On some applications of Orthogonal Polynomials in ...orthonet/orthonet18/notas/ran.pdf · Wigner:[Unreasonable E ectiveness of Mathematics in the Natural Sciences, 1960] The miracle

On some applications of Orthogonal Polynomials

in Mathematical-Physics

Renato Alvarez-Nodarse

There is no branch of mathematics, however abstract, which may not some day

be applied to phenomena of the real world. N.I. Lobachevsky

III Orthonet School, Bilbao, 19-23 October 2018

Renato Alvarez-Nodarse IMUS On some applications of OP in Mathematical-Physics III Orthonet School

Introduction and Motivation


An introduction:

Galileo’s Theorem ...

Von Neumann:[The mathematician, 1947] I thinkthat it is a relatively good approximation totruth that mathematical ideas originate inempirics, although the genealogy is sometimeslong and obscure. But, once they are so concei-ved, the subject begins to live a peculiarlife of its own and is better compared to acreative one [...]

Galileo:[Il Saggiatore, 1623] [Natural] Philosophyis written in that great book which ever lies be-fore our eyes – I mean the universe – but wecannot understand it if we do not first learn thelanguage and grasp the symbols, in which it iswritten. What is that language? Mathematics!


An introduction: Galileo’s Theorem ...


Galileo:[Il Saggiatore, 1623] [Natural] Philosophyis written in that great book which ever lies be-fore our eyes – I mean the universe – but wecannot understand it if we do not first learn thelanguage and grasp the symbols, in which it iswritten.

What is that language? Mathematics!


An introduction: Galileo’s Theorem ...


Galileo:[Il Saggiatore, 1623] [Natural] Philosophyis written in that great book which ever lies be-fore our eyes – I mean the universe – but wecannot understand it if we do not first learn thelanguage and grasp the symbols, in which it iswritten. What is that language? Mathematics!


The mystery of Math & Physics:

Einstein:[Geometry and Experience, 1921] How can itbe that mathematics, being after all a product of hu-man thought which is independent of experience, is soadmirably appropriate to the objects of reality?

Wigner:[Unreasonable Effectiveness of Mathematics inthe Natural Sciences, 1960] The miracle of the ap-propriateness of the language of mathematics for theformulation of the laws of physics is a wonderful giftwhich we neither understand nor deserve.

The real miracle, is that we were able to find a way of quantifying thephenomena that surround us.


The mystery of Math & Physics:



The real miracle, is that we were able to find a way of quantifying thephenomena that surround us.


The mystery of Math & Physics: The two sides of the same coin



The real miracle, is that we were able to find a way of quantifying thephenomena that surround us. As V.I. Arnold says ...


A paradigmatic example: The classical oscillator

We first “arithmetize” the natural phenomenon, and then we usemathematical models to describe it.

m

k

0 x

~F = m · ~a ⇒ mx ′′(t) + kx(t) = 0

x(t) = A cos(ωt + δ), ω =

√k

m,

5 10 15 20 25t

-2

-1

1

2

x

E =m[x(t)′]2

2︸︷︷︸Ec

+k[x(t)]2

2︸︷︷︸V (x)

=1

2kA2 = const.

As a curiosity: The total energy is a continuous function of A and cantake any positive real value.

It is time to show what is the role of SF in all this!




m

k

0 x

~F = m · ~a ⇒ mx ′′(t) + kx(t) = 0


√k

m,

5 10 15 20 25t

-2

-1

1

2

x

E =m[x(t)′]2

2︸︷︷︸Ec

+k[x(t)]2

2︸︷︷︸V (x)

=1

2kA2 = const.






m

k

0 x

~F = m · ~a ⇒ mx ′′(t) + kx(t) = 0


√k

m,

5 10 15 20 25t

-2

-1

1

2

x

E =m[x(t)′]2

2︸︷︷︸Ec

+k[x(t)]2

2︸︷︷︸V (x)

=1

2kA2 = const.






m

k

0 x

~F = m · ~a ⇒ mx ′′(t) + kx(t) = 0


√k

m,

5 10 15 20 25t

-2

-1

1

2

x

E =m[x(t)′]2

2︸︷︷︸Ec

+k[x(t)]2

2︸︷︷︸V (x)

=1

2kA2 = const.






m

k

0 x

~F = m · ~a ⇒ mx ′′(t) + kx(t) = 0


√k

m,

5 10 15 20 25t

-2

-1

1

2

x

E =m[x(t)′]2

2︸︷︷︸Ec

+k[x(t)]2

2︸︷︷︸V (x)

=1

2kA2 = const.


It is time to show what is the role of SF in all this!Renato Alvarez-Nodarse IMUS On some applications of OP in Mathematical-Physics III Orthonet School

A pure mathematical heresy

Special Functions (SF) appear in (almost) all context of Mathematicsand other Sciences.

As Alberto Grunbaum one time said (in a summer school in 2004):“Special functions are to mathematics what pipes are to a house: nobodywants to exhibit them openly but nothing works without them”.

We will show Alberto’s statement in the context of Math-Phys.


A pure mathematical heresy

Special Functions (SF) appear in (almost) all context of Mathematicsand other Sciences.

As Alberto Grunbaum one time said (in a summer school in 2004):“Special functions are to mathematics what pipes are to a house: nobodywants to exhibit them openly but nothing works without them”.

We will show Alberto’s statement in the context of Math-Phys.Renato Alvarez-Nodarse IMUS On some applications of OP in Mathematical-Physics III Orthonet School

The quantum harmonic oscillator

In the microworld we can thing that the HO is a particle moving on apotential V (x) ∼ x2:

In Quantum Mechanics the state of the system is given by the wavefunction Ψ ∈ L2(R) which is the sol. of the stationary Schrodinger Eq.

−~2

2mΨ′′(x) +

1

2mω2x2Ψ(x) = EΨ(x), x ∈ R.

The solution is, n = 0, 1, 2, . . . , x0 =√

~/(mω),

Ψn(x) =

√2n

x0√πn!

e−x2/2x2

0Hn(x/x0), E = ~ω(n + 1/2)

where Hn are the classical Hermite monic polynomials.

The Physical interpretation: |Ψn(x)|2 define de probability density offinding the particle at position x ... and not only.





−~2

2mΨ′′(x) +

1

2mω2x2Ψ(x) = EΨ(x), x ∈ R.


~/(mω),

Ψn(x) =

√2n

x0√πn!

e−x2/2x2

0Hn(x/x0), E = ~ω(n + 1/2)







−~2

2mΨ′′(x) +

1

2mω2x2Ψ(x) = EΨ(x), x ∈ R.


~/(mω),

Ψn(x) =

√2n

x0√πn!

e−x2/2x2

0Hn(x/x0), E = ~ω(n + 1/2)







−~2

2mΨ′′(x) +

1

2mω2x2Ψ(x) = EΨ(x), x ∈ R.


~/(mω),

Ψn(x) =

√2n

x0√πn!

e−x2/2x2

0Hn(x/x0), E = ~ω(n + 1/2)





|Ψ0(x)|2 = e−x2

√π

, |Ψ1(x)|2 = 2x2e−x2

√π

, |Ψ4(x)|2 =(4x4−12x2+3)

2e−x2

24√π

|φ(x

)|2

x

n=0n=1

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

-4 -2 0 2 4

|φ(x

)|2

x

n=0n=4

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

-4 -2 0 2 4

Knowing the properties of Hn is important for applications:

1 How to compute efficiently Ψn(x)?

2 What is the distribution of zeros of Ψn(x)?

3 How to compute integrals like∫R f (x)|Ψn(x)|2dx ?



|Ψ0(x)|2 = e−x2

√π

, |Ψ1(x)|2 = 2x2e−x2

√π

, |Ψ4(x)|2 =(4x4−12x2+3)

2e−x2

24√π

|φ(x

)|2

x

n=0n=1

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

-4 -2 0 2 4

|φ(x

)|2

x

n=0n=4

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

-4 -2 0 2 4

Knowing the properties of Hn is important for applications:

1 How to compute efficiently Ψn(x)?

2 What is the distribution of zeros of Ψn(x)?

3 How to compute integrals like∫R f (x)|Ψn(x)|2dx ?


The classical vs quantum harmonic oscillator

|φ(x

)|2

x

V(x)n=0n=1n=2n=3n=4

0

1

2

3

4

5

6

-4 -2 0 2 4

Why the zero distribution is interesting?

