Are the surface Fermi arcs in Dirac semimetals topologically protected?

Significance

In recent years there has been a surge of interest in quantum materials with topologically protected properties that are robust against the effects of disorder and other perturbations. An interesting example is the newly discovered three-dimensional analog of graphene, called Dirac semimetals. These were expected to have highly unusual conducting states on their surfaces. We show here that the surface states of Dirac semimetals do not exhibit the expected topological robustness and, quite generally, get deformed into states that are of a rather different character. Our theoretical results not only have conceptual importance in the field of topological quantum materials but also make clear predictions that can be tested in several experiments.

Abstract

Motivated by recent experiments probing anomalous surface states of Dirac semimetals (DSMs) Na₃Bi and Cd₃As₂, we raise the question posed in the title. We find that, in marked contrast to Weyl semimetals, the gapless surface states of DSMs are not topologically protected in general, except on time-reversal-invariant planes of surface Brillouin zone. We first demonstrate this finding in a minimal four-band model with a pair of Dirac nodes at $\mathit{k}\mathrm{=}\left(\mathrm{0,0},\mathrm{\pm }\mathit{Q}\right),$ where gapless states on the side surfaces are protected only near $k_z=\mathrm{0}\mathrm{.}$ We then validate our conclusions about the absence of a topological invariant protecting double Fermi arcs in DSMs, using a K-theory analysis for space groups of Na₃Bi and Cd₃As₂. Generically, the arcs deform into a Fermi pocket, similar to the surface states of a topological insulator, and this pocket can merge into the projection of bulk Dirac Fermi surfaces as the chemical potential is varied. We make sharp predictions for the doping dependence of the surface states of a DSM that can be tested by angle-resolved photoemission spectroscopy and quantum oscillation experiments.

Following the theoretical prediction and experimental discovery of topological insulators (1, 2) in the past decade, there has been an explosion of interest in understanding the role of topology in various quantum states of matter. An important set of questions concerns the topological properties of gapless Fermi systems (3–8). Of particular interest are 3D semimetals where the bulk electronic dispersion exhibits point nodes at the Fermi level and the low-energy physics are effectively described by a Weyl or Dirac Hamiltonian (9).

A striking feature of 3D Weyl semimetals (WSMs), which necessarily break either time-reversal or inversion symmetry, is the existence (5) of topologically protected surface “Fermi arcs.” Here the Fermi contour in the surface Brillouin zone breaks up into disconnected pieces, which connect the projection of two Weyl nodes with opposite chirality. The Fermi arcs have been recently observed in angle-resolved photoemission spectroscopy (ARPES) studies of noncentrosymmetric TaAs (10–15).

Here we focus on another outstanding example, the Dirac semimetal (DSM), which is the 3D analog of graphene. In the presence of both time-reversal and inversion symmetries, the electronic excitations near each node are described by a four-component Dirac fermion, and a dispersion that is linear in all directions in $\mathbf{k}$ space (16–18). ARPES has clearly observed linearly dispersing bands near Dirac nodes in two DSM materials, Na₃Bi (19, 20) and Cd₃As₂ (21–23). The signatures of Fermi arcs by ARPES in Na₃Bi (24) and by their peculiar quantum oscillations (25) in Cd₃As₂ (26) have recently been reported. Although DSM can in principle appear in a system with spin rotational symmetry (27), here we focus on DSMs in spin–orbit-coupled systems that are more closely related to material realizations.

We can understand the Dirac fermions as two degenerate Weyl fermions with opposite chirality, where crystal symmetries forbid the two Weyl nodes from hybridizing and opening up a gap at each Dirac point (16, 28). Given this picture of the bulk, it is natural to expect the surface states in a DSM (17, 18) as two copies of the chiral Fermi arc on the WSM surface: i.e., the “double Fermi arcs” shown schematically in Fig. 1A.

Fig. 1.

(A) Schematic k-space picture of a Dirac semimetal showing Dirac nodes along the $k_z$ axis in bulk BZ and possible double Fermi arcs on the surface BZs, shown as blue squares. Note that surfaces perpendicular to the z axis have no arcs. A 2D slice of the bulk BZ perpendicular to the $k_z$ axis is shown as a green square, which projects to a green dashed line on a side surface. (B) Surface spectral density of model in [1], which clearly shows the existence of double Fermi arcs on the (100) surface. (C and D) Continuous deformation of double Fermi arcs on the (100) surface by adding the perturbation $\delta H_4\left(\mathbf{k}\right)$ to [1]. C shows the effect at $m\prime =-0.4t,$ whereas D corresponds to $m\prime =-0.8t,$ showing that the Fermi arcs are progressively destroyed by increasing strength of perturbation. The red solid circles in B–D correspond to the projection of bulk nodes and we set $E_F=0$ to line up the Fermi level with bulk Dirac nodes. (E and F) Electron-doped systems with $m\prime =-0.8t$ (case in D) by raising the Fermi energy to $E_F=0.1t$ and $E_F=0.15t,$ respectively. The large red blobs in E and F mark the projection of bulk states onto the surface BZ. The Fermi contour of the surface states is disconnected from the blobs in E, whereas it merges into the bulk states in F.

In this paper, we address the important question of whether the double Fermi arcs on the surface of Dirac semimetals are topologically protected. If they are, what is the associated topological invariant? The Fermi arcs in WSMs are robustly protected for the following reason: Each plane that lies between a pair of Weyl nodes in momentum space, perpendicular to the separation between them, has an integer Chern number (5, 29, 30) associated with a quantum Hall effect. The chiral edge modes of these momentum-space Chern insulators give rise to robust surface Fermi arcs, which end at the projection of bulk Weyl nodes. Unlike the WSM, the bulk–boundary correspondence in topological phase (8) provides no obvious answer for DSMs, because the stability of bulk Dirac nodes requires crystal rotation symmetries (17, 18) that are explicitly broken on open side surfaces. It was claimed (25), however, that the surface Fermi arcs in DSMs are perturbatively stable against a weak symmetry-breaking surface potential, and a strong surface potential can destroy them.

Here we provide a surprising answer to the question posed in the title. We show that the double Fermi arcs on Dirac semimetal surface are not topologically protected and can be continuously deformed into a closed Fermi contour without any symmetry breaking or bulk phase transition (Fig. 1). The resulting Fermi contours do have topological character, even though they are not as exotic as the Fermi arcs in WSMs. We show that the surface states of a DSM with an odd (even) number of Dirac node pairs are analogous to the surface states of a 3D strong topological insulator (TI) (weak TI or trivial insulator).

To make our results physically transparent, we first focus on the four-band minimal model for DSMs (17, 18, 31) and present simple arguments for lack of topological protection for surface Fermi arcs. We then show numerical results for ARPES spectral functions for the four-band model that demonstrates how a bulk perturbation that respects all symmetries can destroy the Fermi arcs. Next, we go beyond the simple low-energy model and directly address the two DSM materials of experimental interest. We use K theory (4, 32–35) to classify the topological stability of surface states in the presence of time reversal, charge conservation, and space group symmetries in Dirac semimetals Cd₃As₂ and Na₃Bi. This classification indicates that topological protection for gapless surface states on side surfaces (that do not intersect the $\widehat{\mathit{z}}$ axis) of Dirac semimetals exists only in the high-symmetry plane $k_z=0.$ We conclude by examining the experimental implications of our results. We discuss how to understand the existing ARPES and quantum oscillation measurements in light of our results on the lack of topological protection for Fermi arcs, and we make sharp predictions for testing our conclusions.

Minimal Model

We begin with a simple four-band model for a Dirac semimetal with fourfold rotational symmetry along the $\widehat{z}$ axis defined by the Hamiltonian:

$$\begin{array}{c}H\left(\mathbf{k}\right)={\epsilon }_{\mathbf{k}}^0+\left[t\left(\cos k_x+\cos k_y-2\right)+t_z\left(\cos k_z-\cos Q\right)\right]{\tau }_z\\ +\hspace{0.17em}\lambda \sin k_x{\sigma }_x{\tau }_x+\lambda \sin k_y{\sigma }_y{\tau }_x.\end{array}$$ [1]

The Pauli matrices $\overrightarrow{\sigma }$ act on spin and $\overrightarrow{\tau }$ in orbital space, t and $t_z$ are hopping amplitudes, and λ is spin–orbit coupling (SOC). Ignoring ${\epsilon }_{\mathbf{k}}^0$ for the moment, there is a gap everywhere in the Brillouin zone (BZ) except at two points $\mathbf{k}\mathrm{=}\left(\mathrm{0,0},\pm Q\right),$ where the low-energy spectrum is described by linearly dispersing Dirac fermions.

We can always add to Eq. 1 an ${\epsilon }_{\mathbf{k}}^0$ that preserves all of the symmetries and vanishes at the Dirac nodes; e.g., ${\epsilon }_{\mathbf{k}}^0=t_1\left(\cos k_z-\cos Q\right)+t_2\left(\cos k_x+\cos k_y-2\right)$ or ${\epsilon }_{\mathbf{k}}^0=t_1\left(\cos k_z-\cos Q\right)\left(\cos k_x+\cos k_y\right).$ Although this term does not qualitatively change the Dirac spectrum in the bulk, it gives rise to a curvature of the Fermi arc in the surface BZ, as discussed below.

