Quantum Graphs and their generalizations

Transcription

1 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 1/9 Quantum Graphs and their generalizations Pavel Exner Doppler Institute for Mathematical Physics and Applied Mathematics Prague

2 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 2/9 Lecture V Leaky graphs strong coupling, approximation of leaky graphs, eigenvalues and resonances

3 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 3/9 Lecture overview Spectral behaviour of leaky graphs in case of a strong coupling

4 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 3/9 Lecture overview Spectral behaviour of leaky graphs in case of a strong coupling A point-interaction approximation: a method how to find leaky graph spectra numerically

5 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 3/9 Lecture overview Spectral behaviour of leaky graphs in case of a strong coupling A point-interaction approximation: a method how to find leaky graph spectra numerically Geometrically induced spectral bound states of leaky wires and graphs: bent edges

6 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 3/9 Lecture overview Spectral behaviour of leaky graphs in case of a strong coupling A point-interaction approximation: a method how to find leaky graph spectra numerically Geometrically induced spectral bound states of leaky wires and graphs: bent edges Leaky-graph resonances: a solvable model

7 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 4/9 Strong coupling for a compact Γ Let Γ have a single component, smooth and compact Theorem [EY01, 02; EK03, Ex04]: (i) Let Γ be a C 4 smooth manifold. In the limit ( 1) codimγ 1 α we have #σ disc (H α,γ ) = Γ α + O(ln α) 2π for dim Γ = 1, codim Γ = 1,

8 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 4/9 Strong coupling for a compact Γ Let Γ have a single component, smooth and compact Theorem [EY01, 02; EK03, Ex04]: (i) Let Γ be a C 4 smooth manifold. In the limit ( 1) codimγ 1 α we have #σ disc (H α,γ ) = Γ α + O(ln α) 2π for dim Γ = 1, codim Γ = 1, #σ disc (H α,γ (h)) = Γ α2 + O(ln α) 16π2 for dim Γ = 2, codim Γ = 1, and

9 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 4/9 Strong coupling for a compact Γ Let Γ have a single component, smooth and compact Theorem [EY01, 02; EK03, Ex04]: (i) Let Γ be a C 4 smooth manifold. In the limit ( 1) codimγ 1 α we have #σ disc (H α,γ ) = Γ α + O(ln α) 2π for dim Γ = 1, codim Γ = 1, #σ disc (H α,γ (h)) = Γ α2 + O(ln α) 16π2 for dim Γ = 2, codim Γ = 1, and #σ disc (H α,γ ) = Γ ( ǫ α) 1/2 π + O(e πα ) for dim Γ = 1, codim Γ = 2. Here Γ is the curve length or surface area, respectively, and ǫ α = 4 e 2( 2πα+ψ(1))

10 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 5/9 Strong coupling for a compact Γ Theorem, continued: (ii) In addition, suppose that Γ has no boundary. Then the j-th eigenvalue of H α,γ behaves as λ j (α) = α2 4 + µ j + O(α 1 ln α) for codim Γ = 1 and for codim Γ = 2, λ j (α) = ǫ α + µ j + O(e πα )

11 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 5/9 Strong coupling for a compact Γ Theorem, continued: (ii) In addition, suppose that Γ has no boundary. Then the j-th eigenvalue of H α,γ behaves as λ j (α) = α2 4 + µ j + O(α 1 ln α) for codim Γ = 1 and λ j (α) = ǫ α + µ j + O(e πα ) for codim Γ = 2, where µ j is the j-th eigenvalue of S Γ = d ds k(s)2 on L 2 ((0, Γ )) for dim Γ = 1, where k is curvature of Γ, and S Γ = Γ + K M 2 on L 2 (Γ, dγ) for dim Γ = 2, where Γ is Laplace-Beltrami operator on Γ and K, M, respectively, are the corresponding Gauss and mean curvatures

12 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 6/9 Proof technique Consider first the case. Take a closed curve Γ and call L = Γ. We start from a tubular neighborhood of Γ

13 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 6/9 Proof technique Consider first the case. Take a closed curve Γ and call L = Γ. We start from a tubular neighborhood of Γ Lemma: Φ a : [0,L) ( a,a) R 2 defined by (s,u) (γ 1 (s) uγ 2(s),γ 2 (s) + uγ 1(s)). is a diffeomorphism for all a > 0 small enough Γ Λ out a u s Λ in a D, N 2a D, N Σ a constant-width strip, do not take the LaTeX drawing too literary!

14 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 7/9 DN bracketing The idea is to apply to the operator H α,γ in question Dirichlet-Neumann bracketing at the boundary of Σ a := Φ([0,L) ( a,a)). This yields ( N Λ a ) L a,α H α,γ ( D Λ a ) L + a,α, where Λ a = Λ in a Λ out a is the exterior domain, and L ± a,α are self-adjoint operators associated with the forms q a,α[f] ± = f 2 L 2 (Σ a ) α f(x) 2 ds where f W 1,2 0 (Σ a ) and W 1,2 (Σ a ) for ±, respectively Γ

15 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 7/9 DN bracketing The idea is to apply to the operator H α,γ in question Dirichlet-Neumann bracketing at the boundary of Σ a := Φ([0,L) ( a,a)). This yields ( N Λ a ) L a,α H α,γ ( D Λ a ) L + a,α, where Λ a = Λ in a Λ out a is the exterior domain, and L ± a,α are self-adjoint operators associated with the forms q a,α[f] ± = f 2 L 2 (Σ a ) α f(x) 2 ds where f W 1,2 0 (Σ a ) and W 1,2 (Σ a ) for ±, respectively Important: The exterior part does not contribute to the negative spectrum, so we may consider L ± a,α only Γ

16 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 8/9 Transformed interior operator We use the curvilinear coordinates passing from L ± a,α to unitarily equivalent operators given by quadratic forms b + a,α[f] = + L 0 L 0 a a a a j=0 (1 + uk(s)) 2 f s V (s,u) f 2 ds du α 2 du ds + L 0 L a 0 a f u 2 f(s, 0) 2 ds du ds with f W 1,2 ((0,L) ( a,a)) satisfying periodic b.c. in the variable s and Dirichlet b.c. at u = ±a, and 1 b a,α[f] = b + 1 L k(s) a,α[f] 2 ( 1)j 1 + ( 1) j ak(s) f(s, ( 1)j a) 2 ds where V is the curvature induced potential, V (s,u) = k(s) 2 4(1+uk(s)) uk (s) 2(1+uk(s)) 3 5u2 k (s) 2 4(1+uk(s)) 4

