Isochronous classical systems and quantum systems with equally

Journal of Physics Conference Series

OPEN ACCESS

Isochronous classical systems and quantumsystems with equally spaced spectraTo cite this article J F Carintildeena et al 2007 J Phys Conf Ser 87 012007

View the article online for updates and enhancements

You may also likeMultiple island chains in wave-particleinteractionsM C de Sousa I L Caldas A M Ozorio deAlmeida et al

-

Gaseous time projection chambers for rareevent detection results from the T-REXproject I Double beta decayIG Irastorza F Aznar J Castel et al

-

Reconstructing 3D proton dose distributionusing ionoacousticsK W A van Dongen A J de Bleacutecourt ELens et al

-

Recent citationsOn isochronous analytic motions and thequantum spectrumAbd Raouf Chouikha

-

Rational deformations of conformalmechanicsJosamp233 et al

-

Superintegrable classical Zernike systemGeorge S Pogosyan et al

-

This content was downloaded from IP address 22318226118 on 13112021 at 0548

Isochronous classical systems and quantum systems

with equally spaced spectra

Jose F Carinenaa) Askold M Perelomov 1b) and Manuel F Ranadac)

Departamento de Fısica Teorica Facultad de Ciencias Universidad de Zaragoza 50009Zaragoza Spain

E-mail ajfcunizares bperelomodftuzunizares cmfranunizares

Abstract

We study isoperiodic classical systems what allows us to find the classical isochronoussystems ie having a period independent of the energy The corresponding quantumanalog systems with an equally spaced spectrum are analysed by looking for possiblecreation-like differential operators The harmonic oscillator and the isotonic oscillatorare the two main essentially unique examples of such situation

Keywords Isochronous systems ladder operators equispaced spectra

1 Introduction

Inverse problems have been playing a very relevant role in the development of physical theoriesThe physical laws describing the different phenomena are constructed from experimental dataand their symmetries and regularities are used to discard possible laws or to pick up other lawsamong other mathematically possible formulations

In this paper we review some results on two important inverse problems and report somerecent progress on both The first one is the characterisation of possible convex potentials givingrise to isochronous motions ie systems for which the bounded motions have a period which doesnot depend on the energy With this aim we first review in Section 2 the very classical problemof the determination of the potential from the energy dependence of the period a problemwhich dates back to Abel [1] and therefore the characterisation of isoperiodic potentials thosegiving rise to the same dependence The harmonic oscillator is the prototype of an isochronoussystem and all other isochronous systems are isoperiodic with the harmonic oscillator After acharacterisation of such systems we report the recent approach used in [2] to recover a result byChalykh and Veselov [3] among rational potentials only the harmonic oscillator and the isotonicoscillator [4 5] produce isochronous motions There are however other isochronous systemsdescribed by non-rational potentials for instance potentials for which the second derivative hasa discontinuity

A related problem in quantum mechanics is the determination of those potentials with anequispaced energy spectrum Of course only in the best conditions of regularity the solution ofboth problems can be related and this is only possible to order ~ Our approach here is basedon the search for creation- and annihilation-like operators which are assumed to be differential

1 On leave of absence from Institute for Theoretical and Experimental Physics 117259 Moscow Russia

Particles and Fields Classical and Quantum IOP PublishingJournal of Physics Conference Series 87 (2007) 012007 doi1010881742-6596871012007

ccopy 2007 IOP Publishing Ltd 1

operators of arbitrary order Therefore we study the possibility of finding as ladder operatorssome differential operator of order k

Ak =ksum

j=0

vkminusj(x)dj

dxj=

ksum

j=0

vkminusj(x) partj v0(x) equiv 1

The problem is solved for k = 1 2 and once again we find the harmonic oscillator and theisotonic oscillator and we also provide the equations for k = 3 As the solution of such a systemof differential equations cannot be carried out explicitly we attack the problem from a newperspective and reduce the problem to the search of rational solutions of a high-order Kortewegde Vries equation

2 Isochronous systems

The classical harmonic oscillator in one dimension x = minusω2 x is one of most studied modelsand enjoys the property of isochronicity all the solutions of its dynamics are periodic withangular frequency ω q = q0 cosωt+ (v0ω) sinωt = A cos(ωt+ϕ) while A and ϕ are arbitraryMoreover ω is fixed from the own equation and it does not depend on the energy E This canbe directly proved from the expression of the period Recall that for a fixed energy E theperiod T (E) is given by

