Inversion Symmetric Topological Insulators
IInversion Symmetric Topological Insulators
Taylor L. Hughes, Emil Prodan, and B. Andrei Bernevig Department of Physics, University of Illinois, 1110 West Green St, Urbana IL 61801 Department of Physics, Yeshiva University, New York, NY 10016 Department of Physics, Princeton University, Princeton, NJ 08544 (Dated: October 22, 2018)We analyze translationally-invariant insulators with inversion symmetry that fall outside thecurrent established classification of topological insulators. These insulators exhibit no edge or surfacemodes in the energy spectrum and hence they are not edge metals when the Fermi level is in the bulkgap. However, they do exhibit protected modes in the entanglement spectrum localized on the cutbetween two entangled regions. Their entanglement entropy cannot be made to vanish adiabatically,and hence the insulators can be called topological. There is a direct connection between the inversioneigenvalues of the Hamiltonian band structure and the mid-gap states in the entanglement spectrum.The classification of protected entanglement levels is given by an integer n ∈ Z , which is thedifference between the negative inversion eigenvalues at inversion symmetric points in the Brillouinzone, taken in sets of two. When the Hamiltonian describes a Chern insulator or a non-trivialtime-reversal invariant topological insulator, the entirety of the entanglement spectrum exhibits spectral flow . If the Chern number is zero for the former, or time-reversal is broken in the latter, theentanglement spectrum does not have spectral flow, but, depending on the inversion eigenvalues,can still exhibit protected mid-gap bands similar to impurity bands in normal semiconductors.Although spectral flow is broken (implying the absence of real edge or surface modes in the originalHamiltonian), the mid-gap entanglement bands cannot be adiabatically removed, and the insulatoris ‘topological.’ We analyze the linear response of these insulators and provide proofs and examplesof when the inversion eigenvalues determine a non-trivial charge polarization, a quantum Hall effect,an anisotropic 3D quantum Hall effect, or a magneto-electric polarization. In one-dimension, weestablish a link between the product of the inversion eigenvalues of all occupied bands at all inversionsymmetric points and charge polarization. In two dimensions, we prove a link between the productof the inversion eigenvalues and the parity of the Chern number of the occupied bands. In threedimensions, we find a topological constraint on the product of the inversion eigenvalues therebyshowing that some 3D materials are protected topological metals, we show the link between theinversion eigenvalues and the 3D Quantum Hall Effect, and analyze the magneto-electric polarization( θ vacuum) in the absence of time-reversal symmetry. One of the most active fields of research in recent yearshas been the study of non-trivial topological states ofmatter. The paradigm example of such a state is theQuantum Hall Effect, with its Integer (IQHE) and Frac-tional (FQHE) versions. More recently, examples of topo-logical phases that do not require external magnetic fieldshave been proposed, the first being Haldane’s Chern In-sulator model. Although this state has not been experi-mentally realized, a time-reversal invariant (TRI) versionhas been proposed and discovered.
Recent work in the theory of topological insulators showed that an important consideration is not only whichsymmetries the state breaks, but which symmetries mustbe preserved to ensure the stability of the state. A pe-riodic table classifying the topological insulators and su-perconductors has been created. The table organizes thepossible topological states according to their space-timedimension and the symmetries that must remain pro-tected: time-reversal, charge conjugation, and/or chiralsymmetries.
The most interesting entries in this ta-ble, from a practical standpoint, are the 2 and 3D TRItopological insulators which have been already found innature.
These are insulating states classified by a Z invariant that requires an unbroken time-reversal sym-metry to be stable. There are several different methods to calculate the Z invariant, and a non-trivialvalue for this quantity implies the existence of an oddnumber of gapless Dirac fermion boundary states as wellas a non-zero magneto-electric polarizibility in 3D. The current classification of the topological insulatorscovers only the time-reversal, charge conjugation, or chi-ral symmetries and does not exhaust the number of allpossible topological insulators. In principle, for every dis-crete symmetry, there must exist topological insulatingphases with distinct physical properties, and a topologi-cal number that classifies these phases and distinguishesthem from the “trivial” ones. So far, in our discussionwe have used the term “topological” cavalierly so be-fore proceeding, we should ask: what makes an insu-lator topological? We start by first defining a trivialinsulator: this is the insulator that, upon slowly turn-ing off the hopping elements and the hybridization be-tween orbitals on different sites, flows adiabatically intothe atomic limit. In most of the existent literature onnon-interacting topological insulators, it is implicitly as-sumed that non-trivial topology implies the presence ofgapless edge states in the energy spectrum of a systemwith boundaries. However, it is well known from the lit-erature on topological phases that such systems can the-oretically exist without exhibiting gapless edge modes. a r X i v : . [ c ond - m a t . m e s - h a ll ] O c t Hence, the edge modes cannot be the only diagnostic of atopological phase and, consequently, the energy spectrumalone, with or without boundaries, is insufficient to deter-mine the full topological character of a state of matter. Inthe bulk of an insulator, it is a known fact that the topo-logical structure is encoded in the eigenstates rather thanin the energy spectrum. As such, one can expect thatentanglement - which only depends on the eigenstates -can provide additional information about the topologi-cal nature of the system. However we know that topo-logical entanglement entropy (or the sub-leading part ofthe entanglement entropy), the preferred quantityused to characterize topologically ordered phases, doesnot provide a unique classification, and, moreover, van-ishes for any non-interacting topological insulator, be ittime-reversal breaking Chern Insulators or TRI topolog-ical insulators. However, as we will see, careful studies ofthe full entanglement spectrum helps in characterizingthese states. The total entanglement entropy can be continuouslydeformed to zero for trivial insulators, since the atomiclimit to which every trivial insulator can be adiabati-cally continued (by the above definition) is completelylocal and has flat featureless bands. We could there-fore suggest that a non-trivial topological state in a non-interacting translationally invariant insulator should bedefined as having an entanglement entropy which cannotbe adiabatically tuned to zero. However, even this def-inition cannot be entirely correct, as the entanglemententropy strongly depends on the nature of the cut madein the system. For a single particle entanglement spec-trum with eigenvalues { ξ a } the entanglement entropy isdetermined via S ent = − (cid:88) a ( ξ a log ξ a + (1 − ξ a ) log(1 − ξ a )) . (1)For IQH states on the sphere, the many-body wavefunc-tion is a single Slater determinant of occupied Landauorbitals, and hence an orbital cut would result in zeroentanglement entropy since all orbitals are fully occu-pied or unoccupied leading to a set of { ξ a } which are all0’s or 1’s and do not contribute to S ent . Similarly, for atranslationally-invariant Chern or TRI topological insu-lator on a lattice, a momentum space-cut would alwaysgive zero entanglement entropy since the Hamiltonian isdiagonal in this basis. A spatial cut, however, wouldshow mid-gap bands in the entanglement spectrum boththe IQHE and in the topological insulator case( i.e. a setof eigenvalues spanning the ‘gap’ between 0 and 1) simi-lar to the ones in the real energy spectrum.
These mid-gap states give large contributions to the en-tanglement entropy. In fact, for such states, the entan-glement entropy for the spatial cut cannot be made tovanish by any adiabatic changes in the Hamiltonian. Wehence propose that a translationally invariant insulatorcan be classified as topological if it cannot be adiabati-cally connected to a state with zero entanglement entropyfor at least one kind of cut of the system . Explicitly, an insulator should be characterized as topological if it hasprotected mid-gap states in the single-particle entangle-ment spectrum that cannot be pushed to eigenvalues 0or 1 by any adiabatic changes of the Hamiltonian.In the current paper we analyze the physics of insu-lators with inversion symmetry based on the above def-inition. While some of the inversion-symmetric insula-tors exhibit protected edge modes in the energy spectrumwith boundaries ( e.g. a Chern insulator with inversionsymmetry), most do not. However, they can still be topo-logical because their entanglement spectrum for a spa-tial cut exhibits protected mid-gap bands of states. Thiswas first pointed out for 3D strong topological insulatorswith inversion symmetry and soft time-reversal breakingin Ref. 29 . Although it was indicated in Ref. 29 thatthe entanglement spectrum cannot distinguish betweena TRI and inversion invariant topological insulator, andone with TRI slightly broken (compared to the bulk gap),we show that it can distinguish these states. In SectionII we explicitly show that inversion symmetric topologi-cal insulators have two types of entanglement spectrum,both with protected mid-gap states. The characteristicwhich distinguishes the two types of entanglement spec-tra (and the two cases from Ref. 29) is the presence orabsence of spectral flow. For non-trivial TRI or Cherninsulators the entanglement spectrum exhibits spectralflow, very much like their energy spectrum. Heuristi-cally this means that the filled and empty bulk states areconnected via an interpolating set of states localized onthe partition between the two entangled regions. Spec-tral flow in the energy spectrum implies spectral flow inthe entanglement spectrum. If time-reversal is broken forTRI topological insulators, or for T-breaking insulatorswith vanishing Chern number, we show that such spectralflow is interrupted in both the energy and entanglementspectra, and the occupied bands are disconnected fromthe unoccupied ones. One could then assume that, in sys-tems without a continuous spectral connection betweenthe bulk entanglement bands, one could push all the mid-gap entanglement bands to entanglement eigenvalues 0 , , there is a setof protected mid-gap states that form bands which givethe insulator nonzero entanglement entropy (even whenthe spectral flow has been destroyed). For the case oftime-reversal and inversion invariant topological insula-tors, this was shown in Ref. 29 by breaking time-reversalsoftly. We find a formula relating the number of pro-tected mid-gap bands in the entanglement spectrum tothe inversion eigenvalues of the system at inversion sym-metric points. At inversion symmetric k , and for cutswhich separate the system into two equal halves, the en-tanglement spectrum has protected entanglement edgeor surface modes at exactly ξ = 1 / . This means thatthere is no finite-size level repulsion (splitting) betweenthese modes which is a common feature for real bound-ary or interface states in the energy spectrum. Even ifthe original system has other topological invariants, suchas the Chern number, or the TRI Z invariant which areall trivial, the number of entanglement edge modes canbe nonzero.In Section III we analyze the physical response of asubset of these inversion symmetric insulators, and showimportant implications for the charge polarization, theparity of the Chern number, the 3D quantum Hall ef-fect, and the topological magneto-electric polarizability.Namely, given the set of inversion eigenvalues for the oc-cupied bands at all inversion symmetric points in theBrillouin zone, we provide explicit, compact formulas andcomplementary derivations for determining the physicalresponses which only depend on the inversion eigenval-ues. Additionally, as an alternative perspective, we alsoshow that some of the inversion topological invariantsare equivalent to the wavefunction monodromy, which,in principle is an experimentally measurable quantity.In particular, we show the following: in one dimension,the product of inversion eigenvalues over all occupiedbands and over all inversion symmetric points is relatedto the quantized charge polarization. In two dimensions,the product over the inversion eigenvalues determinesthe parity of the Chern number of the occupied bands.In three dimensions, we show several things: first, weprove a topological restriction for the product of inver-sion eigenvalues of any insulator: it must always equal+1. As such, some systems are protected metals, whichcannot be made insulating with weak scattering. Sec-ond, we show that, depending on the product of inver-sion eigenvalues in different inversion symmetric planeswe will have 3D quantum Hall effects on different planesin the sample. We then show that, depending on the in-version eigenvalues, several inversion symmetric systemscan exhibit a quantized magneto-electric polarizability,even though the Hamiltonian my not be adiabaticallycontinuable to a time-reversal invariant topological insu-lator.Finally in Section IV we end the paper with severalexamples of such insulators and with numerical results.While most of the examples we chose have a topologicalresponse connected with an inversion topological invari-ant, we stress that the presence of mid-gap states in theentanglement spectrum is not intrinsically related to thepresence of a non-trivial topological response. For exam-ple, two identical copies of a strong-topological insulatorwith inversion and time-reversal symmetry has a triv-ial Z invariant and thus a trivial response. However,this system will exhibit protected mid-gap entanglementstates. So, while some inversion invariant insulators haveprotected topological responses, some do not. The situa-tion is even more complicated: in several cases, we provethat inversion eigenvalues by themselves cannot uniquelydetermine the response. We can show that a nontrivialquantum spin Hall state and two copies of a Chern in- sulator each with Chern number unity have identical in-version eigenvalues but obviously represent very differentstates of matter. The question of relating all the inver-sion eigenvalues to a response function remains, in thecases where possible, still unsolved. I. PRELIMINARIES
Let us start our discussion with some observations forsimple two- and four-band model Hamiltonians, whichwill be referenced throughout the paper. These proper-ties of the models that we discuss are only dependent ona generic inversion symmetry and not other symmetriespresent “accidentally” in these simple models.
The two band model is H = (cid:80) x (cid:2) c † x α +ˆ σ − i ˆ σ c x +1 + h.c. + c † x (1 + m ) ˆ σ c x (cid:3) , where α, m are two parameters and ˆ σ a are Pauli’s matri-ces. The model is symmetric under the inversion oper-ation c x → ˆ σ c − x . The Bloch representation of H takesthe simple form:ˆ H ( k ) = α cos k + sin( k )ˆ σ + (1 + m − cos k )ˆ σ , (2)and the inversion is implemented by the operator P = ˆ σ :ˆ σ ˆ H ( k )ˆ σ = ˆ H ( − k ) . (3) H is gapped, except when m = − α =0 and m = ∓
1, respec-tively. Throughout this paper, ˆ P k will denote the pro-jector onto the occupied bands at momentum k , whichis a K × K matrix ( K = total number of bands), whoseentries depend on k .The two special points k inv =0 , π where ˆ H ( k ) ismapped onto itself by inversion will play a special rolein the following discussion. We are going to examine the(non-zero) eigenvalues ζ (0) and ζ ( π ) of ˆ P k =0 P ˆ P k =0 andˆ P k = π P ˆ P k = π , respectively, for the 3 different insulatingphases of H . What we find is the following:1. ζ (0) = ζ ( π ) = −
1, for m > ζ (0) = ζ ( π ) = +1, for m < − ζ (0) = − ζ ( π ) = +1, − < m < Z topological invariant: χ P = (cid:89) k inv ,i ∈ occ. ζ i ( k inv ) , (4)which is topologically stable since one cannot change itsvalue without closing the gap of the Hamiltonian. Theexpression of χ P is similar to invariants formed for time-reversal and inversion invariant topological insulators in2D and 3D. For the model of Eq. 2, χ P takes the values χ P =+1for the insulating phases with m / ∈ [ − , χ P = − −101 E ξ −101 E ξ (a) (b)(c) (d) FIG. 1. Energy Spectra for the simple 1D 2-band modelwith open boundary conditions for (a) α = 0 m = − α = 0 m = 1 (trivial) Entanglement spectra for thetwo cases are shown in (c) and (d) respectively for a half-filledFermi sea ground state with periodic boundary conditions. for m ∈ [ − , χ P = − χ P = +1 do not display any end modes. This sim-ple model indicates that, indeed, systems with inversionsymmetry do posses non-trivial topological phases that,apparently, can be classified by the χ P invariant. As weshall see in the following sections, χ P can be linked to aphysical response of the system but it cannot completelyclassify the topological phases of a system with inversionsymmetry, even in 1D. It is instructive to repeat a similar anal-ysis on a 4-band inversion symmetric model. For this weuse the following Hamiltonian, written directly in theBloch representation:ˆ H ( k ) = sin( k )ˆΓ + sin( k )ˆΓ +(1 − m − cos k )ˆΓ + δ ˆΓ + (cid:15) cos( k )(1 + ˆΓ ) (5)where ˆΓ = σ z ⊗ τ x , ˆΓ = 1 ⊗ τ y , ˆΓ = 1 ⊗ τ z , andˆΓ = σ x ⊗ τ z . The Pauli’s matrices τ a , σ a act in theorbital and spin spaces, respectively. ˆ H ( k ) is symmet-ric under inversion, which is implemented by P = ˆΓ ,and is gapped except for a few values of the parameters δ, (cid:15), and m. Note that this system also has an acciden-tal time-reversal symmetry with T = ( iσ y ⊗ K , butthis can be broken without affecting the stability of thetopological state or removing the mid-gap modes in theentanglement spectrum. The two lower energy bands areassumed occupied, and in this case ˆ P P ˆ P and ˆ P π P ˆ P π are 4 × ζ i (0) and ζ i ( π ), i =1 ,
2. We are going to present the inversion eigenvalues forthe insulating phases of the model. There are six suchphases (we discuss only five of them) and their energyspectra with open boundary conditions are shown inFig. 2(a-e). Choosing representative values for the pa-rameters, we find:Case 1) b = δ = (cid:15) = 0 and M < ζ (0) = ζ (0) = +1 ζ ( π ) = ζ ( π ) = +1 . (6)and consequently χ P = +1.Case 2) b = δ = (cid:15) = 0 and M = 0 . ζ (0) = ζ (0) = − ζ ( π ) = ζ ( π ) = +1 . (7)and consequently χ P = +1.Case 3) b = 1, δ = 0 . (cid:15) = 0 and M = < ζ (0) = − , ζ (0) = +1 ζ ( π ) = ζ ( π ) = +1 . (8)and consequently χ P = − b = 1, δ = 1 . (cid:15) = 0 and M = 0 . ζ (0) = − , ζ (0) = +1 ζ ( π ) = +1 , ζ ( π ) = − . (9)and consequently χ P = +1.Case 5) b = 1, δ = 1 . (cid:15) = 0 . M = 0 . ζ (0) = +1 , ζ (0) = − ζ ( π ) = +1 , ζ ( π ) = − . (10)and consequently χ P = +1.The 4-band model reveals a far richer internal struc-ture. Case 1) can be identified with a trivial topologicalphase and, based on the value of χ P and on the pres-ence of end modes seen in Fig. 2(c), one will be inclinedto classify case 3) as a non-trivial topological insulator.But one will have clear difficulties with labeling cases 2),4) and 5). This is a clear indicative that χ P alone isnot enough for a full classification and that additionaltopological invariants are needed for a complete picture.It is instructive to consider the atomic limit of themodel. By the atomic limit we mean the limit of theadiabatic process in which the hopping terms between different sites are tuned to zero. Since the bands aredispersion-less and completely local (disentangled) in thislimit, it makes sense to talk about the parity of an en-tire band (or orbital), since its inversion eigenvalues at k = 0 , π are identical. For a model with 2 occupiedbands, the atomic limit can lead to the following cases,depending of how the occupied atomic orbitals behaveunder inversion: two occupied bands of parity + (labeled++), two occupied bands of parity − (labeled −− ), and −202E 00.51 ξ ξ −202E −202E 00.51 ξ ξ −202E −202E 00.51 ξ (a) (b) (c) (d) (e)(f ) (g) (h) (i) (j) FIG. 2. (a,b,c,d,e) Energy spectra for H in cases 1,2,3,4,5 respectively with open boundary conditions. (f,g,h,i,j) Entanglementspectra for H in cases 1,2,3,4,5 respectively with periodic boundary conditions. one band of parity + and one band of parity − (labeled+ − ). These three options give the complete classifica-tion of the trivial inversion symmetric insulators with 2occupied bands in 1D. Now, a direct calculation will showthat Cases 1), 4) and 5) can be connected to their atomiclimits without closing the insulating gap and that Case1) can be identified with ++ trivial insulator, while bothCases 4) and 5) can be identified with the + − trivial in-sulator. Note that Cases 4 and 5 can be adiabaticallyconnected to each other without closing the bulk insulat-ing gap. The −− trivial insulator occurs in our 4-bandmodel if we take the large m limit.Based on the above discussion and on the ab-sence/presence of the end modes in Fig. 2, we can con-sider the Cases 1, 4, 5 as completely trivial and Case 3 asnon-trivial, but Case 2 is still uncharacterized. It cannotbe continued to the trivial atomic limit without closingthe bulk gap, it displays end modes, yet χ P = +1. To dis-tinguish this phase we must carefully consider the parityeigenvalues. We see that when the k inv points are consid-ered separately, χ P ( k inv ) ≡ (cid:81) i ∈ occ. ζ i ( k inv ) = 1 . Thus, ateach k inv there are an even number of bands with nega-tive parity eigenvalues. In this situation, when for each k inv the local product over the eigenvalues of the occupiedbands is trivial (+1) we can define a second invariant χ (2) P ≡ (cid:81) k inv ,i ∈ occ./ ζ i ( k inv ) , (11)where the product over bands is defined to be the productof half of the bands with negative parity eigenvalues ateach k inv . Note, that we do not require there to be an evennumber of filled bands, just an even number of negativeparity eigenvalues. Out of the five cases discussed above,the χ (2) P invariant can only be defined for cases 1 and 2for which χ (2) P is trivial/non-trivial, respectively. As we shall see, the χ (2) P invariant is relevant and important forinversion symmetric insulators in 3D.The invariant χ = | n − n | , where n and n are thenumber of negative parity eigenvalues at k =0 and π , re-spectively, is also useful for classifying the inversion sym-metric insulators. This invariant generically indicateshow many times the insulating gap must close when onetakes the atomic limit. As we shall see, χ also gives thenumber of robust mid-gap modes in the entanglementspectrum localized on a single cut-boundary. For our 4-band model, χ =0 for Cases 1), 4) and 5); χ =1 for Case3) and χ =2 for Case 2).To conclude, the 2- and 4-band explicit models showthat the systems with inversion symmetry can displaytopologically distinct phases, i.e. they cannot be con-tinuously deformed into one another without closing theinsulating gap. These phases can be characterized by thedifferent values of χ , χ P or χ (2) P (when they can be con-sistently defined). In the remainder of the manuscript,we will try to elucidate the physical signatures of thesetopological phases and to give the physical meaning ofthe invariants mentioned above. For this, we will investi-gate the properties of the entanglement spectrum, whichwe demonstrate to contain robust edge modes, and thephysical responses of the inversion symmetric insulators. II. ENTANGLEMENT SPECTRUM OFTOPOLOGICAL INSULATORS WITHINVERSION SYMMETRY
In this section we discuss the bi-partite single-particleentanglement spectra for inversion symmetric topolog-ical insulators. Previous work on entanglement spec-tra in translationally invariant topological insulators wascarried out in Refs. 23, 29, 30, and 34, where it wasshown that the primary contributions to entanglementarise from states localized near the spatial cut between re-gions A and B. Additionally the entanglement spectrumof disordered Chern insulators has been investigated inRef. 32. The first indication that the presence of in-version symmetry is important for the structure of theentanglement spectrum was presented in Ref. 29. Hereit was shown that, while the physical edge spectrum ofa time-reversal and inversion invariant topological insu-lator is gapped in the presence of an added Zeeman field(which does not close the bulk gap), the entanglementspectrum still contains a gapless mode. The authors ofthat work link the existence of mid-gap states for eachcut in the entanglement spectrum with the existence ofa θ = π vacuum characteristic of a TRI non-trivial topo-logical insulator. Although, as we mentioned before (andwill discuss more in Sec. III), there is not always a di-rect and unique connection between the physical responseand protected states in the entanglement spectrum, thiswas an essential indication that inversion symmetry couldsupport topological states and that the properties of theentanglement spectra were closely connected with inver-sion symmetry.We start this section by detailing how to obtain the en-tanglement spectrum for noninteracting insulators. Wethen look at the entanglement spectrum of topological in-sulators and show that there are two fundamental proper-ties which may be present (i) protected mid-gap states atentanglement eigenvalue ξ = 1 / T -broken case only has protected mid-gap modes.For now we focus solely on insulators with a genericinversion symmetry and show its consequences on theentanglement spectrum. First, we show that if the systemis cut exactly in half, then there can be mid-gap states inthe entanglement spectrum located exactly at a value of1 /
2. This is equivalent to the statement that the mid-gapeigenvalues of the flat band Hamiltonian, for cuts exactlyin half, exhibit no finite-size level repulsion. We thengive an expression for the number of 1 / between inversion symmetric points.In everything presented below, it is very important toclarify that by a spatial cut in a translationally invariant system we mean a cut between primitive unit cells. Thispoint is important when we consider systems with par-tially broken translation symmetry e.g. the dimerizedmodels in Sec. IV. The physics is independent of thechoice of the unit cell. For example, in a multi-orbitalsystem with a one-site unit cell, the cut should not bemade through the orbitals on the same site. A. Obtaining the Entanglement Spectrum of anInsulator
All of the models we study are free fermion Hamil-tonians. To find the single-particle entanglement spec-trum we use Peschel’s method. We begin by assuminga quadratic Hamiltonian of an insulator with α = 1 ...K quantum states per site which is translationally invariant: H = (cid:88) k c † α,k H αβ ( k ) c β,k (12)and the canonical transformation U that diagonalizes it: U † HU = Diag( E n ) . (13) U is the matrix of eigenvectors u n ( k ) of energy E n ( k ): U ( k ) = ( u ( k ) , u ( k ) ....u K ( k )) (14)where each u n ( k ) is a K -component vector. In general,we will use k to denote the wave-vector of components k x , k y , etc..The relationship between the normal mode operators γ βk and the electron creation operators is c αk = U αβ ( k ) γ βk = u nα ( k ) γ nk (15)To calculate the single-particle entanglement spectrumwe simply need the correlation function: C αβij = (cid:104) c † iα c jβ (cid:105) , (16)where c † iα creates an electron in state α at site i . Theexpectation value is taken in the ground state. We canview this correlator as a matrix ˆ C ij , with entries thatdepend on i and j . We have: C αβij = (cid:88) k ,k e ik i − ik j (cid:104) c † αk c βk (cid:105) = (cid:88) k e ik ( i − j ) (cid:88) n ∈ occ. u nα ( k ) ∗ u nβ ( k )= (cid:88) k e ik ( i − j ) P αβk , (17)or more compactlyˆ C ij = (cid:88) k e ik ( i − j ) ˆ P k . (18)We want to make a translationally invariant cut along the y -direction so that k y is still a good quantum number ( k y is a short-hand notation for all the momenta parallel tothe cut, so we are implicitly also treating systems in 2 , C ij ( k y ) = 1 L (cid:88) k x e ik x ( i − j ) ˆ P k x ,k y , (19)where L is the total number of sites along the cut. Follow-ing Peschel, for the entanglement spectrum, we restrict i, j to be in region A, which is an explicit cut in positionspace. There are several physical choices for cuts, but fortopological insulators we will show that a spatial cut candistinguish between topological and trivial insulators.As an aside, note that for an insulator the spec-trally flattened Hamiltonian matrix where the statesabove/below the gap are flattened to energies +1 / − / H flat ( k x , k y ) = − ˆ P k x ,k y . (20)The two above expressions of the correlation functionand of the flat band Hamiltonian explicitly show thatthe entanglement spectrum, i.e. the eigenvalues of therestricted C αβij are identical to the energy levels of theflat band Hamiltonian with open boundaries in region Ashifted by a constant. As such, if the flat band Hamilto-nian is topological (i.e. has protected edge states), thenimmediately we know the entanglement spectrum willhave states localized on the cut. This agrees with theresults of Ref. 30. The interesting thing is that the flat-tened Hamiltonian can have mid-gap states even whenthe un-flattened one does not. B. Properties of the Entanglement Spectrum
We would like to first get an intuitive idea of how theentanglement spectrum of an insulator should look. Theone-body correlation function over the full system (notonly over region A) is a projector. It is the real spacerepresentation of the projector onto the occupied bands,and as such only has the eigenvalues 0 and 1. This isshown explicitly in Appendix A. When we make a cut,the eigenvalues of the one-body correlator deviate from0 , exactly inhalf. A cut exactly in half will enable us to show the ex-istence of exact degeneracies rather than exponential de-generacies only arising in the thermodynamic limit. Ourchoice of cut does not matter in the thermodynamic limit,where it cannot physically matter whether we make a cutexactly in the middle or away from the middle. It is how-ever the case that if we cut the system in two identical halves, we can prove things exactly, otherwise we can justgive arguments.The one-body matrix, computed in the basis c † i,α | (cid:105) ,takes the block form C = (cid:18) C L C LR C RL C R (cid:19) (21)where C L is the matrix of the left-half (the one we di-agonalize for the entanglement spectrum) ( C L ) ij = C αβij , i, j ∈ A ; C R is the matrix of the right-half ( C R ) ij = C αβij , i, j ∈ B ; C LR is the left-right coupling, ( C LR ) ij = C αβij , i ∈ A , j ∈ B ; with C RL = C † LR . Since ˆ C i,j = ˆ C i + n,j + n ,the following extra property is true if the cut is exactlysymmetric (in which case a proper translation of A gives B ): C R = C L . (22)We also find that C LR = C † LR . Moreover, the projectorproperty C = C : (cid:18) C L C LR C † LR C L (cid:19) (cid:18) C L C LR C RL C L (cid:19) = (cid:18) C L C LR C † LR C L (cid:19) (23)gives the following additional identities: C L (1 − C L ) = C † LR C LR ,C LR C † LR = C † LR C LR C L C LR + C LR C L = C LR . (24)Using the last equation, if ψ is an eigenstate of the en-tanglement spectrum matrix C L with eigenvalue (proba-bility) p : C L ψ = pψ (25)then C LR ψ is also an eigenstate with eigenvalue 1 − p : C L C LR ψ = C LR ψ − C LR C L ψ = C LR ψ − pC LR ψ = (1 − p ) C LR ψ. (26)If p = 1 / ψ and C LR ψ have the same 1/2 entanglementprobability, but as we shall see, this does not automati-cally mean that the p = 1 / ψ and C LR ψ are not linearlyindependent, in general.
