Effective masses for zigzag nanotubes in magnetic fields

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1 Effetive masses for zigzag nanotubes in magneti fields Evgeny Korotyaev February 7, 00 arxiv:math-ph/ v1 5 Mar 007 Abstrat We onsider the Shrödinger operator with a periodi potential on quasi-1d models of zigzag single-wall arbon nanotubes in magneti field. The spetrum of this operator onsists of an absolutely ontinuous part (intervals separated by gaps) plus an infinite number of eigenvalues with infinite multipliity. We obtain identities and a priori estimates in terms of effetive masses and gap lengths. 1 Introdution and main results We onsider the Shrödinger operator H B = ( i A ) + V q with a periodi potential V q on the zigzag nanotube Γ N R 3 (1D models of zigzag single-well arbon nanotubes, see [Ha], [SDD]) in a uniform magneti field B = B(0, 0, 1) R 3, B R. The orresponding vetor potential is given by A (x) = 1 [B,x] = B ( x,x 1, 0), x = (x 1,x,x 3 ) R 3. Our model nanotube Γ N is a union of edges Γ ω of length 1, i.e., Γ N = ω Z Γ ω, ω = (n, l, j) Z = Z J Z N, J = {0, 1, }, Z N = Z/(NZ), see Fig. 1 and. Eah edge Γ ω = {x = r ω +te ω, t [0, 1]} is oriented by the vetor e ω R 3 and has starting point r ω R 3. We have the oordinate x = r ω +te ω and the loal oordinate t [0, 1] (length preserving). We define r ω,e ω, ω = (n, l, j) Z by e n,0,j = e 0 = (0, 0, 1), e n,1,j = κ n+j+1 κ n+j + e 0, e n,,j = κ n+j+ κ n+j+1 e 0, κ j = R N ( 0j, s 0j, 0), 0j = os πj N, s 0j = sin πj N, r n,0,j = κ n+j + 3n e 0, r n,1,j = r n,0,j + e 0, r n,,j = r n+1,0,j, where R N = 3. The points r 4 sin π 0,0,j are verties of the regular N-gon P 0. The vertial edge N Γ 0,0,j lies on the ylinder C {x R 3 : x 1 + x = RN }. The starting points r 1,0,j = r 0,,j = κ 1+j + 3e 0, j Z N are the verties of the regular N-gon P 1. P 1 arises from P 0 by the Institut für Mathematik, Humboldt Universität zu Berlin, Rudower Chaussee 5, 149, Berlin, Germany, evgeny@math.hu-berlin.de 1

2 Γ 1,,N Γ 1,,1 Γ 1,, Γ 1, Γ 1,3 Γ 1,1,N Γ 1,1,1 Γ 1,1, Γ 1,0,N Γ 1,0,1 Γ 1,0, Γ 1,0 Γ 0,,N Γ 0,,1 Γ 0,, Γ 0,1 Γ 0, Γ 0,1,N Γ 0,1,1 Γ 0,1, Γ 0,0,N Γ 0,0,1 Γ 0,0, Γ 0,0 Γ 1,,N Γ 1,,1 Γ 1,, Γ 1,1,N Γ 1,1,1 Γ 1,1, Γ 1,1 Γ 1, Γ 1,0,N Γ 1,0,1 Γ 1,0, Γ 1,0 (a) (b) Figure 1: (a) A piee of a nanotube Γ N, (b) a nanotube Γ 1. The fundamental domain is marked by a bold line. Figure : The zigzag nanotube

