Lecture 3: Topological insulators Dimi Culcer UNSW

Outline of this lecture 3D Topological Insulators What is a topological insulator? 𝑘 ∙ 𝑝 theory for Bi2Se3 Surface states of Bi2Se3 Topological Insulators films Thin layer of Bi2Se3 and interlayer tunnelling Quantum anomalous Hall effect in 3D and 2D TIs Extended state picture Edge state picture of 2D QAHE Is transport really dissipationless?

What is a topological insulator? A fancy name for a schizophrenic material All that matters is spin-orbit coupling and time reversal 2D topological insulators Insulating surface, conducting edges – chiral edge states with definite spin orientation 3D topological insulators Insulating bulk, conducting surfaces – chiral surface states with definite spin orientation Many kinds of insulators Band insulator – energy gap ≫ room temperature Anderson insulator – large disorder concentration Mott insulator – strong electron-electron interactions Kondo insulator – localized electrons hybridize with conduction electrons – gap All of these can be topological insulators if spin-orbit strong enough All the materials in this talk are band insulators Also topological superconductors Quasiparticles – Cooper pairs

What is a topological insulator? The first topological insulator was the quantum Hall effect (QHE) QHE is a 2D topological insulator No bulk conduction (except at special points), only edge states Edge states travel in one direction only They cannot back-scatter – have to go across the sample Hall conductivity quantised 𝜎 𝑥𝑦 = 𝑛 𝑒 2 ℎ n is a topological invariant – Chern number This is the integral of the Berry curvature over the Brillouin zone n counts the number of Landau levels below the Fermi energy ~ like the filling factor QHE breaks time-reversal because of the magnetic field The current generation of TIs is time-reversal invariant C.L. Kane & E.J. Mele, Physical Review Letters 95 (2005) 226801. M.Z. Hasan & C.L. Kane, Reviews of Modern Physics 82 (2010) 3045. X.-L. Qi & S.-C. Zhang, Reviews of Modern Physics 83 (2011) 1057. X.-L. Qi, T.L. Hughes & S.-C. Zhang, Physical Review B 78 (2008) 195424.

How do we identify a TI? Kane & Mele found a topological invariant Called Z2 invariant Related to the matrix elements of the time-reversal operator Sandwich this operator between all pairs of bands in the crystal Need the whole band structure – difficult calculation Z2 invariant counts the number of surface states 0 or even is trivial 1 or odd is non-trivial – odd number of Dirac cones Theorem says fermions come in pairs The pair is on the on opposite surface In practice in a TI slab all surfaces have TI states Especially important when looking at Hall transport Bulk conduction Surface states Bulk valence

Most common TI – Bi2Se3 Equivalent layers Se1 equivalent to Se1’ Bi1 equivalent to Bi1’ The system has inversion symmetry Se2 site plays the role of an inversion centre Under spatial inversion Bi1 → Bi1’ Se1 → Se1’ Eigenstates have definite parity Bi2Se3 is slightly non-Bravais 5 atoms per unit cell Zhang et al, Nature Physics 5, 438 (2009)

Bulk Bi2Se3 Start from the atomic energy levels Interested in energy eigenvalues around Γ point What is the effect of crystal field splitting and SO? States near Fermi surface come from p-orbitals Both in Bi (6s26p3) and Se (4s24p4) Crystal field removes degeneracy of p-orbitals Only the pz orbitals are relevant – low-lying This comes from the point-group symmetry of the crystal The lattice has inversion symmetry pz orbitals from Bi, Se near FS have opposite parities Energy-gap between these two controlled by SO Increasing SO may cause a level-inversion See figure In the crystal this corresponds to a band inversion Picture from Shen (2nd edition, 2017) Spin-orbit controls the energy splitting between the orbitals. Increasing the spin-orbit will result in a level inversion.

