-t \sum_{j}\left(c_{2 j}^{\dagger} c_{2 j-1}\left(e^{i k_{x} \frac{a_{1}}{2}}+e^{-i k_{x} \frac{a_{1}}{2}}\right)+\text { h.c. }\right), \] where \(a_{1}=\frac{a \sqrt{3}}{2}\) and \(a\) is the graphene lattice constant. We make a gauge transformation only on the odd sites \(c_{2 j-1} \rightarrow c_{2 j-1} e^{i k_{x} a_{1} 2}\) in the same way that we did before for the periodic Hamiltonian, obtaining \[ H=-t \sum_{j}\left(c_{2 j, k_{x}}^{\dagger} c_{2 j+1, k_{x}}+c_{2 j, k_{x}}^{\dagger} c_{2 j-1, k_{x}}\left(1+e^{-i k_{x} a_{1}}\right)+\text { h.c. }\right) \] Hence the hopping, in ascending order, from odd to even sites \((2 j-1 \rightarrow 2 j)\) is \(t^{\prime}=t(1+\) \(\left.e^{-i k_{x} a_{1}}\right)=t \cdot t_{k}\) with \(t_{k}=1+e^{-i k_{x} a_{1}}\) as in figure 7.6. At \(k_{x}=\pi / a_{1}\), the hopping is \(t^{\prime}=0\), and hence we have two different situations, depending on whether the zigzag chain contains an even or odd number of sites. We analyze these cases separately next.
1 Chains with Even Number of Sites
For a chain with an even number of sites \(2 L_{y}\) (an integer) in the \(y\) -direction, where site 1 is on an \(A\) sublattice (these conditions guarantee that site \(2 L_{y}\) is on an \(B\) sublattice), we have that the two sites at the end of the chain do not couple with the dimerized bulk at \(k_{x}=\pi / a_{1} .\) In this case, we see that the states \[ \left|\psi_{1}\right\rangle=c_{1, \pi / a_{1}}^{\dagger}|0\rangle, \quad\left|\psi_{2}\right\rangle=c_{2 L_{y}, \pi / a_{1}}^{\dagger}|0\rangle \] have \(E=0\) energy in the Hamiltonian and are, hence, zero edge modes. At \(k_{x}=\pi / a_{1}\), these two states are exactly at energy \(0 ;\) the bulk is completely dimerized, as in figure \(7.7\). Away from the point \(k_{x}=\pi / a_{1}\), the edge modes have dispersion, but in the limit \(L_{y} \rightarrow \infty\), the edge modeenergies go to zero. This can be seen by taking the single-particle wavefunction, \(|\psi\rangle=\sum_{i} a_{i}|i\rangle\), and diagonalizing the Hamiltonian to obtain \[ t_{k}^{*} a_{2 j-1}+a_{2 j+1}=\frac{E}{-t} a_{2 j}, \quad t_{k} a_{2 j}+a_{2 j-2}=\frac{E}{-t} a_{2 j-1}, \] Two (quasi) zero-energy modes are present.