Why we need to compute∫R f (x)|Ψn(x)|2dx?


Part I


The N-dimentional isotropic quantum harmonic oscilator (IHO)

For the N-dimensional IHO the stationary Schrodinger Eq. is

(−∆ +

1

2λ2r2

)Ψ = EΨ, ∆ =

N∑

k=1

∂2

∂x2k

, r =

√√√√n∑

k=1

x2k .

Its solution has the form Ψ = R(N)nl (r)Ylm(ΩN), where R

(N)nl (r) is the

so-called radial wave functions

R(N)nl (r) = N (N)

nl r le−12λr2

Ll+ N

2−1

n (λr2), N (N)nl =

√√√√ 2n!λl+N2

Γ(n + l + N

2

) ,

n = 0, 1, 2, . . . , l = 0, 1, 2, . . . , N ≥ 3. Lαn (z) are the Laguerre OP.




(−∆ +

1

2λ2r2

)Ψ = EΨ, ∆ =

N∑

k=1

∂2

∂x2k

, r =

√√√√n∑

k=1

x2k .


(N)nl (r) is the


R(N)nl (r) = N (N)

nl r le−12λr2

Ll+ N

2−1

n (λr2), N (N)nl =

√√√√ 2n!λl+N2

Γ(n + l + N

2

) ,

n = 0, 1, 2, . . . , l = 0, 1, 2, . . . , N ≥ 3.

Lαn (z) are the Laguerre OP.




(−∆ +

1

2λ2r2

)Ψ = EΨ, ∆ =

N∑

k=1

∂2

∂x2k

, r =

√√√√n∑

k=1

x2k .


(N)nl (r) is the


R(N)nl (r) = N (N)

nl r le−12λr2

Ll+ N

2−1

n (λr2), N (N)nl =

√√√√ 2n!λl+N2

Γ(n + l + N

2

) ,

n = 0, 1, 2, . . . , l = 0, 1, 2, . . . , N ≥ 3. Lαn (z) are the Laguerre OP.


Studying the wave functions of quantum systems: the IHO

We are interesting in finding linear relations between the functions Ψn

and its derivatives Ψ(k)n , i.e., A1Ψ

(k1)n1 + A2Ψ

(k2)n2 + A3Ψ

(k3)n2 = 0

Late us take the example of the IHO. Since

Ll+N

2−1

n (λr2) =(N (N)

n,l

)−1r−le

12λr2

R(N)nl (r)

one can look to handbooks of SF (i.e., Abramowitz and Stegun, or theBateman Project by Erdelyi et al.) and from known RRs of Laguerre OP,e.g. the differentation formula or TTRR

d

dxLαn (x) = −Lα+1

n−1 (x),

(n + 1)Lαn+1(x) = (2n + α + 1− x)Lαn (x)− (n + α)Lαn−1(x).

one can obtain RR for the radial function.

That was the usual way that physicists did, but this is not interesting !Why? You can not built the RR that you need. ???





(k1)n1 + A2Ψ

(k2)n2 + A3Ψ

(k3)n2 = 0


Ll+N

2−1

n (λr2) =(N (N)

n,l

)−1r−le

12λr2

R(N)nl (r)


d


n−1 (x),



That was the usual way that physicists did, but this is not interesting !Why? You can not built the RR that you need. ???





(k1)n1 + A2Ψ

(k2)n2 + A3Ψ

(k3)n2 = 0


Ll+N

2−1

n (λr2) =(N (N)

n,l

)−1r−le

12λr2

R(N)nl (r)


d


n−1 (x),



That was the usual way that physicists did, but this is not interesting !Why?

You can not built the RR that you need. ???





(k1)n1 + A2Ψ

(k2)n2 + A3Ψ

(k3)n2 = 0


Ll+N

2−1

n (λr2) =(N (N)

n,l

)−1r−le

12λr2

R(N)nl (r)


d


n−1 (x),



That was the usual way that physicists did, but this is not interesting !Why? You can not built the RR that you need.

???





(k1)n1 + A2Ψ

(k2)n2 + A3Ψ

(k3)n2 = 0


Ll+N

2−1

n (λr2) =(N (N)

n,l

)−1r−le

12λr2

R(N)nl (r)


d


n−1 (x),



That was the usual way that physicists did, but this is not interesting !Why? You can not built the RR that you need. ???Renato Alvarez-Nodarse IMUS On some applications of OP in Mathematical-Physics III Orthonet School

How to obtain RR for the R(N)nl (r) functions?

Theorem (Cardoso & RAN, JPA 2003 RR a la carte)

Let R(N)nl (r), R

(N)n+n1,l+l1

(r) and R(N)n+n2,l+l2

(r) be 3 different radialfunctions of the N-th dimensional I.H.O., n1, n2, l1, l2 ∈ Z

I If mın (n + n1, n + n2, l + l1, l + l2) ≥ 0 =⇒ ∃ non-vanishingpolynomials A0, A1, and A2, such that (General TTRR)

A0(r)R(N)n,l (r) + A1(r)R

(N)n+n1,l+l1

(r) + A2(r)R(N)n+n2,l+l2

(r) = 0.

I If mın (n + n1, l + l1) ≥ 0 and (n1)2 + (l1)2 6= 0 =⇒ ∃ 6= 0polynomials A0, A1, and A2, s.t. (ladder-type RR)

A0(r)R(N)n,l (r) + A1(r)

d

drR

(N)n,l (r) + A2(r)R

(N)n+n1,l+l1

(r) = 0.

Proof: It is based on an Lemma by NU in Special Functions ofMath-Phys (1972 and 1988).




Let R(N)nl (r), R

(N)n+n1,l+l1





(N)n+n1,l+l1

(r) + A2(r)R(N)n+n2,l+l2

(r) = 0.


A0(r)R(N)n,l (r) + A1(r)

d

drR

(N)n,l (r) + A2(r)R

(N)n+n1,l+l1

(r) = 0.

Proof: It is based on an Lemma by NU in Special Functions ofMath-Phys (1972 and 1988).Renato Alvarez-Nodarse IMUS On some applications of OP in Mathematical-Physics III Orthonet School



Let R(N)nl (r), R

(N)n+n1,l+l1





(N)n+n1,l+l1

(r) + A2(r)R(N)n+n2,l+l2

(r) = 0.


A0(r)R(N)n,l (r) + A1(r)

d

drR

(N)n,l (r) + A2(r)R

(N)n+n1,l+l1

(r) = 0.

Let see some examples of how use the above to obtain relations betweenthree different radial functions of the I.H.O.Renato Alvarez-Nodarse IMUS On some applications of OP in Mathematical-Physics III Orthonet School

Example: From the relation between R(N)n,l (r) and L

l+N2−1

n (λr 2)

A0R(N)n,l + A1R

(N)n+n1,l+l1

+ A2R(N)n+n2,l+l2

= 0

mC0L

l+ N2 −1

n + C1L(l+l1)+ N

2 −1n+n1

+ C2L(l+l2)+ N

2 −1n+n2

= 0

• Substituting n1 = −1, n2 = 1, l1 = l2 = 0 and α = l + N2 − 1 ⇒

C0Lαn (s) + C1L

αn−1(s) + C2L

αn+1(s) = 0.