17 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 9/9 Estimates with separated variables We pass to rougher bounds squeezing H α,γ between H ± a,α = U ± a T ± a,α

18 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 9/9 Estimates with separated variables We pass to rougher bounds squeezing H α,γ between H ± a,α = U ± a T ± a,α Here U ± a are s-a operators on L 2 (0,L) U a ± 2 d2 = (1 a k ) ds 2 + V ±(s) with PBC, where V (s) V (s,u) V + (s) with an O(a) error, and the transverse operators are associated with the forms and t + a,α[f] = a a f (u) 2 du α f(0) 2 t a,α[f] = t a,α[f] k ( f(a) 2 + f( a) 2 ) with f W 1,2 0 ( a,a) and W 1,2 ( a,a), respectively

19 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 10/9 Concluding the planar curve case Lemma: There are positive c, c N such that T ± α,a has for α large enough a single negative eigenvalue κ ± α,a satisfying α2 4 (1 + c N e αa/2) < κ α,a < α2 4 < κ+ α,a < α2 4 (1 8e αa/2)

20 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 10/9 Concluding the planar curve case Lemma: There are positive c, c N such that T ± α,a has for α large enough a single negative eigenvalue κ ± α,a satisfying α2 4 (1 + c N e αa/2) < κ α,a < α2 4 < κ+ α,a < α2 4 (1 8e αa/2) Finishing the proof: the eigenvalues of U a ± differ by O(a) from those of the comparison operator we choose a = 6α 1 ln α as the neighbourhood width putting the estimates together we get the eigenvalue asymptotics for a planar loop, i.e. the claim (ii) if Γ is not closed, the same can be done with the comparison operators S D,N Γ having appropriate b.c. at the endpoints of Γ. This yields the claim (i)

21 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 11/9 A curve in R 3 The argument is similar: Σ a Γ D, N

22 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 11/9 A curve in R 3 The argument is similar: Σ a Γ D, N The straightening transformation Φ a is defined by Φ a (s,r,θ) := γ(s) r[n(s) cos(θ β(s)) + b(s) sin(θ β(s))] To separate variables, we choose β so that β(s) equals the torsion τ(s) of Γ. The effective potential is then V = k2 4h 2 + h ss 2h 3 5h2 s 4h 4, where h := 1 + rk cos(θ β). It is important that the leading term is 1 4 k2 again, the torsion part being O(a)

23 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 12/9 A curve in R 3 The transverse estimate is replaced by Lemma: There are c 1, c 2 > 0 such that T ± α has for large enough negative α a single negative ev κ ± α,a which satisfies ǫ α S(α) < κ α,a < ξ α < κ + α,a < ξ α + S(α) as α, where S(α) = c 1 e 2πα exp( c 2 e πα ) The rest of the argument is the same as above

24 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 12/9 A curve in R 3 The transverse estimate is replaced by Lemma: There are c 1, c 2 > 0 such that T ± α has for large enough negative α a single negative ev κ ± α,a which satisfies ǫ α S(α) < κ α,a < ξ α < κ + α,a < ξ α + S(α) as α, where S(α) = c 1 e 2πα exp( c 2 e πα ) The rest of the argument is the same as above Remark: Notice that the result extends easily to Γ s consisting of a finite number of connected components (curves) which are C 4 and do not intersect. The same will be true for surfaces considered below

25 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 13/9 A surface in R 3 The argument modifies easily; Σ a is now a layer neighborhood. However, the intrinsic geometry of Γ can no longer be neglected

26 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 13/9 A surface in R 3 The argument modifies easily; Σ a is now a layer neighborhood. However, the intrinsic geometry of Γ can no longer be neglected Let Γ R 3 be a C 4 smooth compact Riemann surface of a finite genus g. The metric tensor given in the local coordinates by g µν = p,µ p,ν defines the invariant surface area element dγ := g 1/2 d 2 s, where g := det(g µν ). The Weingarten tensor is then obtained by raising the index in the second fundamental form, h µ ν := n,µ p,σ g σν ; the eigenvalues k ± of (h µ ν ) are the principal curvatures. They determine Gauss curvature K and mean curvature M by K = det(h µ ν ) = k + k, M = 1 2 Tr (h µ ν ) = 1 2 (k ++ k )

27 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 14/9 Proof sketch in the surface case The bracketing argument proceeds as before, N Λ a H α,γ H α,γ D Λ a H + α,γ, Λ a := R 3 \ Σ a, the interior only contributing to the negative spectrum

28 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 14/9 Proof sketch in the surface case The bracketing argument proceeds as before, N Λ a H α,γ H α,γ D Λ a H + α,γ, Λ a := R 3 \ Σ a, the interior only contributing to the negative spectrum Using the curvilinear coordinates: For small enough a we have the straightening diffeomorphism L a (x,u) = x + un(x), (x,u) N a := Γ ( a,a) Then we transform H ± α,γ by the unitary operator Ûψ = ψ L a : L 2 (Ω a ) L 2 (N a, dω) and estimate the operators Ĥ± α,γ := ÛH± α,γû 1 in L 2 (N a, dω)

29 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 15/9 Straightening transformation Denote the pull-back metric tensor by G ij, G ij = ( (G µν ) ), G µν = (δ σ µ uh µ σ )(δ ρ σ uh σ ρ )g ρν, so dσ := G 1/2 d 2 s du with G := det(g ij ) given by G = g [(1 uk + )(1 uk )] 2 = g(1 2Mu + Ku 2 ) 2

30 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 15/9 Straightening transformation Denote the pull-back metric tensor by G ij, G ij = ( (G µν ) ), G µν = (δ σ µ uh µ σ )(δ ρ σ uh σ ρ )g ρν, so dσ := G 1/2 d 2 s du with G := det(g ij ) given by G = g [(1 uk + )(1 uk )] 2 = g(1 2Mu + Ku 2 ) 2 Let (, ) G denote the inner product in L 2 (N a, dω). Then Ĥ± α,γ are associated with the forms η α,γ ± [Û 1 ψ] := ( i ψ,g ij j ψ) G α ψ(s, 0) 2 dγ, with the domains W 1,2 0 (N a, dω) and W 1,2 (N a, dω) for the ± sign, respectively Γ

31 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 16/9 Straightening continued Next we remove 1 2Mu + Ku 2 from the weight G 1/2 in the inner product of L 2 (N a, dω) by the unitary transformation U : L 2 (N a, dω) L 2 (N a, dγdu), Uψ := (1 2Mu + Ku 2 ) 1/2 ψ

