Twisting versus bending in quantum waveguides

Transcription

1 Twisting versus bending in quantum waveguides David KREJČIŘÍK Nuclear Physics Institute, Academy of Sciences, Řež, Czech Republic david/ Based on: [Chenaud, Duclos, Freitas, D.K.] Differential Geom. Appl. 23 (2005) [Ekholm, Kovařík, D.K.] Arch. Ration. Mech. Anal., to appear Graph Models of Mesoscopic Systems, Wave-Guides and Nano-Structures Cambridge, April 2007

2 The Problem Hamiltonian 2 2 m Ω D where Ω = twisted and bent tube dependence of spectrum on geometry mathematical model for quantum waveguides due to [Exner, Šeba 1989] Characteristics of the (present) model: unbounded geometry local deformation uniform cross-section

3 Outline 1. Geometry of a twisted and bent tube 2. Strategy 3. Stability of the essential spectrum 4. Effect of bending 5. Effect of twisting 6. Conclusions

4 The Geometry Γ : R R 3 unit-speed curve with curvature κ and torsion τ - possessing an appropriate smooth Frenet frame {e 1, e 2, e 3 } Serret-Frenet formulae: ¼ e 1 e 2 e 3 ½ = ¼ 0 κ 0 κ 0 τ 0 τ 0 ½¼ e 1 e 2 e 3 ½

5 The Geometry Γ : R R 3 ω R 2 unit-speed curve with curvature κ and torsion τ - possessing an appropriate smooth Frenet frame {e 1, e 2, e 3 } Serret-Frenet formulae: ¼ e 1 e 2 e 3 ½ open connected bounded set, a := sup t ω = t ¼ 0 κ 0 κ 0 τ 0 τ 0 ½¼ e 1 e 2 e 3 ½

6 The Geometry Γ : R R 3 unit-speed curve with curvature κ and torsion τ - possessing an appropriate smooth Frenet frame {e 1, e 2, e 3 } Serret-Frenet formulae: ¼ e 1 e 2 e 3 ½ = ¼ 0 κ 0 κ 0 τ 0 τ 0 ½¼ e 1 e 2 e 3 ½ ω R 2 open connected bounded set, a := sup t ω R θ : R SO(2) family of rotation matrices: R θ = - smooth function θ : R R t cosθ sinθ sinθ cosθ

7 The Geometry Γ : R R 3 unit-speed curve with curvature κ and torsion τ - possessing an appropriate smooth Frenet frame {e 1, e 2, e 3 } Serret-Frenet formulae: ¼ e 1 e 2 e 3 ½ = ¼ 0 κ 0 κ 0 τ 0 τ 0 ½¼ e 1 e 2 e 3 ½ ω R 2 open connected bounded set, a := sup t ω R θ : R SO(2) family of rotation matrices: R θ = - smooth function θ : R R t cosθ sinθ sinθ cosθ Ω := L(R ω) tube of cross-section ω 3 L(s, t) := Γ(s) + t µ e θ µ(s) e θ µ := µ=2 3 R θ µν e ν ν=2

8 The Geometry Γ : R R 3 unit-speed curve with curvature κ and torsion τ - possessing an appropriate smooth Frenet frame {e 1, e 2, e 3 } Serret-Frenet formulae: ¼ e 1 e 2 e 3 ½ = ¼ 0 κ 0 κ 0 τ 0 τ 0 ½¼ e 1 e 2 e 3 ½ ω R 2 open connected bounded set, a := sup t ω R θ : R SO(2) family of rotation matrices: R θ = - smooth function θ : R R t cosθ sinθ sinθ cosθ Ω := L(R ω) tube of cross-section ω 3 L(s, t) := Γ(s) + t µ e θ µ(s) e θ µ := µ=2 3 R θ µν e ν ν=2 Assumptions: κ 1 a < 1 and Ω does not overlap itself

9 The motion of the general moving frame e θ 1 e θ 2 e θ 3 = 0 κ cosθ κ sinθ κ cosθ 0 τ θ κ sinθ (τ θ ) 0 e θ 1 e θ 2 e θ 3