1. Properties of Entanglement spectrum with time-reversalsymmetry
The entanglement spectrum maintains the symme-tries of the original Hamiltonian. For example, fortime-reversal symmetry of the original Hamiltonian T ˆ H ( k ) T − = ˆ H ( − k ) (equivalently, T ˆ P k T − = ˆ P − k ): T ˆ C ij ( k y ) T − = (cid:80) k x T e ik x ( i − j ) ˆ P k x ,k y T − == (cid:80) k x e − ik x ( i − j ) T ˆ P k x ,k y T − == (cid:80) k x e − ik x ( i − j ) ˆ P − k x , − k y == (cid:80) k x e ik x ( i − j ) ˆ P k x , − k y = ˆ C ij ( − k y ) (27)so we see that the correlator also has time-reversal sym-metry, and for spin 1 / T = − k and − k .Thus, there are entanglement Kramers’ doublets at time-reversal invariant points where k ≡ − k mod G where G is a reciprocal lattice vector. C. Inversion Symmetric Topological Insulators
In this section we give explicit arguments that the en-tanglement spectrum of an insulator with inversion sym-metry (and without any other symmetry) can have mid-gap states pinned at exactly / without level repulsion when cut exactly in half. An integer number of suchmodes is robust without splitting, so the classificationof the entanglement spectra of insulators with inversionsymmetry is given by an integer Z (compare with the Z case where an even number of modes would be unstable).As an example, in 1D, if a bulk insulator has a number n of filled bands with negative inversion eigenvalues at k = 0 and a number n at k = π , we give explicit ar-guments that the entanglement spectrum for a systemwith periodic boundary conditions (when the system iscut exactly in half) will have 2 | n − n | protected mid-gap modes at exactly 1 /
2. In more than 1D, there willbe conserved momenta parallel to the cut (say k y ), foran insulator cut in the x direction. When cut exactly inhalf, there will be 2 | n − n | zero modes situated at the K y = − K y mod G y for which, in the periodic bulk (be-fore the cut), there were n negative inversion eigenvaluesat ( k x , k y ) = (0 , K y ) and n negative inversion eigenval-ues at ( k x , k y ) = ( π, K y ). We illustrate this explicitlywith several examples in Sec. IV.
1. Properties of Entanglement Spectrum With InversionSymmetry
With inversion symmetry: P ˆ H ( k ) P − = ˆ H ( − k ) , P = 1; P = P − (28)we can define a unitary matrix B ij ( k ) which connects thebands at k and − k : | u i ( − k ) (cid:105) = B ∗ ij ( k ) P | u j ( k ) (cid:105) (29)where the indices i, j run over the occupied bands1 , ..., N . In fact, by performing simple band crossingsbetween the N bands below the gap (which does not in-fluence the physics in the gap which depends only on theground state), we can make the bands non-degenerate inwhich case we can use B ∗ ij ( k ) = e iφ ( k ) δ ij , but we do notneed to choose this gauge here. Since ˆ P k = η ( E F − ˆ H ( k )),where η ( x ) is the Heaviside function, we have: P ˆ P k P = ˆ P − k , (30) which can be used to show: P ˆ C ij P = (cid:88) k e ik ( i − j ) ˆ P − k = ˆ C ji = ˆ C † ij . (31)We now want to relate the appearance of these 1 / k x = 0 , π. In principle only two sites inthe x direction should be enough to reveal the physics.Of course, with just two sites, our cut has to be maderight in the middle of the two-site problem, i.e. we arecomputing the entanglement spectrum of one site vs. theother site. This seems a bit problematic at first because ifwe are looking for the properties of the energy spectrumin a topological insulator phase the wavefunctions of thestates localized on each end will overlap and the degener-acy of these low-energy end states will be lifted becauseof the small size. Crucially, we show that the flat-bandHamiltonian does not have such finite-size eigenvalue re-pulsion between the edge modes even when these modesrest on top of each other on the same site . That is, evenif we bring the ends close to each other e.g. on the samesite (which is the meaning of the one-site entanglementspectrum), it is still true that the end modes do not ex-hibit level repulsion and are degenerate. This statementis true in higher dimensions where the end states be-come propagating edge and surface states. We prove thisstatement for several particular cases, which indicate it istrue in the thermodynamic limit. We first show that forone occupied band (we do not particularize to a specificmodel), there are two mid-gap modes at exactly 1 / k = 0 , π are opposite. Then werepeat this procedure for a chain of four sites cut in half.We then show that for two occupied bands (we do notparticularize to a specific model), there are two mid-gapmodes at exactly 1 / k = 0 , π opposite (while the other two are the same),whereas there are four mid-gap modes at exactly 1 / k = 0 are opposite from theones at k = π (i.e. at one momentum both are negativeand at the other momentum both are positive). We againdo this for a chain of two sites cut in half, then for a chainof four sites cut in half. The main conclusion to be drawnfrom this is that in the flat-band Hamiltonian (entangle-ment spectrum), these mid-gap modes do not experienceeigenvalue repulsion. It is physically clear that, althoughour proofs are only for two and four site flat band Hamil-tonians (cut in half for the entanglement spectrum), levelrepulsion will not set in for larger systems: level repulsionbetween edge modes gets weaker as the distance betweenthem is increased. Finally, at the end, we look at the gen-eral case of N occupied bands for the two-site problemand prove that the number of 1 / | n − n | where n , n are the numberof negative eigenvalues at k = 0 , π . As there is no levelrepulsion when all modes are spatially on top of eachother, we do not expect level repulsion when the num-ber of sites is increased to the thermodynamic limit. Wecheck this numerically for several examples with largersystem sizes ( e.g.
100 sites). Our exercise shows thattime-reversal invariant insulators with inversion symme-try (or even the case with T slightly broken) are not theonly inversion symmetric topological insulators with pro-tected entanglement mid-gap states. These are but oneof a whole series of inversion symmetric insulators withmid-gap entanglement modes.
2. One Occupied Band, Two-Site Problem
First we look at a generic case with one occupied band,two-sites, and periodic boundary conditions. In this case, k -space contains only the points k =0 , π . The wavefunc-tion of the occupied band is | ψ ( k ) (cid:105) :ˆ H ( k ) ψ ( k ) = ε ( k ) ψ ( k ) (32)with inversion eigenvalues: P | ψ (0) (cid:105) = ζ (0) | ψ (0) (cid:105) , P | ψ ( π ) (cid:105) = ζ ( π ) | ψ ( π ) (cid:105) . (33)Since P is a unitary operator which squares to unity, wehave: P † = P (= P − ) and by taking scalar products inthe above we have:( ζ (0) − ζ ( π )) (cid:104) ψ (0) | ψ ( π ) (cid:105) = 0 (34)Hence if ζ (0) = − ζ ( π ) (the eigenvalues can never be zerodue to det P = 1) we have (cid:104) ψ (0) | ψ ( π ) (cid:105) = 0. Noticethat the Hamiltonian ˆ H ( k ) does not impose any restric-tions on the wavefunctions at different momenta k , i.e. at k =0 and k = π we are effectively diagonalizing inde-pendent Hamiltonians. What allows us to relate wave-functions at k = 0 , π is that they are both eigenstates ofthe same matrix P (it is important to recall that P is k -independent).For the two-site problem ( i = 1 , C L = ˆ C = L (cid:80) k ˆ P k = ( ˆ P + ˆ P π ) . (35)The eigenstates of the original Hamiltonian have oppositeinversion eigenvalues then per the above:ˆ P | ψ ( π ) (cid:105) = ˆ P π | ψ (0) (cid:105) = 0 (36)which means that ψ (0) and ψ ( π ), the original Hamil-tonian eigenstates, are also the eigenstates of the entan-glement spectrum, with two eigenvalues at 1 / C L | ψ (0) (cid:105) = | ψ (0) (cid:105) ; C L | ψ ( π ) (cid:105) = | ψ ( π ) (cid:105) . (37)We see that the original Hamiltonian can change, leadingto a change of ψ ( k ), but as long as the inversion eigen-values remain fixed and opposite to each other, and aslong as we can take the flat-band limit (both of whichmean no gap closing), the eigenvalues of the entangle-ment spectrum will be fixed at 1 / . It does not matter what the actual explicit model for ˆ H ( k ) is. If the inver-sion eigenvalues at k = 0 , π are not opposite, there is noreason why (cid:104) ψ (0) | ψ ( π ) (cid:105) = 0, and the 1 /
3. One Occupied Band, Four-Site Problem
With one occupied band with opposite inversion eigen-values, the two-site Hamiltonian has exact 1 / / k -space contains four mo-menta k j = jπ , j = 0 , , ,
3. Call the occupied eigen-state of the Hamiltonian, as above | ψ ( k ) (cid:105) , ˆ H ( k ) ψ ( k ) = ε ( k ) ψ ( k ). For a system cut in half, the entanglementspectrum is given by diagonalizing the matrix C ij = C i − j with i, j = 1 , C L = (cid:18) ˆ C ˆ C ˆ C † ˆ C (cid:19) (38)We have:ˆ C = ˆ C = 14 ( ˆ P + ˆ P π + ˆ P π + ˆ P π ) ≡ ˆ C (39)ˆ C = 14 ( ˆ P − ˆ P π + i ˆ P π − i ˆ P π ) ≡ ˆ C . (40)The eigenstate of C L corresponding to the eigenvalue ξ takes the form ( ψ A , ψ B ), which satisfy the equation:ˆ C ψ A + ˆ C ψ B = ξψ A , ˆ C † ψ A + ˆ C ψ B = ξψ B (41)Due to the presence of inversion symmetry, irrespec-tive of the inversion eigenvalues, we showed before that P ˆ C ij P = ˆ C ji = ˆ C † ij , which renders the second equationof Eq. 41: ˆ C P ψ A + ˆ C P ψ B = ξ P ψ B . (42)We see that this is consistent with the first equation ofEq. 41 if ψ B = m P ψ A , with m = 1. The eigenvalueequation to solve is then:( ˆ C + m ˆ C P ) ψ A = ξψ A (43)0Also because of inversion symmetry, we have that:ˆ P π = P ˆ P π P . (44)For the one-band problem, we know that:ˆ P P = ζ (0) ˆ P ; ˆ P π P = ζ ( π ) ˆ P π (45)where ζ (0) , ζ ( π ) are the inversion eigenvalues at k = 0 , π of the occupied band ψ ( k ). Hence to find the entan-glement spectrum we need to diagonalize the followingoperator:ˆ F = 14 [(1 + mζ (0)) ˆ P + (1 − mζ ( π )) ˆ P π + ˆ P π P ( P + im ) + P ˆ P π ( P − im )] . (46)For the half-mode, we pick an ansatz: ψ A = aψ (0) + bψ ( π ) , (47)which we show can diagonalize ˆ F for an appropriatechoice of a, b : a = − (1 + imζ ( π )) (cid:104) ψ ( π ) | ψ ( π ) (cid:105) b = (1 + imζ (0)) (cid:104) ψ ( π ) | ψ (0) (cid:105) . (48)This choice of a and b makes ( ˆ C + m ˆ C P ) ψ A independentof both ψ ( π ), and P ψ ( π ) in general (i.e. [ ˆ P π P ( P + im ) + P ˆ P π ( P − im )] ψ A = 0), as it should in order forour ansatz to be an eigenstate. With this choice of a, b we find (by taking [(1 + mζ (0)) ˆ P + (1 − mζ ( π )) ˆ P π ] ψ A )that in general the entanglement spectrum eigenvalue isdependent on (cid:104) ψ (0) | ψ ( π ) (cid:105) , and hence the mode is notfixed at 1 /
2. However, if the inversion eigenvalues at k =0 , π are opposite ζ ( π ) = − ζ (0) then (cid:104) ψ (0) | ψ ( π ) (cid:105) = 0and we find the eigenvalue of the entanglement spectrumof our ansatz to be: ξ = 14 (1 + mζ (0)) (49)Recall that have the liberty to choose the values of m = ± , which is equivalent to saying m = ± ζ (0). If wepick m = ζ (0), then our ansatz gives an eigenstate withentanglement eigenvalue equal to exactly 1 /
2. The otherchoice leads to ξ = 0, so is of no interest to us. Theeigenstate at 1 / ψ A , (cid:15) P ψ A ), where: ψ A = i (cid:104) ψ ( π | ψ ( π ) (cid:105) ψ (0) + (cid:104) ψ ( π | ψ (0) (cid:105) ψ ( π ) (50)In the pathological case when (cid:104) ψ ( π ) | ψ ( π ) (cid:105) = (cid:104) ψ ( π ) | ψ (0) (cid:105) = 0, both ψ (0) , ψ ( π ) are 1 / ψ A gives a robust 1 / / m = ζ ( π ) = − ζ (0), but using a different ansatz: ψ (cid:48) A = a (cid:48) ψ ( π ) + b (cid:48) P ψ ( π ) . (51) With this choice for m , the ˆ F matrix to be diagonalizedis: 14 [ ˆ P π P ( P − iζ (0)) + P ˆ P π ( P + iζ (0))] . (52)After straightforward calculations, we find the exact 1 / ψ (cid:48) A = ψ ( π ) + iζ (0) P ψ ( π ) . (53)Thus, for a completely generic Hamiltonian and its eigen-states we have shown that for one occupied band, if theinversion eigenvalues at 0 , π in the bulk are the oppo-site of each other, there are two exact midgap states at1 / ψ A , ζ (0) P ψ A ) , ( ψ (cid:48) A , − ζ (0) P ψ (cid:48) A ) and ψ A , ψ (cid:48) A as above. The standard in-tuition about interface or boundary states is that theeigenvalues repel less as the length of the system is in-creased, and the entanglement spectrum for the case an-alyzed here will have two exact 1 / /
4. Two Occupied bands, Two-Site Problem
On our way to the most general case we now analyzethe two-site, two occupied band problem. This follows inthe same fashion as the previous example, except that wenow have an extra complication. Namely, there are moreoptions for the sets of occupied-band inversion eigenval-ues.We have the two occupied bands of the original Hamil-tonian:ˆ H ( k ) ψ ( k ) = (cid:15) ( k ) ψ ( k ); ˆ H ( k ) ψ ( k ) = (cid:15) ( k ) ψ ( k ) (54)For k = 0 , π we generically have: (cid:104) ψ (0) | ψ (0) (cid:105) = (cid:104) ψ ( π ) | ψ ( π ) (cid:105) = 0 . (55)We denote the inversion eigenvalues for ψ (0), ψ (0), ψ ( π ), ψ ( π ) by ζ (0) , ζ (0) , ζ ( π ) , ζ ( π ) respectively.The order of the occupied inversion eigenvalues at eachinversion symmetric momentum can be changed withoutaffecting the topological structure so we assume that allnegative inversion eigenvalues are listed first. We nowcalculate the number of midgap 1 / k = 0 is the same as the numberof negative inversion eigenvalues at π (which means thenumber of positive inversion eigenvalues is also the same),it is easy to prove that in general there are no protected1 / Case 1:
The number of negative eigenvalues at k = 0differs from the number of negative eigenvalues at k = π by ± i.e. they are both different): ζ (0) ζ ( π ) = ζ (0) ζ ( π ) = ζ (0) ζ ( π ) = ζ (0) ζ ( π ) = − (cid:104) ψ (0) | ψ ( π ) (cid:105) = (cid:104) ψ (0) | ψ ( π ) (cid:105) = 0 (cid:104) ψ (0) | ψ ( π ) (cid:105) = (cid:104) ψ (0) | ψ ( π ) (cid:105) = 0 . (57)The one-site entanglement spectrum obtained by cut-ting the system in half is obtained by diagonaliz-ing the operator C L = ( ˆ P + ˆ P π ) / P = (cid:80) i =1 | ψ i (0) (cid:105) (cid:104) ψ i (0) | , ˆ P π = (cid:80) i =1 | ψ i ( π ) (cid:105) (cid:104) ψ i ( π ) | . Due totheir inversion eigenvalues, eigenstates at π have zeroeigenvalue under the projector at 0 (and vice-versa) butunit eigenvalue under the projector at π. We see that themodes at 1 / ψ (0), ψ (0), ψ ( π ), ψ ( π ). There are exactly 4 ofthem, twice the difference between negative and positiveeigenvalues at the two inversion symmetric points. Case 2
The number of negative eigenvalues at k = 0differs from the number of negative eigenvalues at k = π by ±
1: this implies that at one k point, both inversioneigenvalues are identical. Without loss of generality, letthis point be k = π and let the eigenvalue products be: ζ (0) ζ ( π ) = ζ (0) ζ ( π ) = − ζ (0) ζ ( π ) = ζ (0) ζ ( π ) = 1 (58)which renders the following inner products to be zero: (cid:104) ψ (0) | ψ (0) (cid:105) = (cid:104) ψ (0) | ψ ( π ) (cid:105) = (cid:104) ψ (0) | ψ ( π ) (cid:105) = 0 . Consider the eigenvalue problem:12 ( ˆ P + ˆ P π ) ψ A = αψ A . (59)We expand the state ψ A into the (non-orthogonal) set ofeigenstates ψ (0), ψ (0), ψ ( π ), ψ ( π ): | ψ A (cid:105) = a | ψ (0) (cid:105) + a | ψ (0) (cid:105) + b | ψ ( π ) (cid:105) + b | ψ ( π ) (cid:105) (60)As ψ (0) is orthogonal with all the other eigenstates atboth k = 0 , π since it has a different inversion eigen-value, it is then obvious to see that the first 1 / a , a , b , b ) = (1 , , , / | ψ A (cid:105) = a | ψ (0) (cid:105) + b | ψ ( π ) (cid:105) + b | ψ ( π ) (cid:105) . Thereis a slight complication with this expansion since noth-ing guarantees that the states | ψ (0) (cid:105) , | ψ ( π ) (cid:105) , | ψ ( π ) (cid:105) are orthogonal: in fact, in the generic case, they are not.Moreover, it is not clear that they are even linearly in-dependent. We will assume that the states are linearly independent. This is a perfectly valid procedure since ifthe | ψ (0) (cid:105) , | ψ ( π ) (cid:105) , | ψ ( π ) (cid:105) are not independent, we willsimply get a non-trivial nullspace. However, the non-zeroeigenvalues are still good eigenvalues of the entanglementmatrix. The matrix to diagonalize is: (cid:104) ψ (0) | ψ ( π ) (cid:105) (cid:104) ψ (0) | ψ ( π ) (cid:105)(cid:104) ψ (0) | ψ ( π ) (cid:105) ∗ (cid:104) ψ (0) | ψ ( π ) (cid:105) ∗ (61)with an obvious 1 / a , b , b ) =(0 , −(cid:104) ψ (0) | ψ ( π ) (cid:105) , (cid:104) ψ (0) | ψ ( π ) (cid:105) ) ( a = 0). In the non-generic case when (cid:104) ψ (0) | ψ ( π ) (cid:105) = (cid:104) ψ (0) | ψ ( π ) (cid:105) = 0, ψ (0) is the other 1 / / , π is ± N Occupied bands, Two-Site Problem
We now show that the two-site problem with n neg-ative inversion eigenvalues at k = 0 and n negative in-version eigenvalues at k = π with a total number N ofoccupied bands contains 2 | n − n | zero modes in the en-tanglement spectrum when a real-space cut is made on asystem with periodic boundary conditions ( i.e. there aretwo cuts). The simplest case, which should be obviousfrom our previous examples, is that all the N inversioneigenvalues at k = 0 are identical and are the oppositeof the N eigenvalues at k = π . In this case, the pro-jector at one of the inversion symmetric k ’s annihilatesall the eigenstates at the other inversion symmetric k ,and the 2 N occupied eigenstates of the original two-siteHamiltonian are also the eigenstates of the entanglementspectrum at fixed eigenvalue 1 /
2. Due to their orthog-onality, they are linearly independent. From here it isclear that our formula is correct for this case.Now we will prove the more general formula. Let n and n be the number of eigenvectors for the − P P ˆ P and ˆ P π P ˆ P π , respectively, and assume n > n . Recall that K is the number of orbitals persite, so P is a K × K matrix acting on C K . P has ± H ± ( H − + H + = C K ).Now the subspaces ˆ P C K and ˆ P π C K are invariant un-der the inversion operation P , and ˆ P C K ∩ H − is pre-cisely the subspace spanned by the n eigenvectors ofˆ P P ˆ P corresponding to its negative eigenvalue. Sim-ilarly, ˆ P π C K ∩ H − is precisely the subspace spannedby the n eigenvectors of ˆ P π P ˆ P π corresponding to itsnegative eigenvalue. Since dim[ ˆ P C K ∩ H − ] = n and2dim[ ˆ P π C K ∩ H − ] = n , with n > n , we can always find n − n vectors Ψ n in ˆ P C K ∩ H − that are orthogonal toany vector in ˆ P π C K ∩H − . Since these vectors are in H − ,they are also orthogonal to any vector in ˆ P π C K ∩ H + .In other words, Ψ n ’s are n − n vectors in ˆ P C K per-pendicular to all the vectors in ˆ P π C K . Consequently: C L Ψ n = 12 ( ˆ P + ˆ P π )Ψ n = 12 Ψ n , (62)for all n − n vectors Ψ n . Following the same arguments,we can find n − n vectors in ˆ P π C K ∩ H + that areorthogonal to ˆ P C K , and consequently another set of n − n eigenvectors with eigenvalue . In total, thereare 2( n − n ) robust modes of C L at 1 /
