On a class of self-adjoint operators in Krein space with empty resolvent set

Transcription

1 On a class of self-adjoint operators in Krein space with empty resolvent set Sergii Kuzhel 1,2 1 Institute of Mathematics, NAS of Ukraine, Kiev, Ukraine 2 AGH University of Science and Technology, Krakow, Poland ESF Exploratory Workshop Prague, September 1 S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 1 / 24

2 Outline 1 Elements of the Krein spaces theory. 2 J-self-adjoint operators with empty resolvent set. 3 Exactly solvable models. S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 2 / 24

3 Elements of the Krein spaces theory Definition of Krein spaces. Let H be a Hilbert space with inner product (, ) and let J be a fundamental symmetry in H (i.e., J = J, J 2 = I ). The indefinite metric: [u, v] J := (Ju, v). The Hilbert space H with indefinite metric [, ] J is called the Krein space (H, [, ] J ). Example. H = L 2 (R), (u, v) = R u(x)v(x)dx J = P, Pu(x) = u( x) is the space parity operator in L 2 (R), the indefinite metric [u, v] P = (Pu, v) = R u( x)v(x)dx. The Krein space (L 2 (R), [, ] P ). S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 3 / 24

4 Elements of the Krein spaces theory Fundamental decomposition Let (H, [, ] J ) be a Krein space. Denote P + = 1 (I + J), 2 P = 1 (I J). 2 P ± are orthoprojectors in the Hilbert space H. The fundamental decomposition H = H + H, H + = P + H, H = P H. Example. The decomposition L 2 (R) = L even 2 (R) L odd 2 (R) is fundamental for the Krein space (L 2 (R), [, ] P ). S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 4 / 24

5 Elements of the Krein spaces theory Fundamental decompositions and C-operators. Definition of fundamental decomposition. Let (H, [, ] J ) be a Krein space. An arbitrary decomposition H = L + [ +] J L where L + is positive and L is negative subspaces w.r.t. [, ] J, is called a fundamental decomposition of H. There are many fundamental decompositions of H. Theorem Their description? The collection of operators {C} such that C 2 = I and JC > 0 is in one-to-one correspondence with the set of fundamental decompositions of (H, [, ] J ): L + = 1 (I + C)H, 2 L = 1 (I C)H 2 S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 5 / 24

6 Elements of the Krein spaces theory J-self-adjoint operators An operator A in a Krein space (H, [, ] J ) is called J-self-adjoint if A J = JA. A is J-self-adjoint A is self-adjoint w.r.t. [, ] J. The spectrum of J-self-adjoint operator. In contrast to self-adjoint operators, the spectra of J-self-adjoint ones are just symmetric with respect to the real axis. As to other spectral properties, everything may happen! 1) σ(a) = C is possible; 2) C σ p (A) C + σ r (A) is possible; 3) Jordan blocks for discrete spectra are possible. S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 6 / 24

7 J-self-adjoint operators with empty resolvent set Examples. Let K be a Hilbert space and let be L be a symmetric (non-self-adjoint) operator in K. Consider the operators ( ) ( ) L 0 0 I A := 0 L, J = I 0 in the product Hilbert space H = K K. Then J is a fundamental symmetry in H and A is a J-self-adjoint operator. It is clear that ρ(a) = (since ρ(l) = ). If L is maximal symmetric operator, then one of C ± belongs to σ p (A); another one - belongs to σ r (A). S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 7 / 24

8 J-self-adjoint operators with empty resolvent set Examples Continuous spectra at C \ R Let (H, [, ] J ) be a Krein space and let L + be a maximal positive (but no uniformly positive!) subspace. Then L = L [ ] + is a maximal negative and D = L + +L is a dense subspace in H. The operator A(x + + x ) = x + x, D(A) = D, x ± L ± is J-self-adjoint, J-positive but it has continuous spectrum at C \ R. S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 8 / 24

9 J-self-adjoint operators with empty resolvent set Example Degenerate Sturm-Liouville problems on the finite interval. Operator A in L 2 ( 1, 1) associated with differential expression l(y) = (sgn x)((sgn x)y ) and boundary conditions y( 1) = y(1) = 0. More precisely, Ay = y D(A) = {y W 2 2 ( 1, 0) W 2 2 (0, 1) y(0+) = y(0 ) y (0+) = y (0 ), y(±1) = 0} is P-self-adjoint in (L 2 ( 1, 1), [, ] P ) and it has point spectrum at C \ R. Mingarelli S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 9 / 24

10 J-self-adjoint operators with empty resolvent set WHY ARE THERE J-self-adjoint operators with empty resolvent set??? Hypothesis: In many cases the presence of empty resolvent set means (indicates) the special structure of the corresponding J-self-adjoint operator. Problem: Can one feel better the special structure of J-self-adjoint operators with empty resolvent set? (like Weyl-Kac question: "Can One Hear the Shape of a Drum?") S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 10 / 24