Comparing it with the TTRR ⇒ C0 = s − (2n + α + 1), C1 = n + α and

C2 = n + 1. Coming back to R(N)n,l

√n

(n+l+

N

2−1

)R

(N)n−1,l(r)+

[λr2−

(2n+l+

N

2

)]R

(N)n,l (r)

+

√(n+1)

(n+l+

N

2

)R

(N)n+1,l(r) = 0.



l+N2−1

n (λr 2)

A0R(N)n,l + A1R

(N)n+n1,l+l1

+ A2R(N)n+n2,l+l2

= 0

mC0L

l+ N2 −1

n + C1L(l+l1)+ N

2 −1n+n1

+ C2L(l+l2)+ N

2 −1n+n2

= 0


C0Lαn (s) + C1L

αn−1(s) + C2L

αn+1(s) = 0.


C2 = n + 1.

Coming back to R(N)n,l

√n

(n+l+

N

2−1

)R

(N)n−1,l(r)+

[λr2−

(2n+l+

N

2

)]R

(N)n,l (r)

+

√(n+1)

(n+l+

N

2

)R

(N)n+1,l(r) = 0.



l+N2−1

n (λr 2)

A0R(N)n,l + A1R

(N)n+n1,l+l1

+ A2R(N)n+n2,l+l2

= 0

mC0L

l+ N2 −1

n + C1L(l+l1)+ N

2 −1n+n1

+ C2L(l+l2)+ N

2 −1n+n2

= 0


C0Lαn (s) + C1L

αn−1(s) + C2L

αn+1(s) = 0.


C2 = n + 1. Coming back to R(N)n,l

√n

(n+l+

N

2−1

)R

(N)n−1,l(r)+

[λr2−

(2n+l+

N

2

)]R

(N)n,l (r)

+

√(n+1)

(n+l+

N

2

)R

(N)n+1,l(r) = 0.


Another example of application: ladder operators

A0R(N)n,l (r) + A1

d

d rR

(N)n,l (r) + A2R

(N)n+n1,l+l1

(r) = 0

mB0L

l+N2−1

n (s) + B1Ll+N

2n−1 (s) + B2L

(l+l1)+N2−1

n+n1(s) = 0

For example, set n1 = −1, l1 = 1, and α = l + N2 − 1 ⇒

B0Lαn (s) + B1L

α+1n−1 (s) + B2L

(α+1)n−1 (s) = 0.

No one of the RR of Laguerre pol. helps! But we can use

d


n−1 (x) ⇒ B0Lαn (s)− (B1 + B2)(Lαn )′(s) = 0.

Since Lαn , (Lαn )′ are l.i. ⇒ B0 = 0, B1 = −B2. Thus choosing B2 = 1

[d

dr+ λr − l

r

]R

(N)n,l (r) = −2

√λnR

(N)n−1,l+1(r).



A0R(N)n,l (r) + A1

d

d rR

(N)n,l (r) + A2R

(N)n+n1,l+l1

(r) = 0

mB0L

l+N2−1

n (s) + B1Ll+N

2n−1 (s) + B2L

(l+l1)+N2−1

n+n1(s) = 0


B0Lαn (s) + B1L

α+1n−1 (s) + B2L

(α+1)n−1 (s) = 0.


d


n−1 (x) ⇒ B0Lαn (s)− (B1 + B2)(Lαn )′(s) = 0.


[d

dr+ λr − l

r

]R

(N)n,l (r) = −2

√λnR

(N)n−1,l+1(r).



A0R(N)n,l (r) + A1

d

d rR

(N)n,l (r) + A2R

(N)n+n1,l+l1

(r) = 0

mB0L

l+N2−1

n (s) + B1Ll+N

2n−1 (s) + B2L

(l+l1)+N2−1

n+n1(s) = 0


B0Lαn (s) + B1L

α+1n−1 (s) + B2L

(α+1)n−1 (s) = 0.


d


n−1 (x) ⇒ B0Lαn (s)− (B1 + B2)(Lαn )′(s) = 0.


[d

dr+ λr − l

r

]R

(N)n,l (r) = −2

√λnR

(N)n−1,l+1(r).


Why is so interesting such a method?

Using the NU method we can find recurrences a la carte

Next step: In [Cardoso & RAN, JPA 2003] we have obtained severalRRs. But, which of these RRs are useful for generating numerically the

radial wave functions R(N)n,l (r) ?

The answer: Using a comparative numerical analysis of the obtained RRto numerically obtain the corresponding eigenfunctions we found [RAN,Cardoso & Quintero, ETNA 2006] that:

1 the TTRR obtained from the TTRR of Laguerre polynomials are themost effective way of computing numerically the values of the radialwave functions,

2 the raising-type ladder relation combined with the above RR is themost efficient way to compute the derivatives of the radial wavefunctions.

For other SF (not OP) [Cardoso, Fernandes & RAN, ETNA 2009]















The answer:

Using a comparative numerical analysis of the obtained RRto numerically obtain the corresponding eigenfunctions we found [RAN,Cardoso & Quintero, ETNA 2006] that:







































For other SF (not OP) [Cardoso, Fernandes & RAN, ETNA 2009]Renato Alvarez-Nodarse IMUS On some applications of OP in Mathematical-Physics III Orthonet School

Part II


Factorization method of Schrodinger 1940, Infeld and Hull 1951

Given a Hamiltonian, that is a 2o order diff. operator

Hϕn = λnϕn, H de orden 2

To find 1o order diff operators a and a+ such that

H = a+a, a+ϕn = αnϕn+1, aϕn = βnϕn−1, (a+)∗ = a, a∗ = a+.

Interest: Solving aϕ0 = 0, one gets ϕ0, and a+ϕn generate the others.


Factorization method of Schrodinger 1940, Infeld and Hull 1951

Given a Hamiltonian, that is a 2o order diff. operator

Hϕn = λnϕn, H de orden 2

To find 1o order diff operators a and a+ such that

H = a+a, a+ϕn = αnϕn+1, aϕn = βnϕn−1, (a+)∗ = a, a∗ = a+.

Interest: Solving aϕ0 = 0, one gets ϕ0, and a+ϕn generate the others.


How this work for the quantum harmonic oscillator?

HΨn(x) := −Ψ′′n(x) + x2Ψn(x) = (H−H+ + I )Ψn(x) = λnΨn(x).

a := H+ = x +d

dx, a+ := H− = x − d

dxm

H+Ψ0 = 0, H+Ψn =√

2nΨn−1, H−Ψn =√

2n + 2 Ψn+1.

How it works?

xΨ0(x) + Ψ′0(x) = 0 ⇒ Ψ0(x) =1

4√πe−x

2/2,

and [H−]nΨ0(x) =√

(2n)!!Ψn(x), thus

Ψn(x) =1

π14

√(2n)!!

[H−]ne−x2/2 =

1

π14

√(2n)!!

[xI − d

dx

]ne−x

2/2.

This the classical algebraic realization of the quantum oscillator.




a := H+ = x +d

dx, a+ := H− = x − d

dxm

H+Ψ0 = 0, H+Ψn =√


2n + 2 Ψn+1.

How it works?

xΨ0(x) + Ψ′0(x) = 0 ⇒ Ψ0(x) =1

4√πe−x

2/2,


(2n)!!Ψn(x), thus

Ψn(x) =1

π14

√(2n)!!

[H−]ne−x2/2 =

1

π14

√(2n)!!

[xI − d

dx

]ne−x

2/2.

This the classical algebraic realization of the quantum oscillator.




a := H+ = x +d

dx, a+ := H− = x − d

dxm

H+Ψ0 = 0, H+Ψn =√


2n + 2 Ψn+1.

How it works?

xΨ0(x) + Ψ′0(x) = 0 ⇒ Ψ0(x) =1

4√πe−x

2/2,


(2n)!!Ψn(x), thus

Ψn(x) =1

π14

√(2n)!!

[H−]ne−x2/2 =

1

π14

√(2n)!!

[xI − d

dx

]ne−x

2/2.