T (E) =radic

2m

int x+(E)

xminus

(E)

dxradicE minus U(x)

(1)

where xminus(E) and x+(E) are the roots of the equation U(x) = E U(xplusmn(E)) = E Then whenU(x) = (12)mω2 x2 it turns out to be T (E) = 2πω This suggests us that we can consider aspossible generalisations of the classical harmonic oscillator either systems for which the perioddoes not depend on the energy but whose general solution is not the one given before or simplysystems for which the general solution is also x = A cos(ωt + ϕ) but in which ω can dependon the energy (a nonlinear system) The last possibility has been analysed in [6] (see [7 8] forthe quantum version) and looks as a position-dependent mass oscillator The two-dimensionalversion of this system has been shown to be super-integrable and its HamiltonndashJacobi equationis super-separable [6] and corresponds to the motion of a harmonic oscillator in a surface ofconstant curvature minusλ

The first possible generalisation was however to look for isochronous systems Consider theparticular case of a convex potential well with a minimum in x = 0 for which the potentialfunction is

U(x) =

Uminus(x) if x lt 0U+(x) if x gt 0

where U+(x) is an increasing function and Uminus(x) is a decreasing function xU prime(x) gt 0 if x 6= 0such that then we can invert x as a function of U

x =

φminus(U) if U lt 0φ+(U) if U gt 0

In this case for each energy E there is a oscillatory motion between the two closest turningpoints xminus(E) and x+(E) ie such that φminus(E) = xminus(E) and φ+(E) = x+(E) The period T (E)for a fixed energy E is given by (1) which under the change of variable y = U(x) or the inverseexpression x = φ(y) as U(0) = 0 and U(x+(E)) = E ie φ(0) = 0 and φ(E) = x+(E) becomes

T (E) = 2radic

2m

int E

0

φprime(y)radicE minus y

dy = 2radic

2mE

int 1

0

φprime(E z)radic1 minus z

dz


2

where y = E z The system is said to be isochronous if the period does not depend on theenergy dT (E)dE = 0 which leads to

1

2radicE

int 1

0


dz +radicE

int 1

0

φprimeprime(E z)radic1 minus z

z dz = 0

Therefore the condition for the system to be isochronous is

int 1

0

ζ(E z)radic1 minus z

dz = 0 =1radicE

int E

0

ζ(y)radicE minus y

dy forallE gt 0

where ζ(z) = 2 z φprimeprime(z) + φprime(z) This is only possible when ζ(z) = 0 and the solution of2E z φprimeprime(E z) + φprime(E z) = 0 is φprime(y) = C

radicy and consequently the solution for which φ(0) = 0

is φ(y) = 2Cradicy with inverse function U(x) = (4C2)minus1 x2 ie under the assumed regularity

conditions only the harmonic oscillator is an isochronous systemAn interesting inverse problem is the determination of a function U(x) giving rise to a given

dependence function T (E) As Abel proved in [1] when the potential is convex as indicatedabove such expression can be inverted (see eg Chapter 2 of Landaursquos book [9]) giving rise tothe following integral equation for T (E)

φ+(U) minus φminus(U) =1

πradic

2m

int U

0

T (E)radicU minus E

dE (2)

and therefore the knowledge of T (E) only allows us to determine the difference φ+(U)minusφminus(U)In fact we can split the rhs of the integral (1) as a sum of two integrals between xminus(E) and0 and between 0 and x+(E) respectively and using the same change of variable as before wefind that if ∆ = φ+ minus φminus then

T (E) =radic

2m

int E

0

∆prime(y)radicE minus y

dy

Recall that the convolution product f1 lowast f2 of two functions f1 and f2 is given by

(f1 lowast f2)(E) =

int E

0f1(E minus z)f2(z) dz

and then T (E) isradic

2m times the convolution product of ψ(E) = 1radicE and ∆prime(E) Using

Laplace transformation L[f ](s) =

intinfin

0eminussEf(E) dE in both sides and taking into account that

L[f1 lowast f2] = L[f1]L[f2] we find that L[T ](s) =radic

2mL[ψ](s)L[∆prime](s) But as L[ψ](s) =radicπs

and L[∆prime](s) = sL[∆](s) we can get L[∆](s) from L[T ](s) =radic

2mradicπs sL[∆](s) and we

obtain

L[∆](s) =1radic

2π2m

radicπ

sL[T ](s) =

1radic2π2m

L[ψ](s)L[T ](s)

namely L[∆](s) = (2π2 m)minus12 L[ψ lowast T ](s) and we recover in this way the equation (2)Note that for periodic motions with a period τ = 2πω independent of the energy and m = 1