2. For a moreexplicit proof see Appendix C.
6. Extension to higher dimensions
The extension to higher dimensions is just a matter ofreinserting the extra momenta which are conserved in thepresence of the cut. As an example let us consider 2Dwith an entanglement cut parallel to the y-axis so that k y is a conserved quantum number. The exact mid-gapmodes in the entanglement spectrum will exist only atinversion symmetric points in the momentum parallel tothe cut, i.e. k y = 0 , π . However, since bands becomecontinuous when k y is finely discretized, in the thermo-dynamic limit the existence of 1 / / / , / . This means the system is anon-trivial insulator ( i.e. S ent cannot be made to vanish)if the number of negative (or positive) inversion eigen-values is different between inversion symmetric points.We would like to know at which inversion symmetric k y the exact 1 / | n − n | / k y = 0 when the number of negative inversion eigenval-ues at ( k x , k y ) = (0 ,
0) differs by | n − n | from the num-ber of negative inversion eigenvalues at ( k x , k y ) = ( π, . Additionally there will be exactly 2 | n (cid:48) − n (cid:48) | / k y = π when the number of negative inversion eigenvaluesat ( k x , k y ) = (0 , π ) differs by | n (cid:48) − n (cid:48) | from the numberof negative inversion eigenvalues at ( k x , k y ) = ( π, π ) . Thegeneralization to higher dimensions is the trivial exten-sion of this.
D. Spectral Flow in the Entanglement Spectrum
As mentioned earlier in this section the two importantfeatures of the entanglement spectra of inversion symmet-ric insulators are protected mid-gap modes, and spectralflow. What we mean by spectral flow is a continuousconnection between the valence and conduction bulk en-tanglement bands through the entanglement edge states(an example is seen in Fig. 8h,i ). For a TRI topologicalinsulator in 2D and 3D, or for a Chern insulator in 2D,the entanglement spectrum mirrors the energy spectrumof the open-boundary Hamiltonian. In fact, we have al-ready shown an explicit map between the entanglementspectrum and the energy spectrum of the open bound-ary spectrally flattened Hamiltonian. This implies thatif there is spectral flow between the conduction and va-lence bands in the energy spectrum then such a flow ex-ists in the entanglement spectrum. In fact, this is theonly case where there is true spectral flow in the en-tanglement spectrum. Out of the entire set of inversioninvariant topological insulators only a small subset havespectral flow. Instead most non-trivial systems simplyexhibit protected mid-gap states (or bands) but these donot continuously interpolate between the bulk entangle-ment bands.There is a nice example which illustrates this di-chotomy. Let us consider the 3D strong topological insu-lator with both inversion and time-reversal symmetries.If we preserve P but break T, spectral flow genericallydisappears from the energy spectrum because gaps areopened in the surface state spectrum. The degenera-cies which existed in the entanglement spectrum at thetime-reversal invariant momenta when T is preserved, arealmost all broken, with the exception of the protected de-generacy for states at ξ = 1 / . This degeneracy splittingbreaks the spectral flow in the entanglement spectrum,and opens gaps at the TR-invariant momenta as shownfor a specific model in Fig. 9h,i. As such, the entan-glement spectrum is capable of distinguishing the subtledifference between topological insulators with T and P symmetry from topological insulators with only P sym-metry. III. THE LINEAR RESPONSE
To date, some of the most spectacular features of topo-logical insulators are their responses to external fields.The Chern insulators exhibit a quantized Hall effect andthe 3D TRI topological insulators exhibit a topologicalmagneto-electric effect. The topological invariants whichdistinguish these states from trivial insulators are directlyconnected with the corresponding response coefficient. Infact, a whole ladder of topological responses was uncov-ered in Ref. 9. With this precedent one would hope thatthe inversion invariant topological insulators would alsoexhibit some type of defining physical response. How-ever, this turns out to be true in only a limited set of the3inversion invariant topological insulators. The situationis quite varied (remember we only have generic access tothe information held in the inversion eigenvalues): someinsulators have unique well defined topological responses,some systems can exhibit one of several allowed topolog-ical responses, and for others it is unclear if there is anytopological response at all. We will see examples of allthree cases in Sec. IV. In this section though we focuson the first case where insulators do exhibit a unique re-sponse which is the most interesting physical case. Wediscuss responses in 1,2, and 3 dimensions and then webriefly mention how the general pattern might be ex-tended to higher dimensions to make contact with Refs.9 and 11 in Appendix K.
A. 1D inversion symmetric insulators
The following discussion applies to a generic one di-mensional K -band insulator with N occupied bands, andwith a generic inversion symmetry i.e. P ˆ H ( k ) P − = ˆ H ( − k ) , (63)where the inversion matrix P is unitary and squares tothe identity: P † P = 1 , PP = 1 . (64)As explicitly shown in Appendix D, the charge polariza-tion P of a 1D insulator behaves as: P → − P + je (65)under inversion, where j is a gauge-dependent integer.This shows that the polarization of 1D systems with in-version symmetry can take only two values, 0 and e/ e. Wewill prove that if the χ P invariant, defined as the prod-uct of all inversion eigenvalues of the occupied bands,takes the value 1, then P = 0, and if χ P = −
1, then P = e/
2. More precisely, we will establish that: P = e πi Log (cid:34) N (cid:89) i =1 ζ i (0) ζ i ( π ) (cid:35) . (66)The integer ambiguity of the logarithm is identical to theinteger ambiguity of the polarization.For this, we define the k -dependent N × N unitary ma-trix ˆ B ( k ) B ij ( k ) = (cid:104) u i, − k |P| u j,k (cid:105) , (67)where the indices i and j run only over the occupiedbands. The inversion eigenvalues ζ i (0) and ζ i ( π ) coin-cide with the eigenvalues of the matrix ˆ B ( k ), when eval-uated at the special inversion k -points k =0 and π . It isthen obvious that the determinant of ˆ B ( k ) at these k inv points is the product of the inversion eigenvalues at thatinversion invariant point:det[ ˆ B ( k inv )] = N (cid:89) i =1 ζ i ( k inv ) . (68)We now turn to the calculation of the polarization P = e π (cid:90) π − π dkA ( k ) , (69)where A ( k ) is the adiabatic connection: A ( k ) = − i (cid:88) i ∈ occ (cid:104) u i,k |∇ k | u i,k (cid:105) . (70)We will use the following important relation, which isproven in Appendix E: A ( − k ) = − A ( k ) + i Tr[ ˆ B ( k ) ∇ k ˆ B † ( k )] . (71)The last term can be written in the equivalent form:Tr[ ˆ B ( k ) ∇ k ˆ B † ( k )] = −∇ k Log (cid:2) det[ ˆ B ( k )] (cid:3) . (72)We can now proceed as follows: P = e π (cid:82) π dk [ A ( k ) + A ( − k )]= e πi (cid:82) π dk ∇ k Log (cid:2) det[ ˆ B ( k )] (cid:3) , (73)with the final answer: P = e πi (cid:104) Log (cid:2) det[ ˆ B ( π )] (cid:3) − Log (cid:2) det[ ˆ B (0)] (cid:3)(cid:105) . (74)This, together with Eq. 68 and the fact that the deter-minants can take only the values ±
1, so that det[ ˆ B ] =1 / det[ ˆ B ], prove the statement of Eq. 66.We mention that similar arguments were used in Ref. 9to classify 1D particle-hole symmetric insulators via a Z invariant. In fact, the Z invariant found there is exactlythe value of the charge polarization modulo an integer.For a 1D model with both inversion and particle-holesymmetry, such as the 1D lattice Dirac model, the in-variants coincide. In the non-trivial phase, the 1D Diracmodel exhibits mid-gap energy modes bound to the endsof an open chain. The requirement of particle-hole sym-metry restricts these modes to lie at zero energy if thereare an odd number of them. An even number on eachend is not stable and the degeneracy can be lifted, whichis another manifestation of the Z nature. The mini-mal case is one mode on each end and with particle-holesymmetry at half filling one mode is filled and one isempty. This leads to an excess charge of + e/ − e/ e/ both ends or − e/ both ends. However, because of the gauge-variance of thepolarization this is equivalent to the polarization in theparticle-hole symmetric case. Thus, both insulators havethe same topological electric response. One can form acomplimentary argument by using the effective responseaction for particle-hole symmetric insulators given in Ref.9 and 37: S eff = 12 (cid:90) dxdtP (cid:15) µν F µν (75)where F µν is the field-strength tensor of the externallyapplied electro-magnetic field. The argument for thequantization of P is as follows. In the partition functionthe phase due to this term is e iS eff and under particle-hole symmetry P → − P . Thus if our system is to beparticle-hole symmetric we must have e iS eff = 1 . Forconstant P the integral gives 2 πn for integer n and wehave e πinP = 1 and thus P = 0 , / Z in dimen-sionless units. Since P → − P under inversion symme-try the same argument holds and P is quantized thereas well. A similar argument for magneto-electric polar-izability of 3D insulators with inversion symmetry wasgiven in Ref. 29. As an aside, we recall that in 1Dwe also had an invariant χ (2) P which helped classify the4-band insulator example. We do not know of any re-sponse related to this invariant in 1D inversion invariantinsulators.Additionally, Eq. 66 can be derived using an alterna-tive approach based on a monodromy argument. Theeffect of a magnetic flux Φ through a large one dimen-sional ring can always be gauged away by a transforma-tion: Ψ → e − i Φ Ψ, which is equivalent to a translation in k -space by Φ. The evolution of the states in responseto an adiabatically slowly varying magnetic flux can beunderstood from the evolution of the Bloch states in re-sponse to an adiabatic translation of k -space. The mon-odromy ˆ U ( k, k ) describes the evolution of the occupiedBloch states when the k of the Bloch Hamiltonian ˆ H ( k )is adiabatically varied, and it is the unique solution ofthe equation: ddk ˆ U ( k, k ) = i [ ˆ P k , ∂ k ˆ P k ] ˆ U ( k, k ) , (76)with the initial condition ˆ U ( k , k )= ˆ P k , assuming thatwe start the evolution from k .The monodromy ˆ U ( k , k ) maps the space of occupiedBloch states ˆ P k C K at k into the space of occupiedBloch states ˆ P k C K at k . Since k = ± π are the same,ˆ U ( π, − π ) takes the space ˆ P − π C K into itself, and is a uni-tary operator that we call ˆ U γ , where γ is one of the pathsshown in Fig. 3. U γ gives the change occurring after a fullquantum of magnetic flux has been pumped through thesystem. If bases { ψ i ( k ) } were pre-chosen for all ˆ P k C K spaces, the matrix U ij ( k ) = (cid:104) ψ i ( k ) | ˆ U ( k, k ) | ψ j ( k ) (cid:105) sat-isfies the parallel transport equation ddk ˆ U ( k ) = ˆ A ( k ) ˆ U ( k ) , (77) ! ’ ! - ! ! FIG. 3. The adiabatic transport is carried over γ , . where ˆ A ( k ) is the full non-abelian adiabatic connectiondiscussed by Wilczek and Zee in Ref. 39.There is a direct relation between the determinant ofthe monodromy and the line integral of the abelian con-nection (the trace of the full connection):det[ ˆ U ( k f , k i )] = exp (cid:32) k f (cid:82) k i Tr[ ˆ A ( k )] dk (cid:33) . (78)Indeed, working with predefined basis sets for ˆ P k C K andbreaking the interval k f , k i in small subintervals k f , k n , . . . , k i , we have, up to second order corrections:ˆ U ( k f , k i ) = ˆ U ( k f , k n ) ˆ U ( k n , k n − ) . . . ˆ U ( k , k i )= (cid:0) I + ( k f − k n ) ˆ A ( k n ) (cid:1) . . . (cid:0) I + ( k − k i ) ˆ A ( k i ) (cid:1) . (79)Taking the determinant on both sides and using someelementary identities, we obtaindet[ ˆ U ( k f , k i )] = (cid:0) k f − k n )Tr[ ˆ A ( k n )] (cid:1) . . . (cid:0) k − k i )Tr[ ˆ A ( k i )] (cid:1) = e ( k f − k n ) Tr [ ˆ A ( k n )] . . . e ( k − k i ) Tr [ ˆ A ( k i )] , (80)from which the identity of Eq. 78 follows. The identityis valid in higher dimensions too.Assuming k = − π , a conjugation of Eq. 76 with theinversion operator P gives: ∂ k {P ˆ U ( k, − π ) P − } = i [ ˆ P − k , ∂ k ˆ P − k ] P ˆ U ( k, − π ) P − , with the initial condition P ˆ U ( − π, − π ) P − = ˆ P ( π ). Thisis just the equation for ˆ U ( − k, π ), which shows that P ˆ U ( k, − π ) P − coincides with ˆ U ( − k, π ). Equivalently wecan think that P sends γ into γ (cid:48) in Fig. 3. Now obviouslyˆ U γ ˆ U γ (cid:48) equals the identity, therefore:det[ ˆ U γ P ˆ U γ P − ] = 1 . (81)Using the elementary properties of the determinant, weconclude that det[ ˆ U γ ] =1, hence det[ ˆ U γ ] can take onlytwo values: det[ ˆ U γ ] = ± . (82)5The following calculation will show that the casesdet[ ˆ U γ ]= ± P =0 and P = e (mod Z ), re-spectively. Indeed:det[ ˆ U γ ] = det[ ˆ U ( π,
0) ˆ U (0 , − π )]= det[ ˆ U ( π, P ˆ U (0 , π ) P − ]= det[ ˆ U ( π,
0) ˆ P P ˆ P ˆ U (0 , π ) ˆ P π P − ˆ P π ] . (83)In the last line we have inserted the projectors ˆ P ,π to seeexplicitly the spaces on which P is acting. Using againthe elementary properties of the determinant and the factthat ˆ U ( π,
0) ˆ U (0 , π )=1, we obtain:det[ ˆ U γ ] = det[ ˆ P P ˆ P ] det[ ˆ P π P − ˆ P π ] (84)or det[ ˆ U γ ] = N (cid:81) i =1 ζ i (0) ζ i ( π ) . (85)and thus P = e πi Log det[ ˆ U γ ] (86)(cf. Eq. 78).For a topological insulator, no matter what definitionone uses, it is always the case that, when the hoppingterms between the neighbors are adiabatically turned off,that is, when one takes the atomic limit, the insulatinggap of the system closes at some point in the process. Onecan investigate this issue using the inversion eigenvaluesdirectly, as we have focused on, but here we see how thephysical response enters into the picture. Note that theinversion eigenvalues can be easily computed for simplemodels, but may not always be available, especially forcomplex materials. P or U γ are physically measurable,so they can provide physical signatures of the non-trivialstate. In the atomic limit, the bands have no disper-sion, so in this limit ˆ U γ is just the identity matrix. IfDet[ ˆ U γ ]= −
1, it is obvious that U γ cannot be smoothlyconnected to the identity and the insulator is topological.However, we also must note that if Det[ ˆ U γ ]=1, it is not necessary that the insulator is trivial.We can refine the investigation by asking when can ˆ U γ be smoothly connected to the identity? For this, let uslook again at Eq. 83:ˆ U γ = ˆ U ( π,
0) ˆ P P ˆ P ˆ U (0 , π ) ˆ P π P ˆ P π . (87)The first term, ˆ U ( π,
0) ˆ P P ˆ P ˆ U (0 , π ), is just ˆ P P ˆ P par-allel transported from k =0 to k = π . The parallel trans-port does not alter the eigenvalues of ˆ P (0) P ˆ P (0), whichare pinned at ± P =1). The eigenvalues ofˆ P ( π ) P ˆ P ( π ) are also pinned at ±
1. So what we havein Eq. 87 is a product of two operators with eigenvaluespinned at ± U γ can be deformed intothe identity, then ˆ U ( π,
0) ˆ P P ˆ P ˆ U (0 , π ) can be turnedinto the inverse of ˆ P π P ˆ P π and this requires that ˆ P P ˆ P and ˆ P π P ˆ P π have identical eigenvalues, counting the de-generacy too. The conclusion is: if the set of inversioneigenvalues of ˆ P P ˆ P and ˆ P π P ˆ P π (88)are not identical, then ˆ U γ cannot be connected to theidentity. Since the eigenvalues are restricted to just ± χ = (cid:12)(cid:12)(cid:12) Tr (cid:110) ˆ P P ˆ P − ˆ P π P ˆ P π (cid:111)(cid:12)(cid:12)(cid:12) , (89)tells how many eigenvalues are different for the two ma-trices in Eq. 88. To reach the atomic limit, we need to flip χ inversion eigenvalues, and that will require a minimumof χ gap closings. Unfortunately, we were not able tofind an expression of the topological invariant χ solely interms of the monodromy ˆ U γ , but we know there are pre-cisely χ topological obstructions when trying to connectthe monodromy to the identity. The topological invari-ant χ also gives the number of robust edge modes seenin the entanglement spectrum on a single cut as shownin Sec. II. B. 2D inversion symmetric insulators
The physical response of 2D inversion symmetric in-sulators is much richer than that of 1D. Based solely onthe inversion eigenvalues, one can define several invari-ants, the first of which is the isotropic extension of χ P to2D i.e. χ P = (cid:89) k inv ; i ∈ occ. ζ i ( k inv ) (90)where k inv runs over all four inversion invariant k -points.We show that χ P = ( − C , (91)where C is the first Chern number of the ground state.Thus if χ P is negative the system must be in a quantumHall state, and if it is positive it is in a state with aneven Chern number which can be zero. Thus, only if it isnegative are we sure it is in a topological insulator state.We will prove the statement of Eq. 91 in two ways, firstusing a band crossing argument and then a monodromyargument. Let us begin by assuming we are in a trivialinsulator state in an atomic limit with N occupied bandsand that we have inversion symmetry. We can reachany non-trivial topological insulator state from this limitthrough a series of Hamiltonian deformations that willlead us through band crossings. Our assumption of anatomic limit implies that χ P = +1 initially. If we wantto generate χ P = − !!’
