This model contains two bonds per unit cell (one along the x-direction and the other along y). So we can rewrite the Hamiltonian as:

Transcription

1 1 Problem set # A one-band model on a square lattce Fg. 1 Consder a square lattce wth only nearest-neghbor hoppngs (as shown n the fgure above): H t, j a a j (1.1) where,j stands for nearest neghbors (sold bonds n the fgure above). Here we assume that the hoppng strength t s real. In ths model, all lattce stes are equvalent and thus each unt cell contans only one ste. As a result, the band structure for ths model contans only one band space Ths model contans two bonds per unt cell (one along the x-drecton and the other along y). So we can rewrte the Hamltonan as: H t a a e x r t a a e y r t a a e x t a r a r e y (1.2) Here e x a, 0 and ey 0, a are the lattce vectors (the two green vectors shown n the fgure above) wth a the length of the nearest-neghbor bonds. The frst term n the Hamltonan descrbes hoppngs along the +x drecton. The second one shows hoppngs along +y. The last two terms are the Hermtan conjugate of the frst two, and they are hoppngs along -x and -y drectons. The sum s over all unt cells.

2 2 homewor1.nb Rewrte the Hamltonan n -space as H Ε 1 a a usng the Fourer transform: a 1 N a r and a 1 N a r (1.3) Fnd the dsperson relaton Ε 1 and show that Ε 1 x, y s a perodc functon Ε 1 x 2 n a, y 2 m a Ε 1 x, y (1.4) for any ntegers m and n Complex hoppngs Fg. 2 Let s consder the same lattce, but now we allow the hoppng strength to be complex. Here, we assume that the hoppngs along the y-axs have real coeffcent t. For hoppngs along the x-axs, the hoppng strength s t e Φ f the hoppng s toward the +x drecton (along the arrows shown n Fg. 2). Due to the Hermtan condton H H, the hoppng toward the -x drecton has coeffcent t Φ. The Hamltonan now reads: H t Φ a a e x r t a a e y r t Φ a a e x t a r a r e y (1.5) Rewrte the Hamltonan n -space as H Ε 2 a a usng Fourer transformatons. Fnd the dsperson relaton Ε 2. Show that Ε x, y s also a perodc functon Ε 1 x 2 n a, y 2 m a Ε 1 x, y (1.6) for any ntegers m and n Compare the two models Compare the band structure fnd n problem and 1.1.2,.e.. Ε 1 x, y and Ε 2 x, y. Show that these two dspersons are actually the same functon f one shfts the momentum by some constant: Ε 1 x, y Ε 2 x Α, y Β (1.7) Fnd Α and Β Why the two models have the same band structure (up to a shft of )? As mentoned n the lecture, the phase of the hoppng strength for each ndvdual bond doesn t have any physcal meanng. The only thng matters

3 homewor1.nb 3 here s the flux through each plaquette (sum over all phases around a plaquette). Chec the flux through each plaquette for both these two models and show that they are the same. If two models have the same flux pattern, they are actually the same model, although the phase on each bond may be dfferent. In other words, the two model are connected by a gauge transformaton. Let s consder the Hamltonan of problem H t Φ a a e x r t a a e y r Perform the followng gauge transformaton: t Φ a a e x t a r a r e y (1.8) a r a exp Φ x a a r (1.9) a r a exp Φ x a a r Here x s the x component of r. Now, rewrte the Hamltonan usng a terms of a and a r, the Hamltonan turns nto r (1.10) and a r as our annhlaton and creatons operators. Please show that n H t a a e t a x r a e t a y r a t a e a x e y (1.11) whch s exactly the Hamltonan we studed n problem So ndeed the two models are dentcal A two-band model on a square lattce wth Drac ponts Fg. 3 Let s consder the same lattce. Now we choose the phase of hoppngs accordng to the arrows shown n Fg. 3. H t, j Φj a b j h.c. (1.12) Here, we assume that electrons can hop only between nearest neghbors and we also assume all the nearest-neghbor hoppng strength to have the same absolute value (t, whch s real and postve). For the complex phase Φ j, f the hoppng s along the drecton of the arrow, Φ j 4, so that the hoppng strength s t 4. In the opposte drecton (aganst the arrow), the hoppng strength s t 4. For ths model, there are two dfferent types of stes (red and blac). For the red stes, the arrows at ths ste pont away from the ste along x and towards the ste along y. For the blac stes, the arrows pont towards the ste along x and away from the ste along y. To ndcate that we have two dfferent types of ste, we use a and a to represent fermon creaton and annhlaton operators for the red stes and usng b and b for the blac ste. Because we have two types of

