Outline of Talk. Chern Number in a Band Structure. Contents. November 7, Single Dirac Cone 2

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1 Chern Number in a Band Structure November 7, 06 Contents Single Dirac Cone QAHE in TI. SS Hamiltonian in thick limit SS Hamiltonians with coupling Add magnetic term Haldane Model 5. Basic Graphene Symmetries Inversion TRS Second Nearest Neighbor Haldane Term Supplementary Calculations 8 4. Edge Modes at Mass Inversion Interface Phase From Magnetic Flux Adiabatic Continuity Van Vleck Paramagnetism Mean Field Theory Normal FM Lagrangian for wave-packet Anomalous Velocity QHE and dissipative channels Words Equations Berry connection / vec pot Ak = i nk nk k Berry Phase γ = i d k A n k Berry Curvature F n = k n k n = k Ak Chern Number ν n = π TRS with real spin T = iσ y K = Outline of Talk BZ d kf n Z 0 K 0. Describe what Chern number means and how it is calculated.. Model calculation with a single Dirac cone

2 a Explain qualitatively why this is only half-quantized. b Explain what this means for when a gap closes and re-opens. c On hand:. Haldane Model i. Calculation for state existing at edge a Introduce Hamiltonian i. Full tight-binding rst, without gapping terms. ii. Sublattice asymmetry: show it breaks inversion but not TRS iii. Second-nearest neighbor terms b Add periodic magnetic eld i. Show it breaks TRS, but not inversion necessarily ii. Explain the complex hopping term qualitatively c Topological Phase Diagram i. Plot the gap at the dierent K-points as a function of M > show why chern # changes from 0 to or - ii. Draw the entire phase diagram d On hand: i. Quantiative explanation for complex hopping term 4. QAHE in magnetic TIs a Explain system: i. Two surface states with nite tunneling between them trivial mass gap ii. Write down Hamiltonian iii. Introduce magnetic term, explain what it means b Show via same methods why it can produce nite chern # like in Haldane Model c On hand: i. Van Vleck Paramagnetism ii. Mean eld theory standard iii. Mean eld theory TI? 5. On hand for calculating quantized conductance a Anomalous velocity? b From anomalous velocity up to Hall conductivity c Inuence of Dissipative Channel Single Dirac Cone Hk = i d k kσ i 0 0 i 0 σ =, σ 0 =, σ i 0 = 0 If d = M 0, then there will be a gap. Showing TRS is broken with a mass gap:

3 T HT = d x σ x d y σ y Mσ z H k H k = d x σ x d y σ y + Mσ z Generally we need d i k = d i k to preserve TRS. The normalized, orthogonal wave-functions of this two-level system are the following, incluing a helicity parameter α = ±: ψ + = ψ = dd + d dd d d + d d αid d d d αid 4 5 ψ = = m m m + k + k m + k k m x αik y m m + k B k ke iαθ k 6 7 The Berry connection is A k = i ψ k k ψ k = A x = d k d d k d α dd d 9 αk y k + m k + m m 0 8 A y = αk x k + m k + m m ν = α π F z = αm k + m ˆ 0 F xy d k = signmα This is for innite band width of a Dirac system. However, in a real situation, nite band-width gives two gaps, each of which should roughly contribute ± to the Chern number. QAHE in TI. SS Hamiltonian in thick limit Each surface state has a Hamiltonian Hsf L = v F 0 k k σ = LivF k + 4 where prefactorl = ± is for the top surface and bottom surface, and k ± = k x ± ik y Dispersion is E sf = v F k 5

4 . SS Hamiltonians with coupling Coupling the surfaces results in a mass term. TRS T = H two = iv F k m k 0 iv F k m k m k 0 0 iv F k 0 m k iv F k + 0 K shows that T HT = H., ψ = t t b b 6 In the limit that v F k m k, then m k m 0 + m k and one can rewrite the Hamiltonian in a new basis m k iv F k H two = iv F k + m k m k iv F k +, ψ = iv F k m k which is block diagonal yet again, with each block being an equal superposition of top and bottom surfaces, and they are both gapped: E = ± m k + v F k 8 Note that this Hamiltonian is still maintains TRS, since ipping spin and complex conjugating ipping momentum does not change it.. Add magnetic term A magnetic exchange term to a bath of localized spins inserts itself as a sort of Zeeman coupling for each spin. Therefore gm H M = 0 gm gm 0, ψ = gm which DOES break TRS, since T HT = H. One can re-write the block-diagonal Hamiltonian as hk + m H = k + gm σ z 0 0 h k + m k gm σ z, ψ = which shows two independent sections. When g M 0, then the system has TRS so ν 0. As the exchange eld M is increased, ONE of the gaps closes when g M = m 0 and then reopens with an opposite-sign mass parameter. This means that the Chern number has changed by, so ν = ±, with the sign determined by the sign of M. 4

