3.14. The model of Haldane on a honeycomb lattice

Transcription

1 4 Phys60.n..7. Marginal case: 4 t Dirac points at k=(,). Not an insulator. No topological index...8. case IV: 4 t All the four special points has z 0. We just use u I for the whole BZ. No singularity. C=0..9. case IV: 4 t All the four special points has z 0. We just use u I for the whole BZ. No singularity. C=0..0. the top and? The top and has the opposite Chern numer C C. In fact, one can prove that for any tight-inding models, the total Chern numer for all the ands is always 0. So here, we have C C Summary This model has four phases. Two topological phases with C=+ and two trivial insulator phase with C=0. A topological phase and the trivial insulator phase are separated y a phase transition point, which is known as a topological phase transition. Across the topological phase transition, the topological index changes its value Across a topological transition, the insulating gap closes and then reopens (generically true) This is actually a generic statement: A and can only change its Chern numer y crossing with another and. i.e. At the topological transition point, the system must e a metal or a semi-metal (with no insulating gap). Gap closing is a necessary condition for a topological transition, ut it is not a sufficient one. One may close and reopen the gap without changing the topological index. e.g. the =0 point here..4. The model of Haldane on a honeycom lattice.4.. Honeycom lattice (graphene) Two sites per unit cell (two sulattices) a and (red and lack respectively). The green arrow indicates the lattice vectors v and v. If we shift the lattices y m v n v with m and n eing integers, the lattice is invariant. Vr Vr m v n v (.07) Define a to the length of the nearest-neighor (NN) onds, v a, 0 and v a, a If we only consider hoppings etween the nearest-neighor (NN) sites, the Hamiltonian is H t i, j a i j t i, j i a j (.08) where i,j means nearest neighors.

2 Phys60.n Band structures There are three type of NN onds: () along the y axis with =/, () along =/+/=7/6, () along =/+4/=/6. We need to write them out separately. with H t a i r i rie e 0, a t a i r i rie t a i r i rie h.c. (.09) (.0) e a, a (.) e a, a (.) Here the sum i sums over all unit cells (or say all red sites). Go to the momentum space using a D Fourier transformation a k N i a i k r (.) a i N k a k k r (.4) k N i a i k r (.5) i N k a k k r (.6) The first term in the Hamiltonian t i a ri rie t N i k k' t k k' a k k' k' e N i k r a k k i k' rie k' k' r i t k a k' k k' k' e k,k' t k a k k k' e (.7) This is generically true. By going to the k space, one just need to change a ri rie into a k k and change i (sum over all unit cells) into k (sum over all momentum points), and the coefficient is t exp k r a r with r a and r eing the coordinates of site a and. H t i a ri rie t i a ri rie t i a ri rie h.c. e t a k k k k ke k k a k k 0 k k 0 e a k k t k k a k k e k e k e (.8) k texp k e exp k e exp k e (.9) k k texp k e exp k e exp k e (.0)

3 44 Phys60.n The kernel of the Hamiltonian: k 0 k k 0 (.) The eigenvalues of (k) gives the dispersion relation. k k t cos k x a 4 cos k x a Cos k y a (.) The contour plot of the energy dispersion for the lower and ( as a function of k x and k y ). The red dash lines mark the first BZ, which is a hexagon. The dispersion is a periodic function of k space (the hexagon repeats itself in the figure shown aove). For the lower and, its energy minimum is located at k 0. The maximum of is reached at the corners of the BZ. The first BZ is a hexagon. It has six corners. However, due to the periodic structure in k-space, the three corner points with =0, =/ and =4/ are the same point (their momenta differ y precisely one periodicity). Similarly, the three corners with =, =+/ and =+4/ are the same points. Therefore, there are only two different corner points and they are known as the K and K points, where K 4, 0 a (.) and K ' 4, 0 a (.4) For most momentum points, one of the two ands has positive energy 0 and the other one has negative energy 0. However at the corner of the BZ, K and K, the two ands are degenerate 0.

4 Phys60.n 45 K K t cos k x a 4 cos k x a Cos k y a t cos 4 a 4 cos a 4 a Cos a 0 t cos 4 4 cos t 4 0 Two ands have the same energy at K and K : and crossing points. The energy dispersion for oth ands ( as a function of k x and k y ) in the first BZ (within the red dash lines mark in the figure aove). Near K or K points, the dispersion for the two ands are linear k K. To see this, we expand k near k K and k K ' k K q t cos 4 q x a a 4 cos 4 q x a Cos a q y a (.6) t a q x q y Oq t a q Oq k K ' q t a q Oq (.7) Zoom in near the K points. Q: Fermions with linear dispersion k, what is that? A: A Massless Dirac fermion.

