Supporting Information: Topological Magnon Modes. in Patterned Ferrimagnetic Insulator Thin Films

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1 Supporting Information: Topological Magnon Modes in Patterned Ferrimagnetic Insulator Thin Films Yun-Mei Li,, Jiang Xiao,, and Kai Chang,,, SKLSM, Institute of Semiconductors, Chinese Academy of Sciences, P.O. Box 912, Beijing , China, College of Materials Science and Opto-Electronic Technology, University of Chinese Academy of Sciences, Beijing , China, Department of Physics and State Key Laboratory of Surface Physics, Fudan University, Shanghai , China, Collaborative Innovation Center of Advanced Microstructures, Nanjing , China, and Synergetic Innovation Center of Quantum Information and Quantum Physics, University of Science and Technology of China, Hefei, Anhui , China S1. Band structures of the magnonic crystal The magnetostatic filed in ferrimagnetic thin film is determined by the magnetization: H ms (r)=0, (H ms (r)+m(r))=0 (1) To whom correspondence should be addressed SKLSM, Institute of Semiconductors, Chinese Academy of Sciences, P.O. Box 912, Beijing , China College of Materials Science and Opto-Electronic Technology, University of Chinese Academy of Sciences, Beijing , China Department of Physics and State Key Laboratory of Surface Physics, Fudan University, Shanghai , China Collaborative Innovation Center of Advanced Microstructures, Nanjing , China Synergetic Innovation Center of Quantum Information and Quantum Physics, University of Science and Technology of China, Hefei, Anhui , China S1

2 Figure S1: Title angle of the equilibrium magnetization from the out-of-plane direction. For out-of-plane magnetized thin film with triangle holes, the magnetostatic field is inhomogeneous and has in-plane component, which indicates the equilibrium magnetization is not along out-ofplane direction. The equilibrium magnetization can be obtained by solving LLG equation and Eq. (1) self-consistently. In our system, the equilibrium magnetization has a very small angle with out-of-plane direction. The results are shown in Fig. S1. The maximum tilt angle is as small as rad. The magnetization is divided into two parts: the static one M s (r) M s e z and the dynamical one m(r). The dynamical motion of m(r) is given by dm = γ(m s δh e f f + m H e f f )+βm s dm (2) where H e f f = H ext e z + H ms, δh e f f = (λex 2 )m+h(r,t). h(r,t) is the demagnetization field, given by h(r,t)=0, (h(r,t)+m(r,t))= 0, (3) The tilt angle is so small that m(r,t) is in in-plane direction, we can define m(r,t)=m x (r,t) im y (r,t), and i(1+iβm s ) m(r,t) t = γ(h ext + H z ms)m γm s λ 2 ex 2 m γm s h(r,t), (4) S2

3 air layer Triangle hole PML YIG film Figure S2: Mesh grid of the unit cell. We employ Bloch theorem, m(r,t)=m k (r)e i(k r ωt), h(r,t)= h k (r)e ik r iωt, the secular equation is given by ωm k (r)=γ[h 0 M s λ 2 ex(2ik + 2 )]m k (r) γm s h k (r), (5) where H 0 = H ext +H z ms+m s λ 2 exk 2. For the demagnetization field, we can set h= ψ. The Bloch theorem indicates ψ(r)=e ik r ψ k (r), So we have ( 2 ψ k + 2ik ψ k k 2 ψ k )= m k + ik m k, (6) and h k = (ikψ k + ψ k ). The band structures can be obtained by solving Eq. (1), Eq. (5) and Eq. (6) self-consistently using finite element method. The unit cell and mesh grid are shown in Fig. S2. The total thickness of the air layer and perfectly matched layer (PML) is 80 nm, 20 times of the film thickness. The wih of the out-of-plane mesh is 2 nm for the air layer and PML. The largest size of in-plane triangular mesh 2 nm. For the thin film, the wih of the out-of-plane mesh is 0.8 nm. S3

4 S2. Effective Hamiltonian As mentioned in the main text, the states in the vicinity of K ( ) point can be expanded as: m q (r)= c q A m A(r)+c q B m B(r), where q=k K ( ). The the secular equation becomes (ω q 0 δ i j+ η 0 q p i j sgn(α) K j=a,b 2 σ z i j )c j γm s (1 c q A cq B ) m i h k =ω q c i, (7) where ω q 0 =(ω K c +ω Kv )/2+η 0 (q K+ 1 2 q2 ), m i h k = m i (r)h k(r)d 3 r, p i j = m i (r)( i )m j(r)d 3 r (i, j= A,B) and σ i j z is the Pauli matrix. The integral region is the unit cell. This secular equation should be invariant under C 3 group operation ˆR, which have three independent elements: the identity and the rotations by the angle ϕ = ±2π/3. With a rotation of angle ϕ, cosϕ sinϕ 0 i) h k (r) h k (R 1 r). Here R = sinϕ cosϕ 0 rotates a vector along z axis by an angle ϕ. The absolute value of vector h k (r) should keep invariant under rotation R, and the direction of the vector experience a rotation, which means h k (R 1 r) = R 1 h k (r). So h k (R 1 r) = h x k (R 1 r) ih y k (R 1 r)=e iϕ h k (r). The rotation operator ˆR=exp[ il z ϕ/ h], where L z is the z component of angular momentum operator. For the momentum operator, we have ˆp ˆRˆp ˆR 1 = exp[ il z ϕ/ h]ˆpexp[il z ϕ/ h] = ˆp+ 1 1! [ il zϕ/ h, ˆp]+ 1 2! [ il zϕ/ h,[ il z ϕ/ h, ˆp]] + 1 3! [ il zϕ/ h,[ il z ϕ/ h,[ il z ϕ/ h, ˆp]]]+ Using the commutation commutation relations[l z, ˆp α ]=i h( ˆp y δ αx ˆp x δ αy ), we have ˆp x = ˆp x+ ϕ ˆp y 1 2! ϕ2 ˆp x 1 3! ϕ3 ˆp y + =cosϕ ˆp x + sinϕ ˆp y S4

