Schrödinger operators with non-confining potentials.

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1 Schrödinger operators with non-confining potentials. B. Simon s Non-standard eigenvalues estimates. Fakultät für Mathematik Ruhr-Universität Bochum 21 Sept. 2011

2 Plan of this talk. 1 Part1: Motivation and Examples. A) Basics in geometry. B) Basics for Schrödinger operators. C) A natural question. 2 A) An idea. B) Some nice inequalities. C) Micro-local analysis fails. 3 Dirichlet pb in an unbounded domain. Higher dimension n 3. Open problems.

C) A natural question. 2 A) An idea. B) Some nice inequalities.

3 Then: Let M be a compact (smooth) Riemmanian manifold, equipped with a strictly positive density dx(volume form). n = dim(m). the Laplacian of M. Theorem The spectrum σ( ) of is: positive and discrete accumulates at infinity.

4 In fact let: H(t) = e t. Then H(t) exists for all t > 0 as an unbounded operator and we have: Theorem (Weyl s law) As t tends to 0+ we have: Tr H(t) (2πt) n/2 Vol(M). Proof. Min-Ple formula (parametrix as t 0 + ). Works since M is compact.

5 Consider: regular enough. To simplify, assume that: It is well known that: Theorem The Schrödinger operator: Pf (x) = V : R n R, C R : V (x) > C, x R n. n 2 f (x) + V (x)f (x) = ( + V (x))f (x), j=1 x 2 j is essentially selfadjoint starting from C 0 (Rn ).

6 A classical result is also: Theorem Assume that: lim V (x) = +, x then σ(p) is discrete and accumulates at infinity.

7 A classical result is also: Theorem Assume that: lim V (x) = +, x then σ(p) is discrete and accumulates at infinity. Warning: This does not imply that the heat operator e tp is trace class!!!! Example: V (x) = log( 1 + x 2 ) on R.

8 The "heat kernel" is complicated when V 0. There is a useful classical estimate: Theorem We have: Tr(e tp ) 1 (2π) n R n R n e t( ξ 2+V (x)) dxdξ.

9 The Heat kernel is complicated when V 0. There is a useful classical estimate: Theorem We have: Tr(e tp ) 1 (2π) n R n R n e t( ξ 2+V (x)) dxdξ. Put w = ξ 2 + V (x) then this is a Laplace transform of a surface/volume distribution: 0 e tw S(w)dw.

10 The Heat kernel is complicated when V 0. There is a useful classical estimate: Theorem We have: Tr(e tp ) 1 (2π) n R n R n e t( ξ 2+V (x)) dxdξ. This inequality is a consequence of the Golden-Thompson inequality: Theorem Tr(e A+B ) Tr(e A e B ).

11 We define: and: We have: Theorem Z Cl (t) = 1 (2π) n Z Q (t) = Tr e tp, R n R n When Z Cl is finite for all t > 0 we have: e t( ξ 2+V (x)) dxdξ. Z Q (t) Z Cl (t), as t 0 +.

12 A very natural problem is to know if the condition: Z Cl (t) <, t > 0, is necessary to obtain a bound on Z Q.

13 A very natural problem is to know if the condition: Z Cl (t) <, t > 0, is necessary to obtain a bound on Z Q. Answer is NO since we will see that for n = 2 and V (x, y) = x α y β with α, β > 0 we have: Z Q (t) <, t > 0. Here Z Cl (t) = + and we have asymptotics: Z Q (t) f α,β (t), in 0 +.

Answer is NO since we will see that for n = 2 and V (x, y) = x α y β

14 To see the divergence, just perform integration to get: Z Cl (t) = π t x e α y β dxdy t = 4π Γ( α + 1 t α ) R 2 0 (ty β ) 1 α = +.

15 Trace by parts? One of the first idea I had was to use an operator valued T.F. : Tr x e tp = j=1 e tp e j (x), e j (x) L(L 2 (R y )) using an adapted orthonormal basis e j. Then to do it again with a new adapted basis: Tr y (Tr x e tp ) = j=1 Tr x e tp Ψ j (y), Ψ j (y).