Bulk Bi2Se3 Here z // (111) Near Fermi surface the relevant energy levels are pz orbitals There are called 𝑝 1𝑧 + ↑ , 𝑝 1𝑧 + ↓ , 𝑝 2𝑧 − ↑ , 𝑝 2𝑧 − ↓ Here +/- represents the parity, ↑/↓ represents spin Using these states we can construct the effective 𝑘 ∙ 𝑝 Hamiltonian 𝐻= 𝜀 𝑝 + 𝑀− 𝐵 ∥ 𝑝 ∥ 2 − 𝐵 ⊥ 𝑝 𝑧 2 𝕀 2×2 ℏ 𝑣 ∥ 𝜎 ∙ 𝑝 ∥ + 𝑣 ⊥ 𝜎 𝑧 𝑝 𝑧 𝑣 ∥ 𝜎 ∙ 𝑝 ∥ + 𝑣 ⊥ 𝜎 𝑧 𝑝 𝑧 𝜀 𝑝 − 𝑀− 𝐵 ∥ 𝑝 ∥ 2 − 𝐵 ⊥ 𝑝 𝑧 2 𝕀 2×2 Very much like the Dirac equation Here the energy 𝜀 𝑝 =𝐶 − 𝐷 ∥ 𝑝 ∥ 2 − 𝐷 ⊥ 𝑝 𝑧 2 This breaks the particle-hole symmetry of the system p-linear term due to the different parities of the two sets of basis functions The model is time-reversal invariant as expected

Surface states of Bi2Se3 Surface states can be derived from bulk Hamiltonian Take 𝑘 ∙ 𝑝 Hamiltonian from previous slide Truncate the system at 𝑧=0 𝑝 𝑥 , 𝑝 𝑦 are still good quantum numbers because translationally invariant in xy-plane First set 𝑝 𝑧 →−𝑖ℏ𝜕/𝜕𝑧 and 𝑝 𝑥 , 𝑝 𝑦 →0 Find the solution to the equation 𝐻 𝑧 𝜓 𝑧 =𝜀𝜓 𝑧 where 𝐻(𝑧)= 𝐶+ ℏ 2 𝐷 ⊥ 𝜕 2 𝜕 𝑧 2 + 𝑀+ ℏ 2 𝐵 ⊥ 𝜕 2 𝜕 𝑧 2 𝕀 2×2 −𝑖ℏ𝑣 ⊥ 𝜎 𝑧 𝜕 𝜕𝑧 −𝑖ℏ𝑣 ⊥ 𝜎 𝑧 𝜕 𝜕𝑧 𝐶+ ℏ 2 𝐷 ⊥ 𝜕 2 𝜕 𝑧 2 − 𝑀+ ℏ 2 𝐵 ⊥ 𝜕 2 𝜕 𝑧 2 𝕀 2×2 Here the 𝜀 𝑝 term breaks particle-hole symmetry between conduction & valence bands For 𝐷 ⊥ > 𝐵 ⊥ the band gap closes, the system is no longer an insulator To have a surface state solution we need to focus on 𝐷 ⊥ < 𝐵 ⊥

Surface states of Bi2Se3 We can decouple rows 1&3 from rows 2&4 in H(z) We take two trial solutions 𝜓 1 = 𝑒 𝜆𝑧 𝑎 1 0 𝑏 1 0 and 𝜓 2 = 𝑒 𝜆𝑧 0 𝑎 2 0 𝑏 2 Now we have two independent sets of equations Let 𝐵 ± = 𝐵 ⊥ ± 𝐷 ⊥ 𝑀+ 𝐵 + 𝜆 2 −𝑖 𝑣 ⊥ 𝜆 −𝑖 𝑣 ⊥ 𝜆 −𝑀+ 𝐵 − 𝜆 2 𝑎 1 𝑏 1 =𝜀 𝑎 1 𝑏 1 𝑀+ 𝐵 + 𝜆 2 𝑖 𝑣 ⊥ 𝜆 𝑖 𝑣 ⊥ 𝜆 −𝑀+ 𝐵 − 𝜆 2 𝑎 2 𝑏 2 =𝜀 𝑎 2 𝑏 2