3 following motion: rotate around the axis of the ylinder C by the angle π and translate by N 3 e 0. The non-vertial vetors e 0,1,j and e 0,,j have positive and negative projetions on the vetor e 0. Repeating this proedure we obtain all edges of Γ N. Note that eah non-vertial edge Γ 0,l,j, l = 1, (without the endpoints) lies inside the ylinder C. For eah funtion y on Γ N we define a funtion y ω = y Γω, ω Z. We identify eah funtion y ω on Γ ω with a funtion on [0, 1] by using the loal oordinate t [0, 1]. Our operator H B on Γ N ats in the Hilbert spae L (Γ N ) = ω L (Γ ω ) and is given by (H B f) ω = ωf ω (t) + q(t)f ω (t), ω = d dt ia ω, a ω (t) = (A (r ω + te ω ),e ω ), (1.1) see [Ha], [SDD], where (V q f) ω = qf ω, q L (0, 1) and ω f ω, ω f ω L (Γ N ) satisfies The Kirhhoff Magneti Boundary Conditions: f is ontinuous on Γ N and satisfies ω3 f ω3 (1) + ω f ω (0) ω4 f ω4 (1) = 0, ω1 f ω1 (0) ω f ω (1) + ω f ω (0) = 0, (1.) for all ω 1 = (n+1, 0, j), ω = (n, 1, j), ω = (n,, j), ω 3 = (n, 0, j), ω 4 = (n,, j 1) Z. Condition (1.) means that the sum of derivatives of f at eah vertex of Γ N equals 0 and the orientation of edges gives the sign ±. Suh models were introdued by Pauling [Pa] in 1936 to simulate aromati moleules. They were desribed in more detail by Ruedenberg and Sherr [RS] in For physial models see [Ha], [SDD]. For simpliity we will denote Γ ω,1 Γ 1 by Γ ω, for ω = (n, j) Z 1 = Z J. Thus Γ 1 = ω Z1 Γ ω, see Fig 1. The operator H B is unitarily equivalent to H(a) = N 1 H j(a), a = 3B ot π (see [KL1]), where the self-adjoint operator H 16 N j(a) ats in the Hilbert spae L (Γ 1 ) and is given by (H j (a)f) ω = f ω + qf ω, f = (f ω ) ω Z1 D(H j (a)), where D(H j (a)) onsists of all funtions f = (f ω ) ω Z1, (f ω ) ω Z 1 L (Γ 1 ), that satisfy the Kirhhoff onditions f n,0 (1) = f n,1 (0) = e ia s j f n, (1), f n+1,0 (0) = e ia f n,1 (1) = f n, (0), s = e iπ N, (1.3) f n,0 (1) + f n,1 (0) eia s j f n, (1) = 0, f n+1,0 (0) eia f n,1 (1) + f n, (0) = 0. (1.4) { Define the spae l p = h = {h n } 1 : h n C, } n 1 h n p <, p 1. Let S p, p 1 be the lass of onformal mappings k : C + K(h) = {λ C +, Re λ > 0}\ n 1 [πn, πn+ih n ], where h = (h n ) 1 l p, h n 0 and k(λ) = i λ ( 1 + O(1)) as λ. In this ase we introdue the sets: spetral bands σ n = [λ + n 1, λ n ] = k 1 ([π(n 1), πn]) and gaps γ n = (λ n, λ+ n ), n 1. Reall the needed properties of the Hill operator Hy = y + q(x)y on the real line with a periodi potential q(x + 1) = q(x), x R, see, e.g., [MO]. We introdue the fundamental solutions ϑ(x, λ) and ϕ(x, λ), (x, λ) R C of the equation y + q(x)y = λy satisfying ϑ(0, λ) = ϕ (0, λ) = 1, ϑ (0, λ) = ϕ(0, λ) = 0. The orresponding Lyapunov funtion are given by (λ) = ϕ (1,λ)+ϑ(1,λ), λ C. The spetrum of H is purely absolutely ontinuous and onsists of intervals σ n = [ λ + n 1, λ n], n 1. These intervals are separated by the gaps γ n = ( λ n, λ + n) of length γ n 0. If a gap γ n is degenerate, i.e. γ n = 0, then the orresponding 3