πradic

2

int U

0

τradicU minus E

dE =2 τ

πradic

2

radicU


3

and therefore as the turning points satisfy U(xplusmn(E)) = E namely φplusmn(E) = xplusmn(E)

x+(E) minus xminus(E) =

radic2 τ

π

radicE =

2radic

2

ω

radicE

that in the particular case of a regular symmetric potential for which xminus(E) = minusx+(E) theisochronicity condition becomes

x+(E) =τradic2π

radicE =

radic2

ω

radicE (3)

a condition which in particular the harmonic oscillator holdsThe important point is that the general solution of (2) is of the form

φ+(U) = φ(0)+ (U) + f(U) φminus(U) = φ

(0)minus

(U) + f(U)

where φ(0)+ and φ

(0)minus

are a particular solution of the problem and f(U) is an arbitrary functionof U for which φ+(U) and minusφminus(U) be monotonous increasing functions As an example thechoices U(x) = tanh2(x) and f(U) = 2Artanh

radicU lead to the Morse potential

Note that if we choose f(U) = minus(12)(φ(0)+ (U) + φ

(0)minus (U)) we find a solution φs such

that φsminus(U) = minusφs

+(minusU) and corresponds to a potential that is symmetric under reflectionThis potential is nothing but the Steiner symmetrisation of the potential [10] Of course theambiguity of the choice for the function f(U) disappears when we impose the additional conditionthat the curve U = U(x) be symmetric with respect to the U -axis ie U(x) = U(minusx) becausein this case φ+(U) = minusφminus(U) and therefore

φ(U) =1

2πradic

2m

int U

0

T (E)radicU minus E

dE

Note that a potential U(x) and a shear transformed potential Uf (x) defined by Uf (x +f(U(x))) = U(x) for an arbitrary function f ie φfplusmn(U) = φplusmn(U) + f(U) are isoperiodic [2]

The general solution for an isochronous system with period T is given by

φminus(U) = minusTπ

radicU

2+ f(U) φ+(U) =

T

π

radicU

2+ f(U)

and in particular for f(U) = a we find


radicU

2+ a φ+(U) =

T

π

radicU

2+ a

and we obtain the harmonic oscillator potential

U(x) =ω2

2(xminus a)2 ω =

2π

T

while if the function f is chosen to be

f(U) = αT

π

radicU

2

then

φminus(U) = (minus1 + α)T

π

radicU

2 φ+(U) = (1 + α)

T

π

radicU

2


4

which for |α| 6= 1 corresponds to the potential of two half-oscillators

U(x) =

12 mω2

1 x2 if x le 0

12mω2

2 x2 if x ge 0

with different couplings

ω1 =2π

(1 minus α)T ω2 =

2π

(1 + α)T

glued together at the origin Note that 1ω1 + 1ω2 = 2ω0 where ω0 = 2πT Conversely forsuch a potential with ω1 6= ω2 it suffices to make the choices

α =ω1 minus ω2

ω1 + ω2and ω0 =

2ω1 ω2

(ω1 + ω2)

The cases α = plusmn1 correspond to a half-oscillator and its reflected one either ω1 or ω2 vanishesand ω0 is two times the other one

Another very fundamental identity which characterises all potentials (related by a sheartransformation) having a given dependency T (E) of the period as a function of the energy is [2]

U(x) = U (x+WU (U(x))) where WU(V ) =1

πradic

2m

int U

0

T (E)radicV minus E

dE

Using such relation it can be easily shown (see [2]) that a convex polynomial potential U(x)is isochronous if and only if U(x) = ax2 + bx+ c This results can be generalised for the case ofmeromorphic rational functions for which a rational potential U(x) which does not reduces toa polynomial is isochronous if and only if

U(x) =

(ax2 + bx+ c

x+ d

)2

On the other side a slight modification of Joukowski transformation (Jλ(z) = z + λz with