This the classical algebraic realization of the quantum oscillator.Renato Alvarez-Nodarse IMUS On some applications of OP in Mathematical-Physics III Orthonet School

Again a little of SF theory

The discrete analogues of the QHO have introduced by MacFarlane1989, Biedenharn 1989, Atakishiyev et at 199X, among others. Theywere related with the “discrete” polynomials and are the polynomialsolutions of the Nikiforov & Uvarov (NU) equation:

σ(−s − µ)

∆x(s − 12 )

∆Pn(s)q∆x(s)

− σ(s)

∆x(s − 12 )

∇Pn(s)q∇x(s)

+ λnPn(s)q = 0 ,

∇f (s) = f (s)− f (s − 1), ∆f (s) = f (s + 1)− f (s)

x(s) = c1(qs + q−s−µ) + c3, σ(s) = σ(x(s))− τ(x(s))

2∆x(s − 1

2 )

λn =−[n]q

(q

n−12 + q−

n−12

2τ ′ + [n−1]q

σ′′

2

)= C1q

n + C2q−n + C3

Solutions of the NU Eq.: Askey-Wilson, q-Racah, big q-Jacobi, Hahn,Meixner, etc polynomials, some time called discrete and q- classical


Again a little of SF theory

The discrete analogues of the QHO have introduced by MacFarlane1989, Biedenharn 1989, Atakishiyev et at 199X, among others. Theywere related with the “discrete” polynomials and are the polynomialsolutions of the Nikiforov & Uvarov (NU) equation:

σ(−s − µ)

∆x(s − 12 )

∆Pn(s)q∆x(s)

− σ(s)

∆x(s − 12 )

∇Pn(s)q∇x(s)

+ λnPn(s)q = 0 ,

∇f (s) = f (s)− f (s − 1), ∆f (s) = f (s + 1)− f (s)

x(s) = c1(qs + q−s−µ) + c3, σ(s) = σ(x(s))− τ(x(s))

2∆x(s − 1

2 )

λn =−[n]q

(q

n−12 + q−

n−12

2τ ′ + [n−1]q

σ′′

2

)= C1q

n + C2q−n + C3

Solutions of the NU Eq.: Askey-Wilson, q-Racah, big q-Jacobi, Hahn,Meixner, etc polynomials, some time called discrete and q- classicalRenato Alvarez-Nodarse IMUS On some applications of OP in Mathematical-Physics III Orthonet School

q−Polynomials: In the 1980’s there were two approaches

q-Polynomials

Askey Nikiforov & Uvarov

4ϕ3 σ∆(2)p + τ∆p = λp

Ismail, Gasper, Rah-man, Koorwinder, Koe-koek, Swarttouw ...

Atakishiyev, Suslov,Smirnov, . . .

rφp

(a1, ..., arb1, ..., bp

∣∣∣q ; z) =∞∑k=0

(a1; q)k · · · (ar ; q)k(b1; q)k · · · (bp; q)k

zk

(q; q)k

[(−1)kq

k2

(k−1)]p−r+1

q-Askey tableu [Koekoek, Swarttouw 1996] (recent book with Lesky in 2010)

NU Tableau [Nikiforov, Uvarov 1993]. For a comparison see [RAN, Medem 2001]


Going further: discrete quantum oscillators (DHOs)

Hq(s)ϕn(s) = λnϕn(s),

Hq(s) :=1

∇x1(s)A(s)Hq(s)

1

A(s), ϕn(s) =

A(s)√ρ(s)

dnPn(s;q)

where

Hq(s) := −√σ(−s−µ+1)σ(s)

∇x(s)e−∂s −

√σ(−s−µ)σ(s + 1)

∆x(s)e∂s

+

(σ(−s−µ)

∆x(s)+

σ(s)

∇x(s)

)I , ea∂s f (s) = f (s + a), a ∈ C.

Here ϕn are the q-wave functions and Hq the q-Hamiltonian, andPn(s;q) belong to the NU-tableu.

Problem 1: To find two operators a(s) and b(s) such that

Hq(s) = b(s)a(s)


Going further: discrete quantum oscillators (DHOs)

Hq(s)ϕn(s) = λnϕn(s),

Hq(s) :=1

∇x1(s)A(s)Hq(s)

1

A(s), ϕn(s) =

A(s)√ρ(s)

dnPn(s;q)

where

Hq(s) := −√σ(−s−µ+1)σ(s)

∇x(s)e−∂s −

√σ(−s−µ)σ(s + 1)

∆x(s)e∂s

+

(σ(−s−µ)

∆x(s)+

σ(s)

∇x(s)

)I , ea∂s f (s) = f (s + a), a ∈ C.

Here ϕn are the q-wave functions and Hq the q-Hamiltonian, andPn(s;q) belong to the NU-tableu.

Problem 1: To find two operators a(s) and b(s) such that

Hq(s) = b(s)a(s)


Factorization of the q-Hamiltonian

Solution: Let α ∈ R and A(s) and B(s) are continuous functions. Wedefine a family of α-down and α-up operators by

a↓α(s) :=B(s)√∇x1(s)

e−α∂s

(e∂s

√σ(s)

∇x(s)−√σ(−s − µ)

∆x(s)

)1

A(s),

a↑α(s) :=A(s)

∇x1(s)

(√σ(s)

∇x(s)e−∂s −

√σ(−s − µ)

∆x(s)

)eα∂s

√∇x1(s)

B(s).

Then Hq(s) = a↑α(s)a↓α(s), ∀α ∈ R, and B(s).

Definition: Let ς be a complex number, and let a(s) and b(s) be twooperators. We define the ς-commutator of a and b as

[a(s), b(s)]ς = a(s)b(s)− ςb(s)a(s), ς = qγ 6= 0.


Factorization of the q-Hamiltonian

Solution: Let α ∈ R and A(s) and B(s) are continuous functions. Wedefine a family of α-down and α-up operators by

a↓α(s) :=B(s)√∇x1(s)

e−α∂s

(e∂s

√σ(s)

∇x(s)−√σ(−s − µ)

∆x(s)

)1

A(s),

a↑α(s) :=A(s)

∇x1(s)

(√σ(s)

∇x(s)e−∂s −

√σ(−s − µ)

∆x(s)

)eα∂s

√∇x1(s)

B(s).

Then Hq(s) = a↑α(s)a↓α(s), ∀α ∈ R, and B(s).

Definition: Let ς be a complex number, and let a(s) and b(s) be twooperators. We define the ς-commutator of a and b as

[a(s), b(s)]ς = a(s)b(s)− ςb(s)a(s), ς = qγ 6= 0.


Why we are interested in this?

If we find two operators a(s) and b(s) such that

Hq(s) = b(s)a(s) and [a(s), b(s)]ς = Λ ⇒Proposition 1: we can build three operators K0(s) and K±(s) s.t.

[K0(s),K±(s)] = ±K±(s), [K−(s),K+(s)] = [2K0(s)]q2

i.e. we find the oscillator representation of the q-algebra suq(1, 1)

Proposition 2: Let Hq(s) be an operator, such that ∃a(s), b(s) andς, Λ ∈ C, that Hq(s) = b(s)a(s), and [a(s), b(s)]ς = Λ 6= 0.

Then, if Hq(s)Φ(s) = λΦ(s) ⇒

Hq(s)a(s)Φ(s) = ς−1(λ−Λ) a(s)Φ(s), a(s) lowering op.

Hq(s)b(s)Φ(s) = (Λ + ςλ)b(s)Φ(s). b(s) raising op.

The above propositions means that one can built the dynamical algebrafor Hq(s)





[K0(s),K±(s)] = ±K±(s), [K−(s),K+(s)] = [2K0(s)]q2




Hq(s)a(s)Φ(s) = ς−1(λ−Λ) a(s)Φ(s),

a(s) lowering op.

Hq(s)b(s)Φ(s) = (Λ + ςλ)b(s)Φ(s).

b(s) raising op.






