Lecture notes on topological insulators

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1 Lecture notes on topological insulators Ming-Che Chang Department of Physics, National Taiwan Normal University, Taipei, Taiwan Dated: October 8, 08 Contents I. Review of Berry phase A. Non-degenerate energy level B. Degenerate energy levels 3 C. Magnetic resonance of nuclear quadrupole 4 D. Computation of Berry phase 6 References 7 I. REVIEW OF BERRY PHASE We first review the Berry phase also known as the geometric phase for a non-degenerate energy level Abelian case. The Berry phase for degenerate energy levels non-abelian case is discussed in the second part. Note: For the rest of the lecture notes, a matrix is written in the San-serif font, M = M αβ. A. Non-degenerate energy level Consider a system that varies slowly with parameters λt. One example is a spin in a slowly varying magnetic field Bt. At each instant t, the time-independent Schrödinger equation is H λ n, λ = ε nλ n, λ.. The eigenstates n, λ will be called the snapshot states. According to the quantum adiabatic theorem, if that the energy level ε nλ is gapped from other levels with a minimal separation 0 during the course of evolution, and that the characteristic frequency of the changing parameter Ω 0 0 /, then an initial state n, λ0 would stay at the same level n. After time t, it evolves to the snapshot state at λt, multiplied by a dynamical phase factor, n, λ 0 Ψ nλ t = e i t 0 dt ε nλt n, λ0.. In principle, one can freely assign a λ-dependent phase γ n λt to the snapshot state n, λt, since the new state still satisfies Eq... Thus, in general, Ψ nλ t = e iγnt e i t 0 dt ε nλt n, λt..3 This extra phase had been deemed as removable since the early days of quantum mechanics, until Berry showed the contrary in 984 Berry, 984. The phase γ n is constrained by the time-dependent Schrödinger equation, H Ψ nλ t = i t Ψ nλt..4 Substitute Eq..3 into Eq..4, one has dγ n t = i n, λ n, λ.5 dt t t γ n t = i dt n, λ n, λ..6 t For a cyclic change of the parameters, one has, 0 λ0 λt = λ0,.7 γ n T = i dλ n, λ n, λ,.8 C λ in which C is the loop traversed in the parameter space of λ. γ n T, or γ n C, is the Berry phase acquired by Ψ n after one cycle in energy level n. It depends only on the geometry of the path C, but not on the rate of λ, as long as it is slow enough so that inter-level transition can be neglected. We will call the integrand in Eq..8 the Berry connection, A n λ i n, λ n, λ..9 λ If one re-assigns a λ-dependent phase to the snapshot state, n, λ n, λ = e iχλ n, λ,.0 where χλ is a single-valued function, then A n λ A nλ = A n λ χ λ.. This is similar to the gauge transformation of a vector potential in electromagnetism. Before Berry s discovery, people thought that by choosing an appropriate χλ, A nλ can be set as zero and removed. This is indeed so if the loop integral, γ n C = dλ A n λ,. C

2 x z φ θ FIG. A magnetic field sweeps slowly around the loop C. is zero like the loop integral of an electrostatic field. However, if it is not zero like the loop integral of a vector potential, then the Berry phase cannot be removed by the gauge transformation above since χλ is singlevalued. If the parameter space is three-dimensional, then one can apply the Stokes theorem to write the Berry phase as a surface integral over the area S enclosed by the loop C, γ n C = d a λ A n.3 S = d a F n,.4 S in which F n λ A n is called the Berry curvature. In higher dimension, Stokes theorem remains valid, but needs be written in the language of