λ isin R) which plays a relevant role in aerodynamics applications

Jg(x) =x

2minus 2α

x α ge 0

may be used to prove that if U(x) is a bounded below even convex potential with limxrarrinfinU(x) =infin then for any positive real number α the potential Uα given by

Uα(x) = U(Jα(x)) = U

(x

2minus 2α

x

)

is isoperiodic with U(x) Finally a kind of converse property was also proved in [2] Any non-trivial rational potential Ulowast which is isoperiodic to a given even convex polynomial potential Uis either of the form Uc = U(x+ c) or Uα(x) = U(x2 minus 2αx) for any value of α

Note that the potential corresponding to the isochronous harmonic oscillator Uho(x) =(12)mω2x2 is up to addition of a constant of the form Ug = (18)mω2x2 + gx2 withg = 2mα2ω2


5

3 Quantum systems with equally spaced spectra

The creation and annihilation operators for the quantum harmonic oscillator were introducedby Dirac [11] in 1927 for the description of emission and absorption of radiation They play afundamental role in quantum mechanics and quantum field theory and therefore it is interestingto understand whether other quantum systems admit such creation and annihilation operatorsIn this section we analyse this problem for the particular case of a quantum one-dimensionalsystems with rational potentials with no more than one real pole It appears that under enoughgeneral conditions such operators can only exist for two known systems namely for standardand singular quantum oscillators We guess that this statement is valid probably for a widerclass of potentials not only for rational ones but this more general interesting problem is stillopen

Let us first remind that the Hamiltonian of quantum oscillator is (we use the system ofnatural units with Planck constant ~ = 1 and also choose m = 1)

H =1

2(p2 + ω2x2) p = minusi d

dx

The creation and annihilation operators a+ and a are operators satisfying the equations[Ha] = minusω a and [Ha+] = ω a+ and Dirac defined them by the formulae

a =1radic2

(1radicω

d

dx+

radicω x

) a+ =

1radic2

(minus 1radic

ω

d

dx+

radicω x

)

so that they satisfy the commutation relation [a a+] = 1 It looks more convenient to use similaroperators but with another normalisation of such operators

A =1radic2

(d

dx+ ω x

) A+ =

1radic2

(minus d

dx+ ω x

)

so that this form is still valid for ω = 0 Notice that [AA+] = ωMoreover the fundamental relations

[HA] = minusωA [HA+] = ωA+ (4)

allow us to consider such operators as ladder operators because if ψ is an eigenvector of H witheigenvalue E then if A+ψ 6= 0 A+ψ is an eigenvector of H corresponding to the eigenvalueE + ω while if Aψ 6= 0 Aψ will be an eigenvector of H with eigenvalue E minus ω This impliesthat at least a part of the spectrum of H is equispaced with a equispacing ω Moreover we willrestrict ourselves to the case in which there exists a cyclic eigenstate ψ0 in the sense that theset of orthogonal vectors (A+)kψ0 | k = 0 1 is a complete set Then all the spectrum willbe equispaced

In this case of the harmonic oscillator H prime = Hminus(12)ω = (12)A+A essentially coincides with

its partner H prime = (12)AA+ and as the creation and annihilation operators are of first order

they are also intertwining operators between H prime and its partner H prime ie they satisfy AH prime = H primeAand therefore H primeA+ = A+H prime The Hamiltonian H prime factorised as before is shape-invariant aremarkable property (see eg [12] and references therein)

Note however that as we shall see later on there may be higher-order intertwining operatorsbetween two Hamiltonians and that they give rise to relations among creation and annihilationoperators for both systems when these exist [13] but they can only be formal creation andannihilation operators they satisfy (4) but when applying such operators to an eigenfunctionof one Hamiltonian we can obtain a non-normalisable function and therefore they are not givingrise to an eigenfunction of the partner Hamiltonian


6

Another quantum system admitting creation and annihilation operators and giving rise to aconstant separation ω among neighbour energy levels is the singular harmonic oscillator (alsocalled isotonic oscillator) [4 5] namely the system describing by the Hamiltonian

H =1

2

(p2 +

1

4ω2x2

)+g2

x2

Here only positive values of x are allowed and the spectrum of this system is also equispacedand as indicated in the preceding section the corresponding classical system is isochronousMoreover as indicated above Chalykh and Veselov recently [3] proved that these are the onlytwo cases cases with a rational potential U(x) for which the classical system is isochronous