16 It is clever to use eigenfunctions of the Schrödinger operator: 2 x 2 + y β x α, and to use the scaling properties of this equation. Warning. These eigenfunctions depends also of y. In theory this expresses the partition function as an infinite (countable) family of trace formulae for ODE. Problems: The 1D Scrhrodinger operator is not that easy to solve (appart for α = 2). Tr x e tp is no more a heat operator.

In theory this expresses the partition function as an infinite (countable) family of trace formulae for

17 For n = 2 the operator with perfect scaling is x 2 y 4 or x 4 y 2. We get: Z Q (t) e t(2k+1) 2n+1. It follows that: 0 Z Q (t) e t 2x+1 n,k=0 n=0 e 2t 2x+1 1 dx = 1 t 2 e t 2n+1 e 2t 2n+1 1 t e z e 2z 1 zdz.

18 Performing the last integral yields: Z Q (t) 1 t 2 π 2 8. Remark. Comparing with an abstract value (below), yields: It would follow that: Z Q (t) a t 2 π Γ( 2 ), ( ) a = Tr ( + x 4 ) 3 2. aπ 1 2 Γ( 5 2 ) = 3a 4 = π2 8 a = π2 6.

19 This double trace technique is not correct: Z Q (t) = Tre tp Tr y (Tr x e tp ) = Z sb (t)!!! And the above calculations are: FALSE, surprisingly not totally false, this gives: The correct power of t. An upper bound (the prefactor is too large).

20 In the semi-classical literature: V (x, y) = x 2 y 2. Lead to: D. Robert, J.Math. Pures et Appl. (1982). B. Simon, Annals of Physics (1983). B. Simon, JFA (1983). M. Z. Solomyak, Math USSR (1986). From now I work with: V (x, y) = x α y β, β > α.

21 But if we define the double trace: we have: Theorem Z sb (t) = Tr y (Tr x e tp ), B.Simon s Sliced Bread inequalities. We have: Z Q (t) Z sb (t) Z cl (t). Four our V the r.h.s. is infinite for t > 0 but: Z Q (t) Z sb (t) in t = 0 +.

22 Before going further there is an other interesting inequality for: Z sb (t) = Tr y (Tr x e tp ). Applying the classical inequality to the trace w.r.t. x we get: Z sb (t) Z sgt (t) = 1 e tp2 Tr y (e t( y +V (x,y)) )dxdp. (2π) Where Z sgt stands for "Sliced-Golden-Thompson". We have: Theorem (B.Simon, 83) Z Q (t) Z sb (t) Z sgt (t) Z cl (t).

23 Theorem (B.Simon, 83) Let P = + x α y β. If α < β we put: ν = β + 2 2α, a = Tr(( + y β ) ν ) <. We have: Tr(e tp ) a π Γ(ν + 1)t ( 1 2 +ν). If α = β we have: Tr(e tp ) Γ(2 + 1/α) log(t) t (1+ 1 α ). π

24 This proves: The spectrum of P = + x α y β is discrete. By the Tauberian-Karamata theorem we have estimates on the counting function N P (E). e.g., for α < β we have for E large: N P (E) = #{λ P E} C α,β E ν THIS IS NOT THE WEYL LAW! since: vol(ω e ) = vol R 4({ξ 2 + x α y β e}) = +.

25 Using sliced bread inequality it is enough to study: F (x, t) = Tr L 2 (dy) (exp( t( d 2 ) dy 2 + x α y β )).

26 Using sliced bread inequality it is enough to study: F (x, t) = Tr L 2 (dy) (exp( t( d 2 ) dy 2 + x α y β )). By scaling (w.r.t. y), we have a functional equation: F (xλ ν, tλ 1 ) = F (x, t). Sliced Gordon Thompson (integrating the kinetic energy) gives: Z Q (t) 1 F (x, t)dx. πt 0

27 Using the Feynmann-Kac formula: Z Q (t) = 1 4πt 2t 1 dxdye (x,y),(x,y),2t (exp( 2 b 1(s) α b 2 (s) β ds)) 0 Throwing away paths s.t. x < 1, he got a lower bound: Z Q (t) 1 πt (1 ρ(t)) 1 F (x, t).