5 46 Phys60.n k K q t a 0 q x q y q x q y 0 Oq t a q x x q y y c q (.8) k K ' q t a 0 q x q y q x q y 0 Oq t a q x x q y y c q x x (.9) where c is the speed of light. Each of them is a Weyl fermion. These two Weyl fermions with opposite chirality forms a Dirac fermion. Dirac fermions we learned in quantum mechanics: t 0 c q c q 0 (.0) If one make an unitary transformation (changing the asis, which doesn t change any physics) ' (.) The equation turns into t ' c q x x 0 0 c q ' (.) The Dirac point (and crossing points) in a honeycom lattice is stale as long as the space inversion symmetry r r and the time reversal symmetry t t are preserved. We will come ack to this point latter. No matter how one pertur the systems (e.g. adding longer-range hoppings), the Dirac point is always there as long as the two symmetries mentioned aove are preserved. For graphene, the lower and is filled and the upper and is empty, which is known as half-filling. The half here means that the numer of electrons N e over the numer of sites N s is N e N S. However, it is worthwhile to notice that the ratio etween N e and the numer of unit cells N is, ecause there are two sites in each unit cell. So one of the two ands are totally filled. By gating, one can turn the energy level slightly away from the middle point..4.. Aharonov Bohm effect and complex hopping Q: Can hopping strength t e complex? A: Yes, and the phase can come from the Aharonov Bohm effect in the presence of a magnetic field or spin-orital couplings. The Aharonov Bohm effect:if one moves a particle around a closed contour. The phase difference etween the final and initial states is proportional to the magnetic flux enclosed y the contour e B S e A. l Discrete version on a lattice: Consider three sites a, and c. The hopping strength etween these three sites are t a t c and t ca respectively. If a particle hops from a to and then to c, the hopping strength around this loop is: t a t c t ca t a a t c c t ca ca t a t c t ca acca The phase picked up y the electron is: (.) a c ca e B S (.4) If B is nonzero inside the triangle formed y these three sites, the phase for these hoppings are nonzero. Please notice that: t a and t a has opposite phase, due to the Hermitian condition (One can also prove this using the Fermi s golden rule, which says that the tunneling amplitude from the state f> to i> is the complex conjugate of the tuning amplitude from i> to f>). t a t a (.5)

6 Phys60.n 47 The individual phases for t a, t c and t ca have no physical meaning and they are in fact gauge dependent. Only the total phase on a loop has physical meanings. a e a A l (.6) Underage transformation: A A a a ' e a A l e a A l e a l a a e (.7) Oviously, a is not a physical oservales, since it relies on the gauge choice. However, the total phase around a loop is different. It is a loop integral of A, which is gauge independent and the physics meaning of this Integral is the magnetic flux. a c ca e A l a ' c ' ca ' e A l e A l e l e A l a c ca (.8) The complex hopping strength induced y B fields reaks the time-reversal symmetry ecause B B under time reversal. (in other words, under time-reversal one need to flip the sign of all these phases) Complex next-nearest-neighor (NNN) hoppings (reaking T-symmetry using B fields) Ref: F. D. M. Haldane, Model for a Quantum Hall Effect without Landau Levels: Condensed-Matter Realization of the Parity Anomaly, Phys. Rev. Lett. 6, (988). Now let us add some NNN hoppings and assume their hopping amplitudes are complex. For simplicity, we choose the amplitude and the complex phase to e the same for all NNN onds. If the hopping is along the arrows marked in the figure, the phase of the hopping strength is. If one hops in the opposite direction, the phase is -. This complex phases can e realized (in theory) y applying a staggered B field, which is positive near the center of each hexagon and negative near the edges. The NNN hoppings are from an a-site to an a-site (and from a -site to a -site). For a-to-a hoppings, there are three different types of NNN onds, along =0, / and 4/. Same is true for -to- hoppings. So the Hamiltonian is: H NNN t ' i a ri a riv t ' i a riv a riv t ' i a riv a ri h.c.a and (.9) Here, v and v are marked on the first figure of this section with v a, 0 and v a, a. v a, a