- First, the mode localized close to the \(j=1\) edge: for \(E \approx 0\) we obtain
\[ \frac{a_{2 j+1}}{a_{2 j-1}}=-t_{k}^{*} ; \quad a_{2 j}=0 \text { for all } j \]
with the solution \[ a_{2 j+1}=\left(-t_{k}^{*}\right)^{j} a_{1} \] It is rather clear that this is an edge mode as long as \(t_{k} \neq 1\). Normally, we would say that both \(t_{k}>1\) and \(t_{k}<1\) are edge modes-localized on different edges-but this is not true. The reasoning is the following: we have assumed that the energy of the edge mode is \(E=0\) everywhere in \(k_{x}\) -space, even though, for modes away from \(k_{x}=\pi\), the energy disperses on the order of \(\exp \left(-L_{y}\right)\) (infinitesimal dispersion as we go to the thermodynamic limit but, nonetheless, still dispersion). The reason for this infinitesimal dispersion is the \(2 L_{y}\) site. The site \(2 L_{y}+1\) does not exist, so \(a_{2 L_{y}+1}=0 .\) As such, we have that, for \(E=0\) and \(a_{2 j}=0\) for all \(j\), we get \[ t_{k}^{*} a_{2 L_{y}-1}+a_{2 L_{y}+1}=\frac{E}{-t} a_{2 L_{y}}=0=t_{k}^{*} a_{2 L_{y}-1} \] Unless \(t_{k}=0\) (which happens only at \(k_{x}=\pi\), equation (7.52) requires \(a_{2 L_{y}-1} \rightarrow 0\). This means that the solution \(a_{2 j+1}=\left(-t_{k}^{*}\right)^{j} a_{1}\) works only for \(t_{k}<1\), which is the same as \(\frac{2 \pi}{3} \leq k_{x} a_{1} \leq \frac{4 \pi}{3}\). These are exactly the Dirac points in graphene in terms of \(k_{x} .\) This represents an edge mode localized on the \(j=1\) edge of energy \(E \rightarrow 0\) in the thermodynamic limit. Second, the mode localized on the \(j=2 L_{y}\) edge is obtained if \[ \frac{a_{2 j-2}}{a_{2 j}}=-t_{k} ; \quad a_{2 j+1}=0 \forall j \] which has as solution \[ a_{2 L_{y}-2 j}=\left(-t_{k}\right)^{j} a_{2 L_{y}} \] Again, we have that \(-t_{k} a_{2}=0\), so \(a_{2} \rightarrow 0\), which means \(\left|t_{k}\right|<1\), with identical conditions on \(k\), as before. This is a mode localized at the \(j=2 L_{y}\) edge of energy \(E \rightarrow 0\) in the thermodynamic limit. The discussion of the edge modes just presented is reflected in the form of the Hamiltonian: \[ h=-t\left[\begin{array}{llllllll} 0 & t_{k} & 0 & 0 & 0 & 0 & 0 & 0 \\ t_{k}^{*} & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & t_{k} & 0 & 0 & 0 & 0 \\ 0 & 0 & t_{k}^{*} & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & t_{k} & 0 & 0 \\ 0 & 0 & 0 & 0 & t_{k}^{*} & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & t_{k} \\ 0 & 0 & 0 & 0 & 0 & 0 & t_{k}^{*} & 0 \end{array}\right], \] with many sites in between, whose (almost) zero-energy eigenstates are \[ \psi_{1}=\left(a_{1}, 0, a_{3}, 0, a_{5}, 0, \ldots, a_{2 L_{y}-3}, 0, a_{2 L_{y}-1}, 0\right) \] where \(t_{k}^{*} a_{2 L_{y}-1} \rightarrow 0\) in order to have zero energy and \[ \psi_{1}=\left(0, a_{2}, 0, a_{4}, 0, a_{6}, \ldots, 0, a_{2 L_{y}-2}, 0, a_{2 L_{y}}\right) \] where \(t_{k} a_{2} \rightarrow 0\) to have zero energy.
We now have a picture for the edge modes in the thermodynamic limit. In the bulk, the bands will be the projection onto the \(k_{x}\) -axis of all the band dispersion on \(k_{y}\), so the "bulk" bands will be fat. They will go to zero energy at the Dirac cones, or points. At every point in between the Dirac cones, there will be two, nondispersive edge modes at zero energy, which connect them as in figure \(7.8\). They are localized on one end or the other of the sample, and any local perturbation that could couple the modes on opposite edges cannot lift the degeneracy because it has to go through the whole sample to couple them.
For a chain with an even number of sites \(2 L_{y}\) in the \(y\) -direction, and with site 1 on the \(B\) sublattice, site \(2 L_{y}\) is guranteed to be on the \(A\) sublattice). Going through a similar calculation, we find edge modes from \(\frac{4 \pi}{3}\) to \(2 \pi\) and then continuing from 0 to \(\frac{2 \pi}{3}\). The edge modes will be at asymptotically zero energy and can be seen in figure \(7.9\).