Surface states of Bi2Se3 Focus on the first equation – for a non-trivial solution need det 𝑀+ 𝐵 + 𝜆 2 −𝐸 −𝑖 𝑣 ⊥ 𝜆 −𝑖 𝑣 ⊥ 𝜆 −𝑀+ 𝐵 − 𝜆 2 −𝐸 =0 This gives 4 roots, ± 𝜆 1 ,± 𝜆 2 since it is a quartic in 𝜆 Impose Dirichlet boundary conditions 𝜓 𝑧 =0 at 𝑧=0 and 𝑧=−∞ For 𝑀 𝐵 ⊥ >0 the wave-function takes the form 𝜓 1 = (𝑒 𝜆 1 𝑧 − 𝑒 𝜆 2 𝑧 ) 𝑎 1 0 𝑏 1 0 Similarly from the second equationn we find 𝜓 2 = (𝑒 𝜆 1 𝑧 − 𝑒 𝜆 2 𝑧 ) 0 − 𝑎 1 0 𝑏 1 For the exact coefficients, exponents, normalisation factors – look up Shen

Surface states of Bi2Se3 Effective Hamiltonian for surface states Use 𝜓 1 , 𝜓 2 as the basis states Add the 𝑝 𝑥 , 𝑝 𝑦 terms in the Hamiltonian Work out the matrix elements 𝜓 1 𝐻 𝜓 1 𝜓 1 𝐻 𝜓 2 𝜓 2 𝐻 𝜓 1 𝜓 2 𝐻 𝜓 2 This yields the effective Hamiltonian 𝐻 𝑒𝑓𝑓 = 𝐸 0 − 𝐷 ∥ 𝑝 ∥ 2 + 𝑣 𝑒𝑓𝑓 ( 𝜎 𝑥 𝑝 𝑦 − 𝜎 𝑦 𝑝 𝑥 ) Where 𝑣 𝑒𝑓𝑓 = 𝑣 ∥ 1− 𝐷 ⊥ 2 𝐵 ⊥ 2 The quadratic term does not affect the dispersion much at any transport densities It is however important when we discuss tunnelling later on Note this is exactly Rashba even though the prefactor has a different origin However, here there is only ONE Fermi surface and NO spin precession Bulk conduction Surface states Bulk valence

Helical versus chiral ky kx Spin-momentum locking HgTe quantum well Most of the time we can forget about the quadratic term 𝐻 𝑒𝑓𝑓 = ℏ𝑣 𝑒𝑓𝑓 ( 𝜎 𝑥 𝑘 𝑦 − 𝜎 𝑦 𝑘 𝑥 ) The momentum sets the direction of the spin Direction of spin ⊥ momentum We say the system is chiral HgTe quantum well 𝐻 𝑒𝑓𝑓 = ℏ𝑣 𝑒𝑓𝑓 ( 𝜎 𝑥 𝑘 𝑥 + 𝜎 𝑦 𝑘 𝑦 ) Spin still locked to the momentum but ∥ momentum This system is helical You can rotate one H into the other 𝑘 𝑥 → 𝑘 𝑦 , 𝑘 𝑦 → −𝑘 𝑥 The physics is always the same Graphene similar to HgTe QWs but characterised by pseudospin helicity Same Hamiltonian except 𝜎 is not a real spin but a pseudospin (2 sub-lattices) ky kx