32 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 16/9 Straightening continued Next we remove 1 2Mu + Ku 2 from the weight G 1/2 in the inner product of L 2 (N a, dω) by the unitary transformation U : L 2 (N a, dω) L 2 (N a, dγdu), Uψ := (1 2Mu + Ku 2 ) 1/2 ψ Denote the inner product in L 2 (N a, dγdu) by (, ) g. The operators B ± α,γ := UĤ± α,γ U 1 are associated with the forms b + α,γ [ψ] = ( µψ,g µν ν ψ) g + (ψ, (V 1 + V 2 )ψ) g + u ψ 2 g α ψ(s, 0) 2 dγ, Γ 1 b α,γ [ψ] = b+ α,γ [ψ] + ( 1) j M ( 1)j a(s) ψ(s, ( 1) j a) 2 dγ j=0 for ψ from W 2,1 0 (Ω a, dγdu) and W 2,1 (Ω a,dγdu), respectively Γ

33 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 17/9 Effective potential Here M u := (M Ku)(1 2Mu + Ku 2 ) 1 is the mean curvature of the parallel surface to Γ and V 1 = g 1/2 (g 1/2 G µν J,ν ),µ +J,µ G µν J,ν, V 2 = with J := 1 2 ln(1 2Mu + Ku2 ) K M 2 (1 2Mu + Ku 2 ) 2

34 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 17/9 Effective potential Here M u := (M Ku)(1 2Mu + Ku 2 ) 1 is the mean curvature of the parallel surface to Γ and V 1 = g 1/2 (g 1/2 G µν J,ν ),µ +J,µ G µν J,ν, V 2 = with J := 1 2 ln(1 2Mu + Ku2 ) K M 2 (1 2Mu + Ku 2 ) 2 A rougher estimate with separated variables: squeeze 1 2Mu + Ku 2 between C ± (a) := (1 ± a 1 ) 2, where := max({ k +, k }) 1. Consequently, the matrix inequality C (a)g µν G µν C + (a)g µν is valid

35 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 17/9 Effective potential Here M u := (M Ku)(1 2Mu + Ku 2 ) 1 is the mean curvature of the parallel surface to Γ and V 1 = g 1/2 (g 1/2 G µν J,ν ),µ +J,µ G µν J,ν, V 2 = with J := 1 2 ln(1 2Mu + Ku2 ) K M 2 (1 2Mu + Ku 2 ) 2 A rougher estimate with separated variables: squeeze 1 2Mu + Ku 2 between C ± (a) := (1 ± a 1 ) 2, where := max({ k +, k }) 1. Consequently, the matrix inequality C (a)g µν G µν C + (a)g µν is valid V 1 behaves as O(a) for a 0, while V 2 can be squeezed between the functions C± 2 (a)(k M2 ), both uniformly in the surface variables

36 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 18/9 Concluding the estimate Hence we estimate B ± α,γ by B ± α,a := S ± a I + I T ± α,a with S ± a := C ± (a) Γ + C 2 ± (a)(k M2 ) ± va in the space L 2 (Γ, dγ) L 2 ( a,a) for a v > 0, where T ± α,a are the same as in the case (the same lemma applies)

37 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 18/9 Concluding the estimate Hence we estimate B ± α,γ by B ± α,a := S ± a I + I T ± α,a with S ± a := C ± (a) Γ + C 2 ± (a)(k M2 ) ± va in the space L 2 (Γ, dγ) L 2 ( a,a) for a v > 0, where T ± α,a are the same as in the case (the same lemma applies) As above the eigenvalues of the operators S ± a coincide up to an O(a) error with those of S Γ, and therefore choosing a := 6α 1 ln α, we find λ j (α) = 1 4 α2 + µ j + O(α 1 ln α) as a 0 which is equivalent to the claim (i)

38 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 18/9 Concluding the estimate Hence we estimate B ± α,γ by B ± α,a := S ± a I + I T ± α,a with S ± a := C ± (a) Γ + C 2 ± (a)(k M2 ) ± va in the space L 2 (Γ, dγ) L 2 ( a,a) for a v > 0, where T ± α,a are the same as in the case (the same lemma applies) As above the eigenvalues of the operators S ± a coincide up to an O(a) error with those of S Γ, and therefore choosing a := 6α 1 ln α, we find λ j (α) = 1 4 α2 + µ j + O(α 1 ln α) as a 0 which is equivalent to the claim (i) To get (ii) we employ Weyl asymptotics for S Γ. Extension to Γ s having a finite # of connected components is easy

39 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 19/9 Infinite manifolds Bound states may exist also if Γ is noncompact. The comparison operator S Γ has an attractive potential, so σ disc (H α,γ ) can be expected in the strong coupling regime, even if a direct proof is missing as for surfaces

40 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 19/9 Infinite manifolds Bound states may exist also if Γ is noncompact. The comparison operator S Γ has an attractive potential, so σ disc (H α,γ ) can be expected in the strong coupling regime, even if a direct proof is missing as for surfaces It is needed that σ ess does not feel curvature, not only for H α,γ but for the estimating operators as well. Sufficient conditions: k(s),k (s) and k (s) 1/2 are O( s 1 ε ) as s for a planar curve in addition, the torsion bounded for a curve in R 3 a surface Γ admits a global normal parametrization with a uniformly elliptic metric, K,M 0 as the geodesic radius r

41 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 20/9 Infinite manifolds We must also assume that there is a tubular neighborhood Σ a without self-intersections for small a, i.e. to avoid

42 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 20/9 Infinite manifolds We must also assume that there is a tubular neighborhood Σ a without self-intersections for small a, i.e. to avoid Theorem [EY02; EK03, Ex04]: With the above listed assumptions, the asymptotic expansions (ii) for the eigenvalues derived in the compact case hold again

43 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 21/9 Periodic manifolds One uses Floquet expansion. It is important to choose the periodic cells C of the space and Γ C of the manifold consistently, Γ C = Γ C; we assume that Γ C is connected C Γ C e iθ

44 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 21/9 Periodic manifolds One uses Floquet expansion. It is important to choose the periodic cells C of the space and Γ C of the manifold consistently, Γ C = Γ C; we assume that Γ C is connected C Γ C Lemma: unitary U : L 2 (R 3 ) [0,2π) L 2 (C) dθ s.t. r UH α,γ U 1 = e iθ H α,θ dθ and σ(h α,γ )= σ(h α,θ ) [0,2π) r [0,2π) r

