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Decay of bound states of elliptic PDEs EMSDMF conference, April 2013 E. Skibsted, Institut for Matematiske Fag Aarhus Universitet, Denmark Contents Review of elements of publication (in preparation) Decay of bound states of elliptic


  1. Decay of bound states of elliptic PDE’s EMS–DMF conference, April 2013 E. Skibsted, Institut for Matematiske Fag Aarhus Universitet, Denmark

  2. Contents Review of elements of publication (in preparation) Decay of bound states of elliptic PDE’s , joint with I. Herbst. Introduction, N –body Schrödinger operators. Results. Elements of proof.

  3. One-body Schrödinger operators Consider the operator H = − ∆ + V on L 2 = L 2 ( R d ) for a suitable class of real decaying potentials V = V ( x ) . For any solution φ ∈ L 2 to the eigenvalue equation H φ = E φ , the critical decay rate is: σ c = sup { σ ≥ 0 | e σ | x | φ ∈ L 2 } . Well-known classification: E > 0 ⇒ σ c = ∞ and φ = 0. E < 0 and φ � = 0 ⇒ σ c = ( − E ) 1 / 2 . E = 0 and φ � = 0 ⇒ σ c = 0.

  4. N -body Schrödinger operators The N -body Schrödinger operators (finite masses): N − 1 � � H = ∆ x j + V ij ( x i − x j ) . 2 m j j = 1 1 ≤ i < j ≤ N Hilbert space = L 2 ( R d ( N − 1 ) ) after separation of center of mass motion. Atomic N -body Schrödinger operators on Hilbert space L 2 ( R dN ) : N 1 � � � ∆ x j + V ncl � V elec H = − ( x j ) + ( x i − x j ) . j ij 2 m j j = 1 1 ≤ i < j ≤ N Atomic N -body Schrödinger operators with obstacle: Hamiltonian as above. Particles confined to the exterior of a bounded obstacle Θ ⊂ R d . Hilbert space = L 2 (Ω) , Ω = (Ω 1 ) N = Ω 1 × · · · × Ω 1 , Ω 1 = R d \ Θ . H defined by Dirichlet boundary condition. Example (Coulomb interactions for d ≥ 2) ( y ) = q j q ncl | y | − 1 and V elec V ncl ( y ) = q i q j | y | − 1 j ij

  5. Critical decay rates for N -body Schrödinger operators For any bound state H φ = E φ , 0 � = φ ∈ L 2 , the critical decay rate is: σ c = sup { σ ≥ 0 | e σ | x | φ ∈ L 2 } . Computation of the decay rate? Possibilities? N -body Schrödinger operators take form H = − ∆ + V . Sub-Hamiltonians: H a = − ∆ x a + V a ( x a ) , x = x a ⊕ x a ; a ∈ A . Thresholds of H : � σ pp ( H a ) . T = a ∈A , H a � = H Set T ∋ 0, closed and countable, and (HVZ-theorem) σ ess ( H ) = [ min T , ∞ ) .

  6. N -body theory continued Under decay conditions on pair-potentials: Theorem (FH,IS) For any bound state H φ = E φ , φ � = 0 : σ c < ∞ . 0 < σ c for E �∈ T . E + σ 2 c ∈ T . FH R. Froese, I. Herbst, Exponential bounds and absence of positive eigenvalues for N-body Schrödinger operators , Commun. Math. Phys. 87 no. 3 (1982/83), 429–447. IS K. Ito, E. Skibsted, Absence of positive eigenvalues for hard-core N-body interactions , Institut Mittag-Leffler preprint, fall 2012 no. 31. Absence of positive thresholds and eigenvalues. Example: Theorem (IS, obstacle + Coulomb interactions) For N charged d ≥ 2 dimensional particles confined to the exterior of a bounded obstacle Θ ⊂ R d with 0 ∈ Θ and Ω 1 = R d \ Θ connected, the obstacle Hamiltonian H with Coulomb interactions ( y ) = q j q ncl | y | − 1 and V elec ( y ) = q i q j | y | − 1 , V ncl j ij does not have positive thresholds nor positive eigenvalues.