01 20’3456
FIG. 4. The two paths γ and γ used in the monodromyargument in 2D. of band crossings between bands with opposite inversioneigenvalues. Assume we have an odd number of suchcrossings. This implies that there must be an odd numberof crossings at the inversion invariant points. This is truesince crossings occurring at non-invariant k are accompa-nied by a crossing at − k always giving an even number ofeigenvalue switches which will not affect the value of χ P . Thus we only need to consider the odd number of cross-ings occurring at the invariant momenta. The genericHamiltonian of each crossing between opposite inversioneigenvalue states near an inversion invariant momentumis H = p σ + p σ + m eff σ where ( p , p ) is the mo-mentum away from the inversion invariant point, m eff isa term parameterizing the distance to the band crossing,and σ is the inversion operator projected onto the twocrossing bands. Exactly at the inversion invariant mo-mentum the Hamiltonian reduces to m eff σ as it mustin order to commute with P there. As the crossing occurs m eff switches sign and an inversion eigenvalue of the oc-cupied band is changed. We note that this Hamiltonianis a 2D massive Dirac Hamiltonian which is switching thesign of the mass. At such a crossing, the Chern numberchanges by ± χ P = − i.e. C = 0. The inversion eigenvalues thus give usa rough way to characterize the quantum Hall effect inan insulator.The corresponding monodromy argument proceeds asfollows. We consider the monodromy corresponding tothe path 01234560 in Fig. 4, starting from the middleof the Brillouin zone and continuing on its rim. Thispath can be also be viewed as the composition γ + γ (cid:48) ofthe paths γ =012340 and γ (cid:48) =045610, and the monodromycorresponding to γ + γ (cid:48) can be written as a product ofpartial monodromies:ˆ U γ + γ (cid:48) = ˆ U ˆ U ˆ U ˆ U ˆ U ˆ U ˆ U ˆ U . (92)Since the products ˆ U ˆ U , ˆ U ˆ U , ˆ U ˆ U and ˆ U ˆ U are all equal to the identity, taking the determinant ofEq. 92 leads us to conclude that det[ ˆ U γ + γ (cid:48) ]=1. This isnot surprising and is related to the fact that det[ ˆ U γ + γ (cid:48) ] = e iπC (see Eq. 78) and the first Chern number is aninteger.The next observation is that inversion sends γ into γ (cid:48) and consequently:det[ ˆ U γ + γ (cid:48) ] = det[ ˆ U γ P ˆ U γ P ] = det[ ˆ U γ ] . (93)The conclusion is that the determinant of ˆ U γ can onlytake the values ± . Since the path γ encircles half of theBrillouin zone, and the adiabatic curvature is symmetricwhen k → − k, it is also true that det[ ˆ U γ ] = e iπC (seeEq. 78).We now take a closer look at ˆ U γ . Since inversion sends10 into 40 and 20’ into 30’, we have:ˆ U γ = ˆ U ˆ U ˆ U (cid:48) ˆ U (cid:48) ˆ U ˆ U = P ˆ U P ˆ U P ˆ U (cid:48) P ˆ U (cid:48) ˆ U ˆ U (94)Inserting the appropriate projectors to specify explicitlyon which spaces are the P operators acting, taking thedeterminant and using its elementary properties togetherwith the fact that ˆ U ij ˆ U ji equals the identity, we obtain:( − C = det[ ˆ U γ ] = det[ ˆ P , P ˆ P , ] det[ ˆ P ,π P ˆ P ,π ] × [det { ˆ P π,π P ˆ P π,π ] det[ ˆ P π, P ˆ P π, ] , (95)which completes monodromy argument for Eq. 91.We will briefly mention two other interesting inversioninvariants one can define in 2D. The first is an anisotropicinvariant defined by taking the product of the inver-sion eigenvalues at only two invariant momenta in the2D Brillouin zone. We provide an example in Sec. IVwith χ P = +1 but where this anisotropic invariant isnon-trivial. This invariant has interesting implicationsfor the charge polarization, but there are some subtletiesthat we will illustrate. To begin, assume we have an in-version invariant, insulator Hamiltonian ˆ H ( k x , k y ) withN occupied bands such that (cid:81) Ni =1 ζ i (0 , ζ i (0 , π ) = − . This implies that the 1D Hamiltonian ˆ H (0 , k y ) has acharge polarization P = e/ H (0 , k y ) awayfrom the k y axis. The 1D Hamiltonian ˆ H ( δk x , k y ) is not generically invariant under inversion symmetry becauseit gets mapped onto the Hamiltonian ˆ H ( − δk x , k y ) andthus the polarization does not have to remain quantizedwith value e/ . In fact, by the time we have deformedall the way to the Hamiltonian ˆ H ( π, k y ) the polariza-tion, which must again be quantized since this Hamilto-nian does have inversion, can be completely different. Sowhile we can think of ˆ H ( k x , k y ) as a gapped interpolationbetween ˆ H (0 , k y ) and ˆ H ( π, k y ), inversion symmetry, andthus the value of the polarization, is not preserved alongthe interpolation. Intuitively this makes sense becauseit is exactly when the polarization changes its quantized7value that the system has an odd Chern number. An oddChern number is allowed because a quantum Hall effect isnot forbidden by the requirement of inversion symmetry.Thus, if we have 2D inversion symmetry the anisotropicinvariant does not determine the 2D charge polarization.As an aside, since we clearly know why inversion sym-metry fails, we immediately know how to fix the problem.We fix it by requiring a reflection symmetry ( i.e. paritysymmetry) about an axis, instead of inversion. With-out loss of generality we can have a reflection symmetry M such that M ˆ H ( k x , k y ) M − = ˆ H ( k x , − k y ) . We willsee that the reflection eigenvalues will specify the po-larization in the y-direction (for this choice of reflectionsymmetry). First, we see that the inversion invariantmomenta are also reflection invariant momenta. Thus[ ˆ H ( k inv ) , M ] = 0 , and we can label the states at thesepoints by their reflection eigenvalues which we will alsocall ζ i ( k inv ) . Now let us assume the same setup as theprevious paragraph with (cid:81) Ni =1 ζ i (0 , ζ i (0 , π ) = − . Thepoint is that now when we adiabatically deform awayfrom the k y axis the 1D Hamiltonian ˆ H ( δk x , k y ) is in-variant under reflection. Additionally, the y-componentof the polarization is quantized as long as reflection isa good symmetry. Thus, ˆ H ( k x , k y ) is a gapped inter-polation along which reflection symmetry is preservedand thus the y-component of the polarization is fixedand quantized to e/ (cid:81) Ni =1 ζ i ( π, ζ i ( π, π ) = − . Thus the parity of the Chern number is always even when reflection symmetry is required. In fact, it alwaysvanishes because the quantum Hall effect is incompat-ible with reflection symmetry. Notice that the reflec-tion eigenvalues do not uniquely specify the polariza-tion in the x-direction since the two eigenvalue condi-tions can be satisfied by choosing (cid:81) Ni =1 ζ i (0 , ζ i ( π,
0) = (cid:81) Ni =1 ζ i (0 , π ) ζ i ( π, π ) = ± . Thus, the anisotropic invari-ants in the x-direction can take either value.The other invariant one can define is the isotropic ex-tension of χ (2) P to 2D. If χ P = +1 and the product overthe inversion eigenvalues at every invariant momentum isseparately trivial then χ (2) P is well-defined because thereare an even number of negative inversion eigenvalues ateach invariant momentum and by definition χ (2) P is theproduct over half of those negative eigenvalues. This isthe Z topological invariant which indicates a quantumspin Hall effect when an insulator has both inversion andtime-reversal symmetry. Unfortunately this topologicalinvariant does not yield a unique topological response.We can understand this in several ways. By explicit con-struction take two decoupled copies of the QAHE stateeach with C = 1 to give an IQHE with C = 2 ortake two decoupled copies of QAHE one with C = 1and the other with C = − . The first system breakstime-reversal and still gives an IQHE while the secondpreserves time reversal and will not give a quantum Halleffect. These systems both have χ P = +1 but χ (2) P = − . We can immediately see why this invariant does yield a unique response in the presence of an additional time-reversal symmetry because this requires the total Chernnumber to vanish which only leaves the possibility of aquantum spin Hall state. Thus it is time-reversal symme-try which restricts the allowed physical response. We canunderstand this by a simple symmetry argument as well.The quantum spin Hall effect is even under both inversionsymmetry ( x, y ) → ( − x, − y ) and parity ( x, y ) → ( x, − y )We see that in even space dimensions inversion symmetryis just a rotation. The quantum Hall effect is even underinversion but odd under parity. Thus having a quantumHall effect is compatible with inversion symmetry but notparity (and not time-reversal). This is why the inversion classification does not distinguish between the doubledquantum Hall state and a quantum spin Hall state. Thistype of ambiguity exists in every dimension where onecan define a Chern number invariant. However, as wesaw with the polarization, if we consider the eigenvaluesof a reflection symmetric Hamiltonian we can eliminatethe possibility of a non-zero Chern number thus leavinga quantum spin Hall state as the alternative. C. 3D Inversion Symmetric Insulators
We show that the isotropic extension of χ P in 3D χ P = (cid:89) k inv ; i ∈ occ. ζ i ( k inv ) , (96)where the product runs over all 8 inversion symmetric k -points, is always trivial. We can first see this by usinga band crossing argument. Start from the atomic limitin which bands have identical inversion eigenvalues atall inversion symmetric points. For χ P to be non-trivial(equal to −
1) there has to be an odd number of bandcrossings between bands of opposite inversion eigenval-ues at the inversion invariant momenta when beginningfrom this trivial limit. Without loss of generality let usconsider one crossing between two bands with oppositeinversion eigenvalues at all inversion symmetric points inthe Brillouin zone. The crossing can happen at eitheran inversion symmetric point (by tuning one parameter)or at a non-inversion symmetric point k in the Brillouinzone. In the latter case, there are actually two cross-ings at k and − k because of inversion symmetry. Toinitially close the gap between the two bands one needsa quadratic touching in at least one of the directions,otherwise we would be creating a nonzero Chern num-ber Fermi surface (after gap closing) out of a zero Chernnumber surface(before the closing, in the trivial limit). Ifthe crossing starts at an inversion symmetric point, thequadratic touching and the gap reopening at the inver-sion symmetric point will switch the inversion eigenvaluesof the bands at that point and make χ P = −
1. However,the system will no longer be an insulator: it will havetwo crossings somewhere else in the Brillouin zone. Thatis, although the gap has opened at the inversion sym-metric point, the gapless points have been moved away.8The quadratic touching effectively splits into multiple 3DDirac points. In the generic situation, there are two gap-less points in the Brillouin zone, with relative positionfixed by inversion, and in order to gap the system, weneed to annihilate the two Dirac points. Note that a 3DDirac point is locally stable even if inversion is not pre-served. On a 2D surface surrounding a 3D Dirac point( ˆ H local ( k ) = k i A ij σ j , ( i, j = 1 , , C = sgn(det( A ij )) . The degeneracy points are stableunless two Dirac points of opposite Chern numbers anni-hilate. Inversion symmetry forces the points at k, − k tohave opposite Chern numbers. So, by inversion, annihila-tion of only two Dirac points in the whole Brillouin zonecan only happen at another inversion symmetric point,in which case inversion eigenvalues are switched again togive χ P = 1 for an insulator. If the gap first closes ata non-symmetric point k , we find the same end result:generically two Dirac points are created close to k andtwo close to − k . They can annihilate in pairs, alwaysswitching an even (possibly zero in this case, since the 4dirac points can annihilate two by two at non-inversionsymmetric points in the BZ) number of inversion eigen-values.Another similar way of understanding the 3D bandcrossings is the following: since we are considering atwo-band crossing in 3D there are three varying param-eters and we cannot find a gapped phase with χ P = − , though it is possible to find a gapless phase. If wehave an even number of band crossings then we canopen a gap but this means that two inversion eigen-values are switched leaving χ P = +1 . There is also adeeper reason why this cannot be done. Let us look atthe simplest case of a single two-band crossing at the Γ-point. The effective Hamiltonian can be put in the form H eff = p σ + p σ + p σ which is a (chiral) Weyl-fermion Hamiltonian. From the Nielsen-Ninomiya no-gotheorem this type of Hamiltonian cannot arise withouta partner fermion with the opposite chirality. Thus alltwo-band crossings must generically occur in pairs andthere cannot be an odd number of negative parity eigen-values in an insulating system. Conversely, if no statesat the inversion invariant momenta cross the Fermi-leveland χ P = − H ( k x , k y , k z ) which has N occupiedbands and is invariant under inversion symmetry. Wecan take the plane k z = 0, and think of ˆ H ( k x , k y ,
0) as a2 D inversion symmetric Hamiltonian. We have alreadyproved that the inversion eigenvalues of a 2D inversionsymmetric Hamiltonian are related to the Chern number i. e. : (cid:81) Ni =1 ζ i (0 , , ζ i ( π, , ζ i (0 , π, ζ i ( π, π, − C ( ˆ H ( k x ,k y , . (97) The same thing is true for the inversion symmetric 2DHamiltonian ˆ H ( k x , k y , π ): (cid:81) Ni =1 ζ i (0 , , π ) ζ i ( π, , π ) ζ i (0 , π, π ) ζ i ( π, π, π )= ( − C ( ˆ H ( k x ,k y ,π )) . (98)Hence (cid:81) Ni =1 ζ i (0 , , ζ i ( π, , ζ i (0 , π, ζ i ( π, π, · ζ i (0 , , π ) ζ i ( π, , π ) ζ i (0 , π, π ) ζ i ( π, π, π )= ( − C ( ˆ H ( k x ,k y , C ( ˆ H ( k x ,k y ,π )) . (99)Since ˆ H ( k x , k y , k z ) is gapped due to our assumption ofan insulator, we can think of it as an adiabatic interpo-lation between ˆ H ( k x , k y ,
0) and ˆ H ( k x , k y , π ) by varyingthe parameter k z . Since this interpolation preserves the U (1) charge conservation symmetry i.e. there is no su-perconductivity, it means that the Chern number cannotchange from k z = 0 to k z = π. Thus, C ( ˆ H ( k x , k y , C ( ˆ H ( k x , k y , π )) = C . (100)Hence (cid:89) k inv ; i ∈ occ. ζ i ( k inv ) = ( − C = 1 . (101)
1. Anisotropic Invariants and the 3D Quantum Hall Effect
We have seen that the isotropic invariant in 3D, whichis constructed by multiplying the inversion eigenvalues atall invariant momenta, is always trivial for an insulator.However we can form anisotropic invariants by consider-ing the products of inversion eigenvalues over planes orlines of the Brillouin zone which are mapped onto them-selves under inversion.First consider a plane in the 3D Brillouin zone which ismapped onto itself under inversion. To be explicit, takethe plane k z = π . If the product of inversion eigenvaluesof all the occupied bands in that plane is N (cid:89) i =1 ζ i (0 , , π ) ζ i ( π, , π ) ζ i (0 , π, π ) ζ i ( π, π, π ) = − D QHE with 3D Hall conductance σ xy = odd integer × πc (103)where c is lattice constant in the z direction. From ourabove proof we know that the product of eigenvalues inthe k z = 0 plane must also be −
1. The proof of the3D QHE is simple: ˆ H ( k x , k y , k z ) can be thought of asan adiabatic continuation of ˆ H ( k x , k y , π ) (since we as-sumed it to be an insulator) As such, each k z planeˆ H ( k x , k y , k z = constant) has an odd integer QHE. Mul-tiplying by the momentum, we get the above 3D Hallconductance. It is important to note that this argument9depends crucially on the fact that the Chern number re-mains unchanged when performing adiabatic deforma-tions as long as charge conservation symmetry is pre-served ( i.e. we don’t allow superconducting perturba-tions).In general the 3D quantum Hall effect directions canbe inferred from the eigenvalue formulas. The generalformula for the 3D Hall conductance in terms of the in-version eigenvalues is σ αβ ⊥ γ = G γ (2 n + 1 / − (cid:89) i ∈ occ. (cid:89) k inv ∈ plane ⊥ G γ ζ i ( k inv )) . (104)This expression above gives the 3D Hall conductance as aproduct of the inversion eigenvalues in a plane αβ perpen-dicular to the γ (= x, y, z ) direction. G γ is the reciprocallattice vector in the γ direction and we have left out theunits of e /h . If the product over all the bands of theinversion symmetric eigenvalues in the αβ plane is − plane allows us to specifythe polarization perpendicular to that reflection plane.The argument is basically the same as in 2D so we didnot include it here. The result, however, is explicitlyillustrated with the 3D dimer model shown in Sec. IV.
2. Topological Metal State
As a corollary to the above, when the product over theinversion eigenvalues of several bands at all points in theBrillouin zone is negative: χ P = (cid:89) k inv ; i ∈ occ. ζ i ( k inv ) = − , the system is a metal protected from opening a gap frominfinitesimal perturbations. We can gain some intuitionabout why this metal state exists by looking at the ef-fective Hamiltonian near a band crossing between stateswith opposite inversion eigenvalues at an inversion invari-ant momentum. In the most generic case, in which we donot have any extra point-group symmetries the effectiveBloch Hamiltonian expanded for small k near k inv :ˆ H ( k ) = ( M + A ij k i k j ) σ z + k i B iα σ α (105)where i, j = 1 , , α = 1 ,
2. Notice that since thebands have opposite parity, the mixing elements betweenthem have to be odd in k. The Hamiltonian reduces to
M σ z at k inv and thus the sign of M dictates the occu-pied inversion eigenvalue. When M is tuned thru zero,we have a phase transition with an eigenvalue switch be-tween + and − . For a given A ij matrix, notice that on atleast one side of the switch, we must have a gapless phase:if M >
Det ( A ij ) < M <
Det ( A ij ) > k inv . There are of course two Dirac points,which can then annihilate at another k inv point with ± eigenvalues and can re-open the gap and give χ P = +1 . Because of the different eigenvalues under inversion, weonly need to tune one parameter to get a band crossingin this case.
3. Magneto-electric Polarization and Inversion Symmetry
Although the first 3D isotropic invariant we mentionedis always trivial, Ref. 29 argues that inversion invari-ant insulators in 3 D , which come from strong topologi-cal insulators with softly broken time-reversal invariance,can support an isotropic topological magneto-electric re-sponse ( i.e. a θ = π vacuum). Their argument uses thetransformation properties of the effective response actionwhich we recount here. The topological response actionin 3 D for an insulator coupled to an electromagnetic fieldis S eff [ A µ ] = e πh (cid:90) d xθ E · B (106)where E , B are external applied electric and magneticfields, and θ is an intrinsic quantity which is proportionalto the magneto-electric polarizability. For translation-ally invariant systems the magneto-electric polarizabilityfor time-reversal invariant insulators is P = θ π = 132 π (cid:90) d k(cid:15) ijk Tr (cid:20) ˆ A i ˆ F ij − i A i ˆ A j ˆ A k (cid:21) (107)where ˆ A i ( k ) is the non-abelian adiabatic connection, andˆ F ij ( k ) is the non-abelian adiabatic curvature. Undertime-reversal symmetry P → − P and thus, for time-reversal invariant insulators P = 0 , / π ). P is not a gauge invariant quantity and only definedmodulo an integer. Note that under time-reversal B does not change in the effective action since it is anexternally applied field. Only intrinsic quantities suchas P get acted upon with time-reversal. Using thesetwo values of P we can physically define a non-trivialtime-reversal invariant topological insulator as one with P = 1 / . It has been shown that this definition is equiva-lent to the band structure definition of strong topologicalinsulators.
In the presence of time-reversaland inversion symmetries there is an elegant topologicalinvariant one can define( − P = (cid:89) i ∈ occ./ (cid:89) k α ∈{ k inv } ζ i ( k α ) (108)0which is the product of the inversion eigenvalues at everyinvariant momenta in the 3D Brillouin zone for half theoccupied bands. Since we have time-reversal, half thebands just means one band out of each Kramers’ pair. Weprovide a new physical proof of this equation in AppendixG.The additional insight of Ref. 29 is that P is also oddunder inversion symmetry. This means that the valuesof P are still quantized to be 0 , / D where P = 1 /
2, analogous to the case in 1 D where P = e/ . This argument is an indicator that inver-sion symmetric insulators can support non-trivial topo-logical states with interesting response properties. Nowwe can ask the question, is Eq. 108 still valid when onlyinversion symmetry is preserved? If we only have inver-sion symmetry and no time-reversal symmetry, then thisformula continues to apply if we can adiabatically con-nect the system to the T and P invariant limit withoutbreaking inversion symmetry. However this is not theonly case, and we will show exactly when the inversioneigenvalues in 3D indicate a non-trivial magneto-electricpolarizability protected by inversion symmetry in the fol-lowing section. Our arguments use many of the resultsdiscovered in the previous sections.