45 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 22/9 Comparison operators The fibre comparison operators are S θ = d ds k(s)2 on L 2 (Γ C ) parameterized by arc length for dim Γ = 1, with Floquet b.c., and S θ = g 1/2 ( i µ + θ µ )g 1/2 g µν ( i ν + θ ν ) + K M 2 with periodic b.c. for dim Γ = 2, where θ µ, µ = 1,...,r, are quasimomentum components; recall that r = 1, 2, 3 depending on the manifold type

46 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 23/9 Periodic manifold asymptotics Theorem [EY01; EK03, Ex04]: Let Γ be a C 4 -smooth r-periodic manifold without boundary. The strong coupling asymptotic behavior of the j-th Floquet eigenvalue is λ j (α,θ) = 1 4 α2 + µ j (θ) + O(α 1 ln α) as α for codim Γ = 1 and λ j (α,θ) = ǫ α + µ j (θ) + O(e πα ) as α for codim Γ = 2. The error terms are uniform w.r.t. θ

47 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 23/9 Periodic manifold asymptotics Theorem [EY01; EK03, Ex04]: Let Γ be a C 4 -smooth r-periodic manifold without boundary. The strong coupling asymptotic behavior of the j-th Floquet eigenvalue is λ j (α,θ) = 1 4 α2 + µ j (θ) + O(α 1 ln α) as α for codim Γ = 1 and λ j (α,θ) = ǫ α + µ j (θ) + O(e πα ) as α for codim Γ = 2. The error terms are uniform w.r.t. θ Corollary: If dim Γ = 1 and coupling is strong enough, H α,γ has open spectral gaps

48 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 24/9 Large gaps in the disconnected case If Γ is not connected and each connected component is contained in a translate of Γ C, the comparison operator is independent of θ and asymptotic formula reads λ j (α,θ) = 1 4 α2 + µ j + O(α 1 ln α) as α for codim Γ = 1 and similarly for for codim Γ = 2

49 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 24/9 Large gaps in the disconnected case If Γ is not connected and each connected component is contained in a translate of Γ C, the comparison operator is independent of θ and asymptotic formula reads λ j (α,θ) = 1 4 α2 + µ j + O(α 1 ln α) as α for codim Γ = 1 and similarly for for codim Γ = 2 Moreover, the assumptions can be weakened

50 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 25/9 Soft graphs with magnetic field Add a homogeneous magnetic field with the vector potential A = 1 2 B( x 2,x 1 ). We ask about existence of persistent currents, i.e. nonzero probability flux along a closed loop λ n (ϕ) = 1 ϕ c I n, where λ n (ϕ) is the n-th eigenvalue of the Hamiltonian H α,γ (B) := ( i A) 2 αδ(x Γ) and ϕ is the magnetic flux through the loop (in standard units its quantum equals 2π c e 1 )

51 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 25/9 Soft graphs with magnetic field Add a homogeneous magnetic field with the vector potential A = 1 2 B( x 2,x 1 ). We ask about existence of persistent currents, i.e. nonzero probability flux along a closed loop λ n (ϕ) = 1 ϕ c I n, where λ n (ϕ) is the n-th eigenvalue of the Hamiltonian H α,γ (B) := ( i A) 2 αδ(x Γ) and ϕ is the magnetic flux through the loop (in standard units its quantum equals 2π c e 1 ) B Γ

52 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 26/9 Persistent currents The same technique, different comparison operator, namely S Γ (B) = d ds k(s)2 on L 2 (0,L) with ψ(l ) = e ib Ω ψ(0+), ψ (L ) = e ib Ω ψ (0+), where Ω is the area encircled by Γ

53 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 26/9 Persistent currents The same technique, different comparison operator, namely S Γ (B) = d ds k(s)2 on L 2 (0,L) with ψ(l ) = e ib Ω ψ(0+), ψ (L ) = e ib Ω ψ (0+), where Ω is the area encircled by Γ Theorem [E.-Yoshitomi 03]: Let Γ be a C 4 -smooth. The for large α the operator H α,γ (B) has a non-empty discrete spectrum and the j-th eigenvalue behaves as λ j (α,b) = 1 4 α2 + µ j (B) + O(α 1 ln α), where µ j (B) is the j-th eigenvalue of S Γ (B) and the error term is uniform in B. In particular, for a fixed j and α large enough the function λ j (α, ) cannot be constant

54 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 26/9 Persistent currents The same technique, different comparison operator, namely S Γ (B) = d ds k(s)2 on L 2 (0,L) with ψ(l ) = e ib Ω ψ(0+), ψ (L ) = e ib Ω ψ (0+), where Ω is the area encircled by Γ Theorem [E.-Yoshitomi 03]: Let Γ be a C 4 -smooth. The for large α the operator H α,γ (B) has a non-empty discrete spectrum and the j-th eigenvalue behaves as λ j (α,b) = 1 4 α2 + µ j (B) + O(α 1 ln α), where µ j (B) is the j-th eigenvalue of S Γ (B) and the error term is uniform in B. In particular, for a fixed j and α large enough the function λ j (α, ) cannot be constant Remark: [Honnouvo-Hounkonnou 04] proved the same for AB flux

55 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 27/9 Absolute continuity One is also interested in the nature of the spectrum of H α,γ with a periodic Γ. By [Birman-Suslina-Shterenberg 00,01] the spectrum is absolutely continuous if codim Γ = 1 and the period cell is compact. This tells us nothing, e.g., about a single periodic curve in R d, d = 2, 3.

56 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 27/9 Absolute continuity One is also interested in the nature of the spectrum of H α,γ with a periodic Γ. By [Birman-Suslina-Shterenberg 00,01] the spectrum is absolutely continuous if codim Γ = 1 and the period cell is compact. This tells us nothing, e.g., about a single periodic curve in R d, d = 2, 3. The assumption about connectedness of Γ C can be always satisfied if d = 2 but not for d = 3; recall the crochet curve

57 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 28/9 Absolute continuity Theorem [Bentosela-Duclos-E 03]: To any E > 0 there is an α E > 0 such that the spectrum of H α,γ is absolutely continuous in (,ξ(α) + E) as long as ( 1) d α > α E, where ξ(α) = 1 4 α2 and ǫ α for d = 2, 3, respectively

58 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 28/9 Absolute continuity Theorem [Bentosela-Duclos-E 03]: To any E > 0 there is an α E > 0 such that the spectrum of H α,γ is absolutely continuous in (,ξ(α) + E) as long as ( 1) d α > α E, where ξ(α) = 1 4 α2 and ǫ α for d = 2, 3, respectively Proof: The fiber operators H α,γ (θ) form a type A analytic family. In a finite interval each of them has a finite number of ev s, so one has just to check non-constancy of the functions λ j (α, ) as in the case of persistent currents