differential form see App.??. Example: Consider a spin-/ electron in a magnetic field B = Bˆn see Fig., where ˆn = sin θ cos φ, sin θ sin φ, cos θ. The Hamiltonian is c B H = µ B = µ B σ B, µ B = y e mc..5 The eigen-energies are ε ± = ±µ B B, and the eigenstates are cos θ ˆn, + = e e iφ sin θ, ˆn, = iφ sin θ cos θ..6 Note that e iα±φ ˆn, ± are also eigenstates α ± are not angle-dependent, but they are not single-valued if α ± is not an integer. We emphasize that, when calculating the Berry phase, the snapshot states in Eq..8 need to be single-valued. Even if one adopts only single-valued states, the choice of basis is still not unique. For example, ˆn, ± = e iφ ˆn, ± are also single-valued. You can check that ˆn, ± have φ-ambiguity at θ = π but not at θ = 0, while ˆn, ± have φ-ambiguity at θ = 0 but not at θ = π. The magnetic field B plays the role of the slowly varying parameters λ. The Berry connection can be calculated from Eq..9 with the gradient operator, B = B êr + B θ êθ + B sin θ It is left as an exercise to show that Similarly, φêφ..7 A + B = i B, + B, + B.8 = cos θ ê φ. B sin θ.9 A B = B cos θ ê φ..0 sin θ Both A ± B are singular along θ = π. These vector potentials are the same as that of a magnetic monopole, and the line of singularity corresponds to the Dirac string there. The Berry curvature is, F ± B = B A ± B = ˆB B.. This is the same as the field of a magnetic monopole at the origin with a magnetic charge /. The Berry phase can be calculated using either the line integral of A ±, or the surface integral of F ±. For a loop in Fig., the Berry phase is found to be, γ ± C = ΩC,. where ΩC = π cos θ is the solid angle, as seen from the origin, extended by the loop C. If the loop is lying on the x-y-plane, then the Berry phase can only be π. That is, the state changes sign after a cyclic evolution. The phase in Eq.. has been confirmed by passing neutrons through a tube with a helical magnetic field Bitter and Dubbers, 987. Note that, given ˆn, ± = e iφ ˆn, ±, the Berry connections calculated from the two basis differ by a gradient, Therefore, A ±B = A ± B ± φ B.3 = A ± B ± B sin θ êφ..4 A ±B = ± B + cos θ ê φ..5 sin θ Both A ±B are singular along θ = 0. They generate the same Berry curvatures as those in Eq... The integral of F ± over a closed surface enclosing the origin is quantized, π S B d a F ± B =..6

3 3 TABLE I Analogy between electromagnetism and anholonomy Electromagnetism vector potential Ar magnetic field Br magnetic monopole magnetic charge magnetic flux ΦC Quantum anholonomy Berry connection Aλ Berry curvature Fλ degenerate point Berry index Berry phase γc This integer remains fixed no matter how the surface S B is deformed, as long as it is not torn up. It is sometimes called the Berry index or the topological charge of the degenerate point. Finally, the analogy between the gauge structure of the Berry phase and that of the classical electromagnetism is summarized in Table I. B. Degenerate energy levels One can extend the analysis above to degenerate energy levels Wilczek and Zee, 984. The wave function now has multiple components for a given energy level. As a result, the Berry connection and the Berry curvature become matrix-valued vectors or vector-valued matrices. For simplicity, we consider an energy level ε nλ with only two globally degenerate eigenstates, n,, λ and n,, λ. Again λ are slowly varying parameters. After a time t, the states become compare with Eq..3 Ψ n, t = e i t 0 dt ε nλt.7 n,, λt Γ t + n,, λt Γ t, Ψ n, t = e i t 0 dt ε nλt.8 n,, λt Γ t + n,, λt Γ t. Or, α, β =, Ψ nβ t = e i t 0 dt ε nλt α nαλt Γ αβ t..9 Substitute the states into the Schrödinger equation, one will get, where dγ αβ dt H Ψ nβ t = i t Ψ nβt,.30 = γ = i γ nαλ t nγλ Γ γβ.3 λt A n αγ λγ γβ,.3 A n αβ λ i nαλ nβλ..33 λ The Berry connection now becomes a matrix-valued vector, A n. To simplify the index, we focus only on one degenerate eigenenergy and suppress the index n from now on. To solve for the U Berry rotation matrix Γt, first consider an infinitesimal dt, Γt + dt = Γt + idt λt AtΓt.34 e idt λt At Γt.35 A full path can be divided into small steps, so that Γt = e idλ Aλ e idλ Aλ 0 Γ0.36 P e i λt λ 0 dλ Aλ, Γ0 =,.37 in which P is a path-ordering operator. Because of the path-ordering, Γt usually can only be computed numerically see Sec. I.D. Sometimes one can see non-abelian Berry connections being calculated for energy levels that are not globally degenerate or not degenerate at all. In such cases, the dynamical phase factor in Eq..9 needs be changed to e i t 0 dt ε nαλt, and the right hand side of Eq..3 is multiplied by e i t 0 dt ε nγλ ε nαλ. The Berry connection and Berry curvature can still be defined in the same manner, but the Berry rotation among these energy levels would be mixed with dynamical phase factors. We now consider a gauge transformation, Its dual is Therefore, Or, αλ = γ αλ = γ γλ U γα λ..38 U αγλ γλ..39 A αβλ = i nαλ λ nβλ.40 = U αγa γδ U δβ + iu αγ λ U γβ..4 The phase factor A k = U A k U + iu U..4 λ k e idλ A + idλ A.43 = + idλ U AU dλ U λ U.44 A U + idλ U dλ λ U.45 U λe idλ A Uλ dλ..46

4 dλ dλ Note that F kl is antisymmetric in k, l, and F kl dλ k dλ l = ɛ cabf ab ɛ ckl dλ k dλ l,.58 4 dλ dλ in which ɛ cabf ab = F c is the vector equivalent of the antisymmetric matrix, and ɛ ckl dλ k dλ l = d a c. Therefore, FIG. A small loop in the parameter space of λ. Therefore, after a closed path, Γ [C] = U λ 0 Γ[C]Uλ That is, the non-abelian Berry rotation is gauge covariant. Berry curvature F is defined as the Berry rotation Γ around an infinitesimal loop enclosing an area d a, Γ = e i F d a, or i F Γ d ˆn,.48 a where ˆn is the normal vector of the surface d a. The Berry rotation matrix is composed of 4 steps see Fig., Γ λ.49 = Γλ, λ + dλ Γλ + dλ, λ + dλ + dλ Γλ + dλ + dλ, λ + dλ Γλ + dλ, λ. Using the approximation, one has e ɛa e ɛb = e ɛa+b+ ɛ [A,B] + Oɛ 3,.50 Γλ + dλ + dλ, λ + dλ Γλ + dλ, λ = e ia kλ+dλ dλ k e ia lλdλ l.5 e ia kdλ k +A l dλ l + l A k dλ k dλ l [A k,a l ]dλ k dλ l..5 For the next two steps, it is convenient to adopt instead see p. 39 of Cheng and Li, 988, Γ λ + dλ, λγ λ + dλ + dλ, λ + dλ = e ia kλdλ k e ia lλ+dλ dλ l.53 e ia kdλ k +A l dλ l + k A l dλ k dλ l [A k,a l ]dλ k dλ l..54 Therefore, Γ = e i ka l l A k i[a k,a l ]dλ k dλ l.55 = e if kldλ k dλ l,.56 with the Berry curvature, F kl = k A l l A k i[a k, A l ]..57 F kl dλ k dλ l = F d a,.59 When written in the vectorial form, the Berry curvature is cf. Eq..57, F = λ A i A A..60 Finally, under a gauge transformation, F kl F kl = k A l l A k i[a k, A l].6 = U F kl U..6 That is, the non-abelian Berry curvature is gauge covariant similarly for F. C. Magnetic resonance of nuclear quadrupole The interaction between a nuclear quadrupole with a magnetic field B is Zee, 988, H = J B,.63 where J is the spin operator for spin-j. This coupling is invariant under time reversal. The eigen-energies and eigenstates are H ˆB, ±m = mb ˆB, ±m, m = j,, j..64 Note that ˆB, ±m are degenerate with the same energy mb. Since σα = α = x, y, z for the Pauli matrices, the Hamiltonian is non-trivial only if the spin quantum number j. We are interested in half-integer spins, so the next non-trivial case is j = 3/. In this case, the angular momentum matrices are 0 3/ 0 0 3/ 0 0 J x = 0 0 3/, / 0 0 i 3/ 0 0 i 3/ 0 i 0 J y = 0 i 0 i 3/ 0 0 i,.66 3/ 0 3/ / 0 0 J z = 0 0 / /