The symmetries of $\mathit{H}\left(\mathbf{k}\right)$ and their representations in terms of spin and orbital operators are as follows. Time reversal is implemented by the antiunitary operator $\mathrm{\Theta }=\mathrm{i}{\sigma }_y\cdot \mathcal{K},$ where $\mathcal{K}$ is complex conjugation and ${\mathrm{\Theta }}^2=-1.$ Inversion I corresponds to $U_I\mathrm{=}{\tau }_z,$ with the two orbitals having even and odd parity. Twofold rotation $C_{2,x\left(y\right)}$ about the $\widehat{x}$ (or $\widehat{y}$) axis is implemented by $U_{c_{2,x\left(y\right)}}=\text{i}{\sigma }_{x\left(y\right)},$ and n-fold rotation $C_{n,z}$ by $U_{c_{n,z}}=\exp \left(\text{i}\pi {\sigma }_z\left(1-2{\tau }_z\right)/n\right),$ where $n=4$ here. $H\left(\mathbf{k}\right)$ also has mirror reflection symmetries with respect to the $j=x,y,z$ planes, with $k_j\to \mathrm{-}{k}_j,$ which is implemented by $U_{R_j}=U_{c_{2,j}}{U}_I=\text{i}{\sigma }_j{\tau }_z.$

Instability of Surface Fermi Arcs

Surfaces with a normal not parallel to the $\widehat{z}$ axis are expected to support gapless states with Fermi arcs, shown schematically in Fig. 1A for side surfaces. We address the lack of topological protection for the Fermi arcs of $H\left(\mathbf{k}\right)$ from three different points of view: (i) First, we examine the stability of 1D edge states that exist on 2D slices of BZ at a fixed, generic value of $k_z.$ Here we define a generic $k_z$ to be $k_z\notin \left\{\pm Q,0,\pi \right\},$ so that a fixed-$k_z$ plane will neither contain Dirac nodes nor have the additional symmetry arising from time-reversal invariance. (ii) Next, we prove the existence of an extra mass term in the gapped Dirac Hamiltonian that describes the plane with a generic $k_z.$ (iii) Finally, we show how adding fully symmetric, bulk perturbations to the 3D Hamiltonian $H\left(\mathbf{k}\right)$ can destroy the Fermi arcs, without destroying the bulk Dirac nodes. The projection of these bulk nodes onto the surface does remain gapless, as do the 2D slices located at $k_z=0$ as they have extra (time-reversal) symmetry.

Without loss of generality we consider a (100) surface. This surface preserves only the following symmetries: time reversal $\mathrm{\Theta },$ mirror reflections $R_y,R_z,$ and their combinations. Therefore, a fixed-$k_z$ plane in a system with a (100) surface respects only two symmetry operations: mirror reflection $U_{R_y}=\text{i}{\sigma }_y{\tau }_z$ with respect to (w.r.t.) the $\widehat{y}$ plane and the combination ${\mathrm{\Theta }}_{R_z}=\mathrm{\Theta }{U}_{R_z}=\text{i}{\sigma }_x{\tau }_z\cdot \mathcal{K}$ of time reversal and mirror w.r.t. the $\widehat{z}$ plane. Note that ${\mathrm{\Theta }}_{R_z}^2=1.$ Under these symmetries the Hamiltonian at fixed $k_z,$ denoted by $\tilde{\mathit{H}}\left(k_x,k_y\right),$ transforms as ${\mathrm{\Theta }}_{R_z}^{-1}\tilde{H}\left(k_x,k_y\right){\mathrm{\Theta }}_{R_z}=\tilde{H}\left(-k_x,-k_y\right)$ and $U_{R_y}^{-1}\tilde{H}\left(k_x,k_y\right)U_{R_y}\mathrm{=}\tilde{H}\left(k_x,-k_y\right).$

First (i), consider the surface BZ $\left(k_y,k_z\right)$ for a $\left(100\right)$ surface. If we look at a 1D slice at fixed $k_z,$ we might have expected gapless edge states described by $H_{\text{edge}}=\hslash v_F{\displaystyle {\sum }_{k_y}{\psi }_{k_y}^{\mathrm{†}}\left(k_y{\mu }_z-k_F\right){\psi }_{k_y}}.$ Here ${\psi }_{k_y}^T\mathrm{=}\left({\psi }_{R,k_y},{\psi }_{L,k_y}\right)$ is a two-component spinor of right (R) and left (L) movers and $\overrightarrow{\mathit{\mu }}$ are Pauli matrices in $\left(R,L\right)$ space. Note that the dispersion $v_F$ and the location of the Fermi arc $k_F$ are both, in general, $k_z$ dependent.

The two symmetry operations on the edge states are represented by ${\mathrm{\Theta }}_{R_z}=\text{i}{\mu }_x\cdot \mathcal{K}$ and $U_{R_y}=\text{i}{\mu }_y.$ We now add perturbation $\widehat{M}=m{\displaystyle {\sum }_{k_y}{\psi }_{k_y}^{\mathrm{†}}{\mu }_y{\psi }_{k_y}},$ which preserves both these surface symmetries: $\left[{\mathrm{\Theta }}_{R_z},{\mu }_y\right]=\left[U_{R_y},{\mu }_y\right]=0.$ Clearly $\widehat{M}$ is a mass term for $H_{\text{edge}}$ that destroys the gapless edge states.

Next (ii), we gain insight into the instability of surface states on a fixed-$k_z$ plane within a classification scheme (32–34) of topological insulators. For a fixed, generic $k_z,$ our system is a gapped 2D insulator described by a Dirac Hamiltonian

$${\tilde{H}}_D=k_x{\gamma }_x+k_y{\gamma }_y+m{\gamma }_0,$$ [2]

up to an additive constant term. Here ${\gamma }_x={\sigma }_x{\tau }_x,$ ${\gamma }_y={\sigma }_y{\tau }_x,$ and ${\gamma }_0={\tau }_z$ are Dirac matrices in the minimal model. Let us ask if we can add another mass term $m\prime {\gamma }_0\prime$ to ${\tilde{H}}_D,$ which preserves all of the symmetries and anticommutes with all of the Dirac matrices ${\gamma }_{x,y,0}.$ If we can do this, the $m>0$ phase can be continuously deformed into the $m<0$ phase without closing the gap at $m=0,$ and ${\tilde{H}}_D$ would be topologically trivial. Such a term indeed exists: ${\gamma }_0\prime ={\sigma }_z{\tau }_x$ commutes with ${\mathrm{\Theta }}_{R_z}$ and $U_{R_y}$ and anticommutes with ${\gamma }_{x,y,0}.$ Therefore, the surface states at generic $k_z$ are not topologically protected.

In contrast to generic $k_z,$ the time-reversal invariant planes $k_z=0$ and $k_z=\pi$ are gapped 2D insulators with the well-known ${\mathrm{\mathbb{Z}}}_2$ topological index associated with the quantum spin Hall (QSH) effect (1, 2). In the DSM materials Cd₃As₂ and Na₃Bi, there is a nonsymmorphic glide reflection $g_c$ containing half-translation along the $\widehat{z}$ axis, which serves as a mirror reflection for $k_z=0,\pi$ planes satisfying ${\left(g_c\right)}^2=-1\left(\mathrm{+}1\right)$ for $k_z=0\left(\pi \right).$ Generically such nonsymmorphic glide reflections guarantee that the ${\mathrm{\mathbb{Z}}}_2$ index must be trivial for the $k_z=\pi$ plane (36), which is elaborated below. Therefore, in these materials, only the $k_z=0$ plane can support a nontrivial ${\mathrm{\mathbb{Z}}}_2$ index. A simple calculation (37) shows that the $k_z=0$ plane is indeed a nontrivial QSH insulator, and gapless surface states at $k_z=0$ are topologically protected.

Finally (iii), we directly examine the gapless surface states of the 3D Hamiltonian [1] in a slab geometry with a surface at $x=0,$ which is translationally invariant along y and z. We numerically calculate the spectral density $A\left(\mathbf{k},\omega \right)=-\left(1/\pi \right)\text{ImG}\left(\mathbf{k},\omega +\text{i}{0}^{+}\right)$ associated with the electron Green function $G\left(\mathbf{k},\omega \right).$ Here we measure ω with respect to the zero chemical potential, which coincides with the bulk Dirac nodes. We show in Fig. 1 B–D the surface contributions to $A\left(\mathbf{k},0\right)$ as a function of in-plane momentum $\left(k_y,k_z\right).$ Note that ARPES probes precisely this spectral density multiplied by the Fermi–Dirac function.

In Fig. 1B, we plot the result for $H\left(\mathbf{k}\right)$ of Eq. 1 and clearly see the two Fermi arcs extending from one Dirac node to the other. In the absence of the ${\epsilon }_{\mathbf{k}}^0$ term of the form discussed above, $H\left(\mathbf{k}\right)$ has an (unnatural) “chiral symmetry” and the two Fermi arcs become “degenerate” and collapse to $k_y=0$ for $-Q<k_z<Q$ (Supporting Information). With ${\epsilon }_{\mathbf{k}}^0$ in place, we get the arcs shown in Fig. 1B.