4. Magneto-electric Polarizability for Inversion InvariantInsulators
We begin by reintroducing the unitary matrix B ij ( k ) = (cid:104) u i, − k | P | u j,k (cid:105) (109)where u i,k is a Bloch function with i, j labeling whichoccupied band and k is the Bloch momentum. An im-portant property of the matrix B is B ( − k ) = B † ( k ) (110)which is true because P is unitary and squares to theidentity matrix. This means that at the inversion sym-metric points, the matrix is real and symmetric. Forinversion symmetric insulators, we prove in Appendix Fthat the non-Abelian adiabatic connection satisfies:ˆ A i ( − k ) = − B ˆ A i ( k ) B † + iB ( k ) ∇ i B † ( k ) (111)which implies that the adiabatic curvature is gauge co-variant via: ˆ F ij ( − k ) = B ( k ) ˆ F ij ( k ) B † ( k ) . (112)From this the magneto-electric polarizability is easy, buttedious (see Appendix F) , to obtain:2 P = − π (cid:90) d k(cid:15) ijk T r [( B ( k ) ∂ i B † ( k )) · ( B ( k ) ∂ j B † ( k ))( B ( k ) ∂ k B † ( k ))] . (113) This proves that P is either integer or half-integer de-pending on whether the RHS (which is an integer windingnumber) is even or odd. Since P is itself defined mod 1it means that P defines a Z classification that indicatesa non-trivial topological response.With time-reversal symmetry, states pair up inKramers’ doublets and are degenerate at time-reversalinvariant points (same as inversion symmetric points). Ifwe break all accidental degeneracies in a time-reversal in-variant system we are still left with an even number ofoccupied bands - they are degenerate in pairs at time-reversal symmetric points. The B ( k ) matrix then, atmost, decomposes into diagonal blocks of decoupled U (2)matrices. The winding number can be non-trivial in thiscase because the homotopy group π ( U (2)) = Z (the Z nature appears because P is defined as the windingnumber mod 2). Once we lose time-reversal symmetrythe bands do not have to be degenerate at the k inv . Infact, once time reversal is softly broken, the bands at k inv experience eigenvalue repulsion: before TR was broken,the inversion eigenvalues of the Kramers’ doublets hadto be identical. As such, it seems that once time-reversalis broken we can completely separate the occupied bandsand isolate them from each other at all points in the Bril-louin zone. This would imply that the matrix B could bereduced to an N × N diagonal matrix with U (1) phaseson the diagonal. In other words, B ( k ) will map from 3Dmomentum space into U (1) ⊗ N . Since π ( U (1) ⊗ N ) = 0this implies that the winding number in Eq. 113 alwaysvanishes (note that by making this statement we are im-plicitly assuming that all of the mappings are smoothwhich we will come back to later). This trivial reasoningwould seem to imply that we cannot get a topologicalinsulator with inversion symmetry. Fortunately, this lineof reasoning fails because of the non-trivial global con-straint that the product of all the inversion eigenvaluesmust be positive. Before we deal with the effects of theconstraint we draw several conclusions about a Hamil-tonian with only one occupied band. The non-Abelianform of the winding number implies that with only oneoccupied band P ≡ . Thus, no matter what the inver-sion eigenvalue content, there is no contribution to P from a system with a single occupied band. This alsoholds true for a band that can be completely separatedand untangled from all the other bands at all points inthe Brillouin zone. Such isolated bands do not contributeto a non-trivial P . This is in contrast to the statement atthe end of Ref. 44 which seems to state that a non-trivial P only requires an odd number of pairs of negative inver-sion eigenvalues. As a counter-example, a single occupiedband can have a single pair of negative parity eigenval-ues (which occur at different k inv ) and it has vanishing P . We give such a Hamiltonian in Eq. 138. We will seethat the important thing to consider is pairs of negativeparity eigenvalues at the same k inv . The global constraint on the inversion eigenvalues iscrucial for our discussion. We first give an explicit ex-ample to gain intuition about its importance. Let us1assume that N = 2 and that the occupied bands have ζ (0 , ,
0) = ζ (0 , ,
0) = − k inv are +1 . The naive reasoning fromabove implies that by perturbing this Hamiltonian whilepreserving inversion symmetry we should be able to sep-arate these bands so that they are isolated with no in-termingling degeneracies. However, if we could make thetwo bands non-degenerate everywhere we would contra-dict the constraint χ P = +1 . This is easy to see becauseif we could separate the two bands we could considera different insulating ground state with only one of theprevious two bands occupied and then the product of theinversion eigenvalues of the single occupied band wouldbe negative. But we have proved that this is not possi-ble, so we cannot make the bands fully non-degenerateover the entire Brillouin zone. Alternatively, if we couldseparate the bands, we would have a single band that haseigenvalues − + ++ in the k z = 0 plane and + + ++ onthe k z = π plane. If we consider the 3D Hamiltonian asa gapped interpolation between these two planes then wehave adiabatically connected 2D Hamiltonians with oddand even Chern number respectively. This cannot hap-pen and is thus another contradiction. Hence, we can besure that the two occupied bands are degenerate at atleast two points in the Brillouin zone (by inversion sym-metry). These two points are exactly enough to transferan even Chern number from one plane to the other whichfixes the Chern number issue. Due to this topological ob-struction, the ˆ B ( k ) map cannot be reduced to a map from T into U (1) ⊗ U (1), but instead it is a true map from T into U (2) and therefore its winding can take non-trivialvalues. Additionally, the winding number for each U (2)block is additive due to the trace in the winding number.The conclusion at this step is that, in general, the B ( k )matrix can be decomposed into one and two dimensionaldiagonal blocks. The one dimensional blocks have no con-tribution to the winding number. The two dimensionalblocks come from paired bands that cannot be disentan-gled. Following the previous arguments, one can see thatthis happens only for the following inversion eigenvaluespatterns at the eight k inv (here we restrict the projectorto just these bands): −− ++++++++++++++ , −−−−−− ++++++++++ , −−−−−−−−−− ++++++ , −−−−−−−−−−−−−− ++ Note that the pattern ( + − ) at the same k inv is excludedin all these cases. The reason is because, if the pattern( + − ) shows up at a k inv , then it will necessarily show upat another k inv in order to make χ P =1. In this case, asimple exercise will show that the inversion eigenvaluescan always be separated in two groups with the productof inversion eigenvalues in each group equal to one. Thus,there is no topological obstruction and the bands can beuntangled. The same thing happens for all the otherpossible inversion eigenvalues that are not listed above.We will discuss more about this below and we now returnto the calculation of the winding number.We first discuss the winding number in the continuum(sphere) and then focus on the lattice (torus). We knowfrom Ref. 11 that when considering the isotropic topo- logical invariants such as P it is sufficient to think ofmomentum space as a sphere S instead of the Brillouinzone torus T . Allowing for the full torus structure gives arich set of anisotropic states which we will consider later,but for now we assume that the momentum space topol-ogy is spherical. This effectively reduces the number ofinvariant momenta we need to consider to just two: theorigin and the ‘point at infinity.’ Equivalently we couldthink of the torus with unrestricted inversion eigenval-ues at k inv = (0 , ,
0) but with the inversion eigenval-ues at all the other k inv constrained to be equal bandby band. Now consider a Hamiltonian with N occupiedbands (again N does not have to be even, a crucial dif-ference with the time-reversal case). Since we are in 3Dwe know that the product of all the inversion eigenval-ues must be +1 . This means that there can only be aneven number of inversion eigenvalues that are differentbetween k = 0 and k = ∞ . For example, the number ofnegative eigenvalues at k = ∞ must have the same parity(i.e. even or odd) as the number of negative eigenvaluesat k = 0 . It is clear that, by exchanging parity eigenval-ues between bands, either at k = 0 , ∞ , we can split thebands into two classes: (i) bands with χ P = +1 and (ii)pairs of bands with χ P = − k = 0 and ∞ and case(ii) when they are opposite. For case (i) the bands canbe isolated from each other, but in case (ii) they mustgenerically be in tangled pairs where the χ P = +1 forthe pair. Thus we can understand both cases by consid-ering just 2 occupied bands. For example, a trivial caseis that of two bands with no negative inversion eigenval-ues. This is a realization of case (i) where the inversioneigenvalues of each band separately multiply to +1 . Theglobal constraint does not prevent us from isolating allthe occupied bands and thus, with all eigenvalues posi-tive, P = 0 . All realizations of case (i) are trivial for thesame reason.The first interesting case is that of two bands with asingle pair of negative inversion eigenvalues at the samepoint (say k = 0) but positive inversion eigenvalues at the other point. We consider that case now. The im-portant consequence of the global constraint, as we justsaw, is that the two bands with the negative eigenvaluescan never be completely separated from each other - the(generically) two degeneracy points between them cannotbe lifted or annihilated. B ( k ) restricted to these two-bands is a U (2) matrix and generically takes the form: B ( k ) = e iφ ( k ) ( f ( k ) I + ig a ( k ) σ a ) (114)where ( f ( k )) + g a ( k ) g a ( k ) = 1 . (115)There is a global ± sign ambiguity in the choice of f ( k )but once the sign is chosen at one point, smoothness as-sures the signs at the other points. This ambiguity doesnot have implications for the final result. If we substitutethis form into Eq. 113 all of the dependence on φ ( k ) ( i.e. U (1) part) drops out as long as e iφ ( k ) is smooth (seeAppendix H. Since all loops in S are contractible we cangauge transform B ( k ) to remove the k-dependent phaseso the assumption of smoothness is not an issue. Whatremains is the winding of the SU (2) part which must bean integer. Now, with only the SU (2) part we know thatdue to B ( k ) = B † ( − k ) f ( k ) = f ( − k ) , g a ( k ) = − g a ( − k ) . (116)If we think of S = R ∪ {∞} then it is easy to considerthe general form of B. The function f ( k ) : S → R andits derivative vanishes at k = 0 (and k = ∞ ). This isa Morse function for the sphere and we can expand itaround the origin to find: f ( k ) = N ( k )( M + k a C ab k b + . . . ) (117)where C ab is a 3 × , and N ( k ) is a normaliza-tion factor which is even in k and constrains the matrix B to have unit determinant.Without loss of generality we choose the case whenthe eigenvalues of C ab are all positive (another reasonthe eigenvalues have to have the same sign is to fix theboundary condition at k = ∞ so that it is independentof the path taken to get there). This choice determineswhich sign of M will lead to the non-trivial phase. Sim-ilarly we can expand the function g a ( k ) = N ( k )( D ab k b + . . . ) (118)for a 3 × D ab with no restriction onthe eigenvalues. Generically, D ab will have non-zero de-terminant ( i.e. it will have rank 3). In cases where thedeterminant of D ab is tuned to zero, we have to look usea higher order Taylor-expansion in both f ( k ) and g a ( k ) -maintaining even terms in f ( k ) and odd terms in g a ( k ) . As long as the boundary conditions are fixed, which re-quires us to keep terms in f ( k ) with higher order than g a ( k ), this will not change the result. Without loss ofgenerality we take the case det D (cid:54) = 0 , and by rescalingand rotating we transform to the momentum space ba-sis ( k , k , k ) with C ab = D ab = δ ab . For this choice wehave: N ( k ) = 1 (cid:112) ( M + k ) + k (119)with k = k + k + k . For this B ( k ) we have B (0) = sgn( M ) I × , B ( ∞ ) = I × . (120)By explicit calculation, we find that: P = 1 π (cid:90) ∞ ( M − k )(( M + k ) + k ) k dk = sign( M ) − . (121)We notice that when there is an eigenvalue switch whenpassing from zero to infinity, we have P = 1 / , whereas when there is no eigenvalue switch, we have P = 0. Al-though more terms can be kept in the expansion aroundthe origin this does not influence the result of the wind-ing number, as long as the eigenvalues of B (0) and B ( ∞ )are not changed.We have seen from this simple argument that for twobands which cannot be separated, the contribution to P depends only on the change in inversion eigenvalues. Ifthere is only a single pair of bands which is in case (ii) i.e. cannot be untangled from its partner, then the other N − P and thus thenon-trivial value of P only comes from the two tangledbands. To finish the proof we must consider the casewhen there are more than one set of tangled occupiedbands. If, for example, there are four bands which havenegative eigenvalues at k = 0 (for simplicity we fix all theeigenvalues at k = ∞ to be positive) we can genericallyisolate the four bands into two pairs. The two pairs canbe separated from each other and all other bands, butthe bands making up a single pair must share degenera-cies from the arguments above. Once we have decoupledthe inversion eigenvalues of the two pairs, we can removeall the accidental degeneracies and isolate the pairs fromeach other because a combined pair of bands with nega-tive inversion eigenvalues by itself does not contradict theglobal constraint. Since the pairs can be isolated, theycontribute independently to P . Each pair will contributea half-integer giving P = n = 0 mod Z. We see thereis an even odd effect so that an odd number of pairs ofbands with negative inversion eigenvalues is non-trivialwhile an even number of pairs is trivial.To complete the picture we will discuss how these ar-guments carry over to the lattice case when momentumspace is a torus T . We now have eight invariant mo-menta to consider which can lead to many more differentcombinations of inversion eigenvalues. We will not enu-merate all the possible phases but instead construct thenecessary general principles to classify such states. Weagain consider a set of N occupied bands which does nothave to be an even number. In general the only restric-tion is that the product of all the inversion eigenvaluesof all the occupied bands is equal to one. We can gener-ically perform band crossings only between the occupiedbands to split the bands into three possible classes: (i)a set of n + bands with positive inversion eigenvalues atall k inv , (ii) a set of n e − bands with an even number ofnegative eigenvalues on each band, and a set of n o − bandswhere the product of eigenvalues on each band is − cannot be made equal to +1 via band cross-ings among the other bands in n o − . From the argumentsabove, the sets of n + and n e − bands contribute nothingto P , because they can be isolated one-by-one from allof the occupied bands. Note that the n e − can contributeto non-trivial 3D QHE states in the same manner shownabove. The number of bands in the third class n o − mustbe an even number to satisfy the global constraint. We3will now show that value of P = (1 / n o − mod Z. We callthis process the band decoupling process and we give ex-plicit examples of this band decoupling picture for latticemodels shown in Appendix J.We know that the only bands which can contribute to P are those in n o − . We can consider these bands as a setof n o − / U (2) block to the B ( k ) matrix. We show in AppendixI that once the band decoupling process is finished thenthe U (1) phase in each of the n o − / U (2) blocks is smoothand can thus be completely eliminated from considera-tion. Thus each U (2) block can be contracted to SU (2) . Since this implies that we are really considering mapsfrom T → SU (2), which have the same dimension, wecan connect the winding number of each SU (2) block of B ( k ) to the degree of the map. This argument followsalong the same lines shown in Ref. 43 so we will notinclude all the details. To calculate the degree of themap from T to an SU (2) block of B ( k ) we can chooseany point in SU (2) . For example, we could choose − I × . Since B ( k ) = B † ( − k ) , any k which is not inversion sym-metric contributes to the degree of the map twice, i.e. if − I × occurs at k it will occur at − k , thereby leaving P unchanged. The only contributions, therefore, come fromthe inversion symmetric points, and hence the windingnumber counts the number of inversion symmetric pointsthat have − I × as their inversion eigenvalues. This im-plies that we can simply apply the Kane-Fu formula tothe bands in n o − to determine the value of P . IV. SIMPLE EXAMPLE MODELS
In this section we provide a set of explicit models thatillustrate the majority of the technical details discussed inthe previous sections. For each model we list the interest-ing phases, inversion eigenvalue structure, and describewhat the entanglement spectra should look like. Ad-ditionally we provide figures showing the entanglementspectra for each model which confirms our analytic for-mulae. The models we chose are similar to ones used inmany contexts especially in the field of topological in-sulators. Combined with the models introduced in Sec.I these examples provide valuable intuition about inver-sion invariant topological insulators and the similaritiesand differences between the states protected by inversionsymmetry and those protected by other discrete symme-tries e.g. charge-conjugation, or time-reversal.
A. 1D Models
We have already introduced two illustrative 1D modelsfor which we will analyze the entanglement spectrum.Additionally we will introduce a 1D model of a dimerizedchain. Config ξ Config ξ (a)(b) FIG. 5. Entanglement spectrum for the two-band 1D modelwith (a) random site disorder (b) random site disorder whichis inversion symmetric around the center of the 1D chain.We show the entanglement spectra for many different randomdisorder configurations.
1. Simple two band model
Here we focus on the Hamiltonian given in Eq. 2 andreproduced hereˆ H ( k ) = α cos k + sin( k )ˆ σ + (1 + m − cos k )ˆ σ . This model has one occupied band, and in 1D there aretwo inversion symmetric momenta k = 0 , π. There is aphase transition in this model between two insulatingphases as a function of the parameter m and we havepreviously characterized the two phases of this modelby examining the inversion eigenvalues of the occupiedband. For m > / . Using our inversion criterion we see that this isthe case since the inversion eigenvalues do not changebetween 0 and π. This same characterization applies for m < − − < m < ζ (0) = − ζ ( π ) = + . This cannot be adiabaticallyconnected to an atomic limit. Here the inversion crite-rion implies we should see a pair of entanglement modesat 1 / / χ which takes on values χ = 0when m < − m > χ = 1 when − < m < . The number of required 1 / , H with an onsite disorder term added H = H + (cid:88) i W i c † i c i (122)where the W i are randomly chosen from a uniform distri-bution [ − W/ , W/ . If it is purely random, uncorrelateddisorder the mid-gap entanglement modes are no longerprotected as shown in Fig. 5a. Next we mirror the dis-order around the center of the chain to make a systemwith inversion symmetric disorder. Although unphysical,this helps illustrate the fact that only inversion symme-try is required for the protected mid-gap entanglementstates as we show in Fig. 5b where the cut is along theremaining inversion center.
2. Simple four band model
The Hamiltonian for the simple four band model wasgiven in Eq. 5 and the set of inversion eigenvalues for thedifferent phases were listed in Eqs. 6-10. For conveniencewe reproduce the Hamiltonian hereˆ H ( k ) = sin( k )ˆΓ + sin( k )ˆΓ +(1 − m − cos k )ˆΓ + δ ˆΓ + (cid:15) cos( k )(1 + ˆΓ ) . From the inversion criterion we see that Case 2) and Case3) should have entanglement modes at 1 /
2. In fact, Case2) should have a pair of modes localized on each cut, onefor each occupied band that flips the sign of the inversioneigenvalue. The entanglement spectra for the five casesare shown in Fig. 2f-h and they agree with the analyticprediction. For cases 1-5 the invariant χ = 0 , , , , , , , , / (a)(b) FIG. 6. (a)1d Dimerized Chain (b)2d Dimerized square lat-tice model. Solid and dotted lines represent different hoppingamplitudes. y ξ ξ −π π (a) (b) FIG. 7. Entanglement Spectra for (a)DimerizedChain(b)Dimerized Square Lattice
3. Dimerized Chain
As a final 1D test case we will look at spinless fermionshopping on a dimerized chain. This model is the familiarSu-Schrieffer-Heeger model for electrons in a polyacety-lene chain. Later we will extend this model into 2D and3D to illustrate anisotropic systems with non-trivial en-tanglement spectra. The layout of the chain is shown inFig. 6a along with the choice of two-atom unit cell. TheHamiltonian is given by H = (cid:88) m ∆ (cid:16) c † mA c mA − c † mB c mB (cid:17) + (cid:16) ( − t − δ ) c † mA c mB + ( − t + δ ) c † mB c m +1 A + h.c.) (123)where A, B indicate sublattice A or B and t > . Forour purposes we will set the onsite energy ∆ = 0 . Withthis choice the Hamiltonian has an inversion symmetrywith respect to the middle of a bond with P = σ x i.e. P exchanges sublattices A and B. H = (cid:88) k Ψ † k [( − ( t + δ ) − ( t − δ ) cos k ) σ x + ( t − δ ) sin kσ y ] Ψ k Ψ k = ( c kA c kB ) T . (124)This model has two insulating phases (i) δ < δ > . At the two inversion invariant points we have the BlochHamiltonians ˆ H (0) = − tσ x and ˆ H ( π ) = − δσ x . As ex-pected they both commute with P . For fixed t = 1 wesee that the inversion eigenvalue of the occupied bandat k = 0 is fixed to be + . For k = π the eigenvaluedepends on the sign of δ. So for δ < >
0) we have ζ ( π ) = − . We see that the non-trivial phase occurswhen δ < . In this case the Wannier center of the elec-trons is shifted to the mid-bond center between unit cells.Thus if we cut the system between unit cells there willbe a charge polarization. If δ > within a unit cell. Thus our def-inition of unit cells is important and simply reflects thefact that the polarization is not well-defined absolutelybut is gauge-variant.The entanglement spectra for this model for δ < k =0, and π that there should be 1 / χ = 0 , δ > δ < χ as expected for asystem with periodic boundary conditions and thus twoentanglement cuts.There is a subtlety in the entanglement characteriza-tion of this model which we will now discuss. The real-space Hamiltonian as written in Eq. 123 has brokentranslational symmetry. As stated earlier, our classifi-cation method only applies to translationally invariantHamiltonians since we need to evaluate inversion eigen-values in momentum-space. However, by construction,the Hamiltonian for the dimerized chain only has a verymild breaking of translational symmetry. In fact, as im-plicitly assumed in the analysis, we can just define a unitcell encapsulating two-sites, and in terms of the largerunit cell the Hamiltonian is translationally invariant andour method applies. Our choice for a unit cell has al-ready been implicitly assumed by the time we write theBloch Hamiltonian in Eq. 124. The choice made here isthat sites connected by the hopping term − ( t + δ ) forma single unit cell and the hopping between unit cells is( − t + δ ) . After making this choice we are free to carryout the entanglement analysis by cutting the system be-tween unit cells. Cutting the system within a unit cellis not a position-space cut of a translationally invariantHamiltonian.Now, the important outstanding question is: is theanalysis invariant under the choice of unit cell? It is, andwe explicitly show it for this model. Suppose that we take the unit cell to be sites connected by the hopping( − t + δ ) . Then the Bloch Hamiltonian becomes¯ H = (cid:88) k Ψ † k [( − ( t − δ ) − ( t + δ ) cos k ) σ x + ( t + δ ) sin k σ y ] Ψ k ¯ H (0) = − tσ x ¯ H ( π ) = 2 δσ x . (125)If we fix t > k = 0 is always positive. Thus wesee that for δ > π, and for δ < δ that exhibitsthe non-trivial phase has changed when compared withthe choice of the other unit cell. The physics, however,remains identical. Now when δ > between the new unit cells.This would have lied within the unit cell for our previouschoice and explains why the sign of δ has changed. Forthe choice of unit cell in ¯ H the entanglement spectrumwill have mid-gap modes when δ > i.e. when the in-version eigenvalues are opposite at the two invariant mo-menta. Thus we see that for the new choice of unit cellthe physics and entanglement analysis yields the sameresults. B. 2D Models
1. Dimerized Square Lattice Model
The first 2D model we consider is a trivial extensionof the dimerized chain as illustrated in Fig. 6b. Thisextension has a Hamiltonian which is simply constructedfrom Eq. 124: H = (cid:88) k Ψ † k [( − ( t + δ ) − ( t − δ ) cos k x ) σ x + ( t − δ ) sin k x σ y − t y cos k y ] Ψ k . (126)This model has an inversion symmetry with P = σ x . Atthe inversion invariant points we haveˆ H (0 ,
0) = − t y − tσ x ˆ H ( π,
0) = − t y − δσ x ˆ H (0 , π ) = 2 t y − tσ x ˆ H ( π, π ) = 2 t y − δσ x . We see that although this is more complicated than the1D case everything still commutes with P . For simplic-ity we pick t = 2 t y = 1 and focus on the two gappedphases δ < δ > . In these two phases we have the6following set of eigenvalues δ > ζ (00) = + ζ ( π
0) = + ζ (0 π ) = + ζ ( ππ ) = + δ < ζ (00) = + ζ ( π
0) = − ζ (0 π ) = + ζ ( ππ ) = − . The product of all the eigenvalues in each case is +1so the parity of the first Chern number for these twocases is even. In fact, it is exactly zero for this model.Just as in the 1D case we see that the δ < x -axis to the mid-bond center be-tween each unit cell. If we cut an edge which is per-pendicular to the x-direction there will be a finite chargedensity on the boundary. Although the inversion sym-metry is not enough to determine the polarization wesee that this Hamiltonian also has a reflection symme-try M ˆ H ( k x , k y ) M − = ˆ H ( − k x , k y ) with M = σ x = P . Thus, the charge polarization in the x -direction is quan-tized and equal to P = e/ a where a is the lattice con-stant in the y -direction.Finally we can consider the entanglement spectra ofthese two phases. For δ > / δ < / k x remains a good quantum number and we can ask whetheror not there are 1 / k x = 0 or k x = π. For k x = 0 we look at ζ (0 , ζ (0 , π ) = +1 and for k x = π we consider ζ ( π, ζ ( π, π ) = +1 . Thus for this cut thereare no 1 / y -direction such that k y is a conserved quantum number.For k y = 0 we have ζ (0 , ζ ( π,
0) = − k y = π wehave ζ (0 , π ) ζ ( π, π ) = − / k y = 0 and π. The entanglement spectrumfor a cut parallel to the y -direction is shown in Fig. 7b.In this figure there are clear 1 / k y = 0 , π. In fact, for this simple model there are 1 / k y though our criterion only constrains themodes at the inversion invariant points.