5 5 The spectrum of H is composed of two-fold degenerate energy levels. When B is along the direction in Fig., the eigenstates are m = 3/,, 3/ The Berry connection is ˆB, m = U z φu y θ m.68 = e ijzφ e ijyθ m..69 A mm = i ˆB, m B ˆB, m.70 = i B m U yθu zφ θ êθ + U z φu y θ m..7 sin θ φêφ With the help of the following equations, e ijyθ e ijzφ i θ e ijzφ e ijyθ = J y,.7 e ijyθ e ijzφ i φ e ijzφ e ijyθ = sin θj x + cos θj z,.73 one will get way to remedy this is to use see Chruscinski and Jamiolkowski, 004, ˆB, m = U z φu y θu z φ m.77 = e imφ U z φu y θ m..78 where A = A θ ê θ + A φ ê φ,.74 Also, at θ = 0, ˆB, m m without the φ-ambiguity. Using this new basis, one would get A θ = B J y,.75 A φ = B sin θ sin θj x + cos θj z..76 Note that A φ blows up at θ = 0, π. In the calculation above, the eigenstates ˆB, m = U z φu y θ m are not single-valued if φ φ + π. One 0 i 3/e iφ 0 0 A θ = i 3/e iφ 0 ie iφ 0 B 0 ie iφ 0 i 3/e iφ 0 0 i, 3/e iφ 0.79 and A φ = B sin θ 3 cos θ 3 sin θe iφ 0 0 sin θeiφ cos θ sin θe iφ 0 0 sin θe iφ cos θ sin θeiφ 3 sin θe iφ cos θ..80 Note that A φ now is finite at θ = 0 but θ = π remains a singularity. levels are P 3/ is a projection operator A 3/ = P 3/ A θ ê θ + A φ ê φ P 3/.8 3 = cos θ 0 B sin θ 0 3 ê φ,.8 cos θ The Berry connection for the degenerate m = 3/ which is diagonal Abelian connection. Similarly, the

6 6 Berry connection for the degenerate m = / levels are A / = P / A θ ê θ + A φ ê φ P /.83 = 0 ie iφ ê B ie iφ θ cos θ sin θe iφ B sin θ sin θe iφ ê φ, cos θ which is not diagonal non-abelian connection. The only non-zero component of the Berry curvature F kl k, l = r, θ, φ is F, F 3/ = A 3/ i[a 3/ θ, A 3/ φ ].85 r = 3 B σ It is left as an exercise for the readers to show this, and that F / = 3 B σ These Berry curvatures are the same for the two types of basis in Eqs..68,.77. Note that the Berry rotation is the path-ordered line integral of A, but not the area integral of F, since the Stokes theorem fails in the non- Abelian case. Because of the extra ±3/, ±/ in the diagonal elements of A φ in Eq..80, for a circular path with fixed θ, the Berry phases differ by a sign to those calculated from Eq..76. Eq..80 is to be trusted since it is based on single-valued snapshot states. If you calculate the Berry curvature using Eqs..79,.80 directly without projection, then the result would be zero. This can be explained as follows: The Berry curvature in Eq..57 can be written as, F αβ kl = i k α l β i k α γ γ l β γ k l..88 Without projecting the Berry connection to a subspace, γ runs through bases of the whole Hilbert space. Thus, γ γ γ = and F kl vanishes. Two remarks are in order: First, the Berry phase would result in splitting of the spectra of the nuclear quadrupole resonance. Such a splitting has been confirmed in experiment Tycko, 987. A recent attempt of detecting the non-abelian Berry rotation can be found in Li et al., 06. Second, the quadrupole coupling Hamiltonian is similar to the effective Hamiltonian of heavy holes m = ±3/ and light holes m = ±/ near k = 0 in a semiconductor. Under the four-band approximation, and neglecting anisotropy, the four-band Luttinger Hamiltonian is see, e.g., Winkler, 003, [ H = γ + 5 m γ k γ k J ],.89 where γ, are material-dependent parameters. The