Haydock s recursive solution of self-adjoint problems. Discrete spectrum

Haydock s recursive solution of self-adjoint problems. Discrete spectrum Alexander Moroz Wave-scattering.com wavescattering@yahoo.com January 3, 2015 Alexander Moroz (WS) Recursive solution January 3, 2015 1 / 37

Overview Haydock s recursive solution Alexander Moroz (WS) Recursive solution January 3, 2015 2 / 37

Overview Haydock s recursive solution Novel numerical method Alexander Moroz (WS) Recursive solution January 3, 2015 2 / 37

Overview Haydock s recursive solution Novel numerical method Rabi model Alexander Moroz (WS) Recursive solution January 3, 2015 2 / 37

Overview Haydock s recursive solution Novel numerical method Rabi model Exactly solvable models Alexander Moroz (WS) Recursive solution January 3, 2015 2 / 37

Overview Haydock s recursive solution Novel numerical method Rabi model Exactly solvable models Quasi-exactly solvable (QES) models Alexander Moroz (WS) Recursive solution January 3, 2015 2 / 37

Overview Haydock s recursive solution Novel numerical method Rabi model Exactly solvable models Quasi-exactly solvable (QES) models Conclusions Alexander Moroz (WS) Recursive solution January 3, 2015 2 / 37

The message I m attempting to transmit Tridiagonality is fundamental Alexander Moroz (WS) Recursive solution January 3, 2015 3 / 37

The message I m attempting to transmit Tridiagonality is fundamental Three-term recurrence relations are indispensable Alexander Moroz (WS) Recursive solution January 3, 2015 3 / 37

The message I m attempting to transmit Tridiagonality is fundamental Three-term recurrence relations are indispensable Orthogonal polynomial systems are unavoidable Alexander Moroz (WS) Recursive solution January 3, 2015 3 / 37

The message I m attempting to transmit Tridiagonality is fundamental Three-term recurrence relations are indispensable Orthogonal polynomial systems are unavoidable an efficient numerical method for energy levels characterization of integrable and QES models in terms of intrinsic polynomial properties Alexander Moroz (WS) Recursive solution January 3, 2015 3 / 37

The message I m attempting to transmit Tridiagonality is fundamental Three-term recurrence relations are indispensable Orthogonal polynomial systems are unavoidable an efficient numerical method for energy levels characterization of integrable and QES models in terms of intrinsic polynomial properties confine integrable spectra to four basic classes Alexander Moroz (WS) Recursive solution January 3, 2015 3 / 37

The message I m attempting to transmit Tridiagonality is fundamental Three-term recurrence relations are indispensable Orthogonal polynomial systems are unavoidable an efficient numerical method for energy levels characterization of integrable and QES models in terms of intrinsic polynomial properties confine integrable spectra to four basic classes characterization of chaos? Alexander Moroz (WS) Recursive solution January 3, 2015 3 / 37

Trivial and always valid statement For any linear self-adjoint Ĥ there always exists an orthonormal basis {e n } n=0 such that eigenstates E of Ĥ Ĥ E = E E can be expanded as E = p n (E)e n n=0 Alexander Moroz (WS) Recursive solution January 3, 2015 4 / 37

Haydock s recursive solution (H1) the expansion coefficients {p n (E)} are polynomials with degree p n = n, orthonormal with respect to the density of states (DOS), n 0 (E), p n (E)p m (E)n 0 (E)dE = δ nm (H2) energy eigenstates E are the generating function of the orthogonal polynomials (H3) the orthogonality of the energy eigenstates, E E = δ EE yields a dual orthogonality relation n 0 (E) p n (E)p n (E ) = δ EE n=0 where E and E are both eigenvalues Alexander Moroz (WS) Recursive solution January 3, 2015 5 / 37

Tridiagonality is generic There always exists an orthonormal basis {e n } n=0 self-adjoint operator takes on a tridiagonal form such that a given He n = a n e n + b n+1 e n+1 + b n e n 1 (1) with real recurrence coefficients {a n } and {b n }, where b n 0, n 0. Alexander Moroz (WS) Recursive solution January 3, 2015 6 / 37