Now we ask whether there is a perturbation that has the full symmetry, does not shift the bulk nodes, and yet has a ${\sigma }_{\alpha }\mathrm{\otimes }{\tau }_{\beta }$ structure that anticommutes with each term in $H\left(\mathbf{k}\right)$ (other than the ${\epsilon }_{\mathbf{k}}^0$ term, of course). Such a perturbation does exist and it must be of the form

$$\delta H_4\left(\mathbf{k}\right)=m\prime \left(\cos k_x-\cos k_y\right)\sin k_z{\sigma }_z{\tau }_x.$$ [3]

This perturbation destroys the Fermi arcs near the projections of the Dirac nodes at $k_z=\pm Q$ onto the surface BZ as shown in Fig. 1. We clearly see this in Fig. 1C, where $\delta H_4\left(\mathbf{k}\right)$ with $m\prime =0.4t$ deforms the Fermi arcs into a closed Fermi pocket. An increase in the strength of the perturbation to $m\prime =0.8t$ shrinks the arcs to even smaller pockets as shown in Fig. 1D. Whereas the surface Fermi arcs are progressively destroyed, the bulk Dirac nodes (and bulk states near $E_F=0$) are unaffected by these perturbations. We note that this perturbation of $\mathcal{O}\left(k^3\right)$ is higher order than that retained in usual $\mathbf{k}\cdot \mathbf{p}$ perturbation theory.

The minimal four-band model (31) is equivalent, up to a unitary transformation (Supporting Information), to the well-known effective $\mathbf{k}\cdot \mathbf{p}$ Hamiltonian around the $\mathrm{\Gamma }$ point for DSM materials (17, 18). Despite its simplicity, this model with $n=4$ -fold rotational symmetry about the $\widehat{z}$ axis captures the band inversion near the $\mathrm{\Gamma }$ point of the BZ in Cd₃As₂. A similar model can also be adopted for the case of Na₃Bi with $n=6$ -fold rotational symmetry (Supporting Information).

K Theory

Using K theory we next show that the surface states and associated Fermi arcs in Cd₃As₂ and Na₃Bi are not topologically protected. This mathematical framework has the great advantage of focusing only on the space group, $U\left(1\right)$ charge conservation, and time-reversal symmetry of a material, without the need for a low-energy effective model such as the one used in the analysis above. We thus obtain general and rigorous results that are directly applicable to real materials.

In the K-theory approach (32–34), the classification of distinct gapped symmetric phases is reduced to the following mathematical problem: What is the “classifying space” of a symmetry-allowed mass matrix for a generic Dirac Hamiltonian preserving certain symmetries? Different symmetric phases correspond to disconnected pieces of the classifying space, which cannot be continuously connected to each other without closing the bulk energy gap. Mathematically the group structure formed by these different phases is given by the zeroth homotopy group ${\pi }_0\left(S\right)$ of classifying space S. Here we only sketch the idea behind the calculations; detailed analyses are given in Supporting Information.

Na₃Bi.

The hexagonal DSM material Na₃Bi has a unit cell defined by $\mathbf{a}=a\left(\mathrm{1,0,0}\right),$ $\mathbf{b}=a\left(-1/2,\sqrt{3}/2,0\right),$ and $\mathbf{c}=c\left(\mathrm{0,0,1}\right).$ Its space group is centrosymmetric P6₃/mmc (no. 194) (17), with symmetry operations generated by sixfold screw $s_6,$ $\left(x,y,z\right)\to \left(x-y,x,z+1/2\right);$ glide reflection $g_c,$ $\left(x,y,z\right)\to \left(y,x,z+1/2\right);$ inversion I, $\left(x,y,z\right)\to \left(-x,-y,-z\right);$ and Bravais lattice translations $T_{a,b,c}.$

The two Dirac nodes (17) in Na₃Bi are located at ${\mathbf{Q}}_{\pm }=\pm Q_0\left(2\pi /c\right)\left(\mathrm{0,0,1}\right).$ We consider two different surfaces that can potentially contain nontrivial Fermi arcs: (i) a (010) surface spanned by the a and c axes and (ii) a (110) surface spanned by $\left(\mathbf{a}-\mathbf{b}\right)$ and c. In each case we proceed as follows. We begin by determining the reduced set of symmetries that are preserved in the presence of the surface. By analyzing how these symmetries act on a general Bloch Hamiltonian, we determine those symmetries that preserve a 2D plane in the BZ zone at a fixed value of $k_z\ne 0,\pi /c.$ All such planes, except those passing through the nodes at ${\mathbf{Q}}_{\pm },$ are fully gapped.

The basic strategy is to exploit the classification of gapped insulators described by Dirac Hamiltonians defined on these 2D planes embedded in the 3D Brillouin zone. We use the K-theory classification (32–34) of gapped free-fermion quantum phases by examining the combined algebra for the Dirac γ matrices and the symmetry operations and determining whether the classifying space for symmetry-allowed mass terms is connected or not.

Using this approach we conclude that a generic 2D plane in the BZ at fixed $k_z$ is topologically trivial and does not need to support protected surface states. The plane $k_z=\pi /c$ is also shown to be trivial, due to the presence of the nonsymmorphic glide reflection $g_c.$ Only the $k_z=0$ plane is found to have a nontrivial ${\mathrm{\mathbb{Z}}}_2$ QSH index.

Cd₃As₂.

The space group of the tetragonal DSM Cd₃As₂ is centrosymmetric $\text{I}{4}_1/\text{acd}$ (no. 142) (38). In addition to Bravais lattice translations, its symmetry operations are generated by fourfold screw rotation along the $\widehat{z}$ axis, $\left(x,y,z\right)\to \left(y,1/2-x,1/4+z\right);$ glide reflection $g_c,$ $\left(x,y,z\right)\to \left(x,-y,1/2+z\right);$ and inversion I, $\left(x,y,z\right)\to \left(-x,1/2-y,1/4-z\right);$ where ${\left(s_4\right)}^4={\left(g_c\right)}^2=-T_z.$ The Bravais lattice is expanded by $T_{\pm 1,\pm 1,\pm 1}\mathrm{\equiv }\left(\pm 1,\pm 1,\pm 1\right)/2$ similar to the body centered cubic (BCC) lattice. The two Dirac nodes in Cd₃As₂ are located along the $\widehat{z}$ axis.

The K-theory procedure is used to determine the topological nature of various 2D planes in the BZ at fixed $k_z,$ in a manner completely analogous to that described above, taking into account the symmetries of Cd₃As₂. The calculations are described in Supporting Information, and the conclusions are identical to the ones given above. Only the $k_z=0$ plane has a nontrivial ${\mathrm{\mathbb{Z}}}_2$ QSH index, and all other fixed $k_z$ planes are trivial.

Nature of Surface States

We next show that the closed Fermi pocket on a side surface is essentially the same as the surface Dirac fermion of a 3D strong TI. Consider a weak perturbation that breaks $C_n$ symmetry, but preserves time reversal and inversion. We require that the strength of this perturbation is smaller than the bulk gap at the time-reversal invariant momenta (TRIM) of the 3D BZ. Clearly this perturbation will gap out the Dirac nodes but will not affect parity eigenvalues of filled bands at the TRIM. Now that the bulk is a fully gapped band insulator with time-reversal symmetry, its ${\mathrm{\mathbb{Z}}}_2$ topological invariant is given by the number of odd-parity Kramers pairs in filled bands at TRIM as shown by ref. 37. Because the $k_z=0$ plane is a 2D QSH insulator and $k_z=\pi$ is a trivial 2D insulator, the gapped bulk must be in a 3D strong TI phase. In the process of turning on the perturbation, the surface states at $k_z=0$ remain gapless because time reversal is always preserved. This result demonstrates the equivalence between the closed Fermi pocket for DSMs and surface Dirac fermions of strong TIs.

An important conclusion from our analysis is that if the bulk band touching occurs at an odd number of pairs of Dirac nodes not located at TRIM (such as in Cd₃As₂ and Na₃Bi), then the surface states cannot be fully removed due to time-reversal symmetry. Let $N_p$ Dirac points be located in the top half of the bulk BZ ($0<k_z<\pi$) and their $N_p$ time-reversal counterparts be in the bottom half of the bulk BZ ($-\pi <k_z<0$). Based on our arguments, this will lead to $N_p$ closed Fermi pockets on a generic surface, which are centered around either $k_z=0$ or $k_z=\pi .$ If there are an odd number of Fermi pockets centered around $k_z=0,$ the gapped $k_z=0$ plane in the first BZ must correspond to a 2D QSH insulator with an odd number of helical edge states. Therefore, if $N_p$ is odd, $k_z=0$ and $k_z=\pi$ must have opposite $Z_2$ topological indexes, with one plane being a trivial 2D insulator and the other a QSH insulator. Hence the surface states of an odd-$N_p$ DSM cannot be fully gapped out, in complete analogy with a 3D strong TI. On the other hand, if $N_p$ is even, both the $k_z=0$ and $k_z=\pi$ planes share the same $Z_2$ index; i.e., they are both trivial 2D insulators or both QSH insulators. In the former case there are generally no surface states, whereas gapless surface states in the latter case can be gapped out by translation symmetry-breaking perturbations in analogy to a 3D weak TI. As a result, an “odd DSM” with an odd number of pairs of Dirac points must support time-reversal–protected gapless surface states, in contrast to an “even DSM,” whose surface states can either be fully gapped or need protection from crystal translation symmetry.

Experimental Implications

We have shown above that, in general, the surface states of Dirac semimetals Cd₃As₂ and Na₃Bi have a closed Fermi surface (Fig. 1C), in contrast to the double Fermi arcs that one might naively expect (Fig. 1B). We now address the important question of how one can experimentally distinguish between these two types of surface states.