first term is multiplied by an identity matrix and has no effect on the Berry connection and curvature. The second term is essentially the same as the one in Eq..63, if B is replaced by k. Therefore, this system has exactly the same Berry connection and Berry curvatures listed in Eqs..79,.80,.86, and.87 Murakami et al., 003. D. Computation of Berry phase Numerical computation of Berry rotation using Eq..37 can be implemented as follows: Divide a closed path C into N steps with points λ, λ,, λ N, and λ N+ = λ. For each step, one has [ e idλ Aλ l ] αβ δ αβ + idλ A αβ λ l.90 = αλ l βλ l dλ αλ l λ βλ l = αλ l βλ l + dλ λ αλ l βλ l αλ l+ βλ l..9 Thus with summation convention, Γ αβ C = αλ β N λ N β 3 λ 3 β λ β λ βλ..9 When written in matrices, we have the Berry rotation, ΓC = M N M M,.93 where M l αβ = αλ l+ βλ l..94 A diagonalized unitary matrix is of the form, It follows that, e iγ 0 ΓC = e iγ N det ΓC = e i D α= λα.96 = det Π N l=m l..97 The total Berry phase would be, D λ α = Im ln det Π N l=m l..98 α= Exercise. Replace the single-valued snapshot states n, λ with n, λ = e iγnt n, λ, which are not necessarily single-valued. Show that n, λ t n, λ = 0..99

7 7 A state that evolves under such a restriction is said to be parallel transported. A state that moves under such parallel transport condition acquires a Berry phase γ n C after a cyclic change of λ.. For a spin-/ electron in a magnetic field B = Bˆn, ˆn = sin θ cos φ, sin θ sin φ, cos θ, the eigenstates are, ˆn, + = cos θ e iφ sin θ, ˆn, = a Show that the Berry connections are, A ± B = B cos θ ê φ. sin θ b Show that the Berry curvatures are, F ± B = ˆB B. e iφ sin θ cos θ 3. An electron moving in a magnetic field Br has the Hamiltonian, H = p + Br σ..00 m Suppose the eigenstates of the Zeeman term are n, Br, Br σ n, Br = ε n n, Br..0 Expand the wave function of the moving electron as, Ψ = n=± ψ n r n, Br..0 From the Schrödinger equation, H Ψ = i Ψ /t, show that, [ ] m p A n V n + ε n ψ n = i ψ n t,.03 in which A n r = i n, Br n, Br,.04 r V n r = i n, Br n, Br..05 t The off-diagonal coupling between + and has been ignored this is all right for a smooth magnetic field.. That is, the effective Hamiltonian for the particle motion acquires a Berry potential A n and a similar scalar potential V n. Such potentials would result in forces proportional to Berry curvatures that act on the electron. 4. For a spin-3/ quadrupole coupling with a magnetic field, the projected Berry connections for the m = 3/ sector and the m = / sector are, A 3/ 3 = cos θ 0 B sin θ 0 3 ê φ, cos θ A / = 0 ie iφ ê B ie iφ θ 0 + cos θ sin θe iφ B sin θ sin θe iφ ê φ. cos θ Show that the Berry curvatures are, References = 3 B σ 3, = + 3 B σ 3. F 3/ F / Berry, M. V., 984, Proceedings of the Royal Society of London A: Mathematical, Physical and Engineering Sciences 3980, 45. Bitter, T., and D. Dubbers, 987, Phys. Rev. Lett. 59, 5. Cheng, T.-P., and L.-F. Li, 988, Gauge Theory of elementary particle physics Oxford University Press. Chruscinski, D., and A. Jamiolkowski, 004, Geometric phases in classical and quantum mechanics Birkhauser Boston. Li, T., L. Duca, M. Reitter, F. Grusdt, E. Demler, M. Endres, M. Schleier-Smith, I. Bloch, and U. Schneider, 06, Science 35689, 094. Murakami, S., N. Nagaosa, and S.-C. Zhang, 003, Science , 348. Tycko, R., 987, Phys. Rev. Lett. 58, 8. Wilczek, F., and A. Zee, 984, Phys. Rev. Lett. 5,. Winkler, R., 003, Spin-orbit Coupling Effects in Two- Dimensional Electron and Hole Systems, volume 9 Springer. Zee, A., 988, Phys. Rev. A 38,.