4 4 homewor1.nb stes, we now that the model wll gve us two bands magnetc flux and tme-reversal symmetry Consder a square formed by four nearest-neghbor bonds, whch s the smallest plaquette on ths lattce. If we goes around the square (n the counter-cloc drecton), the Aharonov Bohm phase pced up by the electron s: Φ 1 Φ 2 Φ 3 Φ ± For ths lattce, half of the squares have + and the other half have -. As we learned n the lecture, a ±-flux doesn t brea the tme-reversal symmetry, because. Therefore, ths model s tme-reversally nvarant, smlar to graphene (although we do have complex hoppngs here, they don t brea the tme-reversal symmetry) fnd the Band structure Ths model contans four bonds per unt cell (the four bonds around one blac ste). (1.13) H t 4 a b e x r t 4 a b e x r t 4 a b e y r t 4 a b e y r t 4 b a e x r t 4 b a e x r t 4 b a e y r t 4 b a e y r (1.14) Here, e x a, 0 and ey 0, a wth a beng the length of the nearest-neghbor bonds. The frst term n the Hamltonan descrbes hoppngs from b to a along the +x drecton. The second, thrd and fourth ones are along -x, +y and -y drectons respectvely. The last four terms are the Hermtan conjugate of the frst four, whch descrbes hoppngs from a stes to b stes. The sum here sums over all unt cells. Rewrte the Hamltonan n the -space as H a, b 0 a a 0 b a 0 a b 0 b b a b (1.15) Fnd the two-by-two matrx, whch s the ernel of the Hamltonan 0 0 a a 0 a b 0 b a 0 b b (1.16) Fnd the egenvalues of 0 as a functon of x and y, Ε ± x, y Drac ponts Show that the two bands touch each other at 2 a, 2 a and 2 a, 2 a, Ε 2 a, 2 a Ε 2 a, 2 a Ε 2 a, 2 a Ε 2 a, 2 a (1.17) (1.18) Expand the dsperson Ε ± x, y near the pont, as a power seres: Ε Ε 2 a q x, 2 a q y v F q 2 x q 2 y Oq 2 2 a q x, 2 a q y v F q 2 x q 2 y Oq 2 Fnd the Ferm veloctes v F. Here we fnd two band crossng ponts and near the band crossng ponts, the energy s a lnear functon of the momentum q q x 2 q y 2. In other words, we fnd two Drac ponts n ths model, smlar to graphene. (1.19) (1.20) the Brlloun zone

5 homewor1.nb Fg. 4 Contour plot for Ε as a functon of x and y. Here I set a 1. Darer color means smaller Ε. The two green arrows are the vectors Q 1 and Q 2 defned below. The blac square mars the frst Brlloun zone and the two red dots mars the two Drac ponts. Show that Ε s a perodc functon (the same s true for Ε ) Ε m Q 1 n Q2 Ε (1.21) for any ntegers m and n. Here Q 1, and Q a a 2,. As a result, the Brlloun zone s the square wth four corner ponts located at a a a, 0, 0, a, a, 0 and 0, a. The Brlloun zone and the two vectors Q 1 Drac ponts. and Q2 are shown on the fgure above, as well as the 1.3. two-band model on a square lattce wth a nontrval Chern number Fg. 5