Assume, first, for simplicity that the chemical potential is at the Dirac point (we discuss below the case when it is not). The qualitative differences between the two outcomes are then as follows. The double Fermi arc has a singular change in slope when two arcs meet at the projection of a Dirac node onto the surface BZ. In contrast, a Fermi contour has a smooth curvature everywhere in the surface BZ and it does not pass through the two points that are the projections of the bulk Dirac nodes. In addition, the wave function associated with the surface states merges with the bulk at the tip of the arc, namely the Dirac node; in contrast, the states associated with a regular Fermi contour are exponentially localized at the surface. All of these features could, in principle, be used to distinguish between the two scenarios, using sufficiently high-resolution ARPES data with photon energy dependence used to separate bulk and surface contributions.

We next consider the very interesting quantum oscillation experiments on Cd₃As₂ thin films (26). The observed Shubnikov–de Haas oscillations seem consistent with “Weyl orbits,” which consist of the combination of bulk Landau levels (LLs) and surface Fermi arcs (25). How can we understand these experimental results, if the surface states have a closed Fermi pocket instead of double Fermi arcs?

First, both transport (26, 39, 40) and ARPES (22, 41) experiments confirmed that Cd₃As₂ is slightly n-doped, in the sense that Fermi energy $E_F$ lies above the Dirac points with $k_F\simeq 0.04\hspace{0.17em}{\AA }^{-1}.$ Therefore, the surface projection of the bulk electronic states, near each Dirac point, is in fact a small “blob” denoted by a red oval in Fig. 1 D–F. In this case, the surface states at the Fermi energy may (Fig. 1F) or may not (Fig. 1E) be connected with the bulk Fermi blobs, depending on $E_F.$ When the surface pocket merges into the bulk blobs, the magnetic orbits responsible for quantum oscillations must be the Weyl orbits proposed in ref. 25, which consist of both bulk LLs and surface Fermi arcs. Shubnikov–de Haas oscillations in ref. 26 suggest that this is the situation in Cd₃As₂. The available ARPES results in Na₃Bi (24) are also consistent with surface states merging into bulk projection as $k_z$ increases.

However, if we start to introduce p-type dopants to the system and lower the Fermi level, ultimately the surface pocket must get disconnected from the bulk blobs, as shown in Fig. 1 D and E, when the Fermi level is close enough to the Dirac points. In this case the surface Fermi pocket will provide a closed 2D magnetic orbit for cyclotron motion. This pocket is analogous to the 2D Dirac fermion on the surface of TIs and will lead to quantum oscillations similar to those observed in TIs (42–44), with the electron acquiring a Berry phase of π as it goes around the Fermi contour. As with any 2D Fermi surface, the quantum oscillation frequency $F_s$ has a $1/\cos \theta$ dependence on the angle θ between the magnetic field and surface normal direction. There is an important difference between the quantum oscillations in the two scenarios where the bulk and surface states are mixed and separated : In the latter case, $F_s$ has no dependence on the thickness of the sample, in sharp contrast to the Weyl orbits (25, 26) in the former one.

Therefore, quantum oscillation experiments at different doping levels provide a sharp distinction between double Fermi arcs and TI-like closed surface Fermi pockets, serving as an experimental test of our conclusion. There will be two frequencies in quantum oscillations on DSM thin films (25, 26): $F_b$ associated with bulk Fermi surfaces (due to deviation of $E_F$ from Dirac point) and $F_s$ related to the surface states. As we move $E_F$ toward bulk Dirac points by doping the system, $F_b$ will always decrease monotonically. As demonstrated in ref. 26, a triangle-shaped sample does not exhibit quantum oscillations with frequency $F_s,$ whereas a rectangular sample does, because the Weyl orbits depend on the thickness of the system along the field direction. Thus, in a triangle-shaped sample, double Fermi arcs cannot lead to quantum oscillation at $F_s$ independent of the Fermi level. On the other hand, as the surface pocket is disconnected from the bulk blobs by tuning $E_F$ close to the Dirac point (Fig. 1E), the frequency $F_s,$ with a 2D angular dependence, will show up even in a triangular sample. In particular, when the magnetic length $l_B=\sqrt{\hslash /eB}$ satisfies $k_d\gg l_B^{-1}$ where $k_d$ is the shortest distance between the surface pocket and the bulk blob in surface BZ, the closed surface pocket cannot tunnel into bulk LLs to form a closed Weyl orbit. When the Fermi level lies exactly at the Dirac point, $k_d\sim 0.04\hspace{0.17em}{\AA }^{-1}$ gives an estimate of $B<{10}^{2}T,$ which is within the experimental reach. Of course, very high-resolution ARPES studies can also directly reveal the separation of surface Fermi pocket from the bulk blobs as the photon energy is varied.

Concluding Remarks

Let us conclude by summarizing our main results. Each bulk Dirac node looks like two copies of a Weyl node, which naively suggests that the surface electronic structure of a DSM should look like two copies of that of the WSM, namely a double Fermi arc. We show here—using a simple four-band model and rigorous K-theory calculations—that the double Fermi arcs are not topologically protected, unlike the Fermi arc in a WSM. We find that an arbitrarily small bulk perturbation, which preserves the Dirac nodes and all of the symmetries, can lead to a deformation of the double Fermi arc into a more conventional Fermi contour. Nevertheless, we show that the surface states in a DSM cannot be completely destroyed, because there must be topologically protected states on the $k_z=0$ plane, similar to a 3D TI.

Finally, we explore in detail the experimental implications of our results. We show how we can reconcile our results on the lack of topological protection with existing experiments on Na₃Bi (24) and Cd₃As₂ (26). Although these have been interpreted as being consistent with Fermi arcs, we argue that they are actually a consequence of the mixing of bulk and surface states. We also propose hole-doping the system to disconnect the surface Fermi pocket and bulk states and suggest testable signatures for both ARPES and quantum oscillation experiments.

Appendix A: Symmetries of the Four-Band Hamiltonian

The symmetries of the Hamiltonian [1] are time-reversal $\mathrm{\Theta },$ inversion I, twofold rotations $C_{2,x}$ and $C_{2,y},$ and n-fold rotation $C_{n,z}$ ($n=4$ for Cd₃As₂ and $n=6$ for Na₃Bi), under which

$${\mathrm{\Theta }}^{-1}H\left(\mathbf{k}\right)\mathrm{\Theta }=H\left(-\mathbf{k}\right),$$ [S1]

$$U_I^{-1}H\left(\mathbf{k}\right)U_I=H\left(-\mathbf{k}\right),$$ [S2]

$$U_{c_{2,x}}^{-1}H\left(k_x,k_y,k_z\right)U_{c_{2,x}}=H\left(k_x,-k_y,-k_z\right),$$ [S3]

$$U_{c_{2,y}}^{-1}H\left(k_x,k_y,k_z\right)U_{c_{2,y}}=H\left(-k_x,k_y,-k_z\right),$$ [S4]

$$U_{c_{n,z}}^{-1}H\left(k_{\pm },k_z\right)U_{c_{n,z}}=H\left(e^{\pm \text{i}\left(2\pi /n\right)}{k}_{\pm },k_z\right),$$ [S5]

where $k_{\pm }=k_x\pm \text{i}{k}_y.$

The matrix representations of these operations in spin and orbital basis are described in the main text. The operators involved in the first four are standard.

We explain here why n-fold rotation about the z axis is represented by $U_{c_{n,z}}=\text{exp}\left(\text{i}\pi {\sigma }_z\left(1-2{\tau }_z\right)/n\right).$ On the $k_z$ axis the Hamiltonian commutes with $U_{C_n};$ i.e., $\left[H\left(k_z\right),U_{C_n}\right]=0.$ Hence a common set of states can diagonalize both operators. This result means that the eigenstates of the Hamiltonian can be labeled by eigenvalues α of $U_{C_n},$

$${\left(U_{C_n}\right)}^n=-1\Rightarrow {\alpha }_p=e^{\text{i}\left(2\pi /n\right)\left(p+1/2\right)},\hspace{1em}p=\mathrm{0,1},\mathrm{\cdots },n-1,$$ [S6]

which implies that $U_{C_n}$ is a diagonal matrix as $U_{C_n}=\text{diag}\left({\alpha }_p,{\alpha }_q,{\alpha }_r,{\alpha }_s\right).$ The rotation commutes with time-reversal symmetry $\left[\mathrm{\Theta },U_{C_n}\right]$ = 0, giving rise to ${\alpha }_r={\alpha }_p^{\mathrm{\ast }}$ and ${\alpha }_s={\alpha }_q^{\mathrm{\ast }}.$ Therefore, the rotation can be written as (28)

$$U_{C_n}=e^{\text{i}\left(\pi /n\right){\sigma }_z\left(p+q+1+\left(p-q\right){\tau }_z\right)}.$$ [S7]

Appendix B: Equivalence Between the Four-Band Model and the $k\mathrm{\cdot }p$ Hamiltonian for Cd₃As₂ and Na₃Bi