  7. Bilaplacian ∆ 2 Consider Q ( ξ ) = ( ξ 2 ) 2 , H = Q ( p ) + V , p = − i ∇ . For any bound state H φ = E φ , 0 � = φ ∈ L 2 , the critical decay rate is: σ c = sup { σ ≥ 0 | e σ | x | φ ∈ L 2 } . Possible decay rates? For E � = 0: E > 0 ⇒ σ c = E 1 / 4 . E < 0 ⇒ σ c = ( − E / 4 ) 1 / 4 . Conversely � E 1 / 4 for E > 0 , σ = ( − E / 4 ) 1 / 4 for E < 0 is the critical decay rate for a bound state with energy E for some potential V , possibly for a V with compact support.

  8. General elliptic polynomial Consider any real elliptic polynomial on R d and H = Q ( p ) + V . for a suitable class of real decaying potentials V = V ( x ) . For any bound state H φ = E φ , 0 � = φ ∈ L 2 , the critical decay rate is: σ c = sup { σ ≥ 0 | e σ | x | φ ∈ L 2 } . Possible decay rates? Critical values: If Q ( ξ ) = C and ∇ Q ( ξ ) = 0 for some ξ ∈ R d , C is a critical value of Q . Theorem (HS) For H φ = E φ , φ � = 0 , σ c < ∞ . σ c > 0 if E is a non-critical value of Q. If σ c > 0 there exists ( ω, ξ ) ∈ S d − 1 × R d such that with σ = σ c : Q ( ξ + i σω ) = E , (1) ∇ ξ Q ( ξ + i σω ) = µω ; µ = ω · ∇ ξ Q ( ξ + i σω ) . If the system (1) has a solution for a given σ > 0 we call σ exceptional at E .

  9. Exceptional points, rotationally invariant case Q ( ξ ) = G ( ξ 2 ) Suppose Q ( ξ ) = G ( ξ 2 ) . Note degree ( G ) = q / 2 for q = degree ( Q ) . General facts: There are ≤ q / 2 critical values of Q . There are ≤ q / 2 exceptional points away from ≤ q / 2 − 1 energies, more precisely at any real non-critical value E of G : C → C . Example (Exactly q / 2 exceptional points at non-critical value of Q ) Consider n Q ( ξ ) = G ( ξ 2 ); � ( z + σ 2 G ( z ) = j ) , 0 < σ 1 < · · · < σ n . j = 1 Properties: C = � n j = 1 σ 2 j is the only critical value of Q , and σ ess ( Q ( p ) + V ) = [ C , ∞ ) . The numbers σ 1 < · · · < σ n are the exceptional points at E = 0, and 0 is not a critical value of z → G ( z ) . For each j ∈ { 1 , . . . , n } there exists a compactly supported potential V : σ j is the critical decay rate for some bound state, ( Q ( p ) + V ) φ = 0 .

  10. “Proofs” Theorem (HS) is proved by commutator methods. Two different approaches needed: σ c < ∞ , proved by calculus of differential operators and combinatorics. “Sub-leading symbols” are used. Other statements in Theorem (HS), proved by microlocal analysis. Poisson bracket (i.e. “leading order”) calculations suffice. Results can be generalized to variable coefficient PDE’s.

  11. Why system (1)? Suppose H φ = E φ and σ c > 0. Introduce r ( x ) = r 1 ( x ) = ( 1 + x 2 ) 1 / 2 , ω ( x ) = grad r ( x ) . For σ < σ c put φ σ = e σ r φ, and compute ( Q ( p + i σω ( x )) − E + V ) φ σ = 0 For a decaying potential the principal symbol is x Q ( ξ + i σω ( x )) − E ≈ Q ( ξ + i σ ˆ x ) − E , ˆ x = | x | . Whence by energy bounds 1st system of equations: Q ( ξ + i σω ) = E for some ( ω, ξ ) ∈ S d − 1 × R d .