[K0(s),K±(s)] = ±K±(s), [K−(s),K+(s)] = [2K0(s)]q2











[K0(s),K±(s)] = ±K±(s), [K−(s),K+(s)] = [2K0(s)]q2






The above propositions means that one can built the dynamical algebrafor Hq(s)Renato Alvarez-Nodarse IMUS On some applications of OP in Mathematical-Physics III Orthonet School

The results

Theorem: NECESSARY CONDITION

Let (ϕn)n the eigenfunctions of Hq(s) corresponding to the eigenvalues(λn)n and suppose that the problem 1 has a solution for Λ 6= 0. Then,the eigenvalues λn of the NU q-equation are q-linear or q−1-linearfunctions of n, i.e.,

λn = C1qn + C3 or λn = C2q

−n + C3,

respectively.

This condition is also necessary for the raising andlowering property!

There is also a necessary and sufficient condition but it looksvery technical and we will skip it here. [RAN, Atakishiyev, Costas-Santos,JPA 2005]

Interesting problems: When the α-operators are mutually adjoint? Thereexists a more general dynamical algebra?


The results




−n + C3,

respectively. This condition is also necessary for the raising andlowering property!




The results




−n + C3,





The results




−n + C3,



Interesting problems: When the α-operators are mutually adjoint? Thereexists a more general dynamical algebra?Renato Alvarez-Nodarse IMUS On some applications of OP in Mathematical-Physics III Orthonet School

Example 1: The Al-Salam & Carlitz I

The q-Hamiltonian is, in this case, (x = qs).

Hq(s)=− q2√

a(x−1)(x−a)

(q−1)2x2 e−∂s−√a(1−qx)(a−qx)

x2 e∂s +(√

q(q(x−1)x+a(1+q−qx))

(q−1)2x2

)I

Then, Hq(s)ϕn(s) = q32

1−q−n

(1−q)2ϕn(s) and the operators

a↓(s) ≡ a↓0(s) =q

14 x−1

q1/2 − q−1/2

(√(x − 1/q) (x − a/q) e∂s −

√a I),

a↑(s) ≡ a↑0(s) =q

14 x−1

q1/2 − q−1/2

(√(x − 1) (x − a) e−∂s −

√a/q I

),

are such that



a↑(s)a↓(s) = Hq(s), and [a↓(s), a↑(s)]q−1 =1

kq.

In this case the operators a↑(s) and a↓(s) are mutually adjoint.

Other related cases:

the Al-Salam & Carlitz I polynomials with a = −1 are the discreteq-Hermite I hn(x ; q)

Changing q by q−1, we obtain the factorization and the dynamicalalgebra for the Al-Salam & Carlitz functions II

from where putting a = −1 and x → ix follows the solution for thediscrete Hermite q-polynomials hn(x ; q)




kq.









kq.









kq.







Other examples in the q-Askey Tableau

x(s) Pn(s)q σ(s) + τ(s)∇x1(s) σ(s) λn

qs U(a)n (x ; q) a (x − 1)(x − a) q

32

1−q−n

(1−q)2

q−s V(a)n (x ; q) (1− x)(a− x) a q

12

1−qn

(1−q)2

qs hn(x ; q) −1 x2 − 1 q32

1−q−n

(1−q)2

qs hn(x ; q) 1 + x2 1 q12

1−qn

(1−q)2

q−s vµn (x ; q) µ (1− 1/q)(µ− q/qs) q32

1−qn

(1−q)2

qs Sn(x ; q) x2 q−1x q12

1−qn

(1−q)2

qs pn(x ; a|q) −ax q−1x(x − 1) q12

1−q−n

(1−q)2

qs Lαn (x ; q) ax(x + 1) q−1x q12 a 1−qn

(1−q)2

q−s Cn(x ; a; q) x(x − 1) q−1ax q12

1−qn

(1−q)2

Al-Salam & Carlitz I, II, discrete q-Hermite I, II, q-Charlier-type, Stieltjes-Wigert,

Wall polynomials, discrete q-Laguerre, q-Charlier.

The Askey-Wilson case: Only for some special cases. continuousq-Laguerre and continuous q-Hermite polynomials.


Other examples in the q-Askey Tableau

x(s) Pn(s)q σ(s) + τ(s)∇x1(s) σ(s) λn

qs U(a)n (x ; q) a (x − 1)(x − a) q

32

1−q−n

(1−q)2

q−s V(a)n (x ; q) (1− x)(a− x) a q

12

1−qn

(1−q)2

qs hn(x ; q) −1 x2 − 1 q32

1−q−n

(1−q)2

qs hn(x ; q) 1 + x2 1 q12

1−qn

(1−q)2

q−s vµn (x ; q) µ (1− 1/q)(µ− q/qs) q32

1−qn

(1−q)2

qs Sn(x ; q) x2 q−1x q12

1−qn

(1−q)2

qs pn(x ; a|q) −ax q−1x(x − 1) q12

1−q−n

(1−q)2

qs Lαn (x ; q) ax(x + 1) q−1x q12 a 1−qn

(1−q)2

q−s Cn(x ; a; q) x(x − 1) q−1ax q12

1−qn

(1−q)2

Al-Salam & Carlitz I, II, discrete q-Hermite I, II, q-Charlier-type, Stieltjes-Wigert,

Wall polynomials, discrete q-Laguerre, q-Charlier.

The Askey-Wilson case: Only for some special cases. continuousq-Laguerre and continuous q-Hermite polynomials.Renato Alvarez-Nodarse IMUS On some applications of OP in Mathematical-Physics III Orthonet School

Part III


But now we have some new “discrete” models of QHO

The above results lead to the question of how to generate RR for thecorresponding “discrete” SF?

Theorem (RAN & Cardoso JMAA 2013)

Let x(s) be a linear-type lattice x(s) = qs or x(s) = s. Then, any three

functions y(ki )νi (s), i = 1, 2, 3, are connected by a linear relation

3∑

i=1

Bi (s)y (ki )νi

(s) = 0, y(k)n (s) := ∆(k)yn(s)

where the Bi (s), i = 1, 2, 3, are polynomials.

This is the discrete analog of the NU theorem that we used to obtain theRR for the wave functions of the Schrodinger equation at the begining ofthis talk.


But now we have some new “discrete” models of QHO

The above results lead to the question of how to generate RR for thecorresponding “discrete” SF?

Theorem (RAN & Cardoso JMAA 2013)

Let x(s) be a linear-type lattice x(s) = qs or x(s) = s. Then, any three

functions y(ki )νi (s), i = 1, 2, 3, are connected by a linear relation

3∑

i=1

Bi (s)y (ki )νi

(s) = 0, y(k)n (s) := ∆(k)yn(s)

where the Bi (s), i = 1, 2, 3, are polynomials.

This is the discrete analog of the NU theorem that we used to obtain theRR for the wave functions of the Schrodinger equation at the begining ofthis talk.


The simplest corollaries

Corollary (three-term recurrence relation)

A1(s)yν(s) + A2(s)yν+1(s) + A3(s)yν−1(s) = 0,

where the coefficients Ai (s), i = 1, 2, 3, are polynomials.

Corollary (∆ and ∇-ladder-type relations)

B1(s)yν(s) + B2(s)∆yν(s)

∆x(s)+ B3(s)yν+m(s) = 0, m ∈ Z,

C1(s)yν(s) + C2(s)∇yν(s)

∇x(s)+ C3(s)yν+m(s) = 0, m ∈ Z,

where Bi (s) and Ci (s), i = 1, 2, 3 are polynomials.

Special important cases: m = 1 and m = −1

These operators are usually called raising and lowering operators,respectively.


The simplest corollaries

Corollary (three-term recurrence relation)

A1(s)yν(s) + A2(s)yν+1(s) + A3(s)yν−1(s) = 0,

where the coefficients Ai (s), i = 1, 2, 3, are polynomials.