11 S. Albeverio, U. Günther, and S. Kuzhel, J. Phys. A. 42 (2009) S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 11 / 24 Exactly Solvable Models. Abstract underlying structure. The main ingredients: a symmetric operator S in a Hilbert space H with deficiency indices < 2, 2 >; a fundamental symmetry J which commutes with S in H: SJ = JS. Then the collection of all J-self-adjoint extensions A of S : S A S is determined by the set of unitary matrices ( U = e iφ qe iγ re iξ ) re iξ qe iγ, q 2 + r 2 = 1, φ, γ, ξ [0, 2π). A = A U is simultaneously J-self-adjoint and self-adjoint q = 0, i.e., ( U = e iφ 0 e iξ ) e iξ. 0

12 Exactly Solvable Models. Examples of symmetric operators S H = L 2 (R), J = P, l(f ) = d 2 f dx 2 + x 2 (ix) ɛ f with ɛ = 2 and set [f, g] ± = lim x ± [f (x)g (x) f (x)g(x)], Denote by v 1 and v 2 the solutions of l(v) = 0 with v 1 (0) = v 2 (0) = 1 and v 1 (0) = v 2(0) = 0. The operator Sf = l(f ), f D(S) f, f AC loc (R), l(f ) L 2 (R) with the boundary condition [f, v 1 ] ± = [f, v 2 ] ± = 0 is symmetric in L 2 (R) with deficiency indices < 2, 2 > and SP = PS. S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 12 / 24

13 Examples of symmetric operators S Schrödinger operator with general zero-range potential d 2 dx 2 + ε 11 < δ, > δ + ε 12 < δ, > δ + ε 21 < δ, > δ + ε 22 < δ, > δ, where δ and δ are, respectively, the Dirac δ-function and its derivative (with support at 0) and ε ij are complex numbers. The corresponding symmetric operator S = d 2 dx 2 {u( ) W 2 2 (R) u(0) = u (0) = 0}. has deficiency indices < 2, 2 > and SP = PS. S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 13 / 24

14 Exactly Solvable Models. The concept of stable C-symmetry Let A be a J-self-adjoint extension of S. Then A has the property of stable C-symmetry (i.e., there exists a bounded linear operator C in H such that C 2 = I, JC > 0, SC = CS, and AC = CA. The property of stable C-symmetry depends on the boundary conditions which determine A among other J-self-adjoint extensions of S, i.e depend on the entries of matrix U: U = e iφ ( qe iγ re iξ re iξ qe iγ ), q 2 + r 2 = 1, φ, γ, ξ [0, 2π). S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 14 / 24

15 Exactly Solvable Models. The main abstract result Let S be a symmetric operator with deficiency indices < 2, 2 > and such that SJ = JS. Then the set of J-self-adjoint extensions A of S contains operators with empty resolvent set if and only if there exists an additional fundamental symmetry R such that SR = RS, JR = RJ. In that case (if S is a simple symmetric operator), an arbitrary stable C-symmetry have the form C := C χ,ω = Je χrω = J[(cosh χ)i + (sinh χ)r ω ], where χ 0, ω [0, 2π), and R ω = Re iωj = R[cos ω + i(sin ω)j]. S. Kuzhel, C. Trunk S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 15 / 24

16 The absence of empty resolvent set. If among J-self-adjoint extensions A of a simple symmetric operator S there are no operators with empty resolvent set (i.e. σ(a) = C), then the collection of stable symmetries {C} for S is reduced to the operator J. Therefore, The concept of stable C-symmetry for the set of J-self-adjoint extensions A of S is trivial when there are no operators A with empty resolvent set. Indefinite Sturm-Liouville operators a(y)(x) = (sgn x)( y (x) + q(x)y(x)), x R with a real potential q L 1 loc (R), D is the maximal domain for a(y). Assuming the limit point case of a(y) at both and + S = (sgn x) ( d 2 ) dx 2 + q, D(S) = {y D y(0) = y (0) = 0} with deficiency indices < 2, 2 > and SJ = JS, J = (sgn x). S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 16 / 24

17 Remarks&Conclusions Relationship with the Clifford algebra Cl 2 The operators J and R can be interpreted as basis (generating) elements of the complex Clifford algebra Cl 2 = span{i, J, R, JR}. Thus, S commutes with an arbitrary element of Cl 2 there are J-self-adjoint extensions of S with empty resolvent set. In that case, all operators C which realize the property of stable C-symmetry for J-self-adjoint extensions of S are expressed in terms of Cl 2 Theorem If S commute with elements of the Clifford algebra Cl 2 = span{i, J, R, JR}, then for an arbitrary nontrivial fundamental symmetry J Cl 2 there exists J-self-adjoint extensions of S with empty resolvent set. S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 17 / 24