2. Chern Insulator
Next we move on to study the well-known 2D topologi-cal insulators beginning with the Chern insulator (quan-tum anomalous Hall effect). This is a 2D topologicalinsulator which exhibits a quantum Hall effect and isclassified by an integer invariant, the Chern number. Instead of studying the initially proposed honeycomb lat-tice model we will use the square lattice version which isnothing but massive Dirac fermions on a lattice. The Hamiltonian is H = (cid:88) m,n (cid:26) (cid:104) Ψ † m +1 ,n ( iσ x − σ z )Ψ m,n + Ψ † m,n +1 ( iσ y − σ z )Ψ m,n + h.c (cid:105) + (2 − m )Ψ † m,n σ z Ψ m,n (cid:9) . (127)We can Fourier transform to get the Bloch Hamiltonianˆ H ( k ) = sin k x σ x + sin k y σ y + M ( k ) σ z (128) M ( k ) = 2 − m − cos k x − cos k y . (129)This model exhibits several different phases as a functionof m. For m < , m > < m < , < m < − P = σ z and at the four in-version invariant momenta we haveˆ H (0 ,
0) = − mσ z ˆ H ( π,
0) = 2 − mσ z ˆ H (0 , π ) = 2 − mσ z ˆ H ( π, π ) = 4 − mσ z . For m < , m > < m < ζ (0 ,
0) = − ζ ( π,
0) = ζ (0 , π ) = ζ ( π, π ) = +1 and the systemmust have a Chern number with odd parity. It doessince C = − . For 2 < m < ζ ( π, π ) are negative and again the Chern number musthave odd parity ( C = +1).The location of the 1 / x or y directions the picture remains the same. This indi-cates that we are not dealing with an (weak) anisotropicinsulator as in the dimerized case but a fully 2D topolog-ical insulator state. This is similar to saying that for thequantum Hall effect, no matter what edge we cut in thesystem, there will be edge states. For definiteness assumethat we cut parallel to y so that k y is a good quantumnumber. For 0 < m < / k y = 0 and for 2 < m < / k y = π. In addition to these modes there is ac-tually a dispersing set of modes that are localized on thecut. To clearly see the dispersing modes we look at theentanglement ‘energies’ which are defined to be (cid:15) m = 12 log (cid:20) ξ m − (cid:21) (130)where ξ m are the eigenvalues of the reduced correla-tion matrix C L . The entanglement energies show thefull structure of the entanglement spectrum because theyclearly separate the eigenvalues of C L which are clusterednear zero and one. The energy, entanglement eigenval-ues, and entanglement energies for the Chern insulator inthe m < , < m < < m < k y k y k y k y k y k y k y k y k y −202 E ξ −10010 ε (a)(d)(g) −202 E ξ −10010 ε (b)(e)(h) −202 E ξ −10010 ε (c)(f)(i) - π π - π π - π π - π π - π π - π π - π π - π π - π π FIG. 8. Energy spectrum with an open boundary, entanglement spectrum with a cut parallel to the y-direction, and entangle-ment energies for (Left panel a,d,g)A trivial insulator (Middle panel b,e,h)Non-trivial Chern insulator with negative inversioneigenvalue at ( k x , k y ) = (0 ,
0) implying entanglement mid-gap modes at k y = 0 (Right Panel c,f,i)Chern insulator with negativeinversion eigenvalues at ( k x , k y ) = (0 , , ( π,
0) and (0 , π ) . This implies entanglement mid-gap modes at k y = π. in Fig. 8. In the two non-trivial phases the location ofthe 1 / < m < k y = 0and for 2 < m < k y = π. For the entanglementenergies these become zero modes. These entanglementspectra were cut from a torus geometry so that there aretwo entanglement cuts. This is the reason why there areentanglement modes dispersing in both directions in the C (cid:54) = 0 phases.
3. Quantum Spin Hall Insulator
The quantum spin Hall insulator (QSH) is a time-reversal invariant topological insulator which is mosteasily thought of as two copies of the Chern insulator,one for each spin, with opposite chiralities. The realis-tic material in which this state is realized, HgTe/CdTequantum wells, is best modeled by exactly this type ofHamiltonian. The effective HgTe Hamiltonian is a four-band model on the square lattice with a Hamiltonian given by ˆ H ( k ) = sin k x ˆΓ + sin k y ˆΓ + M ( k )ˆΓ (131) M ( k ) = 2 − m − cos k x − cos k y . (132)where ˆΓ = σ z ⊗ τ x , ˆΓ = 1 ⊗ τ y , ˆΓ = 1 ⊗ τ z where σ a isspin and τ a is the orbital degree of freedom. For this sys-tem the time-reversal operator is T = ( iσ y ⊗ K and theinversion operator is P = ˆΓ . This Hamiltonian is invari-ant under both symmetries. It exhibits phases directlyanalogous to the Chern insulator i.e. it is a trivial insu-lator for m < , m > < m < < m < . The onlydifference with the Chern insulator is that now there aretwo occupied bands which are related by time-reversalsymmetry. The total Chern number of the ground statevanishes but there is a Z invariant given (in the presenceof time-reversal and inversion ) by χ Z = (cid:89) n ∈ occ./ ζ n (0 , ζ n ( π, ζ n (0 , π ) ζ n ( π, π ) (133)where the product runs over half the occupied bands.This invariant has the same formula as χ (2) P defined8 −202 E ξ −10010 ε (a)(d)(g) −202 E ξ −10010 ε (b)(e)(h) −202 E ξ −10010 ε (c)(f)(i) k y k y k y - π π - π π - π π k y k y k y - π π - π π - π π k y k y k y - π π - π π - π π FIG. 9. Energy spectrum with an open boundary, entanglement spectrum with a cut parallel to the y-direction, and en-tanglement energies for (Left panel a,d,g)A trivial insulator (Middle panel b,e,h)Non-trivial quantum spin Hall insulator withtime-reversal symmetry (Right Panel c,f,i)Quantum spin Hall insulator with mildly broken time-reversal symmetry. Comparingh and i one can see that all Kramers’ degeneracies are lifted when time-reversal is broken except the ones at (cid:15) = 0 . By justlooking at e and f it is difficult to tell the difference in the two cases i.e. that spectral flow is broken. above, but we distinguish it here to prevent confusionsince this invariant only implies a non-trivial physical re-sponse when time-reversal is preserved. To specify which half of the occupied bands you just take one from ev-ery Kramers’ pair of bands. If we focus on the transi-tion when m ∼ m < > m > k x , k y ) = (0 , both occupied bands are negative while all others arepositive. The product over inversion eigenvalues of all the occupied bands is trivial, but if we only multiplyover half the Kramers’ pairs we find that χ Z = − C = 2 quantum anomalous Hall state could have the same inversion eigenvalue structure.
4. Quantum Spin Hall Insulator without time-reversalsymmetry
So far the studies of the Chern Insulator and QSHinsulator have just been reconfirmed by recognizing theimportance of inversion eigenvalues when there is an in-version symmetry. The most interesting prospect is whenwe take the QSH effect and break time-reversal but keepinversion. The importance of inversion symmetry for thistype of case in 3D was emphasized in Ref. 29. To breaktime-reversal symmetry we will consider an additionalZeeman term in the QSH Hamiltonian :ˆ H ( k ) = sin k x ˆΓ + sin k y ˆΓ + M ( k )ˆΓ + B x ˆΓ B (134)ˆΓ B = . B x this term lifts the Kramers’ de-generacy of the occupied bands but does not cause anycrossings at the Fermi-level. Although time-reversal isbroken, inversion is still preserved and we can still seethat χ Z = − . This is still well defined because theproduct of all inversion eigenvalues at a particular k inv isstill trivial for all k inv . Thus, this system is an inversioninvariant topological insulator. It was first noted thatsuch states could exist in Ref. 29 where it was suggestedthat as long as inversion symmetry was not broken theentanglement spectra for the time-reversal preserved andbroken cases were the same. However, there is an impor-tant difference between the two. For the time-reversalbroken QSH state we show the entanglement eigenvaluesand entanglement energies in the right panel of Fig. 9.Comparing with the middle panel we see that the entan-glement eigenvalues seemingly show very little differencedue to the fact that most are exponentially close to zeroor one, but the entanglement energies are quite different.The mode at 1 / is protected by inversion symmetry butall of the other “Kramers’” degeneracies are lifted e.g. allthe crossings at k y = − π and k y = π are lifted. This oc-curs because the spectral flow between the bulk valenceand conduction bands is cut-off when time-reversal is bro-ken. The edge states no longer tie together both bandsand there is no “anomaly”-type structure. Thus stateson the left and right half of the system are no longer tiedtogether through the bulk in a topological way.We ask now if there is anything interesting in this sys-tem once time-reversal is broken? As long as inversion ispreserved we cannot connect this state to a trivial atomiclimit while preserving inversion symmetry and there mustalways be a finite entanglement entropy. The finite en-tanglement entropy is due to the fact that the mode at1 / i.e. a field strong enough to re-verse the direction of the Zeeman field in some region ofthe edge then there will be trapped domain wall states.Thus the state is a topological insulator in a very phys-ical sense as well. Unfortunately the inversion invariant( χ (2) P ) in 2D does not imply that this must be the phys-ical response. As mentioned above a C = 2 quantumanomalous Hall effect can have the same value of the Z invariant. However, for this model Hamiltonian we are k y k y k y k y E E ξ ξ − π − π − π − π ππππ (a) (b)(c) (d) FIG. 10. Energy spectrum with open boundary conditionsfor (a)8-band model with T symmetry (b) 8-band modelwith random inversion preserving perturbation. Entangle-ment spectrum for (c) 8-band with T (d) 8-band with randominversion preserving perturbation. saved because there is an additional mirror symmetry M ˆ H ( k x , k y ) M − = ˆ H ( k x , − k y ) with M = σ x ⊗ τ z . Notethat [ M , P ] = [ M , ˆΓ B ] = 0 . Thus, the inversion eigen-values are still valid labels and the Zeeman term doesnot break the mirror symmetry. This symmetry forbidsa non-zero C and thus the QSH response is the uniqueresult.
5. 2D 8-band model
Here we consider a more complicated case of a modelwith 8-bands total and four occupied bands. The modelwe choose is simply two-copies of the QSH model. TheBloch Hamiltonian is given byˆ H ( k ) = sin k x γ x + sin k y γ y + M ( k ) γ z (135)with M ( k ) given in Eq. 132 and γ a = 1 × ⊗ Γ a . This model preserves time-reversal symmetry with T =(1 ⊗ iσ y ⊗ K and inversion symmetry with P = γ z . Inthe presence of time-reversal symmetry this model doesnot yield any non-trivial topological insulators since youalways get an even number of pairs of edge states. If oneadds perturbations which break time-reversal symmetrythen it is possible to generate non-trivial states such asChern insulator states. At half-filling this model willhave four occupied bands, and will exhibit edge statesfor the same range of parameters as the quantum spinHall model above in Eq. 131. These edge states are notprotected generically i.e. one can add a time-reversal andinversion preserving perturbation to the model which willopen a gap in the edge states. We use this model to il-0 ππ − π − π k y k y E ξ (a)(b) FIG. 11. (a)Energy spectrum for the small anisotropic 8-bandHamiltonian with periodic boundary conditions. Difference innegative parity eigenvalues at k y = 0 , π is three. (b) Entan-glement spectrum for the same. At k y = 0 there are six modesat ξ = 1 / . The lines are only filled in as guides to the eye.The only allowed k-values are k y = 0 and k y = π. This is whythe spectra appear to have kinks. lustrate three points (i) even though there is no topolog-ical invariant in the system associated with time-reversalsymmetry, the presence of inversion symmetry predictsthat there will be non-trivial mid-gap states in the en-tanglement spectrum (ii) these mid-gap states are com-pletely stable as long as you do not break inversion sym-metry (iii) the number of mid-gap states is proportionalto the difference in negative parity eigenvalues at the in-variant momenta.In Fig. 10 we show the energy and entanglement spec-tra for two different cases. In Fig. 10a,c we simply diag-onalize Eq. 135 with m = 1 . y -axis. The difference between the number of negative inversion eigenvalues at ( k x , k y ) = (0 ,
0) and( π,
0) is four giving a total number of 2 × / . If we leave out the identity matrix, and P it-self, there are 30 matrices which commute with P . Toillustrate the stability of the 1 / − δ/ , δ/
2] where δ > δ = 0 .
20 that the energy spectrum still has statesin the gap but the crossing-points have been lifted. Theentanglement spectrum, however, still has eight exact 1 / L x = 20 and L y = 2 . We cut the system in themiddle between x = 10 and x = 11 and plot the en-tanglement spectrum vs k y . There are only two allowedvalues for k y = 0 , π. The energy spectrum for such a sys-tem (with the random perturbation included but chosenfrom a uniform distribution [0 , δ ] which is no longer sym-metric about zero) is gapped and has an entanglementspectrum with six mid-gap 1 / δ = 0 . . Counting the occupied states we find numerically thatthere is a difference of three negative inversion eigenval-ues which agrees with the entanglement spectrum. Theenergy and entanglement spectra are shown in Fig. 11.
C. 3D Models
1. Dimerized Cubic Lattice
The 3D dimerized model on a cubic lattice is a trivialextension of the 2D dimerized case into 3D. The Hamil-tonian is given by H = (cid:88) k Ψ † k [( − ( t + δ ) − ( t − δ ) cos k x ) σ x + ( t − δ ) sin k x σ y − t y cos k y − t z cos k z ] Ψ k . (136)For t = 2 t y = 2 t z = 1 there are two different phases δ < δ > . As before, for δ > . For δ < ζ (000) = ζ (00 π ) = ζ (0 π
0) = ζ (0 ππ ) = +1and ζ ( π
00) = ζ ( π π ) = ζ ( ππ
0) = ζ ( πππ ) = − . Again this system has a non-trivial charge polarizationon a surface with ˆ x as a normal vector. However, thisis protected by the reflection symmetry about the yz -plane i.e. M ˆ H ( k x , k y , k z ) M − = ˆ H ( − k x .k y .k z ) with M = σ x = P . Thus, since the product of the reflec-tion eigenvalues is − P = e/ a where a is the area1 ξ ξ (0,0) ( π ,0) ( π , π ) (0,0) (0,0) ( π ,0) ( π , π ) (0,0)(a) (b) FIG. 12. Entanglement Spectra for (a)3DQHE (b)3d WTIwith a cut parallel to the x-y plane plotted along a line in theBrillouin zone. The states in (b) are all doubly degeneratecompared to figure (a). of a plaquette in the yz plane. The product of all theparity eigenvalues is trivial as it must be in 3D, and ad-ditionally the product of the eigenvalues in every plane istrivial. For an entanglement cut such that k y and k z aregood quantum numbers it is clear that there will be 1 / k y , k z ) = (0 , , ( π, , (0 , π ) , and ( π, π ) . Onthe other translationally invariant cuts parallel to the xz or xy planes there will be no 1 /
2. 3D Quantum Hall Effect
The 3D quantum Hall effect state can be thought ofas stacks of 2d quantum Hall states which are connectedtogether. We will use a very simple model for the 3Dquantum (anomalous) Hall effect which is a trivial ex-tension of the 2D Chern insulator. The Hamiltonian isˆ H ( k ) = sin k x σ x + sin k y σ y + M ( k ) σ z (137) M ( k ) = 2 − m − cos k x − cos k y − t ⊥ cos k z . (138)This system has an inversion symmetry with P = σ z . This model exhibits several different phase transitionsbut we will only focus on one, namely the phase transitionthat occurs with a gapless point at ( k x , k y , k z ) = (0 , , . For m < − t ⊥ the system is in a trivial insulating phasewith all inversion eigenvalues positive. At m = − t ⊥ thesystem becomes gapless and stays gapless until m > t ⊥ . For m > t ⊥ the system is in a 3D quantum Hall effectphase with effectively 2D quantum Hall states stackedup in the z -direction. The Bloch Hamiltonians at theinversion invariant points which switch eigenvalues areˆ H (0 , ,
0) = − ( m + t ⊥ ) σ z ˆ H (0 , , π ) = ( − m + t ⊥ ) σ z At m = − t ⊥ the eigenvalue around the Γ-point switchesfrom positive to negative but the system is still gapless.Then at m = + t ⊥ the eigenvalue at (0 , , π ) switches andthe system becomes a gapped insulator. The productover all the eigenvalues is trivial as expected but if we restrict the product to the k z = 0 or k z = π planes theproduct is negative. We proved earlier that this indicatesa non-trivial 3D QHE response.For this arrangement of eigenvalues we can calculatethe location of the 1 / xy -plane there will be no1 / e.g. parallel to the zx -plane then there will be1 / k z , k x ) = (0 ,
0) and ( π, . These nodesare shown in Fig. 12a. The figure shows, in addition tothe 1 / / k z , k z ) = (0 ,
0) and ( π, .
3. 3D Weak Topological Insulator
There are several different classes of the recently pro-posed 3D time-reversal invariant topological insulators.The anisotropic classes, the so-called weak topologicalinsulators, are effectively 2D quantum spin Hall statesstacked into 3D. This is similar to the 3D quantum Halleffect and is essentially just two copies of that system,one for each spin. We use the following model:ˆ H ( k ) = sin k x ˆΓ + sin k y ˆΓ + M ( k )ˆΓ (139) M ( k ) = 2 − m − cos k x − cos k y − t ⊥ cos k z . (140)where the ˆΓ a matrices are the same as in the quantumspin Hall state. This system is time-reversal and inver-sion invariant with an inversion operator P = ˆΓ . Itexhibits phase transitions at the same values of m asthe 3D quantum Hall effect and the only difference isthat there are two occupied bands instead of one. For2 − t ⊥ > m > t ⊥ The system has two pairs of nega-tive inversion eigenvalues, one pair at (0 , ,
0) and oneat (0 , , π ) . The rest of the eigenvalues are all positive.The total product of inversion eigenvalues is trivial, andunlike the 3D quantum Hall case the product of theeigenvalues when restricted to the k z = 0 , π planes isalso trivial. However, there is still something non-trivialhere which arises from taking the eigenvalues from onlyone of the Kramers’ pairs at each invariant momentum.We see that this product is non-trivial and indicates ananisotropic inversion invariant topological insulator. Inthis case, since time-reversal symmetry is preserved, itis a weak topological insulator state. However, with-out time-reversal symmetry the inversion invariant be-ing non-trivial does not require it is a weak topologicalinsulator. As a counter example it could be a 3D quan-tum Hall effect with a Chern number “per layer” thatis an odd multiple of two (unless there is a reflectionsymmetry which requires the Chern number to vanishin each plane). The interesting thing about this model isthat even when time-reversal is softly broken ( i.e. brokenwithout causing a phase transition) the system still is notin a trivial topological state and even though the surfacestates are no longer protected it can exhibit non-trivialbehavior in the entanglement. The entanglement spec-trum for the T-invariant case is shown in Fig. 12b and2 −15−50515 k ε ξ −505 k E −505 k E ξ −15−50515 k ε (0,0) ( π ,0) ( π , π ) (0,0) (0,0) ( π ,0) ( π , π ) (0,0)(0,0) ( π ,0) ( π , π ) (0,0) (0,0) ( π ,0) ( π , π ) (0,0)(0,0) ( π ,0) ( π , π ) (0,0) (0,0) ( π ,0) ( π , π ) (0,0)(a) (b)(c) (d)(e) (f) FIG. 13. (Left column)Strong topological insulator (a)Energy spectrum plotted along a line in the Brillouinzone for open boundary conditions along the z-direction(c)entanglement spectrum with a cut parallel to the x-y planewith periodic boundary conditions (e)entanglement energies.(Right column)Strong topological insulator with time rever-sal breaking (b)energy spectrum (d)entanglement spectrum(f) entanglement energies. In (a) and (b) the mid-gap statesare localized on the surfaces. In (b) there is a gap in the sur-face states due to time-reversal symmetry breaking. In (f) allKramers’ degeneracies are lifted except the one at (cid:15) = 0 . exhibits the same 1/2 mode structure as the 3D quantumHall effect model but with twice as many modes.
4. 3D Strong Topological Insulator
The last class of models we will consider is the 3Dlattice Dirac model which is the minimal model for time-reversal invariant strong topological insulators in 3D. TheBloch Hamiltonian is given byˆ H ( k ) = sin k x ˆΓ + sin k y ˆΓ + sin k z ˆΓ + M ( k )ˆΓ M ( k ) = 3 − m − cos k x − cos k y − cos k z (141)where ˆΓ = σ y ⊗ τ x . As a function of m this model ex-hibits many phase transitions. We will focus on the range m < < m < . There is a phase transition at m = 0 with a band crossing at the Γ-point in k-space.Four bands meet at this point and a pair of inversioneigenvalues is exchanged. For m < k = 0 isˆ H (0) = − m ˆΓ and thus when m switches sign the inver-sion eigenvalues at k = 0 are exchanged. For 0 < m < k = 0 are negative. Inthis phase the product over all inversion eigenvalues istrivial, but if we only keep one of the Kramers’ pairs,then the product of all the eigenvalues of half the occu-pied bands is non-trivial. In the presence of inversionand time-reversal, which is the case here, this invariantis the strong topological Z index. Physically this indexhas two implications: (1) the presence of an odd-numberof massless Dirac cones on any surface (2) a topologicalmagneto-electric effect. To see the topological responseone must apply a time-reversal breaking field on the sur-face to open a gap in the gapless Dirac fermions. Thisinduces a quantum Hall effect confined to the surfacewhich leads to the magneto-electric response. For thetopological phase we picked (0 < m < , the surfacestates are located around the surface Γ-point and the(pair of) entanglement modes will be located at the Γ-point of the conserved momenta parallel to the entangle-ment cut. The energy spectrum, entanglement eigenval-ues, and entanglement energies are shown in Fig. 13a,c,erespectively. The energy spectrum is shown with periodicboundary conditions along x and y and open boundaryconditions along z. The surface states are shown in redand the spectrum is plotted along a 1D path in the Bril-louin zone.
5. 3D Strong Topological Insulator without time-reversalsymmetry
Now we can consider the more interesting question ofthe properties of the system when we break time-reversalbut keep inversion. We add the same Zeeman termshown in Eq. 135 to the bulk of the insulator. We onlybreak time-reversal softly which opens a gap in the sur-face states but does not close the bulk gap. Thus, whilewe can no longer consider the Z invariant protected bytime-reversal, this system will still exhibit a magneto-electric effect since the Zeeman field simply establishesa quantum Hall effect on the surface. This distinctionwas first considered in Ref. 29. Additionally, althoughwe can add surface potentials to push the surface statesinto the bulk bands we still cannot adiabatically con-nect this insulator with an atomic limit. This is clearlyshown in the entanglement spectrum where there are stillmodes protected at 1 / P (cid:54) = 0 for this model. To beexplicit we takeˆ H ( k ) = sin k ˆΓ + sin k ˆΓ + sin k ˆΓ +(5 / − cos k − cos k − cos k )ˆΓ + ˆΓ B (142)This Hamiltonian can be connected through a gappedinterpolation with the trivial Hamiltonian:ˆ H ( k ) = sin k ˆΓ + sin k ˆΓ + sin k ˆΓ + 5 / (143)using the inversion symmetric homotopyˆ H ( k, θ ) = (1 + cos θ ) ˆ H ( k )+ (1 − cos θ ) ˆ H ( k ) + sin θ ˆΓ (144)where ˆΓ = ˆΓ ˆΓ ˆΓ ˆΓ . The second Chern number gen-erated by ˆ H ( k, θ ) is C =1 (odd) and consequently P = mod( Z ) for ˆ H ( k ) since we specifically chose the form ofˆ H ( k ) to be trivial. Note, that as with the cases shown inAppendix J we calculated C numerically using the stan-dard gauge invariant formula in terms of ground-stateprojection operators. V. CONCLUSIONS
The question of what makes an insulator “topologi-cal” has many answers. In this article we presented ananswer which encompasses all of the known topologicalinsulators. The fundamental distinction between an or-dinary band insulator and a topological insulator is theinability to adiabatically connect a topological insulatorto the atomic limit. This distinction can have many man-ifestations including non-trivial topological responses toexternal fields and robust boundary states, however theseproperties are not necessary conditions for a topologicalinsulator. In fact, we have seen examples in this paperwithout protected boundary states, and examples withno topological response. At first sight these insulatorsseem to have no characteristics which distinguish themfrom trivial band insulators. Admittedly these are notthe most interesting systems to consider experimentally,but they still show a striking signature in the entangle-ment spectrum. In fact, all known topological insulators show a signature in the entanglement spectrum when thebi-partition is a position-space cut. Instead of taking thewhole spectrum, one can calculate just the entanglemententropy, which for all topological insulators cannot beadiabatically deformed to zero. This fact is what servesas the basis for our definition and unifies the inversionsymmetric insulators with the ones that are invariantunder other discrete symmetries. The experimental rele-vance of inversion symmetric topological insulators is un-clear, but not out of the question. Although in principledisorder immediately destroys any stability of the topo-logical state (unlike the typical topological insulators ),the robustness of the insulator state is ultimately a ques-tion to be answered in practice. Many of the well-knowntopological insulators simplify when inversion symmetryis required along with the discrete symmetry that stabi-lizes the topological state. Thus it seems like the mostinteresting inversion symmetric insulators are ones whichare derived from parent topological insulator states withweakly broken T or C symmetries. These types of mate-rials would be the first place to search for signatures oftopological protection due to inversion symmetry. Note:
During the last stages of preparation of thismanuscript, we became aware of a paper by A. Turner, YiZhang, R. Mong and A. Vishwanath dealing with similarissues.