4 segments σ n, σ n+1 merge. The sequene λ + 0 < λ 1 λ + 1 <... is the spetrum of the equation y + qy = λy with -periodi boundary onditions, that is y(x + ) = y(x), x R. Here equality λ n = λ + n means that λ ± n is an eigenvalue of multipliity. Note that ( λ ± n ) = ( 1) n, n 1. The lowest eigenvalue λ + 0 is simple, ( λ + 0 ) = 1, and the orresponding eigenfuntion has period 1. The eigenfuntions orresponding to λ ± n have period 1 if n is even, and they are anti-periodi, that is y(x+1) = y(x), x R, if n is odd. The derivative of the Lyapunov funtion has a zero λ n in eah losed gap [ λ n, λ + n ], that is ( λ n ) = 0. Let µ n, n 1, be the spetrum of the problem y +qy = λy, y(0) = y(1) = 0 (the Dirihlet spetrum). It is well-known that µ n [ λ n, λ + n ]. Define the set σ D = {µ n, n 1} and note that σ D = {λ C : ϕ(1, λ) = 0}. Define the quasimomentum k(λ) = aros (λ), λ C +. The funtion k( ) S, where the orresponding vetor (n hn ) 1 l is defined by the equation ( λ n ) = ( 1) n osh h n. If λ + 0 = 0, then k satisfies (here and below 1 = i) k(0) = 0, q 0 + O(1/z) k(λ) = z, z = λ C +, z, (1.5) z k(r ) = ir +, k( σ n ) = [π(n 1), πn], k( γn ) = [πn, πn + i h n ], k(λn ) = πn + i h n, n 1. For a self-adjoint operator H we define the set σ (H) = {λ : λ σ pp (H) is of infinite multipliity}. Reall needed results from [KL1]. Let j = os a j, s j = sin a j, a j = a + πj. If N j 0, then the spetrum σ(h j (a)) = σ (H j (a)) σ a (H j (a)), where σ (H j (a)) = σ D and σ a (H j (a)) = {λ R : ξ j (λ, a) [ 1, 1]} = n 1 σ j,n (a), σ j,n (a) = [λ + j,n 1 (a), λ j,n (a)], where λ + j,0 < λ j,1 λ+ j,1 < λ j,... are zeros of the funtion ξ j 1, and ξ j is the modified Lyapunov funtions given by ξ j = F + s j, F = 9 5, where = ϕ (1, ) ϑ(1, ), j Z N. (1.6) j 4 If j = 0, then the spetrum σ(h j (a)) = σ (H j (a)) = σ D {λ R : F(λ) = 1}. If λ σ a (H j (a)), then the equation y + qy = λy on Γ 1 with onditions (1.3),(1.4) has a solution ψ suh that ψ n+1,0 (0) = e ip j(λ) ψ n,0 (0), ψ n+1,0(0) = e ip j(λ) ψ n,0(0), n Z, where p j (λ) is a quasimonentum. The funtion osp j (λ), j 0 is not entire, and it is define on some Riemann surfae. If we take k j = p j + πj N, then ξ j = osk j (λ) is the entire funtion. Below we onsider only the operator H 0 (a), = osa > 0. Due to (1.6), the results for the operator H 0 (a + πj N ) give the results for the operator H j(a). For simpliity we will write ξ = ξ 0 and λ ± n = λ± n (a) = λ± 0,n (a), n 0. The funtion ξ 1 has only real zeros λ ± n, their labeling is given by λ + 0 < λ 1 λ + 1 < λ λ + <.. and if / { 1, 1}, then λ n < λ+ n. Here σ n = [λ + n 1, λ n ] are the spetral bands and γ n = (λ n, λ+ n ) 4