2 Chains with Odd Number of Sites
For a chain with an odd number of sites, \(2 L_{y}-1\) in the \(y\) -direction, if site 1 is on the \(A\) sublattice, then site \(2 L_{y}-1\) is also on an \(A\) sublattice. This case, is of course, identical to the reflected case in which site 1 is site \(B\) and the site at \(2 L_{y}-1\) is also \(B\). In this case, it is easiest to use the matrix form of the Hamiltonian to immediately see the edge modes. As opposed to the case of an even number of \(y\) sites, the Hamiltonian is missing the last column and row: \[ h=-t\left[\begin{array}{ccccccc} 0 & t_{k} & 0 & 0 & 0 & 0 & 0 \\ t_{k}^{*} & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & t_{k} & 0 & 0 & 0 \\ 0 & 0 & t_{k}^{*} & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & t_{k} & 0 \\ 0 & 0 & 0 & 0 & t_{k}^{*} & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{array}\right], \] with many sites in between. Notice that in this case, the wavefunction \[ \psi_{1}=\left(a_{1}, 0, a_{3}, 0, a_{5}, 0, \ldots, a_{2 L_{y}-3}, 0, a_{2 L_{y}-1}\right) \] with \[ a_{2 j+1}=\left(-t_{k}^{*}\right)^{j} a_{1} \] is always an exact energy solution of the Hamiltonian at any \(k\). For \(\left|t_{k}\right|<1\) -i.e., for \(k\) between \(\frac{2 \pi}{3}\) and \(\frac{4 \pi}{3}\), the edge state is localized on the 1-site, whereas for \(\left|t_{k}\right|>1\) -i.e., in the remainder of the \(BZ\) - the state is localized on the \(\left(2 L_{y}-1\right)\) -site. This is clearly exemplified in figure \(7.10 .\) The edge mode in this case is exactly at zero energy, unlike in the previous case, in which it was split from zero energy by thermodynamically exponential (in \(L_{y}\) ) spitting. Notice that the other state that was available in the \(2 L_{y}\) chain, \[ \psi_{1}=\left(0, a_{2}, 0, a_{4}, 0, a_{6}, \ldots ., a_{2 L_{y}-4}, 0, a_{2 L_{y}-2}, 0\right) \] cannot be a zero-energy state of the system because it would have to satisfy, at the same time, the conditions \[ t_{k} a_{2} \rightarrow 0 ; \quad a_{2 j}=t_{k} a_{2 j+2} ; \quad a_{2 L_{y}-2} \rightarrow 0 \] whose unique solution is the zero wavevector. The degeneracy of the zero mode at each point, including the special \(T\) -invariant points, is \(1 .\) This clearly shows that we are dealing with systems with spinless fermions for which time reversal does not induce Kramers' degeneracy.
3 Influence of Different Mass Terms on the Graphene Edge Modes
We now ask what happens to the edge modes upon opening a gap at the Dirac nodes. In the spinless case, several possibilities appear.
Case 1:
First, we can add a term that gives different energies to lattice sites \(A\) and \(B\), as in the case of BN. As before, this term is \[ h(k)=m \sigma_{z} \] with \(m>0 ; m\) is called an inversion symmetry breaking mass, or Semenoff mass [39]. We ask how the gap opens. The edge modes that terminate (have high amplitude) on sites \(A\) will go up in energy, whereas the edge modes that terminate on sites \(B\) will go down in energy. We hence have the situations in figures 7.11, 7.12, \(7.13\) and 7.14, depending on whether the terminations are \(A-B, B-A, A-A\), or \(B-B\).
Case 2:
We now add a term that breaks time reversal and gaps the system. This is called a Haldane mass; because we do not yet know how to add it on the lattice, we add it at the level of the Dirac fermion mass, which, per sections \(7.3 .2\) and 7.3.1, means we need to add a mass of opposite sign to the Dirac fermions at \(K\) and \(K ^{\prime}\). This mass term is a \(k\) -space-dependent term, which cannot be looked at as a term that adds energy on localized sites. It is not an on-site term. As such, the edge modes will do completely different things than in the case of a BN-type (inversion-breaking) mass. We have seen that in the inversion-breaking case, the edge modes remain linked to the Dirac mass; i.e., the mass is identical at both cones, and the edge modes still connect the cones on the same side of the gap. In the \(T\) -breaking case, a similar thing happens, but because the mass is negative at one cone and positive at the other cone, the edge modes will connect one cone with another by crossing the gap, as in figures 7.15-7.18. The Hall conductance can be inferred if we know where the edge modes in the figure are situated (i.e., on which side of the Laughlin cylinder), but it can be only \(\pm 1\). Thus, it is clear that a topological insulator with nonzero Hall conductance and that maintains lattice translational symmetry exists. We will find one in the next chapter.