  12. 2nd system of equations To obtain the 2nd system of equations of the system (1) write Q ( ξ + i σω ( x )) − E = ( X + i Y )( ξ + i σω ( x )) . Conjugate symbol: a = rY ( ξ + i σω ( x )) . Poisson bracket calculation: σ − 1 { X , Y } = ∂ ξ X ω ′ ( x ) ∂ T ξ X + ∂ ξ Y ω ′ ( x ) ∂ T ξ Y ≥ 0 . (2a) Quantizing: w ( X ) + i Op w ( Y ) ≈ ˜ X + i ˜ Op Y := Q ( p + i σω ( x )) − E , w ( a ) ≈ A := Re ( r ˜ Op Y ) . Commutator identity, virial type argument: e σ r + e σ r i [ V , A ] e σ r . i [ H − E , e σ r A e σ r ] = e σ r � i [ ˜ X , A ] + 2 Re ( ˜ � Y A ) Near energy surface S σ := { X = Y = 0 } : { X , a } + 2 rY 2 = r { X , Y } + 2 rY 2 + { X , r } Y (2b) ≥ rY 2 − Cr − 1 . Improvement: If the system (1) does not have solutions, (2a) is strictly positive (for σ < σ c , but close, and on S σ ), and (2b) is improved as { X , a } + 2 rY 2 ≥ rY 2 + c ≥ c > 0 .

  13. Absence of super-exponentially decaying states Suppose H φ = E φ and φ σ = e σ r φ ∈ L 2 for all σ > 0 . Want to show that φ = 0. Examine possible positivity of [ ˜ X , ˜ Y ] ? Introducing B = ( b 1 , . . . , b d ) = e − σ r p e σ r = p − i σω, we have X − i ˜ ˜ Y = Q ( B ) − λ and ˜ X + i ˜ Y = Q ( B ∗ ) − λ, and 2 i [ ˜ X , ˜ Y ] = [ Q ( B ) , Q ( B ∗ )] . Is [ Q ( B ) , Q ( B ∗ )] positive? Note that p kl := [ b k , b ∗ l ] = 2 σ∂ l ω k , and whence that P := ( p kl ) ≥ 0. Maybe [ Q ( B ) , Q ( B ∗ )] ≈ Q ′ ( B ) PQ ′ ( B ∗ ) T ? Better to replace r = r 1 by r ( x ) = r ǫ ( x ) = � x � − � x � 1 − ǫ + 1 ; ǫ ∈ ( 0 , 1 ) . Then P = ( p kl ) = 2 σω ′ ≥ c ǫ σ r − 1 − ǫ ; ω ( x ) = grad r ǫ ( x ) . Commutator P is strictly positive.

  14. Combinatorial problem for toy model Abstractly, look at commuting operators b 1 , . . . , b d on a Hilbert space H , B ∗ = ( b ∗ 1 , . . . , b ∗ p kl = [ b k , b ∗ B = ( b 1 , . . . , b d ) , d ) , l ]; k , l ≤ d , assuming P := ( p kl ) ≥ 0 and [ b # j , p kl ] = 0 ; j , k , l ≤ d . Representation of commutator: [ Q ( B ) , Q ( B ∗ )] = � ( − 1 ) m + 1 � Q ( α ) ( B ) F αβ Q ( β ) ( B ∗ ); (3a) m ≥ 1 | α | , | β | = m here p m kl � � kl F αβ = m kl ! , ( m kl ) ∈M αβ k , l ≤ d ( m kl ) ∈ N d 2 � � M αβ = � 0 | m kl = β l , m kl = α k � . k l Alternating series. Is the commutator positive? Another representation: [ Q ( B ) , Q ( B ∗ )] = � � Q ( α ) ( B ∗ ) F αβ Q ( β ) ( B ) . (3b) m ≥ 1 | α | , | β | = m Positive series. Answer, yes.

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