Acknowledgements
We acknowledge useful conversa-tions with C.-K. Chiu, E. Fradkin, and F.D.M. Haldane.TLH was supported in part by the NSF DMR 0758462at the University of Illinois, and by the ICMT. EP ac-knowledges a support from the Research Corporation forScience Advancement. BAB was supported by PrincetonStartup Funds, Alfred P. Sloan Foundation, NSF DMR-095242, and NSF China 11050110420, and MRSEC grantat Princeton University, NSF DMR-0819860. TLH andBAB thank the Institute of Physics in Beijing, China forgenerous hosting. BAB thanks Ecole Normale Superieureand Microsoft Station Q for generous hosting.
Appendix A: Proof that the one-body correlationfunction is a projection operator
Here we prove that the one-body correlation functionover the full system (not only over part of the system) isa projector. This can easily be proved: C ij (which is amatrix at each i, j i.e. C αβij ) has the property: (cid:80) j C ij ( k y ) C jk ( k y ) = (cid:80) j N (cid:80) k x e ik x ( i − j ) ˆ P ( k x , k y ) N (cid:80) k (cid:48) x e ik (cid:48) x ( j − k ) ˆ P ( k (cid:48) x , k y ) == (cid:80) k x ,k (cid:48) x (cid:80) j N e i ( k (cid:48) x − k x ) j e ik x i ˆ P ( k x , k y ) e − ik (cid:48) x k ˆ P ( k (cid:48) x , k y ) = (cid:80) k x ,k (cid:48) x N δ k (cid:48) x − k x e ik x i ˆ P ( k x , k y ) e − ik (cid:48) x k ˆ P ( k (cid:48) x , k y ) == N (cid:80) k x e ik x ( i − k ) ˆ P ( k x , k y ) ˆ P ( k x , k y ) = N (cid:80) k x e ik x ( i − k ) ˆ P ( k x , k y ) = C ik ( k y ) . (A1)Note that we have used the notation ˆ P ( k x , k y ) to repre-sent the k-dependent projector onto the occupied states instead of ˆ P k to make the momentum dependence easier4to see. Appendix B: Entanglement eigenvalues for twooccupied bands and four sites
We will analyze only one case of inversion eigenval-ues, i.e. when both inversion eigenvalues at k = 0 dif-fer from the inversion eigenvalues at k = π . For sim-plicity and without loss of generality, we particularize to ζ (0) = ζ (0) = 1, ζ ( π ) = ζ ( π ) = −
1, where the orthog-onality relations between the wavefunctions at the same k and at different inversion symmetric k ’s hold due to theopposite inversion eigenvalues. Let the wavefunctions ofthe two occupied bands be ψ ( k ) and ψ ( k ) . Similar tothe above the arguments in Sec II C 3, the entanglementwavefunctions take the form ( ψ A , m P ψ A ) where ψ A di-agonalizes the operator ˆ C + m ˆ C P = [(1 + m )( ˆ P +ˆ P π ) + ˆ P π P ( P + im ) + P ˆ P π ( P − im )]. We expand thewavefunction ψ A : ψ A = a ψ (0) + a ψ (0) + b ψ ( π ) + b ψ ( π ) . (B1)With this expansion, we have to look at the solutions ψ A which are in the nullspace of ˆ P π P ( P + im ) + P ˆ P π ( P − im ) . Once we have found such solutions, we know theyhave 1 / m = 1 since ( ˆ C + ˆ C P ) ψ A = ( ˆ P + ˆ P π ) ψ A = ψ A due to the fact that wavefunctionsat different inversion symmetric momenta are orthogonalif their inversion eigenvalues are different. We denote theoverlaps: (cid:104) ψ ( π ) | ψ (0) (cid:105) = α , (cid:104) ψ ( π ) | ψ (0) (cid:105) = α (cid:104) ψ ( π ) | ψ ( π ) (cid:105) = α , (cid:104) ψ ( π ) | ψ ( π ) (cid:105) = α (cid:104) ψ ( π ) | ψ (0) (cid:105) = β , (cid:104) ψ ( π ) | ψ (0) (cid:105) = β (cid:104) ψ ( π ) | ψ ( π ) (cid:105) = β , (cid:104) ψ ( π ) | ψ ( π ) (cid:105) = β (B2)These are the only unknown overlaps, as the wavefunc-tion at 3 π/ π/ P are,up to a sign, identical to the ones above. We find thefollowing two exact half modes of the entanglement spec-trum: ( a , a , a , a ) =( i ( β α − β α ) , , β α − β α , β α − β α );( β α − β α , β α − β α , I ( β α − β α ) ,
0) (B3)The other two mid-gap entanglement eigenvalues are ob-tained when m = −
1, in which case we have to diagonal-ize the operator [ ˆ P π P ( P − i ) + P ˆ P π ( P + i )]. In thiscase, we expand the eigenstate: | ψ A (cid:105) = ( c + c P ) (cid:12)(cid:12)(cid:12) ψ ( π (cid:69) + ( c + c P ) (cid:12)(cid:12)(cid:12) ψ ( π (cid:69) (B4)we find 1 / c , c , c , c ) = (0 , , , i ) , (1 , , i,
0) (B5) We then see that we have 4 robust modes at exactly1 / Appendix C: Explicit Proof for the generic two-siteproblem with N occupied bands We now show that the two-site problem with n neg-ative inversion eigenvalues at k = 0 and n nega-tive inversion eigenvalues at k = π out of a number N of occupied bands contains 2 | n − n | zero modesin the entanglement spectrum. Without loss of gen-erality denote the eigenstates of the original Hamilto-nian | ψ (0) (cid:105) . . . | ψ n (0) (cid:105) as the ones with negative in-version eigenvalue at k = 0, | ψ n +1 (0) (cid:105) . . . | ψ N (0) (cid:105) asthe ones with positive inversion eigenvalue at k = 0, | ψ ( π ) (cid:105) . . . | ψ n ( π ) (cid:105) as the ones with negative inversioneigenvalue at k = π , | ψ n +1 ( π ) (cid:105) . . . | ψ N ( π ) (cid:105) as the oneswith positive inversion eigenvalue at k = π . Bands atthe same momentum are all orthogonal, while bands atdifferent momenta are orthogonal if they have oppositeinversion eigenvalues. The simplest case of the above,which should be obvious from our previous examples, isthat of all the N inversion eigenvalues at k = 0 beingidentical and negative of the N eigenvalues at k = π . Inthis case, the projector at one of the inversion symmetric k ’s annihilates all the eigenstates at the other inversionsymmetric k , and the 2 N occupied eigenstates of the orig-inal two-site Hamiltonian are also the eigenstates of theentanglement spectrum at fixed eigenvalue 1 /
2. Due totheir orthogonality, they are linearly independent. Fromhere it is clear that our formula is physically correct:adding the same eigenvalues to both k = 0 , π cannotchange the result.We again expand the eigenstates | ψ A (cid:105) of the correlationfunction C L = ( ˆ P + ˆ P π ) / ψ A = (cid:80) n m =1 a m ψ m (0) + (cid:80) Nm = n +1 a m ψ m (0) ++ (cid:80) n m =1 b m ψ m ( ψ ) + (cid:80) Nm = n +1 b m ψ m ( π ) (C1)In the generic case, we assume that the norms thatare not fixed to vanish by symmetry (such as differ-ent inversion eigenvalues) are all nonzero. In gen-eral, it might be the case that not all the eigenstatesin the expansion above are linearly independent i.e. ψ m (0) might not be linearly independent from a sumof the eigenvalues ψ m ( π ) which have identical inver-sion eigenvalues. In building the matrix to be diago-nalized, we take this into consideration, but when writ-ing the eigenvalue equation, we assume they are lin-early independent - generically, they will be, becausethere are 2 N wavevectors of 2 N components. The vec-tor ( a , . . . , a n , a n +1 , . . . , a N , b , . . . , b n , b n +1 , . . . , b N )has to diagonalize the matrix:5 n × n B n × n N − n × N − n A N − n × N − n B † n × n n × n A † N − n × N − n N − n × N − n (C2)where B ij = (cid:104) ψ i (0) | ψ j ( π ) (cid:105) with i = 1 , . . . , n , j =1 , . . . , n and A ij = (cid:104) ψ i (0) | ψ j ( π ) (cid:105) with i = n + 1 , . . . , N , j = n + 1 , . . . , N . It is easy to see that this matrix has2 | n − n | eigenvalues at exactly irrespective of the A, B matrices. We show it for n = 0, the generalization to n (cid:54) = 0 being straightforward. For n = 0, the matrixreads: n × n N − n × N − n A N − n × N A † N × N − n N × N (C3)Half of the entanglement eigenvalues at 1 / ψ i (0), i = 1 , . . . , n eigen-states. The remaining eigenvalues must then be part ofthe eigenvalues of the matrix: R N − n N = (cid:32) N − n × N − n A N − n × N A † N × N − n N × N (cid:33) (C4)where we have indexed the matrix by the dimension N − n of the upper block-diagonal square matrix and by thedimension N of the lower block diagonal square matrix.We need to compute the determinant of: R N − n N = (cid:18) ( − λ ) N − n × N − n A N − n × N A † N × N − n ( − λ ) N × N (cid:19) (C5)We can prove that this matrix has n eigenvalues inde-pendent of what the matrices A are: as such, we denoteby M αβ a matrix of the form above, but with any ran-dom numbers instead of the matrix made out of normsmatrix A ij = (cid:104) ψ i (0) | ψ j ( π ) (cid:105) . We want to compute thedeterminant of M N − n N . By expanding first on the lastcolumn of the matrix, then immediately after, expandingon the last row of all the matrices obtained, we find therecurrence relation:det( M N − n N ) = ( 12 − λ ) det M N − n N − + x · det( M N − n − N − )(C6)which, applied successively, leads to:det( M N − n N ) = ( − λ ) r det M N − n N − r ++ (cid:80) ri =1 x i · ( − λ ) i − det( M N − n − N − i ) (C7)where r ≤ N − x r are numbers. We choose r = N − M N − n − N − i ) to obtain:det( M N − n − N − i ) = ( − λ ) p − i det M N − n − N − p ++ (cid:80) p − iq =1 x q · ( − λ ) q − det( M N − n − N − i − q ) (C8) where p > i is an integer and the x q are complex numbersnot necessarily equal to the x i ’s in the previous recursion.We again choose p = N −
1, which means i ≤ N − M N − n − N − i ) intoEq[C7], and separating the sum to take into account therestriction i ≤ N − M N − n N ) = O (( − λ ) N − ) ++ (cid:80) N − q =1 (cid:80) N − − q q =1 x q x q (1 − δ q ,N − ) ·· ( − λ ) q + q − det( M N − n − N − ( q + q ) )where O (( − λ ) N − ) means terms proportional to atleast a the ( − λ ) N − . By applying the recursion relationsuccessively, we obtain:det( M N − n N ) = O (( − λ ) N − ( l +1) ) + (cid:80) N − q =1 (cid:80) N − − q q =1 (cid:80) N − − ( q + q ) q =1 ... (cid:80) N − − ( q + ... + q l − ) q l =1 (1 − δ q ,N − )(1 − δ q + q ,N − ) ... (1 − δ q + ... + q l − ,N − ) ·· ( − λ ) ( q + ... + q l − ( l +1) det( M N − n − ( l +1) N − ( q + ... + q l ) ) (C9)when N − n − ( l + 1) = 0, the matrix M N − n − ( l +1) N − ( q + ... + q l ) isfully diagonal and has determinant ( − λ ) N − ( q + ... + q l ) .Hence the term in the sum is:( 12 − λ ) q + ... + q l − l − ( 12 − λ ) N − q − ... − q l = ( 12 − λ ) n (C10)while the lowest order term in O (( − λ ) N − ( l +1) ) isagain ( − λ ) n . We hence see that the determinant ofdet( M N − n N ) ∼ ( − λ ) n , and these are the remaining n eigenvalues. Combined with the first n eigenvalues, wesee we have a total of 2 n eigenvalues in the entanglementspectrum at 1 /
2, just as the formula predicts.
Appendix D: Properties of the 1D adiabaticconnection under inversion symmetry
We begin by recalling some basic facts of inversionsymmetric Hamiltonians. Assume | u i ( k ) (cid:105) is an eigenstateof the Hamiltonian at energyˆ H ( k ) | u i ( k ) (cid:105) = E i ( k ) | u i ( k ) (cid:105) (D1)Then P | u i ( k ) (cid:105) is necessarily an eigenstate at − k of thesame energy: P ˆ H ( k ) P − P | u i ( k ) (cid:105) = ˆ H ( − k ) P | u i ( k ) (cid:105) = P ˆ H ( k ) | u i ( k ) (cid:105) = P E i ( k ) | u i ( k ) (cid:105) = E i ( k ) P | u i ( k ) (cid:105) . We assumed no degeneracies in the spectrum whichmeans | u i ( − k ) (cid:105) = e iα k P | u i ( k ) (cid:105) . (D2)Thus we have: | u i ( k ) (cid:105) = e − iα k P | u m ( − k ) (cid:105) (D3)with E i ( k ) = E i ( − k ) . We assume we are in an insulating state where thecharge polarization is6 P = − ie (cid:90) π − π dk π (cid:88) E i ( k ) < (cid:104) u k,i | ∂ k | u k,i (cid:105) . (D4)Using Eq. D3 we get P = − ie (cid:82) π − π dk π (cid:80) i ∈ occ . (cid:104)P u i ( − k ) | e iα k ∂ k e − iα k P | u i ( − k ) (cid:105) == − ie (cid:82) π − π dk π (cid:80) i ∈ occ . (cid:104)P u i ( − k ) | ∂ k P | u i ( − k ) (cid:105) − ie (cid:82) π − π dk π (cid:80) i ∈ occ . ( − i ) ∂ k α k . (D5)The last term is an integer because it depends only on α π − α − π which can at most be 2 πj , so we will drop it and beleft with: P = − ie (cid:82) π − π dk π (cid:80) i ∈ occ . (cid:104) u i ( − k ) | ∂ k | u i ( − k ) (cid:105) = ie (cid:82) − ππ dk π (cid:80) i ∈ occ . (cid:104) u i ( k ) | ∂ − k | u i ( k ) (cid:105) == − ie (cid:82) π − π dk π (cid:80) i ∈ occ . (cid:104) u i ( k ) | ∂ − k | u i ( k ) (cid:105) = ie (cid:82) π − π dk π (cid:80) i ∈ occ . (cid:104) u i ( k ) | ∂ k | u i ( k ) (cid:105) = − P . (D6)Since P is defined only mod m the two values consistent with inversion symmetry are: P = 0 , e . (D7)We define the trivial insulator as P = 0, while the topological one is P = e . The e/ half a charge on each end. Appendix E: Relation between adiabatic connectionand B ( k ) in 1D We now prove that the matrix B ij = (cid:104) u i, − k |P| u j,k (cid:105) . is unitary and thatˆ A ( − k ) = − ˆ A ( k ) + iT r [ B ( k ) ∇ k B † ( k )]We define the matrix B ij as the matrix connecting thebands at k with the ones at − k : | u i ( − k ) (cid:105) = B ∗ ij ( k ) P | u j ( k ) (cid:105) (E1)where i, j run over the occupied bands 1 , ..., N . Fromthis formula we see thatThat B ij has to be unitary can be easily seen by ex-tending the matrix to belong to all bands, occupied and unoccupied. Since we have a full gap in the system noneof the occupied bands at k can transform to unoccu-pied bands at − k and vice-versa (otherwise we wouldnot have an insulator). This means that the full ma-trix B full = diag ( B occupied , B unoccupied ) is block diagonal.One can prove the full B is unitary by using complete-ness: (cid:88) i ∈ all bands | u i ( k ) (cid:105) (cid:104) u i ( k ) | = I (E2)we have (double index means summation)( B † B ) ij = ( B † ) im B mj = (cid:104) u i,k | P † | u m, − k (cid:105) (cid:104) u m, − k | P | u j,k (cid:105) = (cid:104) u i,k | P † P | u j,k (cid:105) = (cid:104) u i,k | | u j,k (cid:105) = δ ij Hence, since the full matrix B is unitary, so are the B (un)occupied .We now want to express a connection between A ( − k )and A ( k ) . We haveˆ A ( − k ) = − i (cid:104) u i, − k | ∇ − k | u i, − k (cid:105) = i ( u i ( − k )) ∗ α ∇ k ( u i ( − k )) α = iB il ( k ) P ∗ αβ ( u l,k ) ∗ β ∇ k ( B ∗ ij ( k ) P αθ ( u j,k ) θ ) == iB il ( k )[ ∇ k B ∗ ij ( k )] P ∗ αβ ( u l,k ) ∗ β P αθ ( u j,k ) θ + iB il ( k ) B ∗ ij ( k ) P ∗ αβ ( u l,k ) ∗ β ∇ k P αθ ( u j,k ) θ == iB il ( k )[ ∇ k B ∗ ij ( k )]( u l,k ) ∗ β δ βθ ( u j,k ) θ + iB il ( k ) B ∗ ij ( k ) δ βθ ( u l,k ) ∗ β ∇ k ( u j,k ) θ == iB il ( k )[ ∇ k B ∗ ij ( k )] δ jl + iB il ( k ) B ∗ ij ( k ) ( u l,k ) ∗ β ∇ k ( u j,k ) β == iB il ( k )[ ∇ k B ∗ il ( k )] + iδ jl ( u l,k ) ∗ β ∇ k ( u j,k ) β = iB il ( k )[ ∇ k B ∗ il ( k )] + i ( u j,k ) ∗ β ∇ k ( u j,k ) β == iT r [ B ( k ) ∇ k B † ( k )] − ˆ A ( k ) (E3)where repeated indices are summed over. Appendix F: Magneto Electric Polarization asWinding Number of the B ( k ) matrix The Abelian ˆ A ( k ) in the previous section obeys a spe-cial case of the more general non-Abelian transformationˆ A ( − k ) = − B ˆ A ( k ) B † + iB ( k ) (cid:126) ∇ B † ( k ) (F1) where the non-Abelian adiabatic connection is isˆ A αβi ( k ) = − i (cid:104) u α,k | ∇ k i | u β,k (cid:105) . The above implies that7the Berry gauge fields at k and − k are non-abelian gaugetransformed of each other. The field strength gaugetransformation is:ˆ F ij ( − k ) = B ( k ) ˆ F ij ( k ) B † ( k ) (F2) From here the magneto-electric polarizability is easy buttedious to obtain: P = π (cid:82) d k(cid:15) ijk T r [( ˆ F ij ( k ) − i ˆ A i ( k ) ˆ A j ( k )) ˆ A k ( k )] == π (cid:82) d k(cid:15) ijk T r [( ˆ F ij ( − k ) − i ˆ A i ( − k ) ˆ A j ( − k )) ˆ A k ( − k )] == π (cid:82) d k(cid:15) ijk T r [( B ( k ) ˆ F ij ( k ) B † − i ( B ( k ) ˆ A i ( k ) ˆ A j ( k ) B † ( k ) − iB ( k ) ˆ A i ( k ) ∂ j B † ( k ) − + i ( ∂ i B ( k )) ˆ A j ( k ) B † ( k ) + ∂ i B∂ j B † )( − B ( k ) ˆ A k ( k ) B † ( k ) + iB ( k ) ∂ k B † ( k ))] == − π (cid:82) d k(cid:15) ijk T r [( ˆ F ij ( k ) − i ˆ A i ( k ) ˆ A j ( k )) ˆ A k ( k )] − π (cid:82) d k(cid:15) ijk T r [( B ( k ) ∂ i B † )( B ( k ) ∂ j B † )( B ( k ) ∂ k B † )] ++ i π (cid:82) d k(cid:15) ijk ∂ i ( B ( k ) ˆ A j ( k ) ∂ k B † ) = − P − π (cid:82) d k(cid:15) ijk T r [( B ( k ) ∂ i B † )( B ( k ) ∂ j B † )( B ( k ) ∂ k B † )] (F3)which proves the formula in the text. Appendix G: Proof of Fu-Kane Formula In this appendix we provide an alternative proof forthe Fu-Kane formula for the Z invariant of 3D T and P invariant insulators. Consider the Bloch Hamiltonianˆ H ( k ) of an insulator with both inversion P and time-reversal symmetry T . We have P = 1 , T = − , [ P , T ] = 0 T ˆ H ( k ) T − = ˆ H ( − k ) P ˆ H ( k ) P − = ˆ H ( − k ) (G1)and hence: P T ˆ H ( k )( P T ) − = ˆ H ( k ) , ( P T ) = − k. This does not depend onthe dimensionality of the space. The two ingredients wewill use to prove Eq. 108 are (i) band crossing argumentsbetween a 3D insulator in the trivial atomic limit and atopologically non-trivial insulator, and (ii) the fact thata 4 + 1-d Dirac Hamiltonian changes its 4D ‘Hall con-ductance’ by 1 if there is a band crossing of four bands- these are actually two doubly degenerate bands. With-out loss of generality we consider a topological insulatorwith four bands - two occupied bands and two unoccu-pied. As we take a generic T and P invariant insulatorthrough a phase transition, four ( i.e. two doubly degen-erate) bands are generically needed.Assume we have two T and P symmetric Hamiltoniansin 3 D ˆ h ( k ) and ˆ h ( k ) with: T ˆ h , ( k ) T − = ˆ h , ( − k ) P ˆ h , ( k ) P − = ˆ h , ( − k ) . We choose ˆ h ( k ) to be trivial (in the atomic limit withall hoppings taken to vanish). We now construct a time-reversal and inversion invariant interpolation between these two Hamiltonians. We first prove that such agapped interpolation exists. To see this, it is easiest to re-main with our reduced Hamiltonian with 4-bands, whichrepresents the generic effective Hamiltonian of the twodoubly-degenerate bands immediately above and belowthe Fermi level, out of the total N bands in the insu-lator. In this basis, the 3 D effective insulating Hamil-tonian with inversion and time-reversal symmetry hasco-dimension 2 (there are 3 momenta k x , k y , k z and 5Clifford generator Γ a matrices in which a 4-band insu-lating Hamiltonian with doubly degenerate bands can beexpanded). In 4 D , a topological insulator with inversionand time-reversal symmetry has co-dimension 1(there are4 momenta and 5 Γ a matrices), so there is still always away to make it gapped. This shows that generically, a3 D or 4 D insulator with inversion and time-reversal is al-ways gapped. It can be made gapless by tuning 2 and 1parameter(s) respectively. Now, let the gapped interpo-lation between ˆ h ( k ) and ˆ h ( k ) be ˆ h ( k, θ ) which satisfiesthe properties:ˆ h ( k,