In this section we show that in the long-wavelength limit, the four-band model [1] can be related to the following effective Hamiltonian derived from $k\cdot p$ theory (45, 46) around the $\mathrm{\Gamma }$ point,

$${\widehat{H}}_{\mathrm{\Gamma }}\left(\mathbf{k}\right)={\mathit{ϵ}}_{\mathbf{k}}+\left(\begin{array}{cccc}M\left(\mathbf{k}\right)& \lambda k_{+}& 0& 0\\ \lambda k_{-}& -M\left(\mathbf{k}\right)& 0& 0\\ 0& 0& M\left(\mathbf{k}\right)& -\lambda k_{-}\\ 0& 0& -\lambda k_{+}& -M\left(\mathbf{k}\right)\end{array}\right)+O\left({\left|\mathbf{k}\right|}^2\right),$$ [S8]

which is widely adopted to describe topological properties of Dirac semimetal materials Na₃Bi (17) and Cd₃As₂ (18). This effective $k\cdot p$ model is written in the basis of $|S_{J=1/2},J_z=\frac{1}{2}\rangle ,|P_{J=3/2},J_z=\frac{3}{2}\rangle ,|S_{J=1/2},J_z=-\frac{1}{2}\rangle ,|P_{J=3/2},J_z=-\frac{3}{2}\rangle .$ Using Pauli matrices $\overrightarrow{\sigma }$ for the $J_z=\pm J$ index and $\overrightarrow{\tau }$ for the total angular momentum J index, the above effective model can be written as

$${\widehat{H}}_{\mathrm{\Gamma }}\left(\mathbf{k}\right)={\mathit{ϵ}}_{\mathbf{k}}+A\left(k_x{\sigma }_z{\tau }_x-k_y{\sigma }_0{\tau }_y\right)+M\left(\mathbf{k}\right){\sigma }_0{\tau }_z+O\left({\left|\mathbf{k}\right|}^2\right),$$ [S9]

where $M\left(\mathbf{k}\right)=M_0-M_1{k}_z^2-M_2\left(k_x^2+k_y^2\right)+O\left({\left|\mathbf{k}\right|}^2\right)$ with $M_0,M_1,M_2<0.$ Under the unitary rotation

$$\widehat{U}=\text{i}\left(\sin \phi {\sigma }_x+\cos \phi {\sigma }_y\right)\frac{1+{\tau }_z}{2}+e^{\text{i}\phi {\sigma }_z}\frac{1-{\tau }_z}{2},$$ [S10]

we obtain

$$H_4\left(\mathbf{k}\right)={\widehat{U}}^{\mathrm{†}}{\widehat{H}}_{\mathrm{\Gamma }}\left(\mathbf{k}\right)\widehat{U}$$

$$=\lambda \left(k_x{\sigma }_x+k_y{\sigma }_y\right){\tau }_x+M\left(\mathbf{k}\right){\sigma }_0{\tau }_z,$$

$$M\left(\mathbf{k}\right)=t_z\left(1-\cos Q\right)-\frac{t_z}{2}{k}_z^2-\frac{t}{2}\left(k_x^2+k_y^2\right)+O\left({\left|\mathbf{k}\right|}^2\right).$$

Therefore, unitary rotation [S10] indeed transforms effective $k\cdot p$ Hamiltonian [S9], which describes Dirac semimetal materials into the four-band minimal model [1] used for calculations in this paper.

The symmetry operations ${\mathit{\psi }}_{\mathbf{k}}$ for $k\cdot p$ model [S9] are summarized as follows:

$$\widehat{T}=\text{i}{\sigma }_y\cdot \mathcal{K};$$ [S11]

$$\widehat{s_n}=e^{\text{i}\left(\pi /n\right){\sigma }_z\left(2-{\tau }_z\right)},\hspace{1em}n=4,6;$$ [S12]

$$\widehat{I}={\tau }_z,\left(x,y,z\right)\to \left(-x,-y,-z\right);$$ [S13]

$$\widehat{g_c}=\text{i}{\sigma }_x:\left(x,y,z\right)\to \left(-x,y,z+\frac{1}{2}\right).$$ [S14]

After the unitary rotation [S10] we can obtain the symmetry operations for the four-band model [1],

$$\tilde{T}=\text{i}{\sigma }_y\cdot \mathcal{K};$$ [S15]

$$\widetilde{s_n}=e^{\text{i}\left(\pi /n\right){\sigma }_z\left(1-2{\tau }_z\right)},\hspace{1em}n=4,6;$$ [S16]

$$\tilde{I}={\tau }_z:\left(x,y,z\right)\to \left(-x,-y,-z\right);$$ [S17]

$$\widetilde{g_c}=\text{i}{\sigma }_x{\tau }_z:\left(x,y,z\right)\to \left(-x,y,z+\frac{1}{2}\right);$$ [S18]

where we have chosen $\mathit{\phi }=0$ for simplicity in unitary rotation [S10]. For the purpose of simplicity, in most discussions we simply treat the nonsymmorphic n-fold screw rotation $s_n$ as a symmorphic n-fold rotation $C_n^z$ and the glide symmetry $g_c$ as a symmorphic mirror reflection symmetry $R_x.$

Appendix C: Four-Band Tight-Binding Model for Dirac Semimetal Na₃Bi

With $n=6$-fold rotational symmetry along the $\widehat{z}$ axis, Na₃Bi has a hexagonal Brillouin zone in the $\widehat{x}-\widehat{y}$ plane. Choosing ${\overrightarrow{a}}_1=\left(\mathrm{1,0}\right)$ and ${\overrightarrow{a}}_2=\left(1,\sqrt{3}\right)/2$ as the Bravais lattice unit vectors, the primitive vectors of the reciprocal lattice in the $\widehat{x}-\widehat{y}$ plane are ${\overrightarrow{b}}_1=\left(\sqrt{3},-1\right)/\sqrt{3}$ and ${\overrightarrow{b}}_2=\left(\mathrm{0,2}\right)/\sqrt{3}.$ The momentum space is hence parameterized as

$$\mathbf{k}=\left(k_x,k_y,k_z\right)=\left(k_1,\frac{2k_2-k_1}{\sqrt{3}},k_3\right),$$ [S19]

where under sixfold rotation $C_6^z=e^{\text{i}\pi {\sigma }_z\left(1-2{\tau }_z\right)/6}$ we have

$$\left(k_1,k_2,k_3\right)\stackrel{C_6^z}{\to }\left(k_1-k_2,-k_1,k_3\right)$$ [S20]

and under mirror reflection $R_x=\text{i}{\sigma }_x{\tau }_z,$

$$\left(k_1,k_2,k_3\right)\stackrel{R_x}{\to }\left(-k_1,k_2-k_1,k_3\right).$$ [S21]

As a result, the four-band minimal tight-banding model describing Dirac semimetal Na₃Bi is

$$\begin{array}{l}{H}_4\left(\mathbf{k}\right)=\mathit{\lambda }[{\sigma }_x\sin k_1+\frac{{\sigma }_x-\sqrt{3}{\sigma }_y}{2}\sin \left(k_1-k_2\right)+\frac{{\sigma }_x+\sqrt{3}{\sigma }_y}{2}\sin k_2]{\tau }_x\\ +[t\left(\cos k_1+\cos k_2+\cos \left(k_1-k_2\right)-3\right)+\hspace{0.17em}{t}_z\left(\cos k_z-\cos Q\right)]{\tau }_z.\end{array}$$ [S22]

Clearly it preserves time reversal and all space group symmetries, the same as Cd₃As₂ case [1].

Similar to the Cd₃As₂ case with fourfold rotations, Na₃Bi with a sixfold rotation axis allows the following symmetric perturbations to minimal model $H_4\left(\mathbf{k}\right),$

$$\delta H_4\left(\mathbf{k}\right)=\mathrm{\Delta }\sin k_z[{\sigma }_z{\tau }_x\cos k_1+\frac{-{\sigma }_z{\tau }_x+\sqrt{3}{\tau }_y}{2}\cos \left(k_1-k_2\right)-\frac{{\sigma }_z{\tau }_x+\sqrt{3}{\tau }_y}{2}\cos k_2],$$ [S23]

just like [3] in the Cd₃As₂ case. This symmetric term can continuously deform the double Fermi arc on the side surface into a closed pocket.

Appendix D: Details of Numerical Calculation of Spectral Function

In this section we first describe how we calculate the spectral function $A\left(\mathbf{k},\omega \right)$ whose plots at $\mathit{\omega }\mathrm{=}0$ give us information about the Dirac nodes in the bulk and the Fermi arcs on the surface shown in Fig. 1 B–F. An important (quantitative) aspect to the Fermi arcs shown is their curvature in the surface BZ. We explain here the role played by the ${\epsilon }_{\mathbf{k}}^0$ term in $H\left(\mathbf{k}\right),$ without which the Hamiltonian has a special chiral symmetry and the two Fermi arcs lose their curvature and collapse to $k_y=0$ for $-Q<k_z<Q$ (Fig. S1).

Fig. S1.

Surface spectral density for a finite-size system along the x direction described by a four-band model in [1] with ${\epsilon }_{\mathbf{k}}^0=0.$ In A–C, from Left to Right corresponds to adding a perturbation preserving all symmetries of the system with $m\prime =0,$ $m\prime =0.4t,$ and $m\prime =0.8t,$ respectively. In the absence of perturbation the surface states are a line (A) connecting projected bulk nodes. As soon as the perturbation is turned on, the surface states start to disappear from the surface in B and C.

In Fig. S2 we also show the spectral density at the $\omega -k_y$ plane at fixed values of $k_z=0,\pi /6,\pi /3,\pi /2.$ In the absence of symmetry-preserved bulk perturbation $\mathit{\delta }{H}_4$ the gapless edge states extend between projected bulk nodes. Indeed, Fig. S2, Top shows that each fixed $k_z$ plane carries a helical gapless edge state. Upon the inclusion of bulk perturbation the gapless edge states at $k_z=0$ remain intact whereas those off $k_z=0$ become gapped. In the latter the projected bulk nodes also remain gapless. Hence, with perturbation, the only gapless states are limited to the plane located at $k_z=0$ localized merely on the surface and those at the projected bulk nodes with the gapless states extended to the bulk.