It is to be remarked that the creation and annihilation operators for the isotonic oscillator arenot first-order but second-order differential operators The limiting case g rarr 0 is the so-calledlsquohalf-oscillatorrsquo ie a particle moving in the harmonic potential on the lsquohalfrsquo line

U(x) =

18 ω

2 x2 for x gt 0

infin for x lt 0

from all the eigenstates of the Hamiltonian for the harmonic oscillator on the full line only oddsolutions will still be eigenstates in the lsquohalfrsquo oscillator case

Let us remark that given a quantum system described by a Hamiltonian with a potentialU(x) any other potential obtained either by translation ie Ua(x) = U(xminusa) or by reflectionUr(x) = U(minusx) has the same spectrum as the given system When we are interested in theexistence of ladder operators such kind of potentials should be considered as equivalent and itis enough to determine one representative in each equivalence class On the other hand as notonly regular potentials play a role but also singular ones we will analyse the problem also forsingular systems admitting only the case for simplicity in which U is a rational function If ithas k real poles there will be k+1 different quantum problems one in each interval between twoneighbour poles Therefore we will restrict ourselves to the simpler cases in which U is eitherregular or it has a real pole of arbitrary multiplicity In other words we are restricting ourselvesto the case in which either U is regular or it has a real pole assumed to be at x = 0 We aretherefore interested in analysing whether for such a given Hamiltonian there is a realisation ofsuch operators Aminus

k and A+k as differential operators of order k ie

Aminus

k =

ksum

j=0

vkminusj(x)dj

dxj=

ksum

j=0

vkminusj(x) partj v0(x) equiv 1 (5)

with part = ddx We shall denote A+k the adjoint operator of Aminus

k

4 Ladder operators in one-dimensional quantum systems

Let us consider a quantum one-dimensional system described by a Hamiltonian

H =1

2p2 + U(x) = minus1

2

d2

dx2+ U(x) = minus1

2part2 + U(x) (6)

where we assume that the potential U(x) is given either by a rational function free of real poles(and the configuration space is the whole real line) or with exactly one real pole assumed tobe at x = 0 what is enough general because of the invariance of the problem under shift andreflection mentioned before In this last case the configuration space is (0infin) Our aim is todetermine the explicit forms such a function U(x) can take in order for the quantum system


7

to admit creation and annihilation operators and hence the Schrodinger equation for stationarystates

H ψ = E ψ

has at least a part of the discrete spectrum equispaced En = E0 + nω n = 0 1 2 As mentioned before a particularly well-known example is the harmonic oscillator for which

U(x) = (12)ω2 x2 and E0 = ω2 and another example is U1(x) = (18)ω2 x2 + g2x2 definedin the interval (0infin)

The question is when do ladder operators being differential operators of order k exist in thecase under consideration ie when differential operators of order k Aplusmn

k satisfying

[HAplusmn

k ] = plusmnωAplusmn

k (7)

exist where as indicated in (5)

Aminus

k =dk

dxk+ v1(x)

dkminus1

dxkminus1+ middot middot middot + vk(x) = partk + v1(x) part

(kminus1) + middot middot middot + vk(x)

and A+k denotes the adjoint operator of Aminus

k When substituting this expression in the previous equation (7) we obtain a system of

differential equations for the unknown functions U(x) and vj(x) for j = 1 kWe start by analysing the two simplest cases

1 k = 1 Here Aminus

1 is given by Aminus

1 = ddx + v1(x) = part + v1(x) and then the commutationcondition [HAminus

1 ] = minusωAminus

1 having in mind that

[minus1

2part2 v1(x)

]= minus1

2vprimeprime1 minus vprime1 part

leads to the system of differential equations

vprime1 = ω

12 v

primeprime1 + U prime = ω v1

from which we obtain v1(x) = ω x+α and then Aminus

1 = part+ω x+α1 and U(x) = 12 ω

2 x2+α1 x+α2where α1 and α2 are arbitrary constants This potential function can also be written up to anadditive constant in the form

U(x) =1

2ω2(xminus x0)

2 x0 = minusα1

ω2

2 k = 2 In this case

Aminus

2 =d2

dx2+ v1(x)

d

dx+ v2(x) = part2 + v1(x) part + v2(x)