10 The motion of the general moving frame e θ 1 e θ 2 e θ 3 = 0 κ cosθ κ sinθ κ cosθ 0 τ θ κ sinθ (τ θ ) 0 e θ 1 e θ 2 e θ 3 bending : κ 0

11 The motion of the general moving frame e θ 1 e θ 2 e θ 3 = 0 κ cosθ κ sinθ κ cosθ 0 τ θ κ sinθ (τ θ ) 0 e θ 1 e θ 2 e θ 3 bending : κ 0 twisting : { τ θ 0 ω is not circular

12 Equivalent definitions of twisting a tube of non-circular cross-section is not twisted : {e θ 1, e θ 2, e θ 3} is the Tang frame i.e. θ = τ

13 Equivalent definitions of twisting a tube of non-circular cross-section is not twisted : {e θ 1, e θ 2, e θ 3} is the Tang frame i.e. θ = τ parallel transport of the surface normal along generators of the ruled surface generated by e θ 2

14 Equivalent definitions of twisting a tube of non-circular cross-section is not twisted : {e θ 1, e θ 2, e θ 3} is the Tang frame i.e. θ = τ parallel transport of the surface normal along generators of the ruled surface generated by e θ 2 zero curvature of the ruled surface

15 Equivalent definitions of twisting a tube of non-circular cross-section is not twisted : {e θ 1, e θ 2, e θ 3} is the Tang frame i.e. θ = τ parallel transport of the surface normal along generators of the ruled surface generated by e θ 2 zero curvature of the ruled surface no Coriolis acceleration of the (non-inertial) traveller e θ 2

16 The Hamiltonian Ω D Q Ω D : W 1,2 0 (Ω) L 2 (Ω) : { u u 2}

17 The Hamiltonian Ω D Q Ω D : W 1,2 0 (Ω) L 2 (Ω) : { u u 2} Strategy: L : R ω Ω is a diffeomorphism = Ω ( R ω, G ) G = h 2 + h h 2 3 h 2 h 3 h h h(s, t) := 1 [t 2 cos θ(s) + t 3 sinθ(s)] κ(s) h 2 (s, t) := t 3 [τ(s) θ (s)] h 3 (s, t) := t 2 [τ(s) θ (s)]

18 The Hamiltonian Ω D Q Ω D : W 1,2 0 (Ω) L 2 (Ω) : { u u 2} Strategy: L : R ω Ω is a diffeomorphism = Ω ( R ω, G ) G = h 2 + h h 2 3 h 2 h 3 h h h(s, t) := 1 [t 2 cos θ(s) + t 3 sinθ(s)] κ(s) h 2 (s, t) := t 3 [τ(s) θ (s)] h 3 (s, t) := t 2 [τ(s) θ (s)] Ω D H w := G 1/2 i G 1/2 G ij j on L 2( R ω, dvol ) G := det(g) = h 2, (G ij ) := G 1, dvol := h(s, t) ds dt

19 Stability of essential spectrum

20 Stability of essential spectrum Remark. Straight tube (κ = 0 = τ θ ) : E 1 := inf σ( ω D ) σ( R ω D ) = σ ess( R ω D ) = [E 1, ) 0 E 1

21 Stability of essential spectrum Remark. Straight tube (κ = 0 = τ θ ) : E 1 := inf σ( ω D ) σ( R ω D ) = σ ess( R ω D ) = [E 1, ) 0 E 1 Theorem. ( lim κ(s) + τ(s) θ (s) ) = 0 = σ ess ( Ω D ) = [E 1, ) s Proof. Weyl-type criterion adapted to quadratic forms. q.e.d. Classical Weyl s criterion requires to impose additional conditions on derivatives!