59 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 29/9 How can one find the spectrum? The above general results do not tell us how to find the spectrum for a particular Γ. There are various possibilities: Direct solution of the PDE problem H α,γ ψ = λψ is feasible in a few simple examples only

60 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 29/9 How can one find the spectrum? The above general results do not tell us how to find the spectrum for a particular Γ. There are various possibilities: Direct solution of the PDE problem H α,γ ψ = λψ is feasible in a few simple examples only Using trace maps of R k ( k 2 ) 1 and the generalized BS principle R k := R k 0 + αr k dx,m [I αrk m,m] 1 R k m,dx, where m is δ measure on Γ, we pass to a 1D integral operator problem, αr k m,m ψ = ψ

61 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 29/9 How can one find the spectrum? The above general results do not tell us how to find the spectrum for a particular Γ. There are various possibilities: Direct solution of the PDE problem H α,γ ψ = λψ is feasible in a few simple examples only Using trace maps of R k ( k 2 ) 1 and the generalized BS principle R k := R k 0 + αr k dx,m [I αrk m,m] 1 R k m,dx, where m is δ measure on Γ, we pass to a 1D integral operator problem, αr k m,m ψ = ψ discretization of the latter which amounts to a point-interaction approximations to H α,γ

62 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 30/9 2D point interactions Such an interaction at the point a with the coupling constant α is defined by b.c. which change locally the domain of : the functions behave as ψ(x) = 1 2π log x a L 0(ψ,a) + L 1 (ψ,a) + O( x a ), where the generalized b.v. L 0 (ψ,a) and L 1 (ψ,a) satisfy L 1 (ψ,a) + 2παL 0 (ψ,a) = 0, α R

63 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 30/9 2D point interactions Such an interaction at the point a with the coupling constant α is defined by b.c. which change locally the domain of : the functions behave as ψ(x) = 1 2π log x a L 0(ψ,a) + L 1 (ψ,a) + O( x a ), where the generalized b.v. L 0 (ψ,a) and L 1 (ψ,a) satisfy L 1 (ψ,a) + 2παL 0 (ψ,a) = 0, α R For our purpose, the coupling should depend on the set Y approximating Γ. To see how compare a line Γ with the solvable straight-polymer model [AGHH] α n l/n

64 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 31/9 2D point-interaction approximation Spectral threshold convergence requires α n = αn which means that individual point interactions get weaker. Hence we approximate H α,γ by point-interaction Hamiltonians H αn,y n with α n = α Y n, where Y n := Y n.

65 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 31/9 2D point-interaction approximation Spectral threshold convergence requires α n = αn which means that individual point interactions get weaker. Hence we approximate H α,γ by point-interaction Hamiltonians H αn,y n with α n = α Y n, where Y n := Y n. Theorem [E.-Němcová, 2003]: Let a family {Y n } of finite sets Y n Γ R 2 be such that 1 f(y) f dm Y n y Y Γ n holds for any bounded continuous function f : Γ C, together with technical conditions, then H αn,y n H α,γ in the strong resolvent sense as n.

66 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 32/9 Comments on the approximation A more general result is valid: Γ need not be a graph and the coupling may be non-constant; also a magnetic field can be added [Ožanová 06] (=Němcová)

67 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 32/9 Comments on the approximation A more general result is valid: Γ need not be a graph and the coupling may be non-constant; also a magnetic field can be added [Ožanová 06] (=Němcová) The result applies to finite graphs, however, an infinite Γ can be approximated in strong resolvent sense by a family of cut-off graphs

68 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 32/9 Comments on the approximation A more general result is valid: Γ need not be a graph and the coupling may be non-constant; also a magnetic field can be added [Ožanová 06] (=Němcová) The result applies to finite graphs, however, an infinite Γ can be approximated in strong resolvent sense by a family of cut-off graphs The idea is due to [Brasche-Figari-Teta 98], who analyzed point-interaction approximations of measure perturbations with codim Γ = 1 in R 3. There are differences, however, for instance in the 2D case we can approximate attractive interactions only

69 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 32/9 Comments on the approximation A more general result is valid: Γ need not be a graph and the coupling may be non-constant; also a magnetic field can be added [Ožanová 06] (=Němcová) The result applies to finite graphs, however, an infinite Γ can be approximated in strong resolvent sense by a family of cut-off graphs The idea is due to [Brasche-Figari-Teta 98], who analyzed point-interaction approximations of measure perturbations with codim Γ = 1 in R 3. There are differences, however, for instance in the 2D case we can approximate attractive interactions only A uniform resolvent convergence can be achieved in this scheme if the term ε 2 2 is added to the Hamiltonian [Brasche-Ožanová 07]

70 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 33/9 Scheme of the proof Resolvent of H αn,y n is given Krein s formula. Given k 2 ρ(h αn,y n ) define Y n Y n matrix by Λ αn,y n (k 2 ;x,y) = 1 2π [ ( ) ] ik 2π Y n α + ln + γ E δ xy 2 G k (x y) (1 δ xy ) for x,y Y n, where γ E is Euler constant.

71 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 33/9 Scheme of the proof Resolvent of H αn,y n is given Krein s formula. Given k 2 ρ(h αn,y n ) define Y n Y n matrix by Λ αn,y n (k 2 ;x,y) = 1 2π [ ( ) ] ik 2π Y n α + ln + γ E δ xy 2 G k (x y) (1 δ xy ) for x,y Y n, where γ E is Euler constant. Then (H αn,y n k 2 ) 1 (x,y) = G k (x y) + [ Λαn,Y n (k 2 ) ] 1 (x,y )G k (x x )G k (y y ) x,y Y n

72 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 34/9 Scheme of the proof Resolvent of H α,γ is given by the generalized BS formula given above; one has to check directly that the difference of the two vanishes as n

73 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 34/9 Scheme of the proof Resolvent of H α,γ is given by the generalized BS formula given above; one has to check directly that the difference of the two vanishes as n Remarks: Spectral condition in the n-th approximation, i.e. det Λ αn,y n (k 2 ) = 0, is a discretization of the integral equation coming from the generalized BS principle A solution to Λ αn,y n (k 2 )η = 0 determines the approximating ef by ψ(x) = y j Y n η j G k (x y j ) A match with solvable models illustrates the convergence and shows that it is not fast, slower than n 1 in the eigenvalues. This comes from singular spikes in the approximating functions

74 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 35/9 An interlude: scattering on leaky graphs Let Γ be a graph with semi-infinite leads, e.g. an infinite asymptotically straight curve. What we know about scattering in such systems? Not much. First question: What is the free operator? is not a good candidate, rather H α,γ for a straight line Γ. Recall that we are particularly interested in energy interval ( 1 4 α2, 0), i.e. 1D transport of states laterally bound to Γ