Corollary (∆ and ∇-ladder-type relations)

B1(s)yν(s) + B2(s)∆yν(s)

∆x(s)+ B3(s)yν+m(s) = 0, m ∈ Z,

C1(s)yν(s) + C2(s)∇yν(s)

∇x(s)+ C3(s)yν+m(s) = 0, m ∈ Z,

where Bi (s) and Ci (s), i = 1, 2, 3 are polynomials.

Special important cases: m = 1 and m = −1

These operators are usually called raising and lowering operators,respectively.Renato Alvarez-Nodarse IMUS On some applications of OP in Mathematical-Physics III Orthonet School

Application to some discrete systems

We have shown how to built some discrete quantum oscillatorsassociated with several OP on discrete variables.

For these systems the wave functions were

ψn(z) :=

√ρ(z)∆x(s − 1/2)

dnPn(z),

Pn is and discrete OP, dn is the norm of Pn, ρ is the weight function

We can use the same idea we used at the begining of the talk: Combinethe last Eq. with the RR, i.e.,

Pn(z) =dn√

ρ(z)∆x(s − 1/2)ψn(z) +

3∑

i=1

Bi (s)P(ki )ni (s) = 0

to obtain the RR for ψn


Application to some discrete systems

We have shown how to built some discrete quantum oscillatorsassociated with several OP on discrete variables.

For these systems the wave functions were

ψn(z) :=

√ρ(z)∆x(s − 1/2)

dnPn(z),

Pn is and discrete OP, dn is the norm of Pn, ρ is the weight function

We can use the same idea we used at the begining of the talk: Combinethe last Eq. with the RR, i.e.,

Pn(z) =dn√

ρ(z)∆x(s − 1/2)ψn(z) +

3∑

i=1

Bi (s)P(ki )ni (s) = 0

to obtain the RR for ψn


∆-Ladder operadors for Charlier oscillator [Atakishiyev & Suslov 1990]

The wave function is: ψµn (z) =

√e−µµz−n

Γ(z + 1)n!Cµn (z)

[√µB1 +

(√z + 1−√µ

)B2

]ψµn (z) + B2

√z + 1∆ψµn (z)

+ B3

√µm+1(n + 1)m ψ

µn+m(z) = 0.

• Case m = −1

(√

z + 1−√µ)ψµn (z) +

√z + 1∆ψµn (z)−

√nψn−1

µ(z) = 0

• Caso m = 1

[(µ− z)(µ+ n − z)− µ(

√µ−√z + 1)

]ψµn (z)

+ µ√z + 1∆ψµn (z) + µ

√n + 1ψn+1

µ(z) = 0.




√e−µµz−n

Γ(z + 1)n!Cµn (z)

[√µB1 +

(√z + 1−√µ

)B2

]ψµn (z) + B2

√z + 1∆ψµn (z)

+ B3

√µm+1(n + 1)m ψ

µn+m(z) = 0.

• Case m = −1

(√

z + 1−√µ)ψµn (z) +

√z + 1∆ψµn (z)−

√nψn−1

µ(z) = 0

• Caso m = 1

[(µ− z)(µ+ n − z)− µ(

√µ−√z + 1)

]ψµn (z)

+ µ√z + 1∆ψµn (z) + µ

√n + 1ψn+1

µ(z) = 0.




√e−µµz−n

Γ(z + 1)n!Cµn (z)

[√µB1 +

(√z + 1−√µ

)B2

]ψµn (z) + B2

√z + 1∆ψµn (z)

+ B3

√µm+1(n + 1)m ψ

µn+m(z) = 0.

• Case m = −1

(√

z + 1−√µ)ψµn (z) +

√z + 1∆ψµn (z)−

√nψn−1

µ(z) = 0

• Caso m = 1

[(µ− z)(µ+ n − z)− µ(

√µ−√z + 1)

]ψµn (z)

+ µ√z + 1∆ψµn (z) + µ

√n + 1ψn+1

µ(z) = 0.


For more complicated examples

RAN and J. L. Cardoso: Recurrence relations for discretehypergeometric functions. J. Difference Equations. Appl. 11 (2005)829-850

R.Alvarez-Nodarse, J.L. Cardoso. On the Properties of SpecialFunctions on the linear-type lattices. Journal of Mathematical Analysisand Applications 405 (2013) 271–285.

What about the general quadratic lattice x(s) = c1qs + c2q

−s + c3?

R. Sevinik Adıguzel, Recurrence relations of the hypergeometric typefunctions on the quadratic-type lattices, arXiv preprint arXiv:1407.2226

What else?






−s + c3?


What else?






−s + c3?


What else?


Part IV


Why the oscillators are so interesting?

A ”real” problem

m1 m2 mN

R1 R2 RN

Ring of N coupled oscillators ⇒ Chain Model

m1 m2 mN−1 mN

R1 R2 RN

Chain of N coupled quantum oscillators ⇒ Perturbed CM



m1 m2 mN

R1 R2 RN


m1 m2 mN−1 mN

R1 R2 RN

Chain of N coupled quantum oscillators ⇒ Perturbed CMChain of N coupled quantum oscillators ⇒ Perturbed CM

This cover a wide class of quantum hamiltonians

H =1

2

N∑

k=1

(−~2∇2

mk+ κkx

2k

)+ γV (x1, x2, . . . , xN),

V (x1, x2, . . . , xN) =N∑

i=1

∞∑

n=2

Cnix2ni +

N−1∑

i=1

∞∑

m=0

dmi (xi − xi+1)2m.




m1 m2 mN

R1 R2 RN


m1 m2 mN−1 mN

R1 R2 RN




m1 m2 mN

R1 R2 RN


m1 m2 mN−1 mN

R1 R2 RN



H =1

2

N∑

k=1

(−~2∇2

mk+ κkx

2k

)+ γV (x1, x2, . . . , xN),

V (x1, x2, . . . , xN) =N∑

i=1

∞∑

n=2

Cnix2ni +

N−1∑

i=1

∞∑

m=0





m1 m2 mN

R1 R2 RN


m1 m2 mN−1 mN

R1 R2 RN




m1 m2 mN

R1 R2 RN


m1 m2 mN−1 mN

R1 R2 RN



H =1

2

N∑

k=1

(−~2∇2

mk+ κkx

2k

)+ γV (x1, x2, . . . , xN),

V (x1, x2, . . . , xN) =N∑

i=1

∞∑

n=2

Cnix2ni +

N−1∑

i=1

∞∑

m=0




Using the standard perturbation theory in QM we obtain that the firstenergy level of such a system can be described by the followingtridiagonal matrix

(α0 00 AN

), where AN =

α1 β1 0 · · · 0 0β1 α2 β2 · · · 0 00 β2 α3 · · · 0 0...

......

. . ....

...0 0 0 · · · αN βN0 0 0 · · · βN αN+1

.

The eigenproblem

(α0 00 AN

)(x1

X

)= λ

(x1

X

)⇔

An eigenvalue is λ0 = α0 with eigenvector (1, 0, . . . , 0)T .

For the others λ the eigenvec. are (0,X ) s.t. ANX = λX .

I.e., the others eigenval. and eigenvec. are those of AN .



Using the standard perturbation theory in QM we obtain that the firstenergy level of such a system can be described by the followingtridiagonal matrix

(α0 00 AN

), where AN =

α1 β1 0 · · · 0 0β1 α2 β2 · · · 0 00 β2 α3 · · · 0 0...

......

. . ....

...0 0 0 · · · αN βN0 0 0 · · · βN αN+1

.

The eigenproblem

(α0 00 AN

)(x1

X

)= λ

(x1

X

)⇔

An eigenvalue is λ0 = α0 with eigenvector (1, 0, . . . , 0)T .

For the others λ the eigenvec. are (0,X ) s.t. ANX = λX .

I.e., the others eigenval. and eigenvec. are those of AN .Renato Alvarez-Nodarse IMUS On some applications of OP in Mathematical-Physics III Orthonet School

Special but very interesting case ...

(βi)∞i=1 = a1, a2, . . . , ak︸︷︷︸, a1, a2, . . . , ak︸︷︷︸, . . .