5 Haldane Model. Basic Graphene Tight bindng for Graphene. First, lattice vectors, assume A and B are horizontal from each other: r = aˆx But either way: r a b = aˆx = aˆx ± aŷ r b a = aˆx = aˆx ± aŷ 4 H = t e i k r ij c i c j + ɛ i c i c i 5 i,j Find the zero-energy spots when m = 0. We know it is at the K point, which we can nd by setting k x = 0: + 4 cos ak y + 4 cos ak y cos ak x = cos ak y + 4 cos ak y = cos ak y = 0 8 cos ak y = 9 ak y = ± π 0 K ± = ± 4π a ˆk y Now simplify the Hamiltonian: i=a,b 5

6 H = t = t = t 0 e iakx + e i akx e i aky + e i aky e iakx + e i akx e i aky + e i aky 0 0 e iaδx + e i aδx e ± π i+ iaδy + e π i iaδy e iaδx + e i aδx e ± π i+ iaδy + e π i iaδy 0 0 e iaδx + e i e aδx ± π i e iaδy + e π i e e iaδx + e i e aδx ± π i e iaδy + e π i e iaδy 0 0 iaδ t x + + i a δ x + α + iaδ x + i a δ x + α aδ y 0 0 iaδ t x + + α aδ y i a δ x + iaδ x + + α aδ y + i a δ x 0 at 0 αδ y iδ x αδ y + iδ x 0 aδ y iaδy where a at is the fermi velocity v F familiar dispersion relation up to a factor of. Note that α = ± for K/K. This leads to the E = ± v F δ 8. Symmetries.. Inversion r r. This swaps the sublattices. It clearly does not aect the n.n. o-diagonal components, but it does aect the on-site energies. Breaking inversion opens a band gap... TRS ˆT = ikσ y, so T =. The Hamiltonian, as is, has TRS, i.e. A term in the Hamiltonian that breaks TRS may open up a band gap.. Second Nearest Neighbor First, need the vectors: T HT = H 9 The perturbation to the model is r a a = ± aŷ 40 = ± aˆx ± aŷ 4 r b b = ± aŷ 4 H = t = ± aˆx ± aŷ 4 i,j e i k r ij c i c j 44 6

7 which is a diagonal term. As written, this term does not break either symmetry of the lattice, and so does not open a gap. It contributes the following this is at K + : H nnn = t cos aky + cos ak x k y + cos ak x + k y a 4π = t cos 4π = t cos + aδ y a + δ y + cos + cos aδ x aδ x π 4π a + δ y aδ y + cos + cos aδ x + aδ x + π + aδ y 4π a + δ y And the key is that at δ x,y = 0: 4π H nnn 0, 0 = t cos + cos π π + cos 48 4π π = t cos + cos 49 = t 50 which is just a chemical potential shift K ± is the same since any ± happens in front of a δ y and has no consequence. When δ x,y 0, then this term renormalizes the Fermi velocity, and will be dierent for electrons and holes, but doesn't make any fundamental change in the band structure..4 Haldane Term Now introduce a periodic magnetic ux into the unit cell and choose a gauge so that only the nnn hoppings develop a phase. H = t e iφij e i k r ij c i c j 5 i,j Be careful about minus signs! H nnn,a = t cos aky + φ + cos ak x H nnn,b = t cos aky φ + cos ak x k y + φ + cos k y φ + cos ak x + k y + φ ak x + k y φ 5 5 Recall that cosa ± b = cos a cos b sin a sin b 54 7

8 H nnn,a = t cos φ cos aky + cos ak x k y + cos t sin φ sin aky + sin ak x k y + sin H nnn,b = t cos φ +t sin φ So the mass-inducing terms are cos H nnn 0, 0, a = t sin φ H nnn 0, 0, b = t sin φ sin aky + cos ak x sin aky + sin sin ak x aky + sin aky + sin ak x k y + cos k y + sin ak x k y + sin k y + sin ak x + k y ak x + k y ak x + k y ak x + k y ak x + k y k y ak x So 4π H a H b = 4t sin φ sin + sin aδ x aδ y π + sin aδ x + aδ y + π 6 4π = 4t sin sinφ 6 = t sin φ 6 Note that the sign of this term in the Hamiltonian is DEPENDENT on the valley since each valley has a dierent M t sin φ M + t sin φ ν = sign M t sin ν = sign M + t sin φ See that when M goes from + to < t sin φ, the gap at K stays open always and at K it closes and reopens, with the parameter changing sign. This means the total Chern # much change by an integer, and we know that at M t sin φ it is trivial due to the system appearing like an atomic system. 4 Supplementary Calculations 4. Edge Modes at Mass Inversion Interface Hamiltonian of gapped Dirac system with mass that depends on the y coordinate and look for a separable solution Hy = i x σ x i y σ y + myσ z 68 ψx, y = φ x xφ y y 69 8