18 Exactly solvable models. Description of J-self-adjoint extensions with stable C-symmetry. Let S commute with Cl 2. Then ( a J-self-adjoint extension A U has stable qe C-symmetry U = e iφ iγ re iξ ), where one of the following conditions are satisfied: q = 0; re iξ qe iγ 0 < q < cos φ, where q = tanh χ cos φ. In the first case (q = 0 and φ, ξ [0, 2π) are arbitrary), the operator A U has J-symmetry (trivial case). In the second case, (0 < q < cos φ ) the operator A U has the C χ,ω -symmetry, where ω = γ and χ is determined by the relation q = tanh χ cos φ. S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 18 / 24

19 Exactly solvable models The description of domain of stable C-symmetry A J-self-adjoint extension A U has stable C χ,ω -symmetry if and only if the matrix U takes the form U = e iφ tanh χ cos φe iω 1 cosh χ 1 + sinh 2 χ sin 2 φe iξ, 1 + sinh 2 χ sin 2 φe iξ tanh χ cos φe iω 1 cosh χ where φ, ξ [0, 2π). Krein resolvent formula for operators A with stable symmetry? YES! Spectral analysis for operators A with stable symmetry? YES! What happened on the boundary of the domain of stable C-symmetry? The boundary in the formula above when χ. S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 19 / 24

20 Exactly solvable models. The boundary We will say that the operator A U belongs to the boundary of the domain of stable C-symmetry when ( ) ± cos φe iω + sin φe iξ U = e iφ, sin φe iξ ± cos φe iω (the limit of U above when χ ) The operators A U with empty resolvent set are situated at the boundary; A U has empty resolvent set φ = 0 The operators A U which are commutes with an arbitrary stable C-symmetry are situated at the boundary and they correspond to the case φ { π 2, 3π 2 } If ρ(a U ) is non-empty, then ρ(a U ) is real. S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 20 / 24

21 Exactly solvable models. Degenerate Sturm-Liouville problems on [ 1, 1]. The symmetric operator: Sy = y D(S) = {y W 2 2 ( 1, 0) W 2 2 (0, 1) y(0±) = y (0±) = y(±1) = 0} The operator S has the deficiency indices < 2, 2 > and it commutes with the fundamental symmetry Jy(x) = (sgn x)y(x) in H = L 2 ( 1, 1). All possible J-self-adjoint extensions A(= A γ ) of S with empty resolvent set: A γ y = y D(A γ ) = {y W 2 2 ( 1, 0) W 2 2 (0, 1) e iγ y(0+) = y(0 ) e iγ y (0+) = y (0 ), y(±1) = 0}, where γ [0, 2π) is an arbitrary parameter. S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 21 / 24

22 Exactly solvable models. Degenerate Sturm-Liouville problems on [ 1, 1]. The same symmetric operator: Sy = y D(S) = {y W 2 2 ( 1, 0) W 2 2 (0, 1) y(0±) = y (0±) = y(±1) = 0} commutes with span{i, J, P, JP}. Let us choose P as a fundamental symmetry in L 2 ( 1, 1). Then all P-self-adjoint extensions A(= A γ ) of S with empty resolvent set are determined by the formulas A γ y = y, y D(A γ ) y W2 2( 1, 0) W 2 2 (0, 1), y(±1) = 0, and e iγ [y(0+) + y(0 )] = y(0+) y(0 ) e iγ [y (0+) + y (0 )] = y (0+) y (0 ), γ [0, 2π), where γ [0, 2π) is an arbitrary parameter. S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 22 / 24

23 Impulse operator with point perturbation. Consider the symmetric operator S = i d dx, D(S) = {y W 1 2 (R, C 2 ) y(0) = 0} in the Hilbert space L 2 (R, C 2 ) := L 2 (R) C 2. The operator S has deficiency indices < 2, 2 > and it commutes (in L 2 (R, C 2 )) with the fundamental symmetry ( ) ( ) y1 y1 Jy = J =. y 2 y 2 Theorem J-Self-adjoint extensions A U (= A φγ ) of S with empty resolvent set are defined by the formulas: A φγ y = iy D(A φγ ) = { y = ( y1 y 2 ) W2 1 (R, C 2 ) y 2 (0+) = e i(γ+φ) y 1 (0+) y 2 (0 ) = e i(γ φ) y 1 (0 ) S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 23 / 24 }.

24 The End Thank you! S. Kuzhel (Institute of Mathematics) On self-adjoint operators in Krein space with empty resolvent set ss 24 / 24