Fig. S2.

(Top) Spectral density of a four-band model with $\delta H_4=0$ at the $\omega -k_y$ plane. Each panel corresponds to a fixed $k_z$ as shown. (Bottom) The same panels but with $\delta H_4\ne 0.$ The dashed line corresponds to $\omega =0.$ It is clearly seen that the in the absence of bulk perturbation $\delta H_4$ gapless states exist for all values of $k_z$ between nodes, whereas in the presence of perturbation $\delta H_4$ the gapless edge states only at $k_z=0$ and $k_z=\pi /2$ remain. The latter one corresponds to the projection of bulk nodes onto the surface.

We used two methods to calculate the bulk and surface spectral density. Both give the same results. We assume that the system is infinite (periodic) along y and z directions, so both $k_y$ and $k_z$ remain good quantum numbers. (i) We used the iterative Green function method applied to a semiinfinite lattice. In this method as detailed in ref. 47, bulk $G_b\left(\mathbf{k},\omega \right)$ and surface $G_s\left(\mathbf{k},\omega \right)$ with $\mathbf{k}=\left(k_y,k_z\right)$ can be obtained iteratively. (ii) We considered a slab geometry of system with N layers coupled along the x direction. The Hamiltonian of the slab can be easily written as $H\left(k_y,k_z\right)={\displaystyle {\sum }_{i=1}^N{H}_i\left(k_y,k_z\right)+{\displaystyle {\sum }_{i=1}^{N-1}\left(H_{i,i+1}\left(k_y,k_z\right)+h.c.\right)}},$ where $H_i\left(k_y,k_z\right)$ is the intralayer Hamiltonian and $H_{i,i+1}\left(k_y,k_z\right)$ is interlayer coupling. The Hamiltonian H is a $4N\times 4N$ matrix. Having obtained this matrix, the spectral function of the whole system is given as $A\left(\mathbf{k},\omega \right)=-1/\pi \text{Im}\hspace{0.17em}1/\left(\omega -H\left(\mathbf{k}\right)+\text{i}0\right),$ with $\mathbf{k}=\left(k_y,k_z\right).$ Weighted by components of wavefunctions of each layer for all energies, the spectral function of each layer, say the jth one, can be calculated as

$$A_j\left(\mathbf{k},\omega \right)=-\frac{1}{\pi }{\displaystyle \sum _{i=4j-3}^{4j}{\displaystyle \sum _{n=1}^{4N}\text{Im}\frac{{\left|u_{n,i}\right|}^2}{\omega -{\epsilon }_n\left(\mathbf{k}\right)+\text{i}0}}},$$ [S24]

where $H\left(\mathbf{k}\right)u_n={\epsilon }_n\left(\mathbf{k}\right)u_n,$ and $u_{n,i}$ is the ith component of eigenstate $u_n.$ For surface states we put $j=1$ and for states deep in the bulk we can consider the closest integer to $N/2.$

Appendix E: K-Theory Classification of ${\text{Na}}_3\text{Bi}$

The space group of Dirac semimetal Na₃Bi is centrosymmetric P6₃/mmc (no. 194) (17).The space group has 24 symmetry elements generated by sixfold screw $s_6,$ glide reflection $g_c,$ and inversion I as follows,

$$\left(x,y,z\right)\stackrel{s_6}{\to }\left(x-y,x,z+1/2\right),$$ [S25]

$$\left(x,y,z\right)\stackrel{g_c}{\to }\left(y,x,z+1/2\right),$$ [S26]

$$\left(x,y,z\right)\stackrel{I}{\to }\left(-x,-y,-z\right),$$ [S27]

which together with Bravais lattice translations $T_{a,b,c}$ generate all symmetry elements. Here, the unit cell is expanded by $\mathbf{a}=a\left(\mathrm{1,0,0}\right),$ $\mathbf{b}=a\left(-1/2,\sqrt{3}/2,0\right),$ and $\mathbf{c}=c\left(\mathrm{0,0,1}\right),$ and any real space point is represented as $\mathbf{r}=x\mathbf{a}+y\mathbf{b}+z\mathbf{c}.$ The associated reciprocal lattice vectors are ${\mathbf{G}}_a=2\pi /a\left(1,1/\sqrt{3},0\right),$ ${\mathbf{G}}_b=2\pi /a\left(0,2/\sqrt{3},0\right),$ and ${\mathbf{G}}_c=2\pi /c\left(\mathrm{0,0,1}\right).$

The two Dirac nodes in the bulk of Na₃Bi are located along the c axis at ${\mathbf{Q}}_{\pm }=\mathrm{\pm }{Q}_0{\mathbf{G}}_c$ (17). We are interested in side surfaces parallel to node separations that could host possible double Fermi arcs if stable at all. In the following we show that there is no topological protection for the double Fermi arcs from lattice and time reversal symmetries.

Side Surface (100).

The side surface (100) is parallel to a and c. Therefore, it remains invariant under the following symmetries:

$$\left(x,y,z\right)\stackrel{R_b}{\to }\left(x,x-y,z\right),\hspace{1em}{R}_b=T_c^{-1}{g}_c{s}_6$$ [S28]

$$\left(x,y,z\right)\stackrel{R_c}{\to }\left(x,y,1/2-z\right),\hspace{1em}{R}_c=T_c^{2}I{s}_6^3$$ [S29]

$$\left(x,y,z\right)\stackrel{C_2}{\to }\left(x,x-y,1/2-z\right),\hspace{1em}{C}_2=T_c^{2}I{s}_6^2{g}_c.$$ [S30]

Note that $C_2=T_c{R}_c{R}_b.$ The following identities also hold,

$$R_b^2=R_c^2=-1,$$ [S31]

$$R_b{R}_c=-R_c{R}_b,$$ [S32]

where the minus sign comes from $2\pi$ spin rotations associated with spatial rotations due to spin–orbit coupling.

To examine the effect of symmetries on the Bloch Hamiltonian in momentum space, we introduce orthogonal coordinates with unit vectors as follows:

$${\mathbf{k}}_1=\left(\sqrt{3}/2,1/2,0\right),$$ [S33]

$${\mathbf{k}}_2=\left(-1/2,\sqrt{3}/2,0\right),$$ [S34]

$${\mathbf{k}}_3=\left(\mathrm{0,0,1}\right).$$ [S35]

Thus, any vector in the Brillouin zone is represented as $\mathbf{k}={\displaystyle {\sum }_{i=1}^3{k}_i{\mathbf{k}}_i}.$ Under the crystalline symmetry operators above, we obtain

$$\left(k_1,k_2,k_3\right)\stackrel{R_b}{\to }\left(k_1,-k_2,k_3\right),$$ [S36]

$$\left(k_1,k_2,k_3\right)\stackrel{R_c}{\to }\left(k_1,k_2,-k_3\right)\mathrm{,}$$ [S37]

and under time reversal

$$\left(k_1,k_2,k_3\right)\stackrel{\mathrm{\Theta }}{\to }\left(-k_1,-k_2,-k_3\right).$$ [S38]

Assuming $U_b$ and $U_c$ correspond to representations of symmetry operations $R_b$ and $R_c$ acting on orbital and spin degrees of freedom, the symmetric Bloch Hamiltonian transforms as

$$U_{b}H\left(k_1,k_2,k_3\right)U_b^{-1}=H\left(k_1,-k_2,k_3\right),$$ [S39]

$$U_{c}H\left(k_1,k_2,k_3\right)U_c^{-1}=H\left(k_1,k_2,-k_3\right),$$ [S40]

$$\mathrm{\Theta }H\left(k_1,k_2,k_3\right){\mathrm{\Theta }}^{-1}=H\left(-k_1,-k_2,-k_3\right).$$ [S41]

From theses relations it is clear that symmetry operators $U_b$ and ${\mathrm{\Theta }}_c=\mathrm{\Theta }{U}_c$ (a combination of time-reversal $\mathrm{\Theta }$ and reflection $R_c$) preserve a 2D plane at fixed $k_3\ne 0,\pi$ in momentum space, because

$${\mathrm{\Theta }}_{c}H\left(k_1,k_2,k_3\right){\mathrm{\Theta }}_c^{-1}=H\left(-k_1,-k_2,k_3\right).$$ [S42]

Such 2D planes are all gapped except for those passing through nodes at $Q_{\pm }.$ Thus, we can classify gapped 2D insulators embedded in the 3D Brillouin zone. Following the K-theory approach (32–34), we classify gapped free-fermion quantum phases by examining the structure of classifying space for symmetry-allowed mass terms ${\gamma }_0$ in generic Dirac Hamiltonians,

$$H=k_1{\gamma }_1+k_2{\gamma }_2+\text{i}m{\gamma }_0,$$ [S43]

where the ${\gamma }_i$ s satisfy Clifford algebra

$$\left\{{\gamma }_i,{\gamma }_j\right\}=\pm 2{\delta }_{ij},$$ [S44]

where the $+(-)$ sign holds for $i,j=\mathrm{1,2}\left(0\right).$ The symmetry operators satisfy that

$$U_b{\gamma }_1{U}_b^{-1}={\gamma }_1,$$ [S45]

$$U_b{\gamma }_2{U}_b^{-1}=-{\gamma }_2,$$ [S46]

$$U_b{\gamma }_0{U}_b^{-1}={\gamma }_0,$$ [S47]

$${\mathrm{\Theta }}_c{\gamma }_i{\mathrm{\Theta }}_c^{-1}=-{\gamma }_i,\hspace{1em}i=\mathrm{0,1,2.}$$ [S48]