75 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 35/9 An interlude: scattering on leaky graphs Let Γ be a graph with semi-infinite leads, e.g. an infinite asymptotically straight curve. What we know about scattering in such systems? Not much. First question: What is the free operator? is not a good candidate, rather H α,γ for a straight line Γ. Recall that we are particularly interested in energy interval ( 1 4 α2, 0), i.e. 1D transport of states laterally bound to Γ Existence proof for the wave operators is known only for locally deformed line [E.-Kondej 05]

76 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 35/9 An interlude: scattering on leaky graphs Let Γ be a graph with semi-infinite leads, e.g. an infinite asymptotically straight curve. What we know about scattering in such systems? Not much. First question: What is the free operator? is not a good candidate, rather H α,γ for a straight line Γ. Recall that we are particularly interested in energy interval ( 1 4 α2, 0), i.e. 1D transport of states laterally bound to Γ Existence proof for the wave operators is known only for locally deformed line [E.-Kondej 05] Conjecture: For strong coupling, α, the scattering is described in leading order by S Γ := ds d γ(s)2

77 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 35/9 An interlude: scattering on leaky graphs Let Γ be a graph with semi-infinite leads, e.g. an infinite asymptotically straight curve. What we know about scattering in such systems? Not much. First question: What is the free operator? is not a good candidate, rather H α,γ for a straight line Γ. Recall that we are particularly interested in energy interval ( 1 4 α2, 0), i.e. 1D transport of states laterally bound to Γ Existence proof for the wave operators is known only for locally deformed line [E.-Kondej 05] Conjecture: For strong coupling, α, the scattering is described in leading order by S Γ := d2 ds γ(s)2 On the other hand, in general, the global geometry of Γ is expected to determine the S-matrix

78 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 36/9 Something more on resonances Consider infinite curves Γ, straight outside a compact, and ask for examples of resonances. Recall the L 2 -approach: in 1D potential scattering one explores spectral properties of the problem cut to a finite length L. It is time-honored trick that scattering resonances are manifested as avoided crossings in L dependence of the spectrum for a recent proof see [Hagedorn-Meller 00]. Try the same here:

79 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 36/9 Something more on resonances Consider infinite curves Γ, straight outside a compact, and ask for examples of resonances. Recall the L 2 -approach: in 1D potential scattering one explores spectral properties of the problem cut to a finite length L. It is time-honored trick that scattering resonances are manifested as avoided crossings in L dependence of the spectrum for a recent proof see [Hagedorn-Meller 00]. Try the same here: Broken line: absence of intrinsic resonances due lack of higher transverse thresholds Z-shaped Γ: if a single bend has a significant reflection, a double band should exhibit resonances Bottleneck curve: a good candidate to demonstrate tunneling resonances

80 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 37/9 Broken line α = 1

81 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 37/9 Broken line α = E L

82 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 38/9 Z shape with θ = π 2 α = 5 L c = 10

83 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 38/9 Z shape with θ = π 2 α = 5 L c = E L

84 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 39/9 Z shape with θ = 0.32π L c = 10 α = 5

85 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 39/9 Z shape with θ = 0.32π L c = 10 α = E L

86 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 40/9 A bottleneck curve Consider a straight line deformation which shaped as an open loop with a bottleneck the width a of which we will vary

87 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 40/9 A bottleneck curve Consider a straight line deformation which shaped as an open loop with a bottleneck the width a of which we will vary a L L If Γ is a straight line, the transverse eigenfunction is e α y /2, hence the distance at which tunneling becomes significant is 4α 1. In the example, we choose α = 1

88 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 41/9 Bottleneck with a = E L

89 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 42/9 Bottleneck with a = E L

90 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 43/9 Bottleneck with a = E L

91 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 44/9 A caricature but solvable model Let us pass to a simple model in which existence of resonances can be proved: a straight leaky wire and a family of leaky dots.

92 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 44/9 A caricature but solvable model Let us pass to a simple model in which existence of resonances can be proved: a straight leaky wire and a family of leaky dots. Formal Hamiltonian αδ(x Σ) + n i=1 β i δ(x y (i) ) in L 2 (R 2 ) with α > 0. The 2D point interactions at Π = {y (i) } with couplings β = {β 1,...,β n } are properly introduced through b.c. mentioned above, giving Hamiltonian H α,β

93 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 44/9 A caricature but solvable model Let us pass to a simple model in which existence of resonances can be proved: a straight leaky wire and a family of leaky dots. Formal Hamiltonian αδ(x Σ) + n i=1 β i δ(x y (i) ) in L 2 (R 2 ) with α > 0. The 2D point interactions at Π = {y (i) } with couplings β = {β 1,...,β n } are properly introduced through b.c. mentioned above, giving Hamiltonian H α,β Resolvent by Krein-type formula: given z C \ [0, ) we start from the free resolvent R(z) := ( z) 1, also interpreted as unitary R(z) acting from L 2 to W 2,2. Then

94 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 45/9 Resolvent by Krein-type formula we introduce auxiliary Hilbert spaces, H 0 := L 2 (R) and H 1 := C n, and trace maps τ j : W 2,2 (R 2 ) H j defined by τ 0 f := f Σ and τ 1 f := f Π,

95 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 45/9 Resolvent by Krein-type formula we introduce auxiliary Hilbert spaces, H 0 := L 2 (R) and H 1 := C n, and trace maps τ j : W 2,2 (R 2 ) H j defined by τ 0 f := f Σ and τ 1 f := f Π, then we define canonical embeddings of R(z) to H i by R i,l (z) := τ i R(z) : L 2 H i, R L,i (z) := [R i,l (z)], and R j,i (z) := τ j R L,i (z) : H i H j, and

96 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 45/9 Resolvent by Krein-type formula we introduce auxiliary Hilbert spaces, H 0 := L 2 (R) and H 1 := C n, and trace maps τ j : W 2,2 (R 2 ) H j defined by τ 0 f := f Σ and τ 1 f := f Π, then we define canonical embeddings of R(z) to H i by R i,l (z) := τ i R(z) : L 2 H i, R L,i (z) := [R i,l (z)], and R j,i (z) := τ j R L,i (z) : H i H j, and operator-valued matrix Γ(z) : H 0 H 1 H 0 H 1 by Γ ij (z)g := R i,j (z)g for i j and g H j, Γ 00 (z)f := [ α 1 R 0,0 (z) ] f if f H 0, ( ) Γ 11 (z)ϕ := s β (z)δ kl G z (y (k),y (l) )(1 δ kl ) ϕ, with s β (z) := β + s(z) := β + 1 2π (ln z 2i ψ(1))