22 Stability of essential spectrum Remark. Straight tube (κ = 0 = τ θ ) : E 1 := inf σ( ω D ) σ( R ω D ) = σ ess( R ω D ) = [E 1, ) 0 E 1 Theorem. ( lim κ(s) + τ(s) θ (s) ) = 0 = σ ess ( Ω D ) = [E 1, ) s Proof. Weyl-type criterion adapted to quadratic forms. q.e.d. Classical Weyl s criterion requires to impose additional conditions on derivatives! History: [Goldstone, Jaffe 1992]... κ of compact support & ω =disc [Duclos, Exner 1995]... additional vanishing of κ and κ & ω =disc [Dermenjian, Durand, Iftimie 1998]... σ ess of multistratified cylinders [Chenaud, Duclos, Freitas, D.K. 2005]... θ = τ (ω arbitrary)

23 The effect of bending Theorem. κ 0 & θ = τ = inf σ( Ω D ) < E 1 Proof. Trial function based on J 1 ( E 1 ). q.e.d.

24 The effect of bending Theorem. κ 0 & θ = τ = inf σ( Ω D ) < E 1 Proof. Trial function based on J 1 ( E 1 ). q.e.d. Corollary. σ disc ( Ω D ) if in addition lim κ(s) = 0 s 0 bound states! E 1 bending acts as an attractive interacion

25 The effect of bending Theorem. κ 0 & θ = τ = inf σ( Ω D ) < E 1 Proof. Trial function based on J 1 ( E 1 ). q.e.d. Corollary. σ disc ( Ω D ) if in addition lim κ(s) = 0 s 0 bound states! E 1 bending acts as an attractive interacion History: [Goldstone, Jaffe 1992]... κ of compact support & ω =disc [Duclos, Exner 1995]... additional conditions on κ and κ & ω =disc [Chenaud, Duclos, Freitas, D.K. 2005]... ω arbitrary

26 The effect of twisting Theorem([Ekholm, Kovařík, D.K. 2005]). Let κ = 0. Let θ be such that θ 0, θ C 0 (R) and θ L (R). Assume that ω is not circular. Then Ω D E 1 c 1 + Γ( ) Γ(s 0 ) 2 Hardy inequality! where s 0 R is such that θ (s 0 ) 0 and c = c(s 0, θ, ω) > 0. twisting acts as a repulsive interaction

27 The effect of twisting Theorem([Ekholm, Kovařík, D.K. 2005]). Let κ = 0. Let θ be such that θ 0, θ C 0 (R) and θ L (R). Assume that ω is not circular. Then Ω D E 1 c 1 + Γ( ) Γ(s 0 ) 2 Hardy inequality! where s 0 R is such that θ (s 0 ) 0 and c = c(s 0, θ, ω) > 0. twisting acts as a repulsive interaction Proof. Writing ψ(s, t) = J 1 (t) φ(s, t), ψ C 0 (R ω), σ := t 3 2 t 2 3, ( ψ, [H E1 ]ψ ) = J 1 1 φ 2 + J 1 2 φ 2 + J 1 3 φ 2 + θ (J 1 σ φ + φ σ J 1 ) 2 + mixed terms... q.e.d.

28 Theorem([Ekholm, Kovařík, D.K. 2005]). Twisting vs bending Let θ be such that τ θ 0, θ C 0 (R) and θ L (R). Assume that ω is not circular. Assume also that κ C0(R). 1 Then there exists ε > 0 such that κ + κ ε = σ ( D) Ω = [E1, ) where ε = ε(τ, θ, ω). 0 E 1

29 Theorem([Ekholm, Kovařík, D.K. 2005]). Twisting vs bending Let θ be such that τ θ 0, θ C 0 (R) and θ L (R). Assume that ω is not circular. Assume also that κ C0(R). 1 Then there exists ε > 0 such that κ + κ ε = σ ( D) Ω = [E1, ) where ε = ε(τ, θ, ω). 0 E 1 Remark. Mildly curved tubes also studied by [Grushin 2005].

30 Theorem([Ekholm, Kovařík, D.K. 2005]). Twisting vs bending Let θ be such that τ θ 0, θ C 0 (R) and θ L (R). Assume that ω is not circular. Assume also that κ C0(R). 1 Then there exists ε > 0 such that κ + κ ε = σ ( D) Ω = [E1, ) where ε = ε(τ, θ, ω). 0 E 1 Remark. Mildly curved tubes also studied by [Grushin 2005]. Theorem([Bouchitté, Mascarenhas, Trabucho 2006]). Let Ω ε = L(I εω), I bounded. Then graph model Ω ε D ε 2 E 1 (ω) I D κ2 ε C(ω)(τ θ ) 2 + O(ε) where C(ω) > 0 iff ω is not circular.