5 are the gaps. Moreover, λ ± n satisfy ξ(λ ± n) = ( 1) n and λ ± n = λ0,± n + q 0 + ε ± n as n, q 0 = ε ± n = ± q n + o(n 1 ), q n = ε ± n = ± ˆq n ˆq sn 9 see [KL], [KL1]. Here λ 0,± n λ 0,+ 0 = φ 0 [0, π/], 1 + O(1) n, ˆq n = q(t)dt, q(t) osπntdt, n is odd, = 1 q(t)e iπnt dt, ˆq sn = Im ˆq n, n is even, = 1, ε ± n = o(1/n) in other ases, (1.7) are -periodi eigenvalues for the ase q = 0 given by λ 0,± n = πn ± φ 0, os φ 0 = ( 9 λ 0,± n+1 = π(n + 1 ) ± φ 1, φ 1 [0, π/], os φ 1 = ) [ 1, 1], ( 7 ) [ 1, 1]. (1.) The identity ξ(λ) = osk(λ), λ C + defines an analyti funtion (the quasimomentum) k(λ), λ C +, see Theorem 1.. With eah edge of the gap γ n, we assoiate the effetive mass µ + 0, µ± n by (we take some branhes k suh that k(λ) πn as λ λ± n ) λ = λ ± n + (k(λ) πn) (1 + o(1)) as λ λ ± µ n, (1.9) ± n and let µ ± n = 0 if γ n = 0. If q = 0, then the effetive masses µ 0,± n are given by (see Set. ) µ 0,+ 0 = 9 sin φ 0, µ 0,± n φ = ± 9 0 sin φ 0 λ 0,± n, µ 0,± n 1 = ± 9 sin φ 1 λ 0,± n 1, n 1, (1.10) and µ 0,+ n + µ 0, n = O(1/n ) as n. Let F 0 = 9os λ 1. We formulate our first result. Theorem 1.1. Let q L (0, 1). Then the following identities and asymptotis hold true µ ± n = µ ± n = µ 0,± n + ( 1)n+1 F k (λ) = 1 m 1,s=± 0 (λ0,± n n 0,ν=± )ε± n µ ν n + O(1) n 3 as n, (1.11), λ λ ± λ λ n, n 0, (1.1) ν n µ n 1(µ + n + µ n ) =, (1.13) (λ s m 1 λ± n ) 1, µ ± n+1 = m 0,s=± (λ s m λ± n+1 ) 1, n 0, (1.14) where the series onverges absolutely and uniformly on ompat sets in C \ {λ ± n, n 0}. 5

6 Remark. Note that F 0 (λ 0,± n )ε ± n = O(ε ± n/n ) as n. Let λ n, n 1 be the zeros of F. We have ξ (λ n ) = 0. Reall that q 0 = 1 0 q(x)dx. Theorem 1.. i) Let q L (0, 1) and λ + 0 = 0. Then a quasimomentum k(λ) = aros ξ(λ), λ C + belongs to S, (h n ) 1 is defined by the equation ξ(λ n ) = ( 1) n osh h n, and satisfies k(0) = 0, ( ) 9 k(λ) = z + log q 0 + O(1/z), z = i λ 1, λ = z, (1.15) z k(r ) = ir +, k(σ n ) =[π(n 1), πn], k(γ n ) =[πn, πn+ih n ], k(λ n )=πn+ih n, n 1. (1.16) ii) Moreover, for eah n 1 the following estimates hold true h n 3π γn µ ± n 6π n(µ + n µ n ), (1.17) γ n (4πn) (µ + n µ n), (1.1) γ n λ + nµ + n, γ n λ n µ n + 16λ n(µ n), (1.19) h n 4π λ ± n µ± n, h n π γ n µ n, h n π γ n µ + n, (1.0) h n γ n µ + n µ n. (1.1) If γ n is the first non degenerate gap for some n 1, then µ + 0 µ n. iii) Moreover, let a spetral interval σ(n, n 1 ) = [λ + n, λ n 1 ] = n 1 n+1 σ j, where n 1 n is a number of the merged omponents σ j whih are omposed this interval σ(n, n 1 ). Then µ + n(λ n 1 λ + n) 16(n 1 n) (λ + n + λ n 1 ), µ n 1 (λ n 1 λ + n) 3(n 1 n) λ n 1. (1.) Remark. 