In addition to the above symmetries, we also need to consider charge symmetry $U{\left(1\right)}_c$ generated by Q, under which every fermion annihilation operator f is multiplied by the imaginary number: $f\stackrel{\mathbf{Q}}{\to }\text{i}f.$ Therefore, we have

$${\mathbf{Q}}^2={\left(U_b\right)}^2=-1,\hspace{1em}{\left({\mathrm{\Theta }}_c\right)}^2=+1,$$ [S49]

$$\left\{{\mathrm{\Theta }}_c,\mathbf{Q}\right\}=\left[U_b,\mathbf{Q}\right]=\left[{\gamma }_i,\mathbf{Q}\right]=0,$$ [S50]

$$\left\{U_b,U_c\right\}=\left[\mathrm{\Theta },U_{b,c}\right]=0.$$ [S51]

The whole 2D system therefore is characterized by symmetry generators $G=\left\{{\mathrm{\Theta }}_c,U_b,\mathbf{Q}\right\},$ gamma matrices ${\gamma }_{\mathrm{1,2}},$ and mass term ${\gamma }_0.$ Using these operators, we can construct a group of operators all anticommuting with each other, which form the following real Clifford algebra

$$C{l}_{\mathrm{5,0}}=\left\{{\gamma }_1{\mathrm{\Theta }}_c,{\gamma }_2{\mathrm{\Theta }}_c,\mathbf{Q},{\gamma }_1{U}_b{\mathrm{\Theta }}_c,{\gamma }_1{\gamma }_2{\mathrm{\Theta }}_c\mathbf{Q}\right\},$$ [S52]

where every operator squares to be −1. The classification of 2D gapped phases amounts to the extension of this Clifford algebra by adding a mass term

$$C{l}_{\mathrm{5,0}}\to C{l}_{\mathrm{6,0}}=\left\{\mathrm{\dots },{\gamma }_0{\gamma }_1{\gamma }_2{\mathrm{\Theta }}_c\mathbf{Q}\right\},$$ [S53]

because all terms square as −1. The associated classifying space for the mass term in the extension $C{l}_{p,q}\to C{l}_{p+1,q}$ is $R_{p-q+2},$ and the topological classification is given by the zeroth homotopy of classifying space ${\pi }_0\left(R_{p-q+2}\right).$ Thus, we obtain

$${\pi }_0\left(R_7\right)=0,$$ [S54]

implying that the classification is trivial and there are no protected surface states in a generic $k_z\ne 0,\pi$ plane on the side surface (100).

The trivial classification implies that any counterpropagating edge sates on any fixed-$k_z$ ($k_z\ne 0,\pi$) plane can be gapped out by symmetry-preserving perturbations. As mentioned in the previous section, the edge Hamiltonian takes a form $h=v_F{\displaystyle {\sum }_{k}k{\psi }_k^{\mathrm{†}}{\mu }_z{\psi }_k},$ where $\mathit{\psi }={\left({\psi }_R,{\psi }_L\right)}^T.$ In this basis the symmetry operators are $\mathrm{\Theta }=\text{i}{\mu }_y\mathcal{K},$ $U_c=\text{i}{\mu }_z,$ and $U_b=\text{i}{\mu }_y.$ Therefore, a mass term like $m{\mu }_y$ preserving symmetries ${\mathrm{\Theta }}_c=\text{i}{\mu }_{x}K$ and $U_b$ can be added to remove the surface states.

Side Surface (110).

The side surface (110) is perpendicular to ${\mathbf{G}}_a+{\mathbf{G}}_b$ and hence parallel to Bravais vectors $\mathbf{a}-\mathbf{b}$ and c. This plane breaks most crystalline symmetries except for those generated by glide $g_c$ and reflection $R_c$ acting on lattice sites as

$$\left(x,y,z\right)\stackrel{g_c}{\to }\left(y,x,z+1/2\right),$$ [S55]

$$\left(x,y,z\right)\stackrel{R_c}{\to }\left(x,y,1/2-z\right),$$ [S56]

$$\left(x,y,z\right)\stackrel{C_2}{\to }\left(y,x,-z\right),$$ [S57]

where $C_2=R_c{g}_c$ and

$${\left(g_c\right)}^2=-T_c,$$ [S58]

$$g_c{R}_c=-T_c{R}_c{g}_c.$$ [S59]

As before, we introduce orthogonal coordinates with unit vector ${\mathbf{k}}_1$ perpendicular to the (110) plane as follows:

$${\mathbf{k}}_1=\left(1/2,\sqrt{3}/2,0\right),$$ [S60]

$${\mathbf{k}}_2=\left(-\sqrt{3}/2,1/2,0\right),$$ [S61]

$${\mathbf{k}}_3=\left(\mathrm{0,0,1}\right).$$ [S62]

Under the crystalline symmetry operators above, we obtain

$$\left(k_1,k_2,k_3\right)\stackrel{g_c}{\to }\left(k_1,-k_2,k_3\right),$$ [S63]

$$\left(k_1,k_2,k_3\right)\stackrel{R_c}{\to }\left(k_1,k_2,-k_3\right).$$ [S64]

The symmetric Bloch Hamiltonian transforms as

$$U_{g}H\left(k_1,k_2,k_3\right)U_g^{-1}=H\left(k_1,-k_2,k_3\right),$$ [S65]

$$U_{c}H\left(k_1,k_2,k_3\right)U_c^{-1}=H\left(k_1,k_2,-k_3\right),$$ [S66]

$$\mathrm{\Theta }H\left(k_1,k_2,k_3\right){\mathrm{\Theta }}^{-1}=H\left(-k_1,-k_2,-k_3\right),$$ [S67]

where $U_g$ implements glide $g_c$ in spin and orbital basis.

Similar to the side surface (100), here for side surface (110) we can explicitly show that the edge states at fixed $k_z\ne 0,\pi ,$ if any, can be gapped out without breaking any symmetry on the surface. To show this, consider a pair of counterpropagating edge modes described by $h=v_F{\displaystyle {\sum }_{k}k{\psi }_k^{†}{\mu }_z{\psi }_k}$ with symmetry operations represented as

$${\mathrm{\Theta }}_c=\mathrm{\Theta }\cdot U_c=\text{i}{\mu }_y\mathcal{K}\cdot \text{i}{\mu }_z=\text{i}{\mu }_x\mathcal{K},$$ [S68]

$$U_g=\text{i}{e}^{\text{i}{k}_z/2}{\mu }_y.$$ [S69]

A mass term like $\widehat{M}=m{\displaystyle {\sum }_k{\psi }_k^{\mathrm{†}}{\gamma }_0{\psi }_k}$ must satisfy

$$\left\{{\mu }_z,{\gamma }_0\right\}=\left[{\mathrm{\Theta }}_c,{\gamma }_0\right]=\left[U_c,{\gamma }_0\right]=0.$$ [S70]

Clearly a symmetric mass term ${\gamma }_0={\mu }_y$ can destroy the edge states. Therefore, there is no topological protection for surface states at $k_z\ne 0,\pi$ on side surface (110) either.

Surface States at $k_z=0,\pi$.

On both side surfaces (100) and (110) studied above, momentum cuts at $k_z=0,\pi$ are special in the sense that the associated 2D system has more symmetries. In particular, these momenta are time-reversal invariant: i.e., their 2D Hamiltonians are invariant under mirror $U_c$ and time-reversal $\mathrm{\Theta }$ symmetries separately.

With just time reversal ${\mathrm{\Theta }}^2=-1$ and charge conservation $\mathbf{Q},$ the corresponding real Clifford algebra and its extension are given as

$$C{l}_{\mathrm{2,2}}=\left\{{\gamma }_1,{\gamma }_2,\mathrm{\Theta },\mathrm{\Theta }\mathbf{Q}\right\},$$ [S71]

$$C{l}_{\mathrm{2,2}}\to C{l}_{\mathrm{3,2}}=\left\{\mathrm{\dots },{\gamma }_0\right\},$$ [S72]

reproducing the well-known 2D index:

$${\pi }_0\left(R_2\right)={\mathrm{\mathbb{Z}}}_2.$$ [S73]

Remarkably, here nonsymmorphic glide reflection $g_c$ with half translation along the $\widehat{z}$ axis provides extra constraints on these $Z_2$ indexes. This nonsymmorphic glide forbids any nontrivial $Z_2$ index at the $k_z=\pi$ plane (36), as is shown below.

Let us consider crystal symmetries $g_c,R_c$ as well in $k_z=0,\pi$ planes with the open (110) side surface. From [S58] and [S59] we know that $R_c^2=-1$ and

$${\left(g_c\right)}^2=-1,\hspace{1em}\left\{g_c,R_c\right\}=0\hspace{1em}\text{at}\hspace{1em}{k}_z=0;$$ [S74]

whereas

$${\left(g_c\right)}^2=1,\hspace{1em}\left[g_c,R_c\right]=0\hspace{1em}\text{at}\hspace{1em}{k}_z=\pi .$$ [S75]

Therefore, in the $k_z=\pi$ plane, the associated complex Clifford algebra and extension problem are

$$C{l}_5=\left\{{\gamma }_1,{\gamma }_2,\mathrm{\Theta },\mathrm{\Theta }\mathbf{Q},g_c{\gamma }_2\right\}\times R_c,$$ [S76]

$$C{l}_5\to C{l}_6=\left\{\mathrm{\dots },{\gamma }_0\right\},$$ [S77]

because $R_c$ commutes with all other generators. This leads to a classifying space $C_5$ and trivial classification

$${\pi }_0\left(C_5\right)=0.$$ [S78]

This result proves the absence of any protected surface states at $k_z=\pi$ on the side surface of Na₃Bi.