Taking into account that for any function F (x) we have

[part F ] = F prime [part2 F

]= F primeprime + 2F prime part

then from [minus(12)part2 + UAminus

2 ] = minusωAminus

2 we arrive to

[minus1

2part2 + U(x) part2 + v1(x) part + v2(x)

]= minusvprime1 part2 + (minus1

2vprimeprime1 minus vprime2 minus 2U prime) part minus 1

2vprimeprime2 minus v1 U

prime minus U primeprime


8

Therefore the commutation relation [HAminus

2 ] = minusωAminus

2 leads to the following system ofdifferential equations

vprime1 = ω

12 v

primeprime1 + vprime2 + 2U prime = ω v1

12 v

primeprime2 + v1 U

prime + U primeprime = ω v2

(8)

From the first equation we obtain that v1(x) = ω x+ α1 and replacing this value for v1 in thetwo last equations they become

vprime2 + 2U prime = ω (ω x+ α1)

12v

primeprime2 + (ω x+ α1)U


The second equation can be rewritten using the first one as (ω x+α1)Uprime + 1

2 ω2 = ω v2 and

then we can take derivatives in this expression and we obtain vprime2 = U prime +(1ω) (ω x+α1)Uprimeprime and

when we put this in the first equation we arrive at

3U prime +1

ω(ω x+ α1)U

primeprime = ω (ω x+ α1)

Therefore the function w = U prime satisfies the inhomogeneous linear first-order equation

1

ω(ω x+ α1)w

prime + 3w = ω (ω x+ α1)

The general solution of the associated homogeneous linear first-order equation is

w =C

(ω x+ α1)3

while we can see that

w1 =ω2

4x+

1

4ω α1

is a particular solution of the inhomogeneous equation Therefore

U prime(x) =C

(ω x+ α1)3+ω2

4x+

1

4ω α1 =rArr U(x) =

C1

(ω x+ α1)2+ω2

8x2 +

ωα1

4x

which can also be written up to addition of a constant and in the relevant case for whichC1 gt 0 as

U(x) =g2

(x+ α)2+ω2

8(x+ α)2

where α = α1ω and g2 = C1ω2

Finally the value of v2 obtained when we replace U by the previous expression in the relationv2 = (x+ α)U prime + 1

2 ω is

v2 =ω

2minus 2 g2

(x+ α)2+ω2

4(x+ α)2

Therefore Aminus

2 is given by

Aminus

2 =

(d

dx+ω

2(x+ α)

)2

minus 2 g2

(x+ α)2


9


Aminus

3 =d3

dx3+ v1(x)

d2

dx2+ v2(x)

d

dx+ v3(x) = part3 + v1(x) part

2 + v2(x) part + v3(x)

The commutator of H with Aminus

3 is given by

[H Aminus

3 ] = minus vprime1 part3 +

[minus vprime2 minus

1

2vprimeprime1 minus 3U prime

]part2 +


1

2vprimeprime2 minus 2v1U

prime minus 3U primeprime

]part

+[minus1

2vprimeprime3 minus v2U

prime minus v1Uprimeprime minus U primeprimeprime

]

and when we assume that these two operators satisfy the commutation relation [H Aminus

3 ] =minusωAminus

3 we obtain the following system of differential equations

vprime1 = ω

vprime2 + 12 v

primeprime1 + 3U prime = ω v1

vprime3 + 12 v

primeprime2 + 2v1U

prime + 3U primeprime = ω v2

12 v

primeprime3 + v2U

prime + v1Uprimeprime + U primeprimeprime = ω v3

(9)

Unfortunately neither the solution of this system is an easy task nor the computation for thecases k gt 3 is simple and we should look the problem from a more general perspective Ouraim is to point out that the two cases we have studied seems to be the only possible cases

We first remark that the properties of the corresponding classical problem are very useful fordealing with the quantum problem Quantum systems with equispaced spectra are analogousto isochronous systems when the potentials are rational functions [14 15] but there exist othersuch quantum systems whose analogous are not isochronous and only in the WKB approximationthese classes of generalised harmonic oscillators coincide [14 15 16 17 18]

If a rational potential U(x) is such that classical problem is isochronous with an angularfrequency ω then the asymptotic behaviour of U(x) is given by

