Tight binding and emergence of "Dirac" equation in graphene.

Transcription

1 Tight binding and emergence of "Dirac" equation in graphene. A. A. Kozhevnikov 1 1 Laboratory of Theoretical Physics, S. L. Sobolev Institute for Mathematics, and Novosibirsk State University April 22, 2010

2 Outline Introduction 1 Introduction 2 3

3 It is about how Dirac-like equation enter solid state physics. Tight binding model of electronic spectrum in crystal solids Graphene in simplest version of tight binding model Dirac equation near convex points Landau levels and anomalous integer quantum Hall effect

4 Dirac s comb: two approaches 2 d 2 ψ 2m G dx 2 n= δ(x an)ψ = Eψ. mga/ 2 1 limit ( of exact solution: ) E(q) mg2 1+4e mga/ 2 cos qa. 2 2 Find approximate solution: Solution for single δ-potential:e 0 = mg 2 /2 2, ψ 0 (x) = κ 0 e κ 0 x, κ 0 = mg/ 2 Anzatz for approximate solution in periodic potential: ψ(x) = n C nψ 0 (x an)= (E E 0 ) n C n ψ 0 (x na) = G n C n δ(x la)ψ 0 (x na). l n

5 I nn = dxψ 0 (x an )ψ 0 (x an) = (1+κ 0 a n n )e κ 0a n n δ nn dxψ 0 (x an )[ Gδ(x al)]ψ 0 (x an) = Gκ 0 e κ 0( l n + l n ). (E E 0 )C n Gκ 0 C n = αe iqan n n e κ0a n n C n Gκ 0 e κ0a (C n 1+C n +1). E(q) = E 0 2mG2 2 e mga/ 2 cos qa.

6 Tight binding model Simplest version of tight binding model is due to Feynman Lectures on Physics. 1D: Let ε 0, γ, and C n be the energy of an electron in isolated atom, the transition amplitude to the nearest neighbor, and the probability amplitude of sitting at x = an. Then i dc n = ε 0 C n γ(c n 1 + C n+1 ). dt Spectrum: C n e iqan = ε(q) = ε 0 2γ cos qa, γ = Gκ 0 e κ 0a in case of Dirac comb. In 3D and single atom in elementary cell: i dc(rn) dt = E 0 C(R n )+γ l C(R n +δ l ).

7 Crystalline lattice of graphene Two sublattices A and B. Basic vectors: a 1,2 = a 2 (3,± 3). δ `1 a 1 δ 3 A B A B δ 2 a 2 Figure: Crystalline structure of graphene Nearest neighbors: δ 1 = a 2 (1, 3), δ 2 = a 2 (1, 3), δ 1 = a ( 2, 1). 2

8 Reciprocal lattice Reciprocal lattice vectors: a i b j = 2π integer. b 1,2 = 2π 3a (1,± 3). b 1 K K ' b 2 Figure: Reciprocal lattice of graphene Corners of the first Brilloine zone (the Dirac points) ( 1, (1, 3 1 ). K = 2π 3a ) 1, K = 2π 3 3a

9 Electronic spectrum i dc A(n) dt i dc B(n) dt = γ[c B (n+δ 1 )+C B (n+δ 2 )+C B (n+δ 3 )], = γ[c A (n δ 1 )+C A (n δ 2 )+C A (n δ 3 )]. ( CA (n) C B (n) ) = ( α β ) ( exp i Et ) + inq. Spectrum (Wallace 1947): E(q) = ±γ e iqδ 1 + e iqδ 2 + e iqδ = ±γ 1+4 cos 2 q y a cos q ya 2 cos q xa 3 2

10 Spectrum near Dirac points E(K(K )) = 0! Expanding near Dirac points q = p + K(K ) with the help of one obtains cos 3q xa 2 sin q y 3a a2 p 2 x, p ya 3 16 p2 ya 2. E(p x, p y ) = ± 3 2 aγ p 2 x + p 2 y = ± 3 2 aγ p. v F = 3 2 aγ, a = 1.42 Å, γ = 2.8 ev = v F cm/s.

11 "Dirac equation" Expand the eigenvalue equations ( ) Eα = γ e iqδ 1 + e iqδ 2 + e iqδ 3 β, ( ) Eβ = γ e iqδ 1 + e iqδ 2 + e iqδ 3 α and using e ikδ 1 + e ikδ 2 + e ikδ 3 = 0 (the same with K K ) one gets Dirac-like equations in momentum representation.

12 "Dirac" equation in graphene Near K (K ) ( 2E v F ( 3 i)p x ±(1+i 3)p y ( 3+i)p x ±(1 i 3)p y 2E v F ) ( α β ) = 0 Normalized solution near K ( ) α = 1 ( ±ie i(ϑ p/2+π/3) β 2 e i(ϑp/2+π/3) ± ), tanϑ p = py p x = rotation in xy plane by π/6 results in Eψ = v F (σ p)ψ ( near K) Eψ = v F (σ p)ψ ( near K ).

13 Energy levels in magnetic field is given by Eψ = v F ( σ, i e c A )ψ. In the Landau gauge ( A = B( y, ) 0). With ψ(x, y) = e ipx x/ ϕ1 (y) one obtains ϕ 2 (y) Eϕ 1 = v F (p x + eb ) c y y ϕ 2, Eϕ 2 = v F (p x + eb ) c y + y ϕ 1

14 Non-zero Landau energy levels in graphene If E 0 then E 2 ϕ 1 = v 2 F Schrödinger-like equation [ 2 d 2 dy 2 + ( p x + eb )( c y y p x + eb ) c y + y ϕ 1 = ( ) ] eb 2 (y + y 0 ) 2 ϕ 1 = c ( E 2 c 2 F ) + eb ϕ 1 = c ( E n = ±v 2eB F c n ) 1/2, ϕ1n (y) = ψn osc (y + y 0 ), y 0 = cpx eb, degeneracy g = eb 2π c L xl y 2 spinprojection 2 Pseudospin.

15 Non-zero Landau energy levels in graphene 2 spin in magnetic field??? Yes, because at B = 10 T. ω B 0.01 v F (e B/c) 1/2 Lower component of (pseudo)spinor ϕ 2n = v [ F eb E n c (y + y 0)+ d ] ϕ 1n = ±âψn osc = ±ψn 1 osc dy, where â = [ c eb 2e B c (y + y 0)+ d ] dy is destruction operator of oscillator.

16 Zero mode Zero mode (E = 0): [ d dy + eb ] c (y + y 0) ϕ 10 = 0= ϕ 10 (y) = ψ osc 0 (y + y 0), ϕ 20 = 0. Degeneracy g = eb 2π c L xl y 2 spin 1 pseudospin.

17 Resume Introduction Resume: ( E n = ±v 2eB F c n ) 1/2, n = 1, 2,, g = eb 2π c L xl y 2 spin 2 pseudospin, ψ n (x, y) = 1 ( e ipx x/ ψ osc n (y + y 0 ) 2 ±ψn 1 osc (y + y 0) ), Zero mode E = 0, g = eb 2π c L xl y 2 spin 1 pseudospin, ψ 0 (x, y) = e ipx x/ ( ψ osc 0 (y + y 0) 0 ),

18 Figure: Landau levels (Guohong Li and Eva Y. Andrei)

19 Anomalous integer quantum Hall effect Degeneracy L x L y = ( ) 2spin 2 pseudospin N + 2 spin 1 pseudospin eb 2π c σ xy = ± enc B = ± e2 2π c 2(2N + 1), N = 0, 1,.

20 Figure: Integer quantum Hall effect (Castro Neto et al., Rev. Mod. Phys.)