5 and ˆp y = ˆp y ϕ ˆp x 1 2! ϕ2 ˆp y + 1 3! ϕ3 ˆp x + =cosϕ ˆp y sinϕ ˆp x ˆp z = ˆp z. So ˆp ˆRˆp ˆR 1 = R 1 ˆp. ii) m i (r) m i (R 1 r)= j=a,b D R,i j m j (r). D R is the corresponding double group irreducible representation of the C 3 group. If the rotation center is at A point, D R = diag(1,e iϕ ), m A (r) m A (r), m A h k =0. When the rotation center is at B point, D R = diag(e iϕ,1), m B (r) m B (r), m B h k =0. So we know that m i h k =0and p i j = R 1 D R,mi D R,n jp mn. (8) m,n=a,b Solving the equation gives us the relations: p AA = p BB = 0, p AB = p BA and py AB = ipx AB. Choosing proper coordinate system to make p x AB real and taking the TR symmetry into consideration, the effective low-energy Hamiltonian near the corners of FBZ is given by, H eff = ω q 0 + v K(q x σ x + τ z q y σ y ) sgn(α) K 2 σ z, (9) where v K = η 0 p x AB, τ z =± denotes the valley index and σ =(σ x,σ y,σ z ) are the Pauli matrix. S3. Details of Micromagnetic Simulations The dynamics of the magnetization M(r,t) = (M x,m y,m z ) is governed by the Landau-Lifshitz equation of the motion with Gilbert damping, dm(r,t) = γm(r,t) H eff + βm(r,t) dm(r,t), (10) where γ is the electron gyromagnetic ratio and β is the damping coefficient. The effective magnetic field H eff consists of four terms: the external bias field H ext, the magnetic field h 0 (r,t) to excite spin waves, the exchange field (λ 2 ex )M(r,t) and total magnetostatic (including the dynamical S5

6 one) field H ms (r,t), where λ ex = 2A µ 0 M 2 s = 13 nm for yttrium-iron garnets (YIG), A is the exchange stiffness and M s = A/m the saturation magnetization. The boundary condition is adopted as n M=0. The total magnetostatic field H ms (r,t) relates with the magnetization M(r,t) via the Maxwell equations, H ms (r,t)= 0, (H ms (r,t)+m(r,t))= 0. (11) In the main text, we simulate the spin wave dynamics in a 4-nm-thick film by solving Eq. (10) and (11) self-consistently. The external bias field µ 0 H ext = 344 mt. The continous excitation field h 0 (r,t)=h 0 sin(2π f 0 t)e, where µ 0 h 0 = 10 mt and e is a unit vector perpendicular to the antenna. The damping coefficient is adopted as βm s = S4. Phase calculation in Mach-Zehnder interferometer The topological modes emerge in the band gap at K ( ) points. In subsequent text, we use the states located near K point as example to calculate the phase difference between two paths in Mach-Zehnder interferometer. The phase of the topological mode along the interface is given by ϕ =(K+q) r, where r is the space vector along the interface and vector q is always along the interface. It should be noted there are three equivalent K points, marked as K 1, K 2 and K 3 in Fig S2 (a). For the right-going magnons, we should adopt K 1 to calculate the phase, while adopting K 2, K 3 for up-going and down-going magnons, respectively. At the intersecting region of different orientated interfaces, the interface configurations and orientations gives a bent channel for the propagation of the magnons. As shown in Fig S3(b), the path direction from path 0 to path 1 is not a 90 diversion. Because of the bent channel at the crossings, there exists an π-phase difference between path 3 and path 4. The zoom in figure around the interference point is shown in Fig. S3 (d). The interference point is at 3 for magnons converging into path 3, while point 4 for the magnons into path 4. Let the phase difference of the two split beams between the point 1 and 2 be ϕ 0. The S6

7 Figure S3: (color online) (a). Brillouin zone of magnonic crystal. (b). The path direction in the beam splitter. (c) Schematic of Mach-Zehnder interferometer. (d) Zoom in of the interference point. phase difference for the magnons on path 3 is given by ϕ 3 = ϕ 0 + K 1 r 32 K 2 r 51 K 1 r 35 = ϕ 0 + 2π 3 π 2π 3 = ϕ 0 π. The phase difference for path 4 is ϕ 4 = ϕ 0 + K 3 r 52 + K 3 r 45 K 2 r 41 = ϕ 0 + π+ π 3 4π 3 = ϕ 0. An additional π phase for path 3. S7