0) = ˆ h ( k ) ; ˆ h ( k, π ) = ˆ h ( k ) T ˆ h ( k, θ ) T − = ˆ h ( − k, − θ ) P ˆ h ( k, θ ) P − = ˆ h ( − k, − θ ) . (G3)The interpolation between the 3D Hamiltonians is cho-sen this way so that if one interprets θ as a fourth mo-mentum then the resulting 4 D Hamiltonian would re-spect inversion and time-reversal. It was shown in Ref. 9that 2( P (ˆ h ( k )) − P (ˆ h ( k ))) = C (ˆ h ( k, θ )) where C isthe second Chern number of the 4 D Hamiltonian ˆ h ( k, θ ) . Since we chose ˆ h ( k ) to be trivial P (ˆ h ) = 0 mod n ∈ Z. (G4)Hence, if the second Chern number of the 4-D ˆ h ( k, θ ) is odd , we have P (ˆ h ) = 1 / n ∈ Z (G5)8giving rise to the result that if C of the 4 D Hamiltonianis odd, then either ˆ h ( k ) or ˆ h ( k ) is a non-trivial topo-logical insulator. Since we pick ˆ h ( k ) to be our referencetrivial Hamiltonian then this would imply that ˆ h ( k ) isnon-trivial.To understand how to classify the 3 D insulators wefirst need to understand how to get a 4 D insulator withan odd second Chern number. The 4 D trivial Hamil-tonian is simply a momentum-independent interpolationbetween ˆ h ( k ) and itself. This clearly has vanishing C . Since C is a topological invariant, we must have a gap-closing phase transition to change it. As the system isinversion and TR invariant, we have to analyze the cross-ings between two doubly-degenerate bands. This is thegeneric case, even for insulators with an arbitrary numberof bands N, because we can build an “effective” Hamilto-nian close to the transition which will be a 4-band model.With time-reversal and inversion, the Bloch Hamiltonianhas to be of the form ˆ H ( k ) = d a ( k )Γ a . We first con-sider transitions which occur away from the invariantmomenta. Because of inversion (or time-reversal) a gapclosing at k must be matched by one at − k. Such tran-sitions can be tuned by a single parameter. Since thegap closing and re-opening happens away from an in-variant momentum, the inversion eigenvalues of the oc-cupied bands remain unmodified. The non-abelian adia-batic field strengths at the two k-points are equal (up toa gauge transformation, and a minus sign in case of time-reversal). Thus, a gap closing at two points, k and − k makes the total change in second Chern number which is even . Thus a 4D Chern insulator with inversion and timereversal has even second Chern number if the inversioneigenvalues of the occupied bands are the same as in theatomic limit.We now look at the case where the gap closing and re-opening happens at an inversion symmetric point. In thiscase, the Hamiltonian is still 4 × e.g. all positive) then, up to a gauge choice, theinversion matrix is the identity operator. This meansthat [ P , ˆ H (0)] is trivially satisfied and does not providean additional constraint and so the crossing is alwaysavoided with such a large co-dimension. However, if thebands have different inversion eigenvalues,( e.g. bondingand anti-bonding bands), so that the inversion matrix ina specific choice of basis is P = 1 × ⊗ τ z , then we findthat the effective Hamiltonian matrix at the inversionsymmetric points G i / H ( G i /
2) = M i P (G6)where M i is the mass at the inversion symmetric point G i / As such, a gap closing transition whereby the M i changes sign is accompanied by two effects: (i) the second Chern number will change by ± pair of occupied bands change sign as a result of every M i that switches sign. This is the crucial difference be-tween phase transitions which change the first and sec-ond Chern numbers. In two dimensions a change in thefirst Chern number is accompanied by a single inversioneigenvalue switch since the minimal crossing is betweentwo-bands. In 4D, changes in the second Chern num-ber require a four-band crossing and thus two inversioneigenvalues are exchanged. This carries over to 3D wherewe are interested in the Z valued parity of the secondChern number. A change of the parity is effected when a4-band crossing switches two inversion eigenvalues. Wehence proved that a 4D Chern insulator with inversionand time-reversal symmetry has odd second Chern num-ber if the product of half of the inversion eigenvalues (halfmeaning each Kramers’ pair is only counted once) at allTR invariant points is − . By keeping ˆ h ( k ) constant andchanging only ˆ h ( k ) through a gap closing and reopeningwe can make ˆ h ( k, θ ) a 4D insulator with an odd secondChern number. Thus ˆ h ( k ) is a non-trivial Z insulatorwhen Eq. 108 is negative and we have proved the Fu andKane formula. Appendix H: Proof that the winding number isindependent of a smooth U (1) phase We begin with the generic form of a U (2) block of theentire B ( k ) matrix B ( k ) = e iφ ( k ) ( f ( k ) I + ig a ( k ) σ a ) (H1)where ( f ( k )) + g a ( k ) g a ( k ) = 1 . (H2)We drop the argument ( k ) which is implied from now on.By using: σ a σ b = δ ab + i(cid:15) abc σ c (H3)we find: B † ∂ i B = i [ ∂ i φ + ( f ∂ i g c − g c ∂ i f + (cid:15) bac g b ∂ i g a ) σ c ] (H4)Define G ic = ( f ∂ i g c − g c ∂ i f + (cid:15) bac g b ∂ i g a ) (H5)so that: B † ∂ i B = i [ ∂ i φ + G ic σ c ] . (H6)The winding number density is: n W = (cid:15) ijk T r [ B † ∂ i BB † ∂ j BB † ∂ k B ]= − i(cid:15) ijk T r [( ∂ i φ + G ia σ a )( ∂ j φ + G jb σ b )( ∂ k φ + G kc σ c )] == 2 (cid:15) ijk (cid:15) abc G ia G jb G kc . (H7)Thus the U (1) phase does not contribute if it is smooth.9 Appendix I: Proof that the U (1) phase is smooth in3D We have seen that the ˆ B ( k ) matrix decomposes intotrivial one dimensional and non-trivial 2 dimensional di-agonal blocks. Each 2 dimensional block can be seen asmap from T into U (2). Here we show that the U (1)phase of these maps is smooth and for this reason it canbe taken out from winding number calculation. For this,we will show that a discontinuity in the phase impliesthe appearance of the ( + − ) eigenvalue pattern of inver-sion eigenvalues at one k inv point. Since this pattern isexcluded for the U (2) blocks that cannot be reduced to U (1) × U (1), it follows that the phase is actually smooth.Now, a simple example of a map with a non-smoothphase, to keep in mind for intuition is: B ( k ) = e ik (cid:0) cos k + i sin k σ z (cid:1) ≡ e iθ ( k ) U ( k ) (I1)where U ( k ) ∈ SU (2) . Note that this B ( k ) satisfies allthe required constraints and we can see that in Eq. I1everything is smooth as k is continuously varied, exceptwhen we are close to the end of the 2 π cycle, that is whenwe close the loop around the Brillouin torus. If we startfor example from k = − π , one can see that the phase andthe SU (2) factor take different values when k approaches π , but as a whole B ( k ) reaches the same value as the onewe started with at k = π . It is important to note thatthe discontinuity in the phase cannot be removed by anycontinuous deformation of B ( k ).In the following we argue that such a discontinuity inthe phase immediately implies the existence of a ( + − ) pat-tern of inversion eigenvalues at some k inv point. As wealready pointed out, the non-removable discontinuities inthe phase are entirely due to the nontrivial topology ofthe Brillouin torus. The question is what happens when k is taken over a nontrivial loop of the torus? So let usstart from a k inv point and continuously vary k until itadvances by 2 π along one of the k , k or k directions,that is, we are taking k around non-trivial loop of thetorus. As before, we will work with the following expres-sion of the U (2) matrix: B ( k ) = e iθ k ( f k + i(cid:126)g k (cid:126)σ ) . (I2)At the k inv points, θ can take only integer or half-integervalues (in units of π ), and an important observation isthat, if θ is half-integer at a k inv point, then B ( k inv ) nec-essarily takes the form ˆ n(cid:126)σ (ˆ n = unit vector), which has ± θ =0 at k inv (otherwise the ( + − ) pattern will show up). After wecomplete the loop around the torus, in general, we canhave: θ k +2 π = θ k + (cid:26) nπ (a)(2 n + 1) π (b) (I3)In case (a) the phase factor returns to its original value soit is smooth, in which case there is no problem removing it from the winding number calculation. In case (b),the phase factor changes sign. Now let us consider theinversion properties of B ( k ) relative to the k inv = k inv + π point, which also belongs to the loop. We have: e iθ k k (cid:0) f k + k + i(cid:126)g k + k (cid:126)σ (cid:1) = e − iθ k − k (cid:0) f k − k − i(cid:126)g k − k (cid:126)σ (cid:1) . (I4)Since f k and (cid:126)g k take only real values, we can see that if θ k + k is half-integer then necessarily θ k − k is also half-integer. In other words, the half-integer values of θ k comein pairs, unless k is the k inv invariant point itself. Since θ varies from 0 to (2 n +1) π as we vary k by 2 π , there will bean odd number of times when θ k assumes a half-integervalue. It is then evident that one such half-integer valuemust occur at k inv and consequently the ( − +) inversioneigenvalue pattern will show up at k inv . Appendix J: Inversion eigenvalue patterns and thecorresponding magneto-electric polarizability
In this appendix we provide some explicit examples ofour proof of the connection between inversion eigenval-ues and the magneto-electric polarization for inversioninvariant insulators. We focus on 4-band, gapped Hamil-tonians ˆ H ( k ) in 3-dimensions displaying various patternsof inversion eigenvalues. Assuming two occupied bands,we will write these inversion eigenvalue patterns as+ + + + + + ++ − − − − + + ++ , (J1)for example, where one should understand that the twonon-zero eigenvalues of P k inv P P k inv are +1 and − k inv points and +1 and +1 at the remaining k inv points. The order of the k inv is not important for themagneto-electric polarizability, but to be precise we willorder them as { (0 , , , ( π, , , (0 , π, , (0 , , π ) , ( π, π, , ( π, , π ) , (0 , π, π ) , ( π, π, π ) } . For each case, o compute the magneto-electric polariz-ability P , we will begin with a trivial reference Hamil-tonian ( P = 0) and find a gapped and inversion sym-metric interpolation ˆ H ( k, θ ) between this and our ex-ample Hamiltonians. We can compute P as half thesecond Chern number C generated by the interpolationˆ H ( k, θ ) which is considered as a 4+1-d inversion symmet-ric Hamiltonian. An odd C corresponds to a non-trivialinsulator with half-integer magneto-electric polarizabil-ity, while an even C indicates a trivial insulator. Thismethod of calculation is convenient because C can becomputed using a gauge-invariant projector method , sowe can bypass the task of finding a smooth gauge. Allof our numerical observations are in agreement with ourmathematical proofs above.01.) The pattern + + + + + + ++ − − + + + + ++ (J2)is seen with the following gapped Hamiltonian:ˆ H ( k ) = sin k ˆΓ + sin k ˆΓ + sin k ˆΓ +( − . k + cos k + cos k )ˆΓ +0 . k + cos k )(ˆΓ + ˆΓ − ˆΓ − ˆΓ ) . (J3)This Hamiltonian can be connected through a gappedinterpolation with the Hamiltonian:ˆ H ( k ) = sin k ˆΓ + sin k ˆΓ +( − . k + cos k )ˆΓ +0 . k + cos k )(ˆΓ + ˆΓ − ˆΓ − ˆΓ ) . (J4)using the inversion symmetric homotopyˆ H ( k, θ ) = (1 + cos θ ) ˆ H ( k ) + (1 − cos θ ) ˆ H ( k ) . (J5)The second Chern number generated by ˆ H ( k, θ ) is C =0(even) and the P of ˆ H ( k ) is zero because it depends onlyon k and k . Consequently, the P of ˆ H ( k ) is 0 mod( Z ).2.) The pattern + + + + + + ++ − − − − + + ++ (J6)is seen using the following gapped Hamiltonian:ˆ H ( k ) = sin k ˆΓ + sin k ˆΓ + sin k ˆΓ +( − . k + cos k + cos k )ˆΓ − (2 − cos k )(cos k + cos k )(ˆΓ + ˆΓ − ˆΓ − ˆΓ ) . (J7)This Hamiltonian can be connected through a gappedinterpolation with the Hamiltonian:ˆ H ( k ) = sin k ˆΓ + sin k ˆΓ +( − . k + cos k )ˆΓ − k + cos k )(ˆΓ + ˆΓ − ˆΓ − ˆΓ ) . (J8)using the inversion symmetric homotopyˆ H ( k, θ ) = (1 + cos θ ) ˆ H ( k ) + (1 − cos θ ) ˆ H ( k ) . (J9)The second Chern number generated by ˆ H ( k, θ ) is C =0(even) and the P of ˆ H ( k ) is zero because it depends onlyon k and k . Consequently, the P of ˆ H ( k ) is 0 mod( Z ).3.) The pattern − − − − + + ++ − − − − + + ++ (J10) is seen with the following gapped Hamiltonian:ˆ H ( k ) = sin k ˆΓ + sin k ˆΓ + sin k ˆΓ +( − + cos k + cos k + cos k )ˆΓ + ˆΓ . (J11)This Hamiltonian can be connected through a gappedinterpolation with the trivial Hamiltonian:ˆ H ( k ) = sin k ˆΓ + sin k ˆΓ + sin k ˆΓ − ˆΓ , (J12)using the inversion symmetric homotopy of Eq. 144.The second Chern number generated by ˆ H ( k, θ ) is C =2(even) and consequently P = 0 mod( Z ).4.) The pattern + + + + + + ++ − − − − − − −− (J13)is seen using the following gapped Hamiltonian:ˆ H ( k ) = sin k ˆΓ + sin k ˆΓ + sin k ˆΓ − ˆΓ +(2 + cos k + cos k + cos k )ˆΓ +( − k + cos k + cos k )ˆΓ (J14)This Hamiltonian can be connected through a gappedinterpolation with the trivial Hamiltonian:ˆ H ( k ) = sin k ˆΓ + sin k ˆΓ + sin k ˆΓ − ˆΓ (J15)using the inversion symmetric homotopyˆ H ( k, θ ) = sin k ˆΓ + sin k ˆΓ + sin k ˆΓ − ˆΓ + (1 + cos θ )(2 + cos k + cos k + cos k )ˆΓ + (1 + cos θ )( − k + cos k + cos k )ˆΓ + sin θ ˆΓ . (J16)The second Chern number generated by ˆ H ( k, θ ) is C =0(even) and consequently P = 0 mod( Z ).5.) The pattern − − + + + + ++ − − + + + + ++ (J17)is seen using the following gapped Hamiltonian:ˆ H ( k ) = sin k ˆΓ + sin k ˆΓ +( − + cos k + cos k )ˆΓ +( − + 0 . k − cos k ))ˆΓ . (J18)This Hamiltonian can be connected through a gappedinterpolation with the trivial Hamiltonian:ˆ H ( k ) = sin k ˆΓ + sin k ˆΓ + sin k ˆΓ − ˆΓ (J19)1using the inversion symmetric homotopyˆ H ( k ) = sin k ˆΓ + sin k ˆΓ + (1 − cos θ ) sin k ˆΓ +( − + (1 + cos θ )(cos k + cos k ))ˆΓ + (1 + cos θ )( − + 0 . k − cos k ))ˆΓ + sin θ ˆΓ . (J20)The second Chern number generated by ˆ H ( k, θ ) is C =0(even) and consequently P = 0 mod( Z ). Appendix K: Response Theory Argument in HigherDimensions
Although we do not rigorously prove anything in higherdimensions we note that the physical response argumentspresented in this work and Refs. 9 and 29 continue to ap-ply in higher dimensions. This gives us a hint that thereare interesting inversion symmetric topological insulatorsin higher dimensions. For even spacetime dimensions thetopological response actions all take a standard form S eff [ A µ ] = (cid:90) d n xP n (cid:15) a a ...a n − a n F a a . . . F a n − a n − where P n is a response coefficient and F ab is the elec-tromagnetic field-strength tensor. If these were all dy-namical fields then the entire action must transform likea scalar which means that the intrinsic response coeffi-cients P n must transform the same way as the product ofthe electromagnetic fields. Thus, P n must be odd under inversion symmetry for all n. Additionally for even (odd) n, P n is odd under T (C) symmetry. Since we are talkingabout external electromagnetic fields we do not transformthem under the symmetry operation and hence only the P n are changed. Thus in alternating even spacetime di-mensions topological insulators are protected by eitherT or C symmetry. However, in every even spacetimedimension there is a topological insulator protected byinversion symmetry which has a topological response. Inall of these cases P n is not gauge invariant under transfor-mations of the occupied wavefunctions and the oddnessunder the different symmetries quantizes P n to take onlytwo independent values. This yields a Z classificationfor the response coefficient in line with the argumentspresented in Ref. 9.In odd space-time dimensions the generalized Chern-Simons terms describing the electro-magnetic responsesin topological insulators are compatible with inversionsymmetry. The general action is S ( odd ) eff [ A µ ] = C n ( n + 1)!(2 π ) n × (cid:90) d n +1 (cid:15) a a ...a n +1 A a F a a . . . F a n a n +1 . The quantity C n is the n -th Chern number and is evenunder inversion symmetry and thus not restricted by therequirement of preserving inversion symmetry. If n iseven (odd) then the action is also compatible with T ( C ) symmetry. We have not proven it, but our intuitionsuggests that the parity of C n can be characterized by theinversion topological invariants given by the χ ( n ) P definedabove when calculated in (2 n + 1) − d. F. D. M. Haldane, Phys. Rev. Lett. , 2015 (1988). C. L. Kane and E. J. Mele, Phys. Rev. Lett. , 226801(2005). C. L. Kane and E. J. Mele, Phys. Rev. Lett. , 146802(2005). B.A. Bernevig and S.C. Zhang, Phys. Rev. Lett. , 106802(2006). B. A. Bernevig, T. L. Hughes, and S.C. Zhang, Science , 1757 (2006). M. K¨onig, S. Wiedmann, C. Br¨une, A. Roth, H. Buhmann,L. Molenkamp, X.-L. Qi, and S.-C. Zhang, Science ,766 (2007). L. Fu and C. L. Kane, Phys. Rev. B , 045302 (2007). D. Hsieh, D. Qian, L. Wray, Y. Xia, Y. S. Hor, R. J. Cava,and M. Z. Hasan, Nature , 970 (2008). X.-L. Qi, T. Hughes, and S.-C. Zhang, Phys. Rev. B ,195424 (2008). A. P. Schnyder, S. Ryu, A. Furusaki, and A. W. W. Lud-wig, Phys. Rev. B , 195125 (2008). A. Kitaev, Adv. Theo. Phys.: AIP Conf. Proc. , 22(2008). L. Fu and C. L. Kane, Phys. Rev. B , 195312 (2006). J. E. Moore and L. Balents, Phys. Rev. B , 121306(2007). R. Roy, Phys. Rev. B , 195322 (2009). E. Prodan, Phys. Rev. B , 125327 (2009). E. Prodan, J. Phys. A: Math. Theor. , 082001 (2009). A. M. Essin, J. E. Moore, and D. Vanderbilt, Phys. Rev.Lett. , 146805 (2009). M. Freedman, C. Nayak, K. Shtengel, and K. Walker, Ann.Phys. , 428 (2004). A. Hamma, R. Ionicioiu, and P. Zanardi, Phys. Lett. A , 22 (2004). A. Kitaev and J. Preskill, Phys. Rev. Lett. , 110404(2006). M. Levin and X.-G. Wen, Phys. Rev. Lett. , 110405(2006). H. Li and F. D. M. Haldane, Phys. Rev. Lett. , 010504(2008). S. Ryu and Y. Hatsugai, Phys. Rev. B , 245115 (2006). N. Bray-Ali, L. Ding, and S. Haas, Phys. Rev. B ,180504(R) (2009). S. T. Flammia, A. Hamma, T. L. Hughes, and X.-G. Wen,Phys. Rev. Lett. , 261601 (2009). R. Thomale, A. Sterdyniak, N. Regnault, and B. A.Bernevig, Phys. Rev. Lett. , 180502 (2010). R. Thomale, D. P. Arovas, and B. A. Bernevig, Phys. Rev.Lett. , 116805 (2010). F. Pollman and J. E. Moore, New J. Phys. , 025006(2010). A. M. Turner, Y. Zhang, and A. Vishwanath, arxiv:0909.3119. L. Fidkowski, Phys. Rev. Lett. , 130502 (2010). M. Kargarian and G. A. Fiete, Phys. Rev. B , 085106(2010). E. Prodan, T. L. Hughes, and B. A. Bernevig, Phys. Rev.Lett. , 115501 (2010). A. Sterdyniak, N. Regnault, and B. A. Bernevig, arxiv:1006.5435. F. D. M. Haldane, APS 2009 March Meeting Proceeding(unpublished). I. Peschel, J. Stat. Mech , P06004 (2004). J. Zak, Phys. Rev. Lett. , 2747 (1989). J. Goldstone and F. Wilczek, Phys. Rev. Lett. , 986(1981). B. Simon, Phys. Rev. Lett. , 2167 (1983). F. Wilczek and A. Zee, Phys. Rev. Lett. , 2111 (1984). H. B. Nielsen and M. Ninomiya, Nucl. Phys. B , 20(1981). B. Halperin, Japanese Journal of Applied Physics Suppl. , 1913 (1987). L. Fu, C. L. Kane, and E. J. Mele, Phys. Rev. Lett. ,106803 (2007). Z. Wang, X.-L. Qi, and S.-C. Zhang, arxiv: 0910.5954. X. Wan, A. Turner, A. Vishwanath, and S. Y. Savrasov,arxiv: 1007.0016. J. Milnor,
Morse Theory (Princeton University Press,USA, 1963). W. P. Su, J. R. Schrieffer, and A. J. Heeger, Phys. Rev.Lett. , 1698 (1979). D. J. Thouless, M. Kohmoto, M. P. Nightingale, andM. den Nijs, Phys. Rev. Lett. , 405 (1982). M. K¨onig, H. Buhmann, L. W. Molenkamp, T. Hughes,C.-X. Liu, X.-L. Qi, and S.-C. Zhang, J. Phys. Soc. Jpn , 031007 (2008). X.-L. Qi, T. Hughes, and S.-C. Zhang, Nature Physics ,273 (2008). X.-L. Qi and S.-C. Zhang, Phys. Rev. Lett. , 086802(2008). S. Murakami, S. Iso, Y. Avishai, M. Onoda, and N. Na-gaosa, Phys. Rev. B , 205304 (2007). S. Murakami, New J. Phys. , 356 (2007). S. Murakami and S.-I. Kuga, Phys. Rev. B78