9 Now, φ x = e ik X x, whereas φ y = e y 0 my dy 70 Plugging in φ y φ x, one gets a function of φ x and m to nd the exact spinor for φ x. Then you can solve for the plane-wave part which gives a chiral mode. 4. Phase From Magnetic Flux Phase acquired from a path with a vector potential: φ = e ˆ h A d r 7 Which is Gauge dependent if not on a closed loop: A = A + χ φ φ 7 But if you have a closed loop, then you have a xed quantity: ϕ = e A d r = e h h B d s 7 And the complex phase around the loop is related to this phase. You can put this phase on whatever hopping term you like, due to the exibility of gauge transformations. 4. Adiabatic Continuity General Hamiltonian and solution to id: Ĥtψ n t = E n tψ n t 74 Ψt = n θ n t = c n tψ n te iθnt 75 ˆ t 0 E n t dt 76 Subtitute Ψ into Ĥt and you get i ċ n ψ n + c n ψ n + ic n ψ n θn e iθn = c n Ĥψ n e iθn 77 n n but θ n = E so ċ n ψ n e iθn = c n ψ n e iθn 78 n n now inner product with an arbitrary eigenfunction ψ m and you get ċ m t = c n ψ m ψ n e iθn θm 79 n then calculating the dierentiated inner product you get the exact form ċ m t = c m ψ m ψ ψ m m Ĥ ψ n c n e iθn θm 80 E n E m n m but if tĥ, then the second term drops out. This means that c m t = c m 0e t 0 ψ m ψ m dt = c m 0e iγmt 8 so that there is a dynamical phase factor known as the geometrical phase, but the amplitudes remain constant. Thus, a particle remains in the n th eigenstate at all times if the Hamiltonian varies slowly. 9

10 4.4 Van Vleck Paramagnetism Assume an insulator with no net magnetism, i.e. µ z B = 0 8 so that there is no rst-order correction to the system. But then let's say that c S z v 0 8 This means that with perturbation theory in a eld µb E G, one gets a wave-function ψ v = ψ v + B E G c S z v ψ c 84 And now we have a moment 0 S z 0 B c S z v 85 E G So that in the presence of an applied eld the magnetization is And for a band insulator with more than one quantum number k then M N = B c S z v 86 E G χ = c,v,k c S z v E c,k E v,k 87 In typical systems, these terms are very small. What happens in a TI? Well, there is a point in the TI band structure where the gap comes close to closing and at the same time this causes a mazimization of the second order eects. 4.5 Mean Field Theory 4.5. Normal FM H = J i,j s i s j h i s i 88 so dene the mean eld m i = s i and rewrite H = J i,j m i + δs i m j + δs j h i s i 89 H H MF 90 H MF = J m i m j + m i δs j + m j δs i h s i 9 i,j i = J i,j m + ms i m h i s i 9 and i,j = i j nni and thus j nni = x which is the number of nearest-neighbor spins. Therefore H MF = Jm Nx h + mjx i s i 9 This gives us a partition function Z = e Jm Nx β [ h + mjx cosh k B T ] N 94 0

11 which gives a Free Energy [ ] h + mjx F = k B T ln Z Nk B T ln cosh k B T M = F h + mjx H tanh k B T Lagrangian for wave-packet Marder, section 6.4, page 45. L = w i t w w H ev w, assuming no magnetic elds so H = p and w kkc = e ik kc A berry and Hψk = E k ψ k. The Lagrangian after math simplies to m +Ur and w = N k w kk c e i k r c ψ k L = k c r c + kc A berry E kc ev r c 97 L = d L r c dt r and L c = d L k c dt 98 kc k c ee = d dt r c + E k = d k c dt kc and eventually with a bunch of vector identities, etc. kc A berry k = e E Anomalous Velocity r = E k k k F 0 Semiclassical calculations or Kubo formula, or other means, give the following two equations of motion for a wave-packet when the external magnetic eld is zero: k = e E 0 r = E k k k F 04 Ek while if the system is in a band gap then BZ k 0, e.g. when integrating over completely lled bands. As such, I pre-emptively remove that term from the calculation reader can verify for him/herself that this is ok. So, substituting, we get r = vk = e E F 05 For a D system, F = F z. To calculate the conductivity one needs j = ˆσ E 06 so j y = σ xy E x 07

12 then, noting that the distribution function g is roughly the Fermi function g f in this context. j y = e d k vkg 08 4π = e d ke h π F f 09 = e h π E x d kf z 0 Normally, σ xy = dj y de x = e d kf z h π = e h ν ν = d kf z Z 4 π g f τe f µ v k E 5 and when you take the derivative dj β de α the rst term drops out since there is no electric eld there. However, when inside a gap and with nite Berry curvature, vk f µ = 0 and the only remaining term is the one with the Berry curvature multiplying the general Fermi function. 4.8 QHE and dissipative channels σ xy = σ QH + σ dis = σ QH 6 σ xx = σ QH + σ dis σ dis 7 so so σ = ρ = ρ xx = ρ xy = σd σ Q σ Q σ d + σ Q σ d σd σ Q σ Q σ d σ d σd + h ξ σ Q e ξ + σ Q σd + h σ Q e ξ So as the dissipative channel's conductance goes away, the dissipating resistance goes to zero, and the Hall resistance approaches the quantized value.