1) There is a big differene beween the quasimomentum for the operator H 0 and the Hill operator H: h l and (n hn ) 1 l, and, in partiular, the unperturbed vetor h 0 l. ) Note that n 1 n in (1.). Below we will sometimes write λ ± n(a), µ ± n(a), ξ(λ, a),.., instead of λ ± n,µ ± n, ξ(λ),.., when several magneti fields are being dealt with. Theorem 1.3. Let q L (0, ) and a [0, π ), n 0. Then h 0 n(a) h n (a), µ 0,± n (a) µ ± n(a), σ 0 n(a) σ n (a), (1.3) h n (a) h n (a 1 ), µ ± n(a) µ ± n(a 1 ), σ 0 n(a) σ n (a 1 ), all π 3 a a 1 π. (1.4) In the present paper we obtain only loal estimates, i.e., estimates for fix n 0. For the Hill operator there exist a priori two sided estimates [K1], [K]. In order to obtain similar results for the zigzag nanotubes, where h l, we have to study arefully the quasimomentum k( ) as onformal mapping. In our paper we only touh this problem. 6

7 Identities and asymptotis Reall that F = 9 5 4, where = ϕ (1, ) ϑ(1, ) and, F satisfy q 0 = 1 0 q(t)dt, (λ) = os λ + q 0 sin λ λ F(λ) = F 0 (λ) + 9q 0 sin λ λ F (λ) = F 0 (λ) + 9q 0 λ + O(e Im ) λ, (λ) = o(e Im ), (.1) λ λ 1 ( e Im ) λ + O, F 0 (λ) = 9 os λ 1, λ (.) os ( λ e Im ) λ + O λ λ 3 (.3) as λ, uniformly on bounded sets of q L (0, 1) (see [KL]). The identity µ ± n = ( 1)n ξ (λ ± F+s n ), n 0 from [KK1] together with ξ = yields µ ± n = ( 1)n ξ (λ ± n ) = (λ ± ( 1)nF n ), n 0. (.4) In the unperturbed ase q = 0 the modified Lyapunov funtion is given by ξ 0 = F 0 + s, F 0 = 9 osz 1 = osa > 0, s = sin a, z = λ. (.5) The funtion ξ 0 (z ) is π-periodi and on the period [0, π] has a maximum at z 0, π and a minimun at z = π : max x R ξ0 (x ) = ξ(0) = 1 + s = > 1 if 1 and max x R ξ0 (x ) = 1 if = 1, (.6) min x R ξ0 (x ) = ξ(π /4) = 1 4 < 1 if 1 and ξ 0 (π/) = 1 if = 1. (.7) We have gaps γn 0 = (λ 0, n, λ 0,+ n ), where ξ 0 (λn 0,± ) = ( 1) n. Using ξ 0 (λn 0,± ) = ( 1) n we have the following equetion for zn 0,± = λn 0,± > 0: 9 os z 1 + s = ( 1) n, os z = ( + ( 1) n 7 ) [ 1, 1]. 9 Then -periodi eigenvalues λ 0,± n = (zn 0,± ) have the form (1.), i.e., z 0,+ 0 = φ 0 [0, π ], z0,± n = πn ± φ 0, os φ 0 = 9 ( + 7 ) [ 1, 1], z 0,± n 1 = π(n 1 )±φ 1, φ 1 [0, π/], os ( π φ 1) = os φ 1 = ( 7 ) [ 1, 1] 9 7

8 for n 1. The funtion k 0 = aros ξ 0 (λ) is a onformal mapping from the upper half-plane C + onto a quasimomentum domain K(h 0 ) = C\ n 1 [πn, πn+ih 0 n ], where h = (h0 n ) 1, h n 0 is defined by the equation osh h 0 n = ( 1)n ξ(λ n ) 1 and satisfies osh h 0 0 = 1 + s 1, h 0 n = h0 0, and osh h0 1 = Using (.4), we dedue that effetive masses for q = 0 are given by 1, h 0 n+1 = h0 1. (.) Thus µ 0,± n ) n = ( 1) n+1f (λ 0,± = 9( 1)n sin zn 0,± zn 0,±, λn 0,± = (zn 0,± ), n 0. (.9) µ 0,+ 0 = 9 sin φ 0, µ 0,± n φ 0 = ± 9 sin φ 0 ( πn ± φ 0), µ0,+ n +µ0, n = 9φ 0 sin φ 0 4[ (πn) φ 4 0] and µ 0,± n = ± 9 sin φ 1 ( πn ± φ 1), µ0,+ n + µ n 0, = 9φ 1 sin φ 1 < 0, n is odd. 4( (πn) φ 4 1 ) Proof of Theorem 1.1. Using asymptotis (1.7) we obtain F (λ ± n ) = F 0 (λ± n ) + O(1/n3 ), F 0 (λ± n ) = F 0 (λ0,± n ) + F 0 (λ0,± n )ε± n + O(1/n3 ) < 0, n is even, as n. Combine last asymptotis with µ ± n = ( 1)n F (λ ± n ) (see (.4)) we get (1.11). Using the Cauhy theorem about residues for the funtion k = ξ ξ 1 1 k (ρ) πi ρ =t ρ λ dρ = k (λ) 1 The identity ξ = F+s and b ± = ± s gives λ ν n <t,ν=± µ ν n λ λ ν n, we dedue that, all t λ ± n, n 0. k = ξ 1 ξ = F ( (F + s ) = F (F b )(F b + ). (.10) Using os φ ± = 1+b ± [ 1, 1], φ 9 ± [0, π ], we obtain F 0 (ρ) b ± = 9 (os z os φ ±) = 9 4 sin(z φ ±) sin(z + φ ± ), z = ρ. (.11) We need the simple estimates (for eah r (0, π ]) sinz e Im z (1 e r ), any z C r = {z C : z πn r, n Z}. (.1) Using (.11),(.1) we obtain for ζ 1 = z φ ±, ζ = z + φ ± C r (we take small r << 1) F 0 (ρ) b ± = 9 4 sinζ 1 sin ζ 9 4 e Im z (1 e r ) as ρ. (.13)

9 Substituting (.),(.3) and (.13) into (.10) we dedue that k (ρ) = F (ρ) (F(ρ) b )(F(ρ) b + ) = O(ρ 1 )e 4 Im ρ e 4 Im ρ (1 + o(1)) = O(ρ 1 ) as t, for all ρ {µ C : µ = t} C r, where t φ ±, t φ ± C r. Thus we obtain k (ρ) dρ = O(1/n) as t, whih yields (1.1). ρ =t ρ λ In order to show (1.13) we need the following identities µ + n + µ n = A λ λ + n λ λ n + B n, A n = (µ+ n + µ n ) ( ) 1 1 +, n λ λ + n λ λ n B n = (µ+ n µ n ) ( 1 1 ) = (µ+ n µ n )(ε+ n ε n ) λ λ + n λ λ n (λ λ + n )(λ λ n ). Asymptotis (1.11) give µ + n + µ n < and (µ + n µ n )(ε+ n ε n ) <, whih implies ( µ + λ 0 λ λ + + ) A n µ (µ + n + µ n), and λ B n 0 0 n 1 n 1 n 1 as λ. These asymptotis together with (1.1), (3.3) yield (1.13). The Hadamard fatorization ξ(λ) + 1= ( ) 1 λ ξ gives (λ) = λ s n+1 ξ(λ)+1 Then the identity (.4) implies n 0,s=± µ ± n = ξ (λ ± n ) = ξ (λ ± n) ξ(λ ± n) + 1 = whih yields (1.14). The proof for µ ± n+1 is similar. n 0,ν=± 1 λ ±, n λ ν n+1 n 0,s=± 1. λ λ s n+1 3 Conformal mappings and estimates Proof of Theorem 1.. i) We need some results from [MO]. Let a funtion f be entire and f(z ), z C of exponential type and f(λ) be real on the real line and f(λ) = O(1) and f( λ) = C f e λ 1 +o(1) as λ + for some onstant C f > 0. Assume that all zeros of the funtion f 1 are real and their labeling is given by ζ 0 + < ζ1 ζ 1 + < ζ ζ + <... Then there exists a onformal mapping k : C + K(h) for some sequene h = (h n ) 1 l suh that f(λ) = osk(λ) and k(λ) = λ( + O(1)) as λ and if ζ 0 + = 0, then k satisfies k(r ) = ir +, k([ζ + n 1, ζ n ]) = [π(n 1), πn], k([ζ n, ζ + n ]) = [πn, πn + ih n ], k(ζ n ) = πn + ih n, n 1, where ζ 1 < ζ < ζ 3 <.. are zeros of f and ζ n [ζ n, ζ + n ] for all n 1. 9