97 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 46/9 Resolvent by Krein-type formula To invert it we define the reduced determinant D(z) := Γ 11 (z) Γ 10 (z)γ 00 (z) 1 Γ 01 (z) : H 1 H 1,

98 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 46/9 Resolvent by Krein-type formula To invert it we define the reduced determinant D(z) := Γ 11 (z) Γ 10 (z)γ 00 (z) 1 Γ 01 (z) : H 1 H 1, then an easy algebra yields expressions for blocks of [Γ(z)] 1 in the form [Γ(z)] 1 11 = D(z) 1, [Γ(z)] 1 00 = Γ 00 (z) 1 + Γ 00 (z) 1 Γ 01 (z)d(z) 1 Γ 10 (z)γ 00 (z) 1, [Γ(z)] 1 01 = Γ 00 (z) 1 Γ 01 (z)d(z) 1, [Γ(z)] 1 10 = D(z) 1 Γ 10 (z)γ 00 (z) 1 ; thus to determine singularities of [Γ(z)] 1 one has to find the null space of D(z)

99 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 47/9 Resolvent by Krein-type formula With this notation we can state the sought formula: Theorem [E-Kondej 04]: For z ρ(h α,β ) with Im z > 0 the resolvent R α,β (z) := (H α,β z) 1 equals R α,β (z) = R(z) + 1 i,j=0 R L,i (z)[γ(z)] 1 ij R j,l(z)

100 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 47/9 Resolvent by Krein-type formula With this notation we can state the sought formula: Theorem [E-Kondej 04]: For z ρ(h α,β ) with Im z > 0 the resolvent R α,β (z) := (H α,β z) 1 equals R α,β (z) = R(z) + 1 i,j=0 R L,i (z)[γ(z)] 1 ij R j,l(z) Remark: One can also compare resolvent of H α,β to that of H α H α,σ using trace maps of the latter, R α,β (z) = R α (z) + R α;l1 (z)d(z) 1 R α;1l (z)

101 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 48/9 Spectral properties of H α,β It is easy to check that σ ess (H α,β ) = σ ac (H α,β ) = [ 1 4 α2, )

102 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 48/9 Spectral properties of H α,β It is easy to check that σ ess (H α,β ) = σ ac (H α,β ) = [ 1 4 α2, ) σ disc given by generalized Birman-Schwinger principle: dim ker Γ(z) = dim ker R α,β (z), 1 H α,β ϕ z = zϕ z ϕ z = R L,i (z)η i,z, i=0 where (η 0,z,η 1,z ) ker Γ(z). Moreover, it is clear that 0 σ disc (Γ(z)) 0 σ disc (D(z)); this reduces the task of finding the spectrum to an algebraic problem

103 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 49/9 Spectral properties of H α,β Theorem [E-Kondej 04]: (a) Let n = 1 and denote dist (σ, Π) =: a, then H α,β has one isolated eigenvalue κ 2 a. The function a κ 2 a is increasing in (0, ), lim a ( κ2 a {ǫ ) = min β, 14 } α2, where ǫ β := 4e 2( 2πβ+ψ(1)), while lim a 0 ( κ 2 a) is finite.

104 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 49/9 Spectral properties of H α,β Theorem [E-Kondej 04]: (a) Let n = 1 and denote dist (σ, Π) =: a, then H α,β has one isolated eigenvalue κ 2 a. The function a κ 2 a is increasing in (0, ), lim a ( κ2 a {ǫ ) = min β, 14 } α2, where ǫ β := 4e 2( 2πβ+ψ(1)), while lim a 0 ( κ 2 a) is finite. (b) For any α > 0, β R n, and n N + the operator H α,β has N isolated eigenvalues, 1 N n. If all the point interactions are strong enough, we have N = n

105 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 49/9 Spectral properties of H α,β Theorem [E-Kondej 04]: (a) Let n = 1 and denote dist (σ, Π) =: a, then H α,β has one isolated eigenvalue κ 2 a. The function a κ 2 a is increasing in (0, ), lim a ( κ2 a {ǫ ) = min β, 14 } α2, where ǫ β := 4e 2( 2πβ+ψ(1)), while lim a 0 ( κ 2 a) is finite. (b) For any α > 0, β R n, and n N + the operator H α,β has N isolated eigenvalues, 1 N n. If all the point interactions are strong enough, we have N = n Remark: Embedded eigenvalues due to mirror symmetry w.r.t. Σ possible if n 2

106 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 50/9 Resonance for n = 1 Assume the point interaction eigenvalue becomes embedded as a, i.e. that ǫ β > 1 4 α2

107 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 50/9 Resonance for n = 1 Assume the point interaction eigenvalue becomes embedded as a, i.e. that ǫ β > 1 4 α2 Observation: Birman-Schwinger works in the complex domain too; it is enough to look for analytical continuation of D( ), which acts for z C \ [ 1 4 α2, ) as a multiplication by d a (z) := s β (z) ϕ a (z) = s β (z) 0 µ(z, t) t z 1 dt, 4α2 µ(z,t) := iα 16π (α 2i(z t) 1/2 ) e 2ia(z t)1/2 t 1/2 (z t) 1/2 Thus we have a situation reminiscent of Friedrichs model, just the functions involved are more complicated

108 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 51/9 Analytic continuation Take a region Ω of the other sheet with ( 1 4 α2, 0) as a part of its boundary. Put µ 0 (λ,t) := lim ε 0 µ(λ+iε,t), define I(λ) := P 0 µ 0 (λ,t) t λ 1 dt, 4α2 and furthermore, g α,a (z) := iα 4 e αa (z+ 1 4 α2 ) 1/2.

109 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 51/9 Analytic continuation Take a region Ω of the other sheet with ( 1 4 α2, 0) as a part of its boundary. Put µ 0 (λ,t) := lim ε 0 µ(λ+iε,t), define I(λ) := P 0 µ 0 (λ,t) t λ 1 dt, 4α2 e αa (z+ 1 4 α2 ) 1/2. and furthermore, g α,a (z) := iα 4 Lemma: z ϕ a (z) is continued analytically to Ω as ( ϕ 0 a(λ) = I(λ) + g α,a (λ) for λ 1 ) 4 α2, 0, ϕ a (z) = 0 µ(z, t) t z 1 4 α2 dt 2g α,a(z), z Ω

110 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 52/9 Analytic continuation Proof: By a direct computation one checks lim ε 0 ϕ± + a (λ ± iε) = ϕ 0 a(λ), 1 4 α2 < λ < 0, so the claim follows from edge-of-the-wedge theorem.