No back-scattering ky kx To flip the momentum you must flip the spin Ordinary Coulomb scattering cannot do this Need time-reversal breaking – magnetic impurities Eigenstates of Rashba Hamiltonian In 𝑘 ∙ 𝑝 language 𝑢 𝑘 = 1 2 𝑒 −𝑖𝜃 ±1 One represents electrons, the other holes Fermi’s Golden Rule 𝑆𝑐𝑎𝑡𝑡𝑒𝑟𝑖𝑛𝑔 𝑟𝑎𝑡𝑒= 2𝜋 ℏ 𝑘′ 𝜓 𝑘 𝑈 𝑑𝑖𝑠𝑜𝑟𝑑𝑒𝑟 𝜓 𝑘′ 2 𝛿 𝜀 𝑘 − 𝜀 𝑘′ Substitute Rashba eigenstates (for electrons) You get a factor of 1+ cos 𝜃− 𝜃 ′ Vanishes when 𝜃 ′ =− 𝜃 i.e. no back-scattering This provides topological protection against localisation Which is coherent back-scattering ky kx

Spin-polarized current 𝐻 𝑠𝑜 =𝛼 𝜎 ∙ 𝑘 × 𝑧 Current operator proportional to spin Charge current = spin polarization Spin polarization exists throughout surface Not in bulk because Bi2Se3 has inversion symmetry This is a signature of surface transport Smoking gun for TI behavior Detection – Dan Ralph group, Nature 511, 449 (2014) Conducting surface  Insulating bulk

Spin-polarized current ky ky kx kx No E E // x

Thin film of Bi2Se3 – non-magnetic Formally it is exactly like a single surface Except the system is ow truncated at 𝑧= 𝑧 𝑡𝑜𝑝 and 𝑧 𝑏𝑜𝑡𝑡𝑜𝑚 Results in tunnelling between top & bottom layers Because of wave function overlap Important in ultra-thin films < 5 nm Now the effective Hamiltonian is 4 × 4 𝐻 𝑇𝐼𝑇𝐹 = 𝑅𝑎𝑠ℎ𝑏𝑎 𝑡 0 0 𝑡 𝑡 0 0 𝑡 −𝑅𝑎𝑠ℎ𝑏𝑎

Thin film of Bi2Se3 – non-magnetic The energy eigenvalues are 2-fold degenerate This is because time-reversal symmetry is still preserved We have Kramers degeneracy 𝜀 ± = ± 𝑣 𝑒𝑓𝑓 2 𝑝 2 + 𝑡 2 But now the eigenstates contain spins from both layers When tunnelling is strong we can no longer think of individual surfaces The asymptotic limits are easy to understand Expand the eigen-energies in the small parameter Small t, like 2 TI surfaces Large t, like ordinary parabolic semiconductor

Thin film of Bi2Se3 – magnetic We need to add a magnetisation 𝑀 ∥ 𝑧 Same form for top and bottom Let M have units of energy 𝐻 𝑇𝐼𝑇𝐹 = 𝑀 𝑖𝑣 𝑒𝑓𝑓 𝑝 − −𝑖𝑣 𝑒𝑓𝑓 𝑝 + −𝑀 𝑡 0 0 𝑡 𝑡 0 0 𝑡 𝑀 −𝑖𝑣 𝑒𝑓𝑓 𝑝 − 𝑖𝑣 𝑒𝑓𝑓 𝑝 + −𝑀 Rotate this to get rid of the off-diagonal blocks 𝜓 1 = 1 2 𝜓 1 + 𝜓 3 𝜓 2 = 1 2 𝜓 2 − 𝜓 4 𝜓 3 = 1 2 𝜓 1 − 𝜓 3 𝜓 4 = 1 2 𝜓 2 + 𝜓 4 Liu et al, arXiv:1508.07106