7 48 Phys60.n H NNN t ' i a ri a riv t ' i a riv a riv t ' i a riv a ri h.c.a and t ' k a k a k kv kv kv h.c. a and t ' k a k a k cosk v cosk v cosk v (.40) t ' k k k cosk v cosk v cosk v H H NN H NNN k a k k k k k k a k k (.4) k t k' e k' e k' e (.4) k k t k' e k' e k' e k t ' cosk v cosk v cosk v k t ' cosk v cosk v cosk v (.4) (.44) (.45) If one computes the eigenvalues of (k), one find that the two ands never cross with each other for any k (if t is non-zero and is NOT an integer times ). The energy dispersion for oth ands ( as a function of k x and k y ) in the first BZ (within the red dash lines mark in the figure aove). Using Pauli matrices: 0 k I x k x y k y z k z 0 k k t ' cos cosk v cosk v cosk v z k k t ' sin sink v sink v sink v x Re k tcosk e cosk e cosk e (.46) (.47) (.48) (.49) y Im k tsink e sink e sink e (.50) The energy dispersions: k 0 k x k y k z k (.5) Without the complex NNN hoppings, 0 z 0, so the dispersion k x k y k (.5)

8 Phys60.n 49 At K and K, x y 0, so the two ands degenerate (without the complex NNN hoppings) k 0 (.5) which gives the Dirac points. Now, with these complex NNN hoppings, At K and K points, x y 0, ut z 0. The gap etween the two ands: k k x k y k z k (.54) At K or K, the gap is: k k x k y k z k z k K 6 t ' sin (.55) In fact, at the K point, z at the K point z t ' sin t ' sin (.56) (.57) They have opposite signs (as long as is not n ). Based on what we learned last time, this means that one cannot define the wavefunction in the whole BZ. We need to cut the systems into two regions. The region I contains the K point, the region II contains the K points. And we need to use different eigenvectors for these two regions. Using the same method we learned in the last lecture, one finds that the Chern numer here is ±. If one choose the oundary etween region I and II to e very close to the K or K points, one find that: k z k xk yk z k x k y k x k y k q x q y q x q y (.58) xk yk.4.5. Potential energy (reaking inversion symmetry) Let s keep NNN hoppings to e zero for now and add some potential energy to the Hamiltonian to see whether we can open a gap to get an insulator. H Potential V M i a i a i V M i i i V N M i a i a i M i i i (.59) The V part (average potential etween a and sites) just adds a constant term to the energy, since the total particle numer N is conserved. So we can drop the V term and only consider the difference etween the potential energies at a and sites (M). In k-space H Potential M i a i a i M i i i M k a k a k M k k k k a k k M z a k k (.60) It adds a z component to the Hamiltonian. Same as the NNN complex hopping, this term also opens a gap at the Dirac point. If we add this term to the Hamiltonian without the complex NNN hoppings, the gap is M at K and K, and the system is a trivia insulator at half-filling. Q: What will happen if we have oth H Potential and H NNN? A: We just need to look at the signs of z at K and K points. If they have the same sign, we can use one wavefunction to cover the whole BZ, so C=0 (trivial insulator). If they have opposite signs, the system is a topological insulator with C=± At the K point, z M at the K point t ' sin (.6)

9 50 Phys60.n z M t ' sin (.6) Therefore, as long as M t ' sin, the system is an topological insulator ( z flips sign). If M t ' sin, z is always positive (or negative) and thus the system is topologically trivial. The marginal case M t ' sin is a topological transition. Here, z 0 at either the K point or the K point. Because x y 0 at these two points, the gap must e zero at one of the two points. So there is a and crossing in the system (either at K o K, depending on the sign of M ant t sin ) and the system is not an insulator. The insulating gap is closed at the topological transition point. Remarks: The model of Haldane is the first example of a topological insulator eyond quantum Hall effect. It demonstrates that topological insulator is a generic concept, which may appear in any insulating systems (NOT just quantum Hall). It also demonstrates that as long as the topological index is nonzero, one will oserve all the topological phenomena expected for a quantum Hall state, including the quantized Hall conductivity and the existence of the edge states. The key differences etween the model of Haldane and the quantum Hall effects are () the B field is on average zero in the model of Haldane while the QHE has a uniform B field and () there is a very strong lattice ackground in the model of Haldane while the QHE requires weak lattice potential. Systems similar to the Haldane s model are known as topological Chern insulators or Chern insulators (average B is 0 and have a strong lattice potential). But sometimes, Chern insulator are also used to refer to the quantum Hall effect. The model of Haldane is also the foundation to explore more complicated and exotic topological states. For example, the time-reversal invariant topological insulators was first proposed using a modified Haldane s models, which we will study latter in the semester.