28 The main result follows if: (b) (a) lim t 0 + tν 1 lim tν t F (x, t)dx = 0, F (x, t)dx = aγ(ν + 1). (a) contains 2 facts: Decay w.r.t. t (y scaling gives Ct αν β ). Existence of the integral (singularity x α β in x = 0). Recall that if V is homogenous of degree p 2: σ( + cv ) = c 2 2+p σ( + V ).

29 (b) Follows from the functional equation (scales t out): F (x, t)dx = F (xt ν, 1)dx = t ν F (x, 1)dx If A = d 2 dy 2 + y β, we scale A out via: F (x, 1)dx = F (1, x 1 ν )dx = ν s ν 1 F (1, s)ds 0 = νtr 0 s ν 1 e sa ds = νγ(ν)tr(a ν ). 0 0

30 It is tempting to try to understand the spectrum of: P h = h 2 + x α y β. For semi-classical eigenvalues λ j (h): λ j (h) E Ch, C > 0.

31 It is tempting to try to understand the spectrum of: P h = h 2 + x α y β. For semi-classical eigenvalues λ j (h): λ j (h) E Ch, C > 0. Motivation? When h 0 +, counting these e.v. should provide information on a density dn such that: N(E 2 ) N(E 1 ) = E 2 E 1 dn(s). I show now that a powerful method (WKB parametrix) fails for these operators.

32 F.I.O. for U h (t) = exp( it h P h) (WKB Method). Search the propagator U h as: i K t (x, y) = e h φ(t,x,y,ξ) a(h, t, x, y, ξ)dξ. R n φ(t, x, y, ξ) = S(t, x, ξ) y, ξ. If K t solves the Schrödinger equation: { t S(t, x, ξ) = (HJ) x S(t, x, ξ) 2 + V (x), S(0, x, ξ) = x, ξ. a(h, t, x, y, ξ) h j a j (t, x, y, ξ).

33 F.I.O. for exp( it h P h) (WKB Method) t small. { t S(t, x, ξ) = (HJ) x S(t, x, ξ) 2 + V (x), S(0, x, ξ) = x, ξ. (HJ) is not complete. Our flow is complicated. So we stay near t = 0!!! We get: S(t, x, ξ) = x, ξ + tp(x, ξ) + O(t 2 ).

34 Usual method. We can estimate (2πh) n Υ(E, h) (regularized s.d.) via i e h (S(t,x,ξ) x,ξ te) ˆϕ(t)a(t, x, ξ)dtdxdξ. R T R n Like always... So WHERE is the problem?

35 Usual method. We can estimate (2πh) n Υ(E, h) (regularized s.d.) via i e h (S(t,x,ξ) x,ξ te) ˆϕ(t)a(t, x, ξ)dtdxdξ. R T R n a / C 0!!!!!!!! a 1 near Σ E = {ξ 2 + V = E}!!!!!!!! The method of analysis collapses. If not: For t = 0 : the full surface Σ E contributes. This surface has infinite volume: ABSURD...

36 The previous strategy works for Dirichlet pb in domains: D E = { x y γ E} R Figure: Some domains D E (with γ = 1) Proof: a) By scaling change E into 1. b) Pick a sequence x αn y βn with: α n + and αn β n = γ..

37 With Nils Schippkus (Bochum) we have extented B. Simon s result to n-dimension. Current state of research: The (generic) case: V (x) = We got recently better, e.g. Open PB is: n x j α k, j=1 α 1 > α 2 >... > α n. α 1 >... > α n 1 = α n. α n k = α n+1 k =... = α n, k 3.

38 Exact trace formulae for power of resolvant: Tr( + x α ) ν. Duality with classical mechanics. If there is a dynamics (α i 2): spectrum and length of periodic orbits? Validity of classical theorems? e.g., Egorov s theorem: P h = h 2 + V α (x) e it h P h Op(a)e it h P h = Op(a Φ t ) + R t?? Eigenvectors estimates?

MATH 425, PRACTICE FINAL EXAM SOLUTIONS.

MATH 425, PRACTICE FINAL EXAM SOLUTIONS. MATH 45, PRACTICE FINAL EXAM SOLUTIONS. Exercise. a Is the operator L defined on smooth functions of x, y by L u := u xx + cosu linear? b Does the answer change if we replace the operator L by the operator