Periodic Schrödinger Operators ∞ (IR2 /Γ). Let Γ be a lattice in IR2 . Let V ∈ CIR Proposition S.9 The spectrum of the self–adjoint boundary value problem h i 2 i∇ ∇ + V (x) φ = λφ φ(x + γ ) = eik·γγ φ(x) ∀γγ ∈ Γ or, equivalently, the spectrum of the self–adjoint boundary value problem h i 2 i∇ ∇x −k +V (x) ψ = λψ ψ(x+γγ ) = ψ(x) ∀γγ ∈ Γ consists of a sequence of eigenvalues e1 (k) ≤ e2 (k) ≤ e3 (k) ≤ · · · with, for each n, en (k) continuous in k and periodic with respect to Γ# and limn→∞ en (k) = ∞. Theorem S.10 The spectrum of −∆ + V (x) is d # en (k) k ∈ IR /Γ , n ∈ IN FC 1 Fermi Curves Defined The (real “lifted”) Fermi curve for energy λ is k ∈ IR en (k) = λ for some n ∈ IN 2 2 = k ∈ IR i∇ ∇x − k + V (x) − λ ψ = 0 Fbλ,IR (V ) = 2 2 2 2 i∇ ∇x − k + V (x) ψ = 0 k∈C 2 for some 0 6= ψ ∈ H (IR /Γ) We may absorb the λ in V . FbIR (V ) = Fb0,IR (V ) FIR (V ) = FbIR (V )/Γ# The complexifications Fb(V ) = 2 F(V ) = Fb(V )/Γ# 2 for some 0 6= ψ ∈ H (IR /Γ) FC 2 The Free Fermi Curve First set V = 0. Then 2 2 e i x·b b ∈ Γ# 2 is a basis of L (IR /Γ) of eigenfunctions of i∇ ∇x − k . 2 i x·b = (−b − k)2 ei x·b i∇ ∇x − k e = Nb (k) ei x·b = Nb,1 (k)Nb,2 (k) ei x·b where Nb (k) = (k1 + b1 )2 + (k2 + b2 )2 Nb,ν (k) = (k1 + b1 ) + i(−1)ν (k2 + b2 ) Hence 2 # b F (V = 0) = k ∈ C ∃ b ∈ Γ with Nb (k) = 0 [ [ Nb = Nb,ν = b∈Γ# where b∈Γ# ν=1,2 2 2 Nb = k ∈ C (k1 + b1 ) + (k2 + b2 ) = 0 2 ν Nb,ν = k ∈ C (k1 + b1 ) + i(−1) (k2 + b2 ) = 0 2 FC 3 −ik1 N0,2 N−b,2 Nb,1 Nb,2 N0,1 N−b,1 k2 Free Fermi Curve: Fb(V = 0) FC 4 −ik1 k2 Free Fermi Curve: F(V = 0) FC 5 Theorem. Let r > 1. There exists an entire function F on C2 × Lr (IR2 /Γ) such that 2 ker i∇ ∇x − k + V (x) 6= {0} ⇐⇒ F (k, V ) = 0 Idea of Proof. Write i∇ ∇x − k 2 + V (x) = 1l − ∆ + W (k, x) . 2 ker i∇ ∇x − k + V (x) 6= {0} ⇐⇒ ker 1l − ∆ + W (k, x) 6= {0} 1 1 ⇐⇒ ker 1l + √1l−∆ W (k, x) √1l−∆ 6= {0} 1 1 √ √ ⇐⇒ det 1l + 1l−∆ W (k, x) 1l−∆ = 0 c = Denote the eigenvalues of W λi = λi (k, V ), i = 1, 2, 3, · · ·. ∞ X |λi | 2r <∞⇒ i=1 ∞ Y √ 1 √ 1 W (k, x) , 1l−∆ 1l−∆ P 1 Since 1+|n|2r < ∞ n∈ZZ2 (1 + λi )e −λi + 12 λ2i converges i=1 Set 2 b b −tr W + 12 tr W c c) F (k, V ) = lim det(1l + W )e = det3 (1l + W finite rank FC 6 “Theorem”. Outside of a compact set F(V ) is very close to F(0) = Fb(0)/Γ# except that each can open up to z1 z2 = t z1 z2 = 0 Generically all the double points open up and then we have a Riemann surface (a one dimensional complex manifold). Idea of Proof. Fix any k. This k ∈ Fb(V ) iff there is a nonzero ψ obeying ψ(x + γ ) = ψ(x) ∀γγ ∈ Γ and h i 2 i∇ ∇x − k + V (x) ψ = 0 Expand ψ(x) = to 2 P (−b − k) ψ̃b + c∈Γ# X (1) ψ̃c eic·x . Then (1) is equivalent Ṽ (b − c)ψ̃c = 0 ∀b ∈ Γ# (2) c∈Γ# FC 7 Recall that b+k 2 = Nb (k) = Nb,1 (k)Nb,2 (k) where Nb,ν (k) = (k1 + b1 ) + i(−1)ν (k2 + b2 ) Nb (k) vanishes on Nb = Nb (k) = 0 . Fix k∈C 2 small and define the (-)tube about Nb by Then Tb = T1 (b) ∪ T2 (b) 2 Tν (b) = k ∈ C |Nb,ν (k)| < k∈ / Tb =⇒ |Nb (k)| ≥ 1+|Im k|1− |Im k| 1 + |Im k|1− and • T b ∩ T b0 is compact whenever b 6= b0 • T ν (b) ∩ T ν (b0 ) = ∅ if b 6= b0 • T b ∩ T b0 ∩ T b00 = ∅ for all distinct b, b0 , b00 in Γ] FC 8 −ik1 T2 (0) T2 (b) T2 (−b) T1 (b) T1 (0) T1 (−b) k2 FC 8a To study the part of F̂ (V ) in the intersection of ∪d∈B Td and C2 \ ∪b6∈B Tb for some finite subset B of Γ] it is natural to split (2) Nb (k)ψ̃b + P Ṽ (b − c)ψ̃c = 0 ∀b ∈ B (2B ) Ṽ (b − c)ψ̃c = 0 ∀b ∈ B 0 (2B0 ) c∈Γ# Nb (k)ψ̃b + P c∈Γ# 0 # where B = Γ \ B. Set φ̃b = Then (2B ) is φ̃b + P c∈B 0 Ṽ (b−c) Nc (k) φ̃c Using Mb,c b,c∈B 0 ≤ max n =− P Nb (k)ψ̃b 0 Ṽ (b − c)ψ̃c if b ∈ B 0 . if b ∈ B ∀b ∈ B 0 c∈B o P P Mb,c Mb,c , sup sup c∈B 0 b∈B 0 b∈B 0 c∈B 0 one can easily see that, for k in the region under con Ṽ (b−c) sideration, Nc (k) b,c∈B 0 1. FC 9 Hence Rb,c b,c∈B 0 Ṽ (b−c) Nc (k) , with Rb,c = δb,c + bounded inverse and P −1 0 φ̃b = − Rb,d 0 Ṽ (d − c)ψ̃c has a ∀b ∈ B 0 c∈B d0 ∈B 0 or ψ̃c0 = − P c∈B d0 ∈B 0 −1 1 R 0 ,d0 Ṽ c Nc0 (k) (d0 − c)ψ̃c ∀c0 ∈ B 0 Substituting this into (2B ) yields, for all b ∈ B, Nb (k)ψ̃b + P Ṽ (b − c)ψ̃c c∈B P − c∈B c0 ,d0 ∈B 0 Ṽ (b−c0 ) −1 Nc0 (k) Rc0 ,d0 Ṽ (d0 − c)ψ̃c = 0 This has a nontrivial solution if and only if the |B|×|B| determinant h det Nb (k)δb,c + Ṽ (b − c) − B (k) Sb,c i b,c∈B =0 where B Sb,c (k) = P c0 ,d0 ∈B 0 Ṽ (b−c0 ) −1 Nc0 (k) Rc0 ,d0 Ṽ (d0 − c) FC 10 Proposition 19.5,6 Let k ∈ C2 with |Im k| > const . a) Let k ∈ C2 \ ∪b Tb . Then k 6∈ F̂ (V ). b) Let k ∈ T0 \ ∪b6=0 Tb . Then k ∈ F̂ (V ) if and only if k2 = N0 (k) = A(k) {0} where A(k) = −Ṽ (0) + S0,0 (k) obeys |A(k) + Ṽ (0)| ≤ ∂ n+m n m A(k) ≤ ∂k ∂k 1 2 const |Im k|2− const |Im k| if m + n = 1 c) Let k ∈ T0 ∩ Td . Then k ∈ F̂ (V ) if and only if N0 (k) + Ṽ (0)−D(k)1,1 Nd (k) + Ṽ (0)−D(k)2,2 = Ṽ (−d) − D(k)1,2 Ṽ (d) − D(k)2,1 where {0,d} D(k)i,j = Sdi ,dj (k), d(1) = 0, d(2) = d and obeys |D(k)i,j | ≤ ∂ n+m n m D(k)i,j ≤ ∂k ∂k 1 2 const |Im k|2− const |Im k| if m + n ≥ 1 FC 11 Theorem 19.9 Let V ∈ L2 (IR2 /Γ) with kbṼ (b)k1 < ∞. Then F(V ) is a reduced one dimensional complex analytic variety, which consists of at most two components. If F(V ) is smooth, then it is irreducible. Theorem 20.2 Let q ∈ C ∞ (IR2 /Γ) be such that F(q) is smooth. Then reg han ∪ F(q) F(q) = F(q)com ∪ F(q)reg 2 ∪ F(q) 1 where F(q)com = a compact Riemann surface with boundary ∼ F(q)reg = C \ ({some disks} ∪ {compact}) 1 ∼ F(q)reg = C \ ({some disks} ∪ {compact}) 2 [ han ∼ F(q) Yj = j≥g=1 Here g is the genus of F(q)com , Yj is biholomorphic to 2 z1 z2 = tj , |z1 |, |z2 | ≤ 1 (z1 , z2 ) ∈ C CN and, for all N , |tj | ≤ N for all j . j FC 12 Planar Parts of the Fermi Curve FC 13 Interacting Fermi Curve: F(V ) FC 14