Proof of He n = a n e n + b n+1 e n+1 + b n e n 1 n = 0: find a 0, b 1, e 1 He 0 = a 0 e 0 + b 1 e 1 a 0 = e 0, He 0 b 2 1 = (H a 0)e 0, (H a 0 )e 0 e 1 = (H a 0 )e 0 /b 1 Alexander Moroz (WS) Recursive solution January 3, 2015 7 / 37

Proof of He n = a n e n + b n+1 e n+1 + b n e n 1 n = 0: find a 0, b 1, e 1 He 0 = a 0 e 0 + b 1 e 1 a 0 = e 0, He 0 b 2 1 = (H a 0)e 0, (H a 0 )e 0 e 1 = (H a 0 )e 0 /b 1 Induction step to determine a n, b n+1, e n+1 e n 1, He n = e n, He n 1 b n a n = e n, He n b n+1 e n+1 = (H a n )e n b n e n 1 b 2 n+1 = (H a n)e n b n e n 1, (H a n )e n b n e n 1 e n+1 = [(H a n )e n b n e n 1 ]/b n+1 By construction e n+1 normalized to one and orthogonal to e n and e n 1. Alexander Moroz (WS) Recursive solution January 3, 2015 7 / 37

Final check: e n+1 orthogonal to e n 2,...,e 0 Tridiagonality of He n, self-adjointness, and normalizability b n+1 e m, e n+1 = e m, He n = He m, e n Alexander Moroz (WS) Recursive solution January 3, 2015 8 / 37

Final check: e n+1 orthogonal to e n 2,...,e 0 Tridiagonality of He n, self-adjointness, and normalizability b n+1 e m, e n+1 = e m, He n = He m, e n Tridiagonality for m < n 1 He m is a linear combination of e m 1, e m, and e m+1, all of which have zero overlap with e n if m + 1 < n. Alexander Moroz (WS) Recursive solution January 3, 2015 8 / 37

Relation to orthogonal polynomials I Wave function expansion E = p n (E)e n n=0 Alexander Moroz (WS) Recursive solution January 3, 2015 9 / 37

Relation to orthogonal polynomials I Wave function expansion E = p n (E)e n n=0 Three-term recurrence relation (TTRR) Expansion coefficients satisfy Ep n (E) = a n p n (E) + b n+1 p n+1 (E) + b n p n 1 (E) (2) with b k 0 and an initial condition p 0 = 1 and p 1 = 0. {p n (E)} are by the very definition orthogonal polynomials. Alexander Moroz (WS) Recursive solution January 3, 2015 9 / 37

Favard s theorem (Theorem I-4.4 of Chihara s book) For given arbitrary sequences of complex numbers {c n } and {λ n } in the TTRR xp n = P n+1 + c n P n + λ n P n 1 there always exists a moment functional, that is a linear functional L acting in the space of (complex) monic polynomials C[E], such that the polynomials P n defined by the TTRR are orthogonal under L: L(P k P l ) = 0 k l N The functional L is unique if we impose the normalization condition L(P 0 ) = L(1) = µ 0, where µ 0 is a chosen positive constant. Alexander Moroz (WS) Recursive solution January 3, 2015 10 / 37

Relation to orthogonal polynomials II Shorthand for basis e n = p n (H)e 0 Alexander Moroz (WS) Recursive solution January 3, 2015 11 / 37

Relation to orthogonal polynomials II Shorthand for basis e n = p n (H)e 0 Orthogonality relations e m, e n = p m (H)e 0, p n (H)e 0 = δ mn Alexander Moroz (WS) Recursive solution January 3, 2015 11 / 37

Relation to orthogonal polynomials III Change to the orthonormal basis of eigenstates e 0 = k ω k ψ(e k ) Alexander Moroz (WS) Recursive solution January 3, 2015 12 / 37

Relation to orthogonal polynomials III Change to the orthonormal basis of eigenstates e 0 = k ω k ψ(e k ) Orthogonal polynomials of a discrete variable e m, e n = k ω k 2 p m (E k )p n (E k ) = δ mn Alexander Moroz (WS) Recursive solution January 3, 2015 12 / 37