U(x) sim 1

2ω2 x2 at xrarr plusmninfin

if U(x) has not singularities on the real axis and

U(x) sim 1

8ω2 x2 at xrarr infin

in x gt 0 if U(x) has a singularity on the real axis This fact is a direct consequence of theresult in [3] We give here a simpler proof for this less general result which furthermore showsthe reason of factor 18 instead of 12 If the potential U(x) has not singularities on the realaxis then the asymptotic behaviour of U(x) at x = plusmninfin should be of the form U(x) sim αx2nwith n isin N+ because if the leading term is an odd power the motion at sufficiently high energywould be unbounded Therefore we can assume that the asymptotic behaviour is given by aneven function

Recall that the expression of the period as a function of the energy in a one-dimensionalbounded and therefore periodic motion of a particle of mass m = 1 under the action of apotential U(x) is (1) which gives rise to (2) from which we derived the isochronicity condition


radic2 τ

π

radicE =

2radic

2

ω

radicE


10

that in the particular case of a regular symmetric potential reduces to (3) When T (E) is notconstant but it is asymptotically constant for big enough E ie if we assume that

T (E) = τ(1 +

α1

E+α2

E2+ middot middot middot

)

we obtain that


πradic

2

int U

0

T (E)radicU minus E

dE =2

πradic

2

radicU

(τ +

β1

U+β2

U2+ middot middot middot

)

with βk = αk Ik where

Ik =

int 1

0

2 dζ

(1 minus ζ2)k

and then


radic2

π

radicE

(τ +

β1

E+β2


) (10)

On the other side given a rational potential with a real pole assumed to be at x = 0 theclassical motion takes place in the open interval (0infin) and then as limErarrinfin xminus(E) = 0 we seethat for sufficiently high energy x+(E) behaves as x+(E) sim (

radic2 τπ)

radicE

In the case of a potential U(x) sim k x2n the isochronicity condition for high energy leads to

k

(2

ω

)2n (E

2

)n

= E if U is regular

and

k 2n

(2

ω

)2n (E

2

)n

= E if U has a pole

with ω = 2πτ and therefore n = 1 and then either k = ω22 for regular U or k = ω28 if Uhas a pole

Of course in the asymptotic behaviour of the potential only the leading term is determinedand therefore the above asymptotic dependence can be replaced by any second order polynomialin x with the same leading term as the above potential

As a corollary using the semi-classical approach to quantum mechanics we can concludethat if the energy spectrum of the quantum Hamiltonian (6) with a rational function U(x)is equispaced with difference ω among neighbour eigenstates then the asymptotic behaviourfor x rarr infin of the potential is U(x) sim αω2 x2 or any second order polynomial P2(x) of seconddegree with the same leading term because the classical limit should be a periodic motion Hereα = 12 when U is regular whereas α = 18 when there is one real pole

This suggests us to study the case of potentials of the form U(x) = P2(x) +U1(x) or simplyU(x) = αω2 x2 + U1(x) with U1(x) decreasing at xrarr plusmninfin

Let assume that the Hamiltonian for such a potential admits ladder operators Aminus(ω)and A+(ω) for which Aplusmn(0) exist Then the condition [H(ω) Aminus(ω)] = minusωAminus(ω) whenparticularised for ω = 0 leads to the commutativity of A(0) with H(0) = H1 = minus(12)d2dx2 +U1(x)

In the fundamental papers by Burchnall and Chaundy the theory of commuting differentialoperators has been developed [19] More specifically in the case we are considering there are twodifferent possibilities depending on k being either even or odd When k is even k = 2m the onlypossibility is that the differential operator A2m be a polynomial function of order m in H1 Ifon the contrary k is odd then the results of Burchnall and Chaundy (see also [20 21]) establish


11

that U1(x) should be a solution of a high-order Korteweg de Vries equation [22] Therefore thepotential function U1(x) takes the form [20 21]

U1(x) =lsum

j=0

mj(mj + 1)

(xminus xj)2(11)

where mj are non-negative integers and xj are complex numbers ie U1(x) only can have(maybe complex) poles of second order at points xj Note that as we have assumed that thepotential is real with each complex pole its conjugate value is also a pole

There are only two possibilities1 There is no real pole but all poles xj are complex numbersHence there will be an even number of poles l = 2r and with each pole its conjugate is

also a poleThe assumed non-existence of real poles implies that the potential function U1 in the domain

of the integral (2) is bounded and when there exist terms in (11) they will destroy the precedingisochronicity condition [1] and therefore in such a case the analogous quantum case cannot havean equispaced spectrum