Thin film of Bi2Se3 – magnetic After rotation to the new basis 𝐻 𝑇𝐼𝑇𝐹 = 𝑀+𝑡 𝑖𝑣 𝑒𝑓𝑓 𝑝 − − 𝑖𝑣 𝑒𝑓𝑓 𝑝 + −𝑀−𝑡 0 0 0 0 0 0 0 0 −𝑀+𝑡 − 𝑖𝑣 𝑒𝑓𝑓 𝑝 + 𝑖𝑣 𝑒𝑓𝑓 𝑝 − 𝑀−𝑡 The eigenvalues are no longer degenerate because breaking TR Top block 𝜀 ± = ± 𝑣 𝑒𝑓𝑓 2 𝑝 2 + 𝑀+𝑡 2 Bottom block 𝜀 ± = ± 𝑣 𝑒𝑓𝑓 2 𝑝 2 + 𝑀−𝑡 2 Notice this is just 2 copies of Rashba + magnetisation Instead of M we have 𝑀±𝑡 some effective magnetisation The two blocks do not represent the two surfaces But the structure of the Hamiltonian is the same Liu et al, arXiv:1508.07106

Surface QAHE in Bi2Se3 TI thin/ultra-thin film Hamiltonian ~ 2 blocks In both cases it is Rashba + effective magnetisation The magnetisation opens a gap in the spectrum We can place the Fermi energy in this gap For now assume everything is happening at 𝑇=0 Valence band full, conduction band empty There is definitely no 𝜎 𝑥𝑥 because there is no Fermi surface Consider the contribution to the AHE due to one block We already calculated this in Lecture 2 The Berry curvature is exactly the same This time we only have one band – the valence band We can extend the k-integral from −∞ to +∞ This is cheating but doing it formally correctly takes a lot of work – need full 3D model One block gives 𝜎 𝑥𝑦 = 𝑒 2 2ℎ Surface states εF Surface states

Surface QAHE in Bi2Se3 The other block contributes 𝜎 𝑥𝑦 = 𝑒 2 2ℎ These add up because the normal to the other surface is flipped, total 𝜎 𝑥𝑦 = 𝑒 2 ℎ Easiest to understand for 𝑡=0 when we can think of independent surfaces All surfaces are topological and the Hall current flows around the sample This is the famous quantised anomalous Hall effect, the Chern number is 1 Notice we get the same result regardless of tunnelling Because the structure of the Hamiltonians is the same t appears as a renormalisation of the magnetisation But we do not care because we are extending k from −∞ to +∞ The magnitude of the effective magnetisation becomes irrelevant In the two cases the mechanism for current flow must be different When t is large the electrons live on both surfaces – picture below not valid Cross-sectional view Electric field into the page

Edge states of Bi2Se3 film See Shen for edge states – same demonstration as for surface states For the edge state to be well-defined We need non-magnetic tunnelling term t to be significant So the film needs to be 3 nm thick and certainly less than 5 nm Some sort of edge states exist even if t negligible But very complicated and poorly understood We understand the maths We don’t fully understand the physics E → Top view

Is transport really dissipationless? Not completely dissipationless for two reasons The system is coupled to leads Otherwise you cannot pass a current through it So the system talks to the outside world The leads are metallic and interact with phonons etc Also, there is a potential difference across the sample We can think of it as transport through an island There is no dissipation in the island itself But there is dissipation in the leads and ACROSS the island Jared Cole is currently working on this There is a top gate – needed to switch a transistor on/off This is like a parallel-plate capacitor (conductor – insulator – conductor) Actually most of the energy dissipated inn transistors comes from the gate There is also the issue of room temperature operation Bi2Se3 bulk gap 0.3 eV compared to Si 1.1 eV

Summary 3D Topological Insulators 2D Topological Insulators Bi2Se3 surface states follow Rashba-like dispersion Spin-orbit is everything: no kinetic-energy term Each surface exhibits a quantum anomalous Hall effect adding up to 𝑒 2 ℎ 2D Topological Insulators Thin layer of Bi2Se3 QAHE in 2D TIs can be explained by edge states Only one state on each edge → transport without scattering like in QHE → dissipationless Not truly dissipationless Because of leads and top gate But can still save a lot of energy Thanks to Haizhou Lu for many explanations