10 The funtion ξ satisfies these onditions, then the statement i) have been proved and we need only to show (1.15). Identities (λ) = os k(λ) and asymptotis k(λ) = z q 0+o(1), z = z iy = λ, y, see [MO], and (.1) give F(λ) = 9 (os k(λ) + o(λ 1 e y )), ξ(λ) = 9 os k(λ) + o(z 1 e y ) = 9 16 e ie k(λ)+o(z 1). Then osk(λ) = e ik(λ) (1 + O(e 4y )) yields (1.15). ii) We have proved the existene of the onformal mapping k : C + K(h) for some h l. For suh onformal mapping the estimates (1.17), (1.1) were proved in [K1]. Moreover, if γ n is the first non degenerate gap for some n 1, then µ + 0 µ n [K1]. Let n 1. We need the following estimates from [KK] g n h n π g n m ± n π m± n, where z± n m± n = µ± n, z± n = λ ± n > 0 (3.1) h n g n m + n m n, g n m ± n, where (z + n + z n ) g n = γ n. (3.) Consider µ + n. Using the estimate g n m + n we obtain γ n 4(z + n + z n )z + n µ + n λ + nµ + n, whih yields the first estimate in (1.19). Consider µ n. Using the estimate g n m n and identites from (3.1), (3.) we obtain γ n 4(z + n + z n )z n µ n λ n µ n + 4 g n z n µ n, g n z n 4λ n µ n, whih yields (1.19). The estimate from (3.1) givees h n π m ± n = 4π λ ± n µ ± n, whih gives the first estimate in (1.0). Using estimates and (3.1), (3.), we dedue that g n m n γ n µ n, g n m + n γ n µ + n h n π γ n µ n, h n π γ n µ + n, whih yields the last two estimates in (1.0). The identities from (3.1), (3.) give g n m + n m n = 4z+ n z n g n µ + n µ n γ n µ + n µ n sine 4z + n z n (z+ n + z n ). This implies (1.1). iii) We need the estimate (see Theorem.4 from [KK]) µ ± σ(n, n 1 ) 16(n 1 n) λ ± ( λ + λ + ), µ + = µ + n, λ + = λ + n, µ = µ n 1, λ = λ n 1, where n 1 n is the number of the merged omponents whih are omposed the band (λ +, λ ). This estimate and λ + < λ yields (1.). 10

11 Proof of Theorem 1.3. Using the estimates F(λ n ) F 0 (λ 0 n) 1 (see Lemma 3.1 from [KL]) we dedue that h 0 n h n for all n 1. Applying these fats to the qusimomentums k, k 0 and using (3.4) we get (1.3). We show (1.4). If n is even, then F(λ n ) 1 and f n () = osh h n = F(λn)+s = + F(λn)+1 and f F(λn)+1 n () = 1 < 1. If n is odd, then F(λ n ) 5 (see Lemma 3.1 from [KL]) and f 4 n() = osh h n = F(λn)+s = F(λn)+1 and f n() = 1+ F(λn)+1. Assume that < 1, then f n() 1 4 < 0. Thus the funtion h n (a) on the interval [ π, π] is inreasing and h 3 n(a) < h n (a 1 ) for all π a < a 3 1 π. Then estimate (3.4) yields (1.4). Lemma 3.1. The asymptotis (1.15) and following one hold true Proof. Using (.1)-(.3), we obtain k (λ) = 1 λ + q 0 + o(1) λ as λ. (3.3) k (λ) = ξ (λ) ξ(λ) 1 = (λ) ξ ξ(λ) (1 + O(ey )), ξ (λ) ξ(λ) = F (λ) F(λ) (1 + O(ey )), where y = λ > 0. Identities (λ) = os k(λ), z = iy = λ and asymptotis k(λ) = z q 0+o(1), z k (λ) = 1 (1 + q 0+o(1) ), (see [MO]) and (.1) give z λ F(λ) = 9 (os k(λ) + o(λ 1 e y )), F (λ) = 9 ( sin k(λ)) k (λ) + o(z 3 e y ), F (λ) F(λ) whih yields (3.3). k(λ) = sin os k(λ) ( k (λ) + o(λ 1 )) = i( k (λ) + o(z 3 ) = i z (1 + q 0 + o(1) ), λ Lemma 3.. Let k 1, k S and let h 1,n h,n for all