2 There is one real pole at the point x0As in this paper we consider to be equivalent two potentials obtained one from the other by

means of shift and reflection and we are restricting ourselves to the case of potentials having atmost one real pole at the point x0 we can choose x0 to be x0 = 0 Consequently the functionU1(x) is a regular function on the semi-axis 0 lt x lt +infin and once again as the potential isreal with each complex pole its conjugate value is also a pole ie there are l = 2r + 1 polesand only one x0 = 0 is real

In this case the function U(x) should have the following behaviour

U(x) sim

g2

x2 xrarr 0

18 ω

2 x2 +g21

x2 xrarr infin

where

g2 = m0(m0 + 1) g21 =

2r+1sum

j=0

mj(mj + 1)

ie g1 ge g gt 0 The equation U(x) = E has two roots xminus(E) lt x+(E) such that for big enoughvalues of the energy are E rarr infin

x+(E) simradicE

ω+

g1radicE

xminus(E) sim gradicE

Therefore the difference between both turning points as a function of the energy is

x+(E) minus xminus(E) simradicE

ω+g1 minus gradic

E

and as indicated above in order for the spectrum to be equidistant and then the analogousclassical system be isoperiodic we should have g1 = g Hence we can have only one pole (or allmj = 0 except m0) We arrive to the known case of the isotonic oscillator

U(x) =1

8ω2 x2 +

g2

x2


12

Note however that the derivation of this property is not fully rigourous because one can admitthat the classical system be such that T (E) is not constant but it is asymptotically constant

So we come to the following resultsIf we consider the family of rational potentials U(x) of the form

U(x) = αω2 x2 + U1(x) α =1

2or

1

8

where ω is a positive constant and u1(x) is a rational function having at most a real poleand vanishing at x rarr plusmninfin Then the quantum problem admits creation and annihilationoperators (which include the eigenfunctions of the harmonic oscillator in their domains) andhas an equispaced spectrum with distance ω only for the two cases considered before

It is also worthy of mention that in multidimensional case we know some examples of quantumsystems admitting creation and annihilation operators B+

k and Bk see papers [23 24] and [25]for details However the analog of the previous result for multidimensional rational potentialsremains still unknown

Acknowledgements

Support of projects BFM-2003-02532 FPA-2003-02948 and SAB-2003-0256 is acknowledged

References[1] NH Abel ldquoAuflosung einer mechanischen Aufgaberdquo J Reine Angew Math 1 153ndash157 (1826)[2] M Asorey JF Carinena G Marmo and AM Perelomov ldquoIsoperiodic classical systems and their quantum

counterpartsrdquo Ann Phys 322 1444ndash65 (2007)[3] OA Chalykh and AP Veselov ldquoA remark on rational isochronous potentialsrdquo J Nonl Math Phys 12

Suppl 1 179ndash83 (2005)[4] F Calogero ldquoSolution of a three body problem in one dimensionrdquo J Math Phys 10 2191ndash96 (1969)[5] Z Dongpei ldquoA new potential with the spectrum of an isotonic oscillatorrdquo J Phys A 20 4331ndash4336 (1987)[6] JF Carinena MF Ranada M Santander and M Senthilvelan ldquoA non-linear oscillator with quasi-harmonic

behaviour two- and n-dimensional oscillatorsrdquo Nonlinearity 17 1941ndash1963 (2004)[7] JF Carinena MF Ranada and M Santander ldquoOne-dimensional model of a quantum nonlinear harmonic

oscillatorrdquo Rep Math Phys 54 285ndash293 (2004)[8] JF Carinena MF Ranada and M Santander ldquoA quantum exactly solvable nonlinear oscillator with quasi-

harmonic behaviourrdquo Ann Phys 322 434ndash59 (2007)[9] LD Landau and EM Lifshitz Mechanics Pergamon Press (1981)

[10] R Subramanian and KV Bhagwat ldquoA lower bound for ground-state energy by Steiner symmetrisation ofthe potentialrdquo J Phys A Math Gen 20 69-78 (1987)

[11] PAM Dirac ldquoThe quantum theory of the emission and absortion of radiationrdquo Proc Roy Soc A (London)114 243ndash265 (1927)

[12] JF Carinena and A Ramos ldquoRiccati equation Factorization Method and Shape Invariancerdquo Rev MathPhys 12 1279ndash304 (2000)

[13] A