6 6 homewor1.nb Let s add next-nearest-neghbor hoppngs to the model we consdered above n Problem 1.2. Here we eep the nearest-neghbor hoppngs the same as n Problem 1.2. H H NN H NNN (1.22) Here H NN descrbes the nearest-neghbor hoppngs and t s the same as the Eq. (1.12). H NN t, j Φj a b j h.c. (1.23) H NNN are the next-nearest-neghbor hoppngs whch are mared by the dagonal lnes n Fg. 5. Here we assume that all the next-nearest-neghbor hoppngs are real. Along the sold dagonal lnes, the hoppng strength s t. Along dashed lnes, the strength s -t. (They have opposte sgns) H NNN, j t j ' a a j, j t j ' b b j (1.24) magnetc flux and tme-reversal symmetry Consder a trangle formed by two nearest-neghbor bonds and one next-nearest-neghbor bond, whch s the smallest plaquette on ths lattce. If we goes around the square (n the counter-cloc drecton), the Aharonov Bohm phase pced up by the electron s: Φ 1 Φ 2 Φ 3 The phase for nearest-neghbor hoppngs are ± 4 (Φ 1 Φ 2 ± 4). For next-nearest-neghbor bonds, the phase s 0 f t j ' 0 and f t j ' 0. Therefore the total phase s (1.25) Φ 1 Φ 2 Φ 3 ± 2 or Φ 1 Φ 2 Φ 3 ± 2 (1.26) These fluxes are NOT nteger tmes. As we learned n the lecture, they brea the tme-reversal symmetry. So ths model s NOT tme-reversally nvarant Band structure H NN has been taen care of n the prevous problem. Now we convert H NNN nto -space. For red-red hoppngs, there are two next-nearestneghbor bonds per unt cell, one along the 45-degree lne and the other along 135-degree lne. In addton, there are two other NNN bonds for blac-blac hoppngs. As a result the Hamltonan s: H NNN t ' a v1 a r t ' a v2 a r t ' b v1 b r t ' b v2 b r h.c. (1.27) Here, v 1 a, a and v2 a, a wth a beng the length of the nearest-neghbor bonds. The sum here sums over all unt cells. Rewrte H NNN n the -space as H NNN a, b 1 a a 1 b a 1 a b 1 b b a b (1.28) Fnd the two-by-two matrx, whch s the ernel of the Hamltonan for H NNN 1 1 a a 1 a b 1 b a 1 b b (1.29) As a result, the full Hamltonan H H NN H NNN tae the followng form n -space H a, b a a b a a b b b a b (1.30) where 0 1. Fnd the matrx and fnd ts egenvalues Ε ±. Show that gaps open up at the two Drac ponts by computng the gap: Ε 2 a, 2 a Ε 2 a, 2 a (1.31)

7 homewor1.nb 7 Ε 2 a, 2 a Ε 2 a, 2 a Fnd and show >0 as long as t s nonzero. (1.32) Egenvectors To fnd the egenvectors of 0 1, we frst rewrte the two two matrx n terms of dentty and Paul matrces. 0 I x Σ x y Σ y z Σ z (1.33) where I and Σ x and Σ y 0 0 and Σ z (1.34) Fnd the four coeffcents 0, x, y and z and show that u I 1 I z x y (1.35) s one egenvector for the lower band. In addton, please show that u II 1 II x y z (1.36) s also a egenvector for the lower band Can one use one wavefuncton to cover the whole Brlloun zone? Compute u I and u II at, and show that one of them s sngular. Compute u I and u II at, and show that one of them s sngular. Can one use a sngle wavefuncton u I or u II to cover the whole Brlloun zone? Compute the Chern number Draw a small crcle around the pont, to separate the Brlloun zone nto two regons: nsde and outsde. Insde the crcle, use wavefuncton u I and outsde use u II. Now there s no sngular pont. As we have proved n class, at the boundary of these two regons (the crcle), the two wavefunctons are connected by a gauge transformaton Χ x y x y (1.37) Expand near, as 2 a q x, 2 a q y At small q, show that (1.38) Χ x y x y q x q y 1 34 q x q y Θ 3 4 (1.39) where Θ s the polar angle. As a result, the Chern number s

8 8 homewor1.nb C 1 ΧΘ 0 ΧΘ So ths s a topologcal nsulator, f we fll the lowest band. (1.40) 1.4. (optonal) two-band model on a square lattce: topologcally trval nsulator Let s consder the same model as n Problem 1.2 (wth only nearest-neghbor hoppngs). Now we add dfferent potental energy for a stes and b stes H t, j Φj a b j h.c.v a a b b (1.41) Fnd the ernel of the Hamltonan for ths model, whch s a two-by-two matrx. Fnd ts egenvalues and show that gaps are opened at the two Drac ponts. Expand the ernel n terms of dentty and Paul matrces. 0 I x Σ x y Σ y z Σ z (1.42) Compute u I and u II at, and show that one of the s sngular. Compute ui and u II at, and show that one of the s sngular. Can one use a sngle wavefuncton u I or u II to cover the whole momentum space?