111 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 52/9 Analytic continuation Proof: By a direct computation one checks lim ε 0 ϕ± + a (λ ± iε) = ϕ 0 a(λ), 1 4 α2 < λ < 0, so the claim follows from edge-of-the-wedge theorem. The continuation of d a is thus the function η a : M C, where M = {z : Im z > 0} ( 1 4 α2, 0) Ω, acting as η a (z) = s β (z) ϕ l(z) a (z), and our problem reduces to solution if the implicit function problem η a (z) = 0.

112 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 53/9 Resonance for n = 1 Theorem [E-Kondej 04]: Assume ǫ β > 1 4 α2. For any a large enough the equation η a (z) = 0 has a unique solution z(a) = µ(b) + iν(b) Ω, i.e. ν(a) < 0, with the following asymptotic behaviour as a, µ(a) = ǫ β + O(e a ǫ β ), ν(a) = O(e a ǫ β )

113 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 53/9 Resonance for n = 1 Theorem [E-Kondej 04]: Assume ǫ β > 1 4 α2. For any a large enough the equation η a (z) = 0 has a unique solution z(a) = µ(b) + iν(b) Ω, i.e. ν(a) < 0, with the following asymptotic behaviour as a, µ(a) = ǫ β + O(e a ǫ β ), ν(a) = O(e a ǫ β ) Remark: We have ϕ a (z) 0 uniformly in a and s β (z) as Im z. Hence the imaginary part z(a) is bounded as a function of a, in particular, the resonance pole survives as a 0.

114 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 54/9 Scattering for n = 1 The same as scattering problem for (H α,β,h α ) α β a

115 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 54/9 Scattering for n = 1 The same as scattering problem for (H α,β,h α ) α β a Existence and completeness by Birman-Kuroda theorem; we seek on-shell S-matrix in ( 1 4 α2, 0). By Krein formula, resolvent for Im z > 0 expresses as where v z := R α;l,1 (z) R α,β (z) = R α (z) + η a (z) 1 (,v z )v z,

116 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 55/9 Scattering for n = 1 Apply this operator to vector ω λ,ε (x) := e i(λ+α2 /4) 1/2 x 1 ε 2 x 2 1 e α x 2 /2 and take limit ε 0+ in the sense of distributions; then a straightforward calculation give generalized eigenfunction of H α,β. In particular, we have

117 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 55/9 Scattering for n = 1 Apply this operator to vector ω λ,ε (x) := e i(λ+α2 /4) 1/2 x 1 ε 2 x 2 1 e α x 2 /2 and take limit ε 0+ in the sense of distributions; then a straightforward calculation give generalized eigenfunction of H α,β. In particular, we have Proposition: For any λ ( 1 4 α2, 0) the reflection and transmission amplitudes are R(λ) = T (λ) 1 = i 4 αη a(λ) 1 e αa (λ α2 ) 1/2 ; they have the same pole in the analytical continuation to Ω as the continued resolvent

118 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 56/9 Resonances from perturbed symmetry Take the simplest situation, n = 2 α β 0 a a β 0 + b

119 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 56/9 Resonances from perturbed symmetry Take the simplest situation, n = 2 α β 0 a a β 0 + b Let σ disc (H 0,β0 ) ( 1 4 α2, 0 ), so that Hamiltonian H 0,β0 has two eigenvalues, the larger of which, ǫ 2, exceeds 1 4 α2. Then H α,β0 has the same eigenvalue ǫ 2 embedded in the negative part of continuous spectrum

120 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 56/9 Resonances from perturbed symmetry Take the simplest situation, n = 2 α β 0 a a β 0 + b Let σ disc (H 0,β0 ) ( 1 4 α2, 0 ), so that Hamiltonian H 0,β0 has two eigenvalues, the larger of which, ǫ 2, exceeds 1 4 α2. Then H α,β0 has the same eigenvalue ǫ 2 embedded in the negative part of continuous spectrum One has now to continue analytically the 2 2 matrix function D( ). Put κ 2 := ǫ 2 and s β (κ) := s β ( κ 2 )

121 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 57/9 Resonances from perturbed symmetry Proposition: Assume ǫ 2 ( 1 4 α2, 0) and denote g(λ) := ig α,a (λ). Then for all b small enough the continued function has a unique zero z 2 (b) = µ 2 (b) + iν 2 (b) Ω with the asymptotic expansion µ 2 (b) = ǫ 2 + ν 2 (b) = κ 2 b s β (κ 2)+K 0 (2aκ 2) + O(b2 ), κ 2 g(ǫ 2 )b 2 2( s β (κ 2)+K 0 (2aκ 2)) s β (κ 2) ϕ 0 a(ǫ 2 ) + O(b3 )

122 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 58/9 Unstable state decay, n = 1 Complementary point of view: investigate decay of unstable state associated with the resonance; assume again n = 1. We found that if the unperturbed ev ǫ β of H β is embedded in ( 1 4 α2, 0) and a is large, the corresponding resonance has a long halflife. In analogy with Friedrichs model [Demuth, 1976] one conjectures that in weak coupling case, the resonance state would be similar up to normalization to the eigenvector ξ 0 := K 0 ( ǫ β ) of H β, with the decay law being dominated by the exponential term

123 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 58/9 Unstable state decay, n = 1 Complementary point of view: investigate decay of unstable state associated with the resonance; assume again n = 1. We found that if the unperturbed ev ǫ β of H β is embedded in ( 1 4 α2, 0) and a is large, the corresponding resonance has a long halflife. In analogy with Friedrichs model [Demuth, 1976] one conjectures that in weak coupling case, the resonance state would be similar up to normalization to the eigenvector ξ 0 := K 0 ( ǫ β ) of H β, with the decay law being dominated by the exponential term At the same time, H α,β has always an isolated ev with ef which is not orthogonal to ξ 0 for any a (recall that both functions are positive). Consequently, the decay law (ξ 0,U(t)ξ 0 ) 2 ξ 0 2 has always a nonzero limit as t

124 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 59/9 Summarizing Lecture V A strong coupling turns leaky wires into essentially one-dimensional objects as far as the discrete spectrum is concerned

125 Summer School Lectures; Les Diablerets, June 6-10, 2011 p. 59/9 Summarizing Lecture V A strong coupling turns leaky wires into essentially one-dimensional objects as far as the discrete spectrum is concerned The analogous problem for scattering remains open