In the $k_z=0$ plane on the other hand, we found the Clifford algebra as follows:

$$C{l}_{\mathrm{4,2}}=\left\{{\gamma }_1,{\gamma }_2,\mathrm{\Theta },\mathrm{\Theta }\mathbf{Q},g_c{\gamma }_2,R_c{g}_c{\gamma }_2\right\}$$ [S79]

$$C{l}_{\mathrm{4,2}}\to C{l}_{\mathrm{5,2}}=\left\{\dots ,{\gamma }_0\right\}.$$ [S80]

The corresponding classifying space is $R_{2-4+2}=R_0,$ leading to an integer classification

$${\pi }_0\left(R_0\right)=\mathrm{\mathbb{Z}}.$$ [S81]

This integer index ν simply labels how many pairs of counterpropagating edge modes appear in the $k_z=0$ plane on the side surface. In the case of Na₃Bi, both experiments and effective model [S22] indicate $\nu =1;$ i.e., the $k_z=0$ plane is a 2D topological insulator.

Appendix F: K-Theory Classification of ${\text{Cd}}_3{\text{As}}_2$

The space group of Dirac semimetal Cd₃As₂ is centrosymmetric $\text{I}{4}_1/\text{acd}$ (no. 142) (38). In addition to Bravais lattice translations, it is generated by fourfold screw rotation along the $\widehat{z}$ axis, glide reflection $g_c,$ and inversion I defined as

$$\left(x,y,z\right)\stackrel{s_4}{\to }\left(y,1/2-x,1/4+z\right),$$ [S82]

$$\left(x,y,z\right)\stackrel{g_c}{\to }\left(x,-y,1/2+z\right),$$ [S83]

$$\left(x,y,z\right)\stackrel{I}{\to }\left(-x,1/2-y,1/4-z\right),$$ [S84]

where ${\left(s_4\right)}^4={\left(g_c\right)}^2=-T_z.$ The Bravais lattice is expanded by $T_{\pm 1,\pm 1,\pm 1}\equiv \left(\pm 1,\pm 1,\pm 1\right)/2$ similar to the BCC lattice. Two Dirac nodes in Cd₃As₂ are located along the $\widehat{z}$ axis. There are two inequivalent side surfaces parallel to node separation, i.e., (100) and (110) surfaces. Completely similar to the Na₃Bi case, here nonsymmorphic glide reflection $g_c$ dictates that the $k_z=\pi$ plane must have a trivial $Z_2$ index with no protected surface states on side surfaces, whereas the $k_z=0$ plane can support a nontrivial $Z_2$ QSH index. Indeed, experiments and minimal models [1] suggest that the $k_z=0$ plane is a 2D insulator, leading to protected $k_z=0$ surface states on side surfaces of Cd₃As₂. In the following we analyze stability of generic surface states on two representative side surfaces.

Side Surface (100).

This side surface preserves a symmetry subgroup generated by two glides

$$\left(x,y,z\right)\stackrel{g_c}{\to }\left(x,-y,1/2+z\right),$$ [S85]

$$\left(x,y,z\right)\stackrel{g_b}{\to }\left(x,1/2+y,1/4-z\right),$$ [S86]

where

$$g_b=T_{\overline{1}\overline{1}\overline{1}}I{\left(s_4\right)}^2\Rightarrow {\left(g_b\right)}^2=-T_y,$$ [S87]

$$g_b{g}_c=-T_y{T}_z^{-1}{g}_c{g}_b.$$ [S88]

In an orthogonal basis the Bloch Hamiltonian transforms as

$$U_{g_c}H\left(k_x,k_y,k_z\right)U_{g_c}^{-1}=H\left(k_x,-k_y,k_z\right),$$ [S89]

$$U_{g_b}H\left(k_x,k_y,k_z\right)U_{g_b}^{-1}=H\left(k_x,k_y,-k_z\right).$$ [S90]

We define

$${\mathrm{\Theta }}_{g_b}=\mathrm{\Theta }{U}_{g_b},$$ [S91]

which transforms the Bloch Hamiltonian as

$${\mathrm{\Theta }}_{g_b}H\left(k_x,k_y,k_z\right){\mathrm{\Theta }}_{g_b}^{-1}=H\left(-k_x,-k_y,k_z\right).$$ [S92]

In momentum space we have

$$U_{g_c}^2=-e^{\text{i}{k}_z},$$ [S93]

$${\mathrm{\Theta }}_{g_b}^2=e^{\text{i}{k}_y},$$ [S94]

$${\mathrm{\Theta }}_{g_b}{U}_{g_c}{\mathrm{\Theta }}_{g_b}^{-1}=-e^{\text{i}\left(k_y-k_z\right)}{U}_{g_c}.$$ [S95]

As before we focus on planes with fixed $k_z\ne 0,\pi .$ At $k_y=0,\pi$ the effective time-reversal symmetry squares ${\mathrm{\Theta }}_{g_b}^2=\pm 1,$ which leads to a one-dimensional classification for the Hamiltonian:

$$H\left(k_x\right)=k_x{\gamma }_x+m{\gamma }_0.$$ [S96]

At $k_y=0,$ the corresponding Clifford algebra and its extension become as follows:

$$C{l}_{\mathrm{0,3}}=\left\{{\gamma }_x,{\mathrm{\Theta }}_{g_b},{\mathrm{\Theta }}_{g_b}\mathbf{Q}\right\},$$ [S97]

$$C{l}_{\mathrm{0,3}}\to C{l}_{\mathrm{1,3}}=\left\{\dots ,{\gamma }_0\right\}.$$ [S98]

As a result the classification is given by

$${\pi }_0\left(R_{0-3+2}\right)=0.$$ [S99]

At $k_y=\pi ,$ the corresponding Clifford algebra and its extension remain the same as above, and because ${\mathrm{\Theta }}_{g_b}^2=-1,$ at $k_y=\pi ,$ the classification to be trivial is as follows:

$${\pi }_0\left(R_{2-1+2}\right)=0.$$ [S100]

We can also understand the robustness of surface states in terms of a low-energy effective Hamiltonian. Consider a pair of counterpropagating edge modes at fixed $k_z\ne 0,\pi ,$ described by $h\left(k_y\right)=v_F{k}_y{\mu }_z.$ In the following we show that a mass term $\delta h$ is allowed by symmetry to gap out these surface states at $k_z\ne 0,\pi .$ The transformation of the mass term is given as follows. A minimal representation of symmetry operators is

$$U_{g_c}\left(\mathbf{k}\right)=\text{i}{\mu }_y{e}^{\text{i}{k}_z/2},$$ [S101]

$${\mathrm{\Theta }}_{g_b}\left(\mathbf{k}\right)=\text{i}{\mu }_x{e}^{\text{i}{k}_y/2}\mathcal{K}.$$ [S102]

The symmetric mass term is found to be $\delta h=m{\mu }_y,$ which can destroy gapless surface states at fixed $k_z\ne 0,\pi$ on side surface (100).

Side Surface (110).

There are two glides

$$\left(x,y,z\right)\stackrel{g_d}{\to }\left(y+1/2,x,z+1/4\right),\hspace{1em}{g}_d=T_{1\overline{1}\overline{1}}{g}_c{s}_4,$$

$$\left(x,y,z\right)\stackrel{g_b}{\to }\left(x,y+1/2,1/4-z\right),\hspace{1em}{g}_b=T_{\overline{1}11}I{\left(s_4\right)}^2,$$

satisfying

$${\left(g_d\right)}^2=-T_{111},\hspace{1em}{\left(g_b\right)}^2=-T_y,$$

$$g_b{g}_d{g}_b^{-1}=-T_{\overline{1}1\overline{1}}{g}_d.$$

Although both glides are broken on side surface (110), there is a twofold rotation, the combination of two glides, preserved on the (110) surface:

$$\left(x,y,z\right)\stackrel{C_2}{\to }\left(y,x,1/2-z\right),\hspace{1em}{C}_2=T_x^{-1}{g}_d{g}_b.$$

The combination ${\mathrm{\Theta }}_{C_2}$ of time reversal $\mathrm{\Theta }$ and $C_2$ rotation satisfying ${\left({\mathrm{\Theta }}_{C_2}\right)}^2=1$ preserves each surface momentum, but the classification of the 1D system at fixed $k_x-k_y$ and $k_z$ is trivial,

$${\pi }_0\left(R_{0-3+2}\right)=0,$$ [S103]

from Clifford algebra and extension problem

$$C{l}_{\mathrm{0,3}}=\left\{{\gamma }_1,{\mathrm{\Theta }}_{C_2},{\mathrm{\Theta }}_{C_2}\mathbf{Q}\right\},$$ [S104]

$$C{l}_{\mathrm{0,3}}\to C{l}_{\mathrm{1,3}}=\left\{\mathrm{···},{\gamma }_0\right\}.$$ [S105]

Therefore, generally surface states at $k_z\ne 0,\pi$ are not robust against perturbations on side surface (110).