n 1. Then the orresponding spetral bands σ 1,n, σ,n and effetive masses µ ± 1,n, µ±,n satisfy σ 1,n σ,n, µ ± 1,n µ±,n, all n 1. (3.4) Proof. Let r j (λ) = kj(λ), λ = ζ + iη C +, j = 1,, where k j = u j + iv j. The funtion r j is a onformal mapping from C + onto R j = {r = k, k K(h j )}. Let λ j (r), r R j be the inverse mapping λ j = r 1 j. The funtion s j = Im r j = u j v j is harmoni, nonnegative in C + and s j C(C + ) and r j satisfies ( 1 t r j (λ) = λ+c j + 1 π λ j,1 s j (t) t λ 1 + t ) dt, C j = 1 π λ j,1 s j (t)dt t(1 + t ), λ C +, (3.5) where λ j,1 > 0 and s j (t)dt <, see [K1]. In the domain D λ j,1 (1+t ) ε = {λ C + : ε arg λ π ε}, 0 < ε < π there is an estimate t λ t sin ε for all t > 0. This and 11

12 (3.5) yields r j (λ) = λ(+o(1)) as λ D ε, λ. But for any κ there exists a onstant ρ = ρ(κ) > 0 suh that {λ : λ > ρ} D κ r j (D ε ), j = 1, for some ε < κ < π. Then λ j (r) = r(1 + o(1)), r = t + is D κ as r, and λ 1 (r (iη)) iη = λ 1(r (iη)) r (iη)) 1 as η, (r (iη)) iη whih yields f(iη) = η(1 + o(1)) as η. Then the Herglotz Theorem yields f(λ) = Imλ 1 (r (λ)) Imλ (r (λ)) = η, λ = ζ + iη C +. (3.6) Then Im λ 1 Im λ 0 in the domain r (C + ) and Im λ 1 (r) = Im λ (r) = 0, r R give λ 1(r) = s Imλ 1(r) s Imλ (r) = λ (r), r R, r t n = (πn), (reall that r = t + is) whih implies σ 1,n = Moreover, we dedue that tn t n 1 λ 1(r)dr tn t n 1 λ (r)dr = σ,n, n 1. r λ j (r) λ + j,n = λ j (r)dr = (r t n) t n (πn) µ + (1 + o(1)) as r t n + 0, j,n whih yelds µ + 1,n µ+,n. The proof for µ 1,n is similar. Referenes [Ha] Harris P. Carbon Nanotubes and Related Strutures, Cambridge Univ. Press., Cambridge, [KK] Kargaev, P.; Korotyaev, E. Effetive masses and onformal mappings. Comm. Math. Phys. 169(1995), [KK1] Kargaev P.; Korotyaev E. The inverse problem for the Hill operator, a diret approah. Invent. Math. 19(1997), [K1] Korotyaev, E. The estimates of periodi potential in terms of effetive masses. Commun. Math. Phys., 13(1997), [K] Korotyaev, E. Estimates of periodi potentials in terms of gap lengths. Comm. Math. Phys. 197 (199), no. 3, [KL] Korotyaev, E.; Lobanov, I. Shrödinger operators on zigzag nanotubes, will be published in Ann. H. Poinaré. 1

13 [KL1] Korotyaev, E.; Lobanov, I. Magneti Shrödinger operators on zigzag nanotubes, preprint 006. [MO] Marhenko, V.; Ostrovski, I. A haraterization of the spetrum of the Hill operator. Math. USSR Sb. 6 (1975), [Pa] Pauling, L. The diamagneti anisotropy of aromati moleules, Journal of Chemial Physis, 4(1936), [RS] Ruedenberg, K.; Sherr, C. Free-eletron network model for onjugated systems. I. Theory, The Journal of Chemial Physis, 1(1953), [SDD] Saito, R.; Dresselhaus, G.; Dresselhaus, M. Physial properties of arbon nanotubes, Imperial College Press,