(γi)∞i=1 = b2

1, b22, . . . , b

2k︸︷︷︸, b

21, b

22, . . . , b

2k︸︷︷︸, . . .

i.e, k-periodic sequences ⇒ k-Toeplitz matrices

Motivation: Quasicrystals, periodic chain of oscillators, ...

The key matrix: Jkn+k−1 and the key interval Σk

For solving the problem we made an alliance with an expert on this topic!


Special but very interesting case ...

(βi)∞i=1 = a1, a2, . . . , ak︸︷︷︸, a1, a2, . . . , ak︸︷︷︸, . . .

(γi)∞i=1 = b2

1, b22, . . . , b

2k︸︷︷︸, b

21, b

22, . . . , b

2k︸︷︷︸, . . .

i.e, k-periodic sequences ⇒ k-Toeplitz matrices

Motivation: Quasicrystals, periodic chain of oscillators, ...

The key matrix: Jkn+k−1 and the key interval Σk

For solving the problem we made an alliance with an expert on this topic!


Where are the eigenvalues of Jn: Notation

πk(x) := β + Dk(x) + (−1)k(b2k + b2

1b22 · · · b2

k−1

), k ≥ 2

Dk(x) :=

∣∣∣∣∣∣∣∣∣∣∣∣∣

x − a1 1 0 · · · 0 1b2

1 x − a2 1 · · · 0 00 b2

2 x − a3 · · · 0 0...

......

. . ....

...0 0 0 · · · x − ak−1 1b2k 0 0 · · · b2

k−1 x − ak

∣∣∣∣∣∣∣∣∣∣∣∣∣

.

∆k−1(x) :=

∣∣∣∣∣∣∣∣∣∣∣∣∣

x − a1 1 0 · · · 0 0b2

1 x − a2 1 · · · 0 00 b2

2 x − a3 · · · 0 0...

......

. . ....

...0 0 0 · · · x − ak−2 10 0 0 · · · b2

k−2 x − ak−1

∣∣∣∣∣∣∣∣∣∣∣∣∣

.

Set α := 2b1 · · · bk , β := b21 + · · ·+ b2

k , and Σk := π−1k ([β−α, β+α]).


The polynomial mappings [β − α, β + α] 7→ π−13 ([β − α, β + α])

5

and

Mj :=

j if 0 ≤ j ≤ ⌊k/2⌋ ,

k − j + 1 if ⌊k/2⌋+ 1 ≤ j ≤ k .(2.9) ?Mj?

Proof. The sequences βss≥0 and γss≥1 satisfying the periodicity conditions(2.2) generate a MOPS, (Pn)n, defined by the three-term recurrence relation

xPn(x) = Pn+1(x) + βnPn(x) + γnPn−1(x) , n = 0, 1, 2, · · ·

with initial conditions P−1 = 0 and P0 = 1. It follows from very well knownfacts in the Theory of Orthogonal Polynomials that the zeros of Pn are theeigenvalues of the matrix Jn, which are all real and simple, and the zeros ofPn interlace with those of Pn−1. Under the given hypothesis, it is known (seee.g. [12, 16]) that the support of the measure with respect to which the MOPS(Pn)n is orthogonal consists of an union of k intervals such that any two ofthese intervals may intersect at a single point, plus at most k − 1 isolatedpoints between them. Furthermore (cf. e.g. [13, 16]) these k intervals aredefined by the inverse polynomial mapping [β−α, β+α] 7→ π−1

k ([β−α, β+α])and they are separated by the points ξ1, · · · , ξk−1 which are the solutions ofthe algebraic equation ∆k−1(x) = 0 (see Figure 1 for the case k = 3, wherethe inverse polynomial mappings [β − α, β + α] 7→ π−1

3 ([β − α, β + α]) and[−α, α] 7→ (π3 − β)−1([−α, α]) are illustrated). This justifies statement (i) in

β−α

α+β

α

−α

π3(x)

π3(x)− βπ−13 ([β − α, β + α]) = (π3 − β)−1([−α, α])

ξ1 ξ2

Figure 1: Inverse polynomial mapping

the Proposition. In order to prove (ii) notice first that (cf. e.g. [13])

Pnk+k−1(x) = ∆k−1(x)Un

(πk(x) − β

α

), n = 0, 1, 2, · · · (2.10) zerosPnk+k-1

where Un is the Chebyshev polynomial of the second kind of degree n,

Un(x) :=sin(n+ 1)θ

sin θ, x = cos θ. (2.11) che-pol

Thus the zeros of Pnk+k−1 (hence the eigenvalues of Jnk+k−1) are the k− 1 realnumbers ξ1, · · · , ξk−1, that are located between the k intervals I1, · · · , Ik, andthe kn real numbers x such that Un

(πk(x)−β

α

)= 0, i.e.,

πk(x) = β + α cosjπ

n+ 1, j = 0, 1, · · · , k − 1 . (2.12) trigEq1


Where are the eigenvalues of Jn:

ξ1ξ2

Σ3 I1 2I 3I

I The set Σk is an union of k intervals such that any two of theseintervals intersect at most at a single point

Σk = π−1k ([β − α, β + α]) = I1 ∪ · · · ∪ Ik .


Where are the eigenvalues of Jn: The exact case Jkn+k−1

ξ1ξ2

Σ3 I1 2I 3I

I Except for at most k − 1 ones, all the eigenvalues of Jkn+k−1 arelocated in the set Σk , for all n = 0, 1, 2, · · · .

I In fact, each I1, . . . , Ik contains exactly n eigenvalues of Jkn+k−1 inits interior, and the remaining k − 1 are located in the gaps betweenthem

I Between I` and I`+1 (` = 1, · · · , k − 1) ∃ exactly one eigenvalue ξ`of Jkn+k−1. These k − 1 eigenvalues are the k − 1 solutions of∆k−1(x) = 0.


Where are the eigenvalues of Jn: The general case Jkn+j−1

ξ1ξ2

Σ3 I1 2I 3I

I For each j = 0, 1, · · · , k , all the eigenvalues of Jkn+j−1

(n = 1, 2, · · · ) are contained in the convex hull of the set Σk and inbetween two consecutive intervals I` and I`+1 (` = 1, · · · , k − 1) thenumber of eigenvalues of Jkn+j−1 is at most (Nj important number)

Nj :=

j + 1 if 0 ≤ j ≤ bk/2c ,

k − j + 1 if bk/2c < j ≤ k .

The number of eigenvalues of Jkn+j−1 in each I` is at leastn − 1− Nj .


In a serious math talk should be at least 1 Proof (W. Van Assche)


The Proof: 1. We started discussing a particular example


The Proof: 2. We then started to “fight” with a general proof.


The Proof: 3. We solved it!. Time for beer!


Interlacing Properties for Jµ,λn : The result ∀µ, λ ∈ R

The QM model leads to a perturbed matrix:

Jµ,λn :=

β0 + µ 1 0 . . . 0 0γ1 β1 1 . . . 0 00 γ2 β2 . . . 0 0...

...... . . . ...

...0 0 0 . . . βn−2 10 0 0 . . . γn−1 βn−1 + λ

being the original tridiagonal matrix a k-periodic Toplitz matrix.



I The eigenvalues of Jµ,λn are real and simple.

I Between two consecutive eigenvalues of Jn there exists atmost two eigenvalues of Jµ,λn , and conversely.

I At most two eigenvalues of Jµ,λn can live out the interval of theconvex hull of Σk .

If we concentrate our attention in the Jµ,λnk+j−1 matrix we can be moreprecise:

I At most two eigenvalues of the perturbed matrix Jµ,λnk+j−1 lie out ofthe convex hull of Σk .

I At most Nj + 2 eigenvalues of Jµ,λnk+j−1 are in the gap between twoconsecutive intervals I` and I`+1 (` = 1, · · · , k − 1).