Relation to orthogonal polynomials III Change to the orthonormal basis of eigenstates e 0 = k ω k ψ(e k ) Orthogonal polynomials of a discrete variable e m, e n = k ω k 2 p m (E k )p n (E k ) = δ mn Weight function is the local density of function (DOS) n(e) = k ω k 2 δ(e E k ) Alexander Moroz (WS) Recursive solution January 3, 2015 12 / 37

Weight function more rigorously (pp. 62-63 of Chihara s book) lim n x nl = ξ l σ lim j ξ j Alexander Moroz (WS) Recursive solution January 3, 2015 13 / 37

Weight function more rigorously (pp. 62-63 of Chihara s book) lim n x nl = ξ l σ lim j ξ j (a) ξ l = σ = (l 1) Alexander Moroz (WS) Recursive solution January 3, 2015 13 / 37

Weight function more rigorously (pp. 62-63 of Chihara s book) lim n x nl = ξ l σ lim j ξ j (a) ξ l = σ = (l 1) (b) < ξ 1 < ξ 2 <... < ξ l = σ for some l 1 Alexander Moroz (WS) Recursive solution January 3, 2015 13 / 37

Weight function more rigorously (pp. 62-63 of Chihara s book) lim n x nl = ξ l σ lim j ξ j (a) ξ l = σ = (l 1) (b) < ξ 1 < ξ 2 <... < ξ l = σ for some l 1 (c) < ξ 1 < ξ 2 <... < ξ l <... < σ = Alexander Moroz (WS) Recursive solution January 3, 2015 13 / 37

Weight function is the set of limit points of flows of zeros Alexander Moroz (WS) Recursive solution January 3, 2015 14 / 37

Numerical recipe Choose N c N 0 and determine the first N 0 zeros x Ncl, l N 0, of P Nc (x). Usually a good starting point is to take N c N 0 + 20. Because P Nc (x) has N c simple zeros, any omission of a zero could be easily identified. Gradually increase the cut-off value of N c. The latter is what drives the incessant flows of polynomial zeros x Ncl, wherein each flow is characterized by the parameter l. Monitor convergence of the respective flows induced by the very first n zeros of P Nc (x). Each flow is a monotonically decreasing sequence having necessary a fixed limit point. Terminate your calculations when the N 0 -th zero of P Nc (x) converged to ξ N0 within predetermined accuracy. Then as a rule all other flows x Ncl with l < N 0 have converged, too. Alexander Moroz (WS) Recursive solution January 3, 2015 15 / 37

Rabi model Hamiltonian ĤR Ĥ R = ω1â â + gσ 1 (â + â) + µσ 3 ω 0... TLS resonance frequency, ω... cavity mode frequency µ = ω 0 /2, [â, â ] = 1, g... a coupling constant 1... the unit matrix σ j... the Pauli matrices Alexander Moroz (WS) Recursive solution January 3, 2015 16 / 37

Rabi model Hamiltonian ĤR Ĥ R = ω1â â + gσ 1 (â + â) + µσ 3 ω 0... TLS resonance frequency, ω... cavity mode frequency µ = ω 0 /2, [â, â ] = 1, g... a coupling constant 1... the unit matrix σ j... the Pauli matrices In what follows: κ = g/ω, = µ/ω, = 1 Alexander Moroz (WS) Recursive solution January 3, 2015 16 / 37

Convergence toward the 1000th eigenvalue of the Rabi model in positive parity eigenspace n,999-999 10 0,1 1E-3 1E-5 1E-7 1E-9 1E-11 κ=0.2, =0.4 κ=1.0, =0.7 1E-13 0 50 100 150 (n-1000)/ ɛ 999 = 998.907883759510, 997.950425260357, 973.989087026621 for (κ, ) = (0.2, 0.4), (1, 0.7), (5, 0.4), respectively Alexander Moroz (WS) Recursive solution January 3, 2015 17 / 37