31 ω = ( a, a) ( b, b) ( a, a) b 0 ( π 2 E 1 = 2a) Twisted strips

32 ω = ( a, a) ( b, b) ( a, a) b 0 ( π 2 E 1 = 2a) Twisted strips Theorem([D.K. 2006]). Assume that κ cosθ = 0 and 0 < τ θ a 2. Then Ω D E 1 c 1 + Γ( ) Γ(s 0 ) 2 where s 0 R is such that (τ θ )(s 0 ) 0 and c = c(s 0, τ θ, a) > 0.

33 ω = ( a, a) ( b, b) ( a, a) b 0 ( π 2 E 1 = 2a) Twisted strips Theorem([D.K. 2006]). Assume that κ cosθ = 0 and 0 < τ θ a 2. Then Ω D E 1 c 1 + Γ( ) Γ(s 0 ) 2 where s 0 R is such that (τ θ )(s 0 ) 0 and c = c(s 0, τ θ, a) > 0. negative curvature of the ambient space acts as a repulsive interaction

34 Twist via boundary conditions N y D ( π 2 E 1 = 4a) 2a x D N

35 Twist via boundary conditions N y D E 1 = ( π 4a) 2 2a x D N Theorem([Kovařík, D.K. 2006]). DN E 1 c 1 + x 2 where c = c(a) > 0.

36 Conclusions Moral: bending acts as an attractive interaction twisting acts as a repulsive interaction 0 0 E 1 E 1 Hardy inequalities in twisted tubes Ω D E 1 ρ( ) > 0

37 Conclusions Moral: bending acts as an attractive interaction twisting acts as a repulsive interaction 0 0 E 1 E 1 Hardy inequalities in twisted tubes Ω D E 1 ρ( ) > 0 Open problems: higher-dimensional generalisations?

38 Conclusions Moral: bending acts as an attractive interaction twisting acts as a repulsive interaction 0 0 E 1 E 1 Hardy inequalities in twisted tubes Ω D E 1 ρ( ) > 0 Open problems: higher-dimensional generalisations? effect of twisting on the essential spectrum?

39 Conclusions Moral: bending acts as an attractive interaction twisting acts as a repulsive interaction 0 0 E 1 E 1 Hardy inequalities in twisted tubes Ω D E 1 ρ( ) > 0 Open problems: higher-dimensional generalisations? effect of twisting on the essential spectrum? other physical motivations?

40 Mourre s theory for twisted tubes? Theorem([D.K., Tiedra de Aldecoa 2004]). Assume that θ = τ (plus some fast decay of κ, τ at infinity). Then A := i 2 (s s + s s) is strictly conjugate to H on R \ {E n } n=1, i.e. P H i[h, A]P H c P H with some c > 0. Corollary. Eigenvalues of H are countable and can accumulate at E n only.

41 Mourre s theory for twisted tubes? Theorem([D.K., Tiedra de Aldecoa 2004]). Assume that θ = τ (plus some fast decay of κ, τ at infinity). Then A := i 2 (s s + s s) is strictly conjugate to H on R \ {E n } n=1, i.e. P H i[h, A]P H c P H with some c > 0. Corollary. Eigenvalues of H are countable and can accumulate at E n only. Commutator for the straight tube: i[h 0, A] = 2( 2 s) = ψ n := 1 J n represent the bad functions.

42 Mourre s theory for twisted tubes? Theorem([D.K., Tiedra de Aldecoa 2004]). Assume that θ = τ (plus some fast decay of κ, τ at infinity). Then A := i 2 (s s + s s) is strictly conjugate to H on R \ {E n } n=1, i.e. P H i[h, A]P H c P H with some c > 0. Corollary. Eigenvalues of H are countable and can accumulate at E n only. Commutator for the straight tube: i[h 0, A] = 2( 2 s) = ψ n := 1 J n represent the bad functions. However, ( ψ n, i[h, A] ψ n ) = θ σ J n 2 > 0 for a straight but twisted tube! Conjecture. The result of Mourre theory can be improved for twisted tubes.