I There are at most (k − 1)Nj + 2k eigenvalues of the perturbed

matrix Jµ,λnk+j−1 out of the set Σk .



I The eigenvalues of Jµ,λn are real and simple.

I Between two consecutive eigenvalues of Jn there exists atmost two eigenvalues of Jµ,λn , and conversely.

I At most two eigenvalues of Jµ,λn can live out the interval of theconvex hull of Σk .

If we concentrate our attention in the Jµ,λnk+j−1 matrix we can be moreprecise:

I At most two eigenvalues of the perturbed matrix Jµ,λnk+j−1 lie out ofthe convex hull of Σk .

I At most Nj + 2 eigenvalues of Jµ,λnk+j−1 are in the gap between twoconsecutive intervals I` and I`+1 (` = 1, · · · , k − 1).

I There are at most (k − 1)Nj + 2k eigenvalues of the perturbed

matrix Jµ,λnk+j−1 out of the set Σk .


How close are the EV of Jn and Jµ,λ?

Theorem: Let z`,1 < · · · < z`,n be the n eigenvalues of Jkn+k−1 that liein the (topological) interior of the interval I` (` = 1, 2, · · · k). Then, ifπk(ξi ) 6= β ± α for all i = 1, · · · , k − 1,

|z`,ν+1 − z`,ν | ≤%`

n + 1, %` :=

απ

mınx∈I` |π′k(x)| ≤ % =απ

mınx∈Σk|π′k(x)|

for all ` = 1, 2, · · · , k and ν = 1, 2, · · · , n − 1.

As a corollary we have our main result:

Theorem: Denote by z(j ,µ,λ)`,1 < · · · < z

(j ,µ,λ)`,n′j (`)

the eigenvalues of Jµ,λkn+j−1

that lie in the interior of the interval I` (` = 1, 2, · · · k). Then, for almostall ν = 1, 2, . . . , n − 1

∣∣∣z(j ,µ,λ)`,ν − z`,ν

∣∣∣ ≤ 5k%

n + 1.


How close are the EV of Jn and Jµ,λ?

Theorem: Let z`,1 < · · · < z`,n be the n eigenvalues of Jkn+k−1 that liein the (topological) interior of the interval I` (` = 1, 2, · · · k). Then, ifπk(ξi ) 6= β ± α for all i = 1, · · · , k − 1,

|z`,ν+1 − z`,ν | ≤%`

n + 1, %` :=

απ

mınx∈I` |π′k(x)| ≤ % =απ

mınx∈Σk|π′k(x)|

for all ` = 1, 2, · · · , k and ν = 1, 2, · · · , n − 1.

As a corollary we have our main result:

Theorem: Denote by z(j ,µ,λ)`,1 < · · · < z

(j ,µ,λ)`,n′j (`)

the eigenvalues of Jµ,λkn+j−1

that lie in the interior of the interval I` (` = 1, 2, · · · k). Then, for almostall ν = 1, 2, . . . , n − 1

∣∣∣z(j ,µ,λ)`,ν − z`,ν

∣∣∣ ≤ 5k%

n + 1.


Numerical experiments: Tridiagonal 1-Toeplitz matrices

Aµ,λN =

a1 + µ b1 0 · · · 0 0b1 a1 b1 · · · 0 00 b1 a1 · · · 0 0...

......

. . ....

...0 0 0 · · · a1 b1

0 0 0 · · · b1 a1 + λ

AN := A0,0N =

a1+0 b1 0 · · · 0 0b1 a1 b1 · · · 0 00 b1 a1 · · · 0 0...

......

. . ....

...0 0 0 · · · a1 b1

0 0 0 · · · b1 a1+0



0 20 40 60 80 100number of eigenvalue

1

2

3

4

5

λ Α (∗

); λ

∼ Α (ο

)

0 20 40 60 80 100number of eigenvalue

−0.04

−0.03

−0.02

−0.01

0.00

λ∼ Α −

λ Α

Left panel: the EV λ(AN) (stars) and λ(Aµ,λN ) (open circles) for N = 101. In the

right panel λ(Aµ,λN )− λ(AN) for N = 101. a1 = 3, λ = µ = 1 and b1 = 1

3.50 3.85 4.20

EV around the eigenvalue 66 λA (with stars) and λA (using open circles).



From the above example one can think that the interlacing property canbe reformulated as follows:

Conjeture

Between two eigenvalues of Jn there exists one and only one eigenvalueof Jµ,λn , and conversely.

The following example shows that it is, in general, not true.

4.1 4.3 4.6

We plot the 12–15 EV the EV λ(AN) (stars) and λ(Aµ,λN ) (open circles). Notice

that between the eigenvalues 13 and 14 of the perturbed matrix AN

there are two eigenvalues of the matrix Aµ,λN . The parameters of the

numerical simulations are b1 = 1/2, a1 = 4, µ = 1/2, λ = −3/2, and N = 21.




Conjeture



4.1 4.3 4.6








Conjeture



4.1 4.3 4.6






Numerical experiments. ⇑ 3-periodic ⇓ 5 & 7 periodic

0 20 40 60 80 100 120 140 160−4

−2

0

2

4

6

8

10

Eigenvalue numbers

Eig

enva

lues

0 20 40 60 80 100 120 140 160−4

−2

0

2

4

6

8

10

12

Eigenvalue numbers

Eig

enva

lues

0 20 40 60 80 100 120 140 160−6

−4

−2

0

2

4

6

8

10

12

Eigenvalue numbers

Eig

enva

lues

0 50 100 150−6

−4

−2

0

2

4

6

8

10

12

Eigenvalue numbers

Eig

enva

lues


Part V


Part V


To end: As it begun ... J. von Neumann wrote

I think that it is a relatively good approximation to truth thatmathematical ideas originate in empirics, although the genealogy issometimes long and obscure. But, once they are so conceived, thesubject begins to live a peculiar life of its own and is better comparedto a creative one [...]

As a mathematical discipline travels far from itsempirical source [...] it is beset with very grave dangers. [...]

[...] is a grave danger that the subject will develop along the line ofleast resistance, that the stream, so far from its source, will separateinto a multitude of insignificant branches, and that the discipline willbecome a disorganized mass of details and complexities. In other words,at a great distance from its empirical source, or after much“abstract” inbreeding, a mathematical subject is in danger ofdegeneration [...] whenever this stage is reached, the only remedyseems to me to be the rejuvenating return to the source: there-injection of more or less directly empirical ideas.



I think that it is a relatively good approximation to truth thatmathematical ideas originate in empirics, although the genealogy issometimes long and obscure. But, once they are so conceived, thesubject begins to live a peculiar life of its own and is better comparedto a creative one [...] As a mathematical discipline travels far from itsempirical source [...] it is beset with very grave dangers. [...]







Conclusions

There are other interesting problem (chain models, perturbation ofTopelitz matrices, random and quantum walks, etc.) that can bediscussed in this talk that are directely motivated by real empiricalproblems.

The SF theory and methods is very useful in solving applied problems(and not only) and there a lot of interesting open problems for young(and not so young) mathematicians interested on them.

Questions?

That’sall

Thanks


Conclusions



Questions?

That’sall

Thanks


Conclusions



Questions?

That’sall

Thanks


You promised!!!!


Part IV


Part IV



I think that it is a relatively good approximation to truth thatmathematical ideas originate in empirics, although the genealogy issometimes long and obscure. But, once they are so conceived, thesubject begins to live a peculiar life of its own and is better comparedto a creative one [...]

As a mathematical discipline travels far from itsempirical source [...] it is beset with very grave dangers. [...]











Conclusions



Questions?

That’sall

Thanks


Conclusions



Questions?

That’sall

Thanks


Conclusions



Questions?

That’sall

Thanks


Documents

On some applications of Orthogonal Polynomials in ...orthonet/orthonet18/notas/ran.pdf · Wigner:[Unreasonable E ectiveness of Mathematics in the Natural Sciences, 1960] The miracle