An approximation of the spectrum ɛ l = l κ 2 of displaced harmonic oscillator by the zeros x nl nl - l 10 0,1 1E-3 1E-5 1E-7 10 0,1 1E-3 1E-5 1E-7 10 0,1 1E-3 1E-5 1E-7 1 2 50 60 70 80 90 100 1 2 400 420 440 460 480 500 1 2 860 880 900 920 940 960 980 1000 l κ=2 n=100 n=500 n=1000 Alexander Moroz (WS) Recursive solution January 3, 2015 18 / 37

I m looking for somebody to analyze level statistics. My input: Thousands of energy levels per parity subspace. Alexander Moroz (WS) Recursive solution January 3, 2015 19 / 37

Comparison with Braak s PRL 107, 100401 (2011) A transcendental function is constructed [ G ± (ζ) = K n (ζ, κ) 1 ] κ n ζ n n=0 and one determines eigenvalues as zeros of G ± (ζ). Alexander Moroz (WS) Recursive solution January 3, 2015 20 / 37

Comparison with Braak s PRL 107, 100401 (2011) A transcendental function is constructed [ G ± (ζ) = K n (ζ, κ) 1 ] κ n ζ n n=0 and one determines eigenvalues as zeros of G ± (ζ). The expansion coefficients K n are determined by a TTRR φ n+1 f n(ζ) (n + 1) φ n + 1 n + 1 φ n 1 = 0 where f n (ζ) = 2κ + 1 ) (n ζ 2. 2κ n ζ Alexander Moroz (WS) Recursive solution January 3, 2015 20 / 37

Shortcomings of Braak s approach Relies heavily on unproven hypotheses that G ± (ζ) takes on a zero value between its subsequent poles at ζ = n and ζ = n + 1 once - by implicitly presuming that at one of the poles G ± (ζ) goes to + and at the neighboring pole goes to, with a monotonic behavior from + to between the poles; twice - implicitly presuming that G ± (ζ) goes to one of ± at both subsequent poles, and in between the poles it has rather a featureless behavior, e.g., similar to a cord hanging on two posts; none - occurs under the similar circumstances as described in the previous item, if the cord is too short, e.g., it does not stretch sufficiently up or down so as to cross the abscissa. Alexander Moroz (WS) Recursive solution January 3, 2015 21 / 37

Shortcomings of Braak s approach Relies heavily on unproven hypotheses that G ± (ζ) takes on a zero value between its subsequent poles at ζ = n and ζ = n + 1 once - by implicitly presuming that at one of the poles G ± (ζ) goes to + and at the neighboring pole goes to, with a monotonic behavior from + to between the poles; twice - implicitly presuming that G ± (ζ) goes to one of ± at both subsequent poles, and in between the poles it has rather a featureless behavior, e.g., similar to a cord hanging on two posts; none - occurs under the similar circumstances as described in the previous item, if the cord is too short, e.g., it does not stretch sufficiently up or down so as to cross the abscissa. Numerically no advantage over the classical Schweber s solution [Ann. Phys. (N.Y.) 41, 205 (1967)]. Alexander Moroz (WS) Recursive solution January 3, 2015 21 / 37

Comparison with Braak s PRL 107, 100401 (2011) - II The TTRR for the coefficients K n is not a fundamental one, i.e. not of Haydock s type, because of f n (ζ) is not a linear function of energy. Any relation with the orthogonal polynomials {p n } is eluded. Alexander Moroz (WS) Recursive solution January 3, 2015 22 / 37

Comparison with Braak s PRL 107, 100401 (2011) - II The TTRR for the coefficients K n is not a fundamental one, i.e. not of Haydock s type, because of f n (ζ) is not a linear function of energy. Any relation with the orthogonal polynomials {p n } is eluded. Instead at looking straight at the zeros of p N of sufficiently large degree, a complicated composite object, G ± (ζ), is formed. Alexander Moroz (WS) Recursive solution January 3, 2015 22 / 37

Is Rabi model integrable or solvable? Alexander Moroz (WS) Recursive solution January 3, 2015 23 / 37

Structure relation and bispectral condition Divided-difference operator or discrete derivative D x f (x) = f (ι +(x)) f (ι (x)) ι + (x) ι (x) Alexander Moroz (WS) Recursive solution January 3, 2015 24 / 37

Structure relation and bispectral condition Divided-difference operator or discrete derivative Structure relation D x f (x) = f (ι +(x)) f (ι (x)) ι + (x) ι (x) D x p n (x) = B n (x)p n (x) + A n (x)p n 1 (x) where D x : Π n [x] Π n 1 [x], Π n [x] being the linear space of polynomials in x over C with degree at most n Z 0 Alexander Moroz (WS) Recursive solution January 3, 2015 24 / 37

Consequences of a structure relation if there is one structure relation there is another - simply apply the fundamental TTRR Alexander Moroz (WS) Recursive solution January 3, 2015 25 / 37

Consequences of a structure relation if there is one structure relation there is another - simply apply the fundamental TTRR a pair of mutually adjoint raising and lowering ladder operators orthogonal polynomials satisfy in general a second-order difference equation = bispectral property Alexander Moroz (WS) Recursive solution January 3, 2015 25 / 37

Consequences of a structure relation if there is one structure relation there is another - simply apply the fundamental TTRR a pair of mutually adjoint raising and lowering ladder operators orthogonal polynomials satisfy in general a second-order difference equation = bispectral property allows one to introduce a discrete analogue of the Bethe Ansatz equations Alexander Moroz (WS) Recursive solution January 3, 2015 25 / 37

Consequences of a structure relation if there is one structure relation there is another - simply apply the fundamental TTRR a pair of mutually adjoint raising and lowering ladder operators orthogonal polynomials satisfy in general a second-order difference equation = bispectral property allows one to introduce a discrete analogue of the Bethe Ansatz equations the polynomials are hypergeometric of Askey-Wilson type Alexander Moroz (WS) Recursive solution January 3, 2015 25 / 37

D x requires four primary classes of special non-uniform lattices 1 the linear lattice Alexander Moroz (WS) Recursive solution January 3, 2015 26 / 37

D x requires four primary classes of special non-uniform lattices 1 the linear lattice 2 the linear q-lattice Alexander Moroz (WS) Recursive solution January 3, 2015 26 / 37

D x requires four primary classes of special non-uniform lattices 1 the linear lattice 2 the linear q-lattice 3 the quadratic lattice Alexander Moroz (WS) Recursive solution January 3, 2015 26 / 37

D x requires four primary classes of special non-uniform lattices 1 the linear lattice 2 the linear q-lattice 3 the quadratic lattice 4 the q-quadratic lattice Alexander Moroz (WS) Recursive solution January 3, 2015 26 / 37

D x requires four primary classes of special non-uniform lattices 1 the linear lattice 2 the linear q-lattice 3 the quadratic lattice 4 the q-quadratic lattice Either Λ = {x x = u 2 n 2 + u 1 n + u 0, n N} Alexander Moroz (WS) Recursive solution January 3, 2015 26 / 37

D x requires four primary classes of special non-uniform lattices 1 the linear lattice 2 the linear q-lattice 3 the quadratic lattice 4 the q-quadratic lattice Either or Λ = {x x = u 2 n 2 + u 1 n + u 0, n N} Λ = {x x = u 2 q n + u 1 q n + u 0, n N} where u j are real constants and the real parameter q satisfies 0 < q < 1 Alexander Moroz (WS) Recursive solution January 3, 2015 26 / 37

Is Rabi model integrable or solvable? Alexander Moroz (WS) Recursive solution January 3, 2015 27 / 37

Characterization of tridiagonal He n 1 = a n 1 e n 1 + b n e n + b n 1 e n 2 (A) b n = 0 for some n = N > 0 Alexander Moroz (WS) Recursive solution January 3, 2015 28 / 37

Characterization of tridiagonal He n 1 = a n 1 e n 1 + b n e n + b n 1 e n 2 (A) b n = 0 for some n = N > 0 (B) b n 0 for all n 0 Alexander Moroz (WS) Recursive solution January 3, 2015 28 / 37

He n 1 = a n 1 e n 1 + b n e n + b n 1 e n 2 with b N = 0 for some N > 0 p N (E) enters the TTRR first for n = N. One has p n+n (E) = p N (E)Q n (E) (n 0) i.e. p N +1 (E) is proportional to p N (E). The quotient polynomials Q n (E) form a new orthogonal sequence of polynomials. The polynomials {p n (E)} N 1 n=0 decouple from {p n(e)} n=n. Alexander Moroz (WS) Recursive solution January 3, 2015 29 / 37

He n 1 = a n 1 e n 1 + b n e n + b n 1 e n 2 with b N = 0 for some N > 0 p N (E) enters the TTRR first for n = N. One has p n+n (E) = p N (E)Q n (E) (n 0) i.e. p N +1 (E) is proportional to p N (E). The quotient polynomials Q n (E) form a new orthogonal sequence of polynomials. The polynomials {p n (E)} N n=0 1 decouple from {p n(e)} n=n. The operator Ĥ has necessary a finite dimensional invariant subspace V N = {e n } N n=0 1. Alexander Moroz (WS) Recursive solution January 3, 2015 29 / 37

Quasi-exact solvabality of a linear differential operator Ĥ Ĥ preserves a finite dimensional subspace V N of a Hilbert space L 2 (S) (S R), ĤV N V N, dim V N = N <, on which the operator Ĥ is naturally defined Alexander Moroz (WS) Recursive solution January 3, 2015 30 / 37

Quasi-exact solvabality of a linear differential operator Ĥ Ĥ preserves a finite dimensional subspace V N of a Hilbert space L 2 (S) (S R), ĤV N V N, dim V N = N <, on which the operator Ĥ is naturally defined The basis of V N admits an explicit analytic form V N = span {e 0 (x),..., e N 1 (x)} Alexander Moroz (WS) Recursive solution January 3, 2015 30 / 37

Basic properties of finite OPS defined by xp n = P n+1 + c n P n + λ n P n 1 If λ k > 0 and c k are real for k = 0, 1,..., N 1 then: Alexander Moroz (WS) Recursive solution January 3, 2015 31 / 37

Basic properties of finite OPS defined by xp n = P n+1 + c n P n + λ n P n 1 If λ k > 0 and c k are real for k = 0, 1,..., N 1 then: the polynomials of the resulting finite orthogonal polynomial sequence {P k } N k=1 have real and simple zeros the zeros of {P k } N k=1 interlace (the Sturm property) ν P (E) = N 1 k=0 w k θ(e E k ), where θ(x) is Heaviside s step function, E k are the N zeros of P N, and the coefficients w k can be found by solving N 1 l=0 P k (E l ) w l = δ k0, k = 0, 1,..., N 1. Alexander Moroz (WS) Recursive solution January 3, 2015 31 / 37

Basic properties of finite OPS defined by xp n = P n+1 + c n P n + λ n P n 1 If λ k > 0 and c k are real for k = 0, 1,..., N 1 then: the polynomials of the resulting finite orthogonal polynomial sequence {P k } N k=1 have real and simple zeros the zeros of {P k } N k=1 interlace (the Sturm property) ν P (E) = N 1 k=0 w k θ(e E k ), where θ(x) is Heaviside s step function, E k are the N zeros of P N, and the coefficients w k can be found by solving N 1 l=0 w k > 0 for 0 k < N P k (E l ) w l = δ k0, k = 0, 1,..., N 1. Alexander Moroz (WS) Recursive solution January 3, 2015 31 / 37

Basic properties of finite OPS defined by xp n = P n+1 + c n P n + λ n P n 1 If λ k > 0 and c k are real for k = 0, 1,..., N 1 then: the polynomials of the resulting finite orthogonal polynomial sequence {P k } N k=1 have real and simple zeros the zeros of {P k } N k=1 interlace (the Sturm property) ν P (E) = N 1 k=0 w k θ(e E k ), where θ(x) is Heaviside s step function, E k are the N zeros of P N, and the coefficients w k can be found by solving N 1 l=0 w k > 0 for 0 k < N dν P (E) is unique P k (E l ) w l = δ k0, k = 0, 1,..., N 1. Alexander Moroz (WS) Recursive solution January 3, 2015 31 / 37

Normalization paradox Orthogonality L[π(x)p n (x)] = 0 for every polynomial π(x) of degree k < n. Hence the norm of p n (E) for n N vanishes, i.e. 0 1 b N +1 L[E N p N (E)] Paradox One cannot satisfy (H1), i.e. that {p n (E)} are orthonormal with respect to the density of states (DOS), n 0 (E), p n (E)p m (E)n 0 (E)dE = δ nm Alexander Moroz (WS) Recursive solution January 3, 2015 32 / 37

Resolution Assume that the respective spectra obtained in V N and L 2 (S)\V N form two disjoint intervals on the real axis, S = S qes S, sup S qes < inf S. Alexander Moroz (WS) Recursive solution January 3, 2015 33 / 37

Resolution Assume that the respective spectra obtained in V N and L 2 (S)\V N form two disjoint intervals on the real axis, S = S qes S, sup S qes < inf S. On S qes is n 0 (E)dE represented by the discrete measure dν P (E) Alexander Moroz (WS) Recursive solution January 3, 2015 33 / 37

Resolution Assume that the respective spectra obtained in V N and L 2 (S)\V N form two disjoint intervals on the real axis, S = S qes S, sup S qes < inf S. On S qes is n 0 (E)dE represented by the discrete measure dν P (E) On S is n 0 (E)dE represented by dν PQ (E) = 1 p 2 N (E) dν Q(E) (E S ) Alexander Moroz (WS) Recursive solution January 3, 2015 33 / 37

Conclusions tridiagonality is fundamental Alexander Moroz (WS) Recursive solution January 3, 2015 34 / 37

Conclusions tridiagonality is fundamental TTRR are indispensable Alexander Moroz (WS) Recursive solution January 3, 2015 34 / 37

Conclusions tridiagonality is fundamental TTRR are indispensable OPS are unavoidable Alexander Moroz (WS) Recursive solution January 3, 2015 34 / 37

Conclusions tridiagonality is fundamental TTRR are indispensable OPS are unavoidable an efficient numerical method for energy levels Alexander Moroz (WS) Recursive solution January 3, 2015 34 / 37

Conclusions tridiagonality is fundamental TTRR are indispensable OPS are unavoidable an efficient numerical method for energy levels characterization of integrable and QES models in terms of intrinsic polynomial properties Alexander Moroz (WS) Recursive solution January 3, 2015 34 / 37

Conclusions tridiagonality is fundamental TTRR are indispensable OPS are unavoidable an efficient numerical method for energy levels characterization of integrable and QES models in terms of intrinsic polynomial properties confine integrable spectra to four basic classes Alexander Moroz (WS) Recursive solution January 3, 2015 34 / 37

Conclusions tridiagonality is fundamental TTRR are indispensable OPS are unavoidable an efficient numerical method for energy levels characterization of integrable and QES models in terms of intrinsic polynomial properties confine integrable spectra to four basic classes characterization of chaos? Alexander Moroz (WS) Recursive solution January 3, 2015 34 / 37

R. Haydock, p. 217 of his review: The generality of this result suggests a new way of looking at quantum mechanics. Since any quantum system can be transformed into a chain model, we need only investigate such chain models in order to see the varieties of quantum phenomena that are possible. To understand a particular physical system, we need only find the chain model appropriate to it. Alexander Moroz (WS) Recursive solution January 3, 2015 35 / 37

The End Alexander Moroz (WS) Recursive solution January 3, 2015 36 / 37

Further reading This talk has been based on excerpts from my arxiv:1205.3139; EPL 100, 60010 (2012); AP 338, 319-340 (2013); AP 340, 252-266 (2014); J. Phys. A: Math. Theor. 47(49), 495204 (2014); AP 351, 960-974 (2014) Visit me at http://www.wave-scattering.com. F77 source codes and numerical data files for the Rabi model can be freely downloaded from http://www.wave-scattering.com/rabi.html My all F77 scattering codes are available at http://www.wave-scattering.com/codes.html For list of my unfinished projects see http://www.wave-scattering.com/projects.html Alexander Moroz (WS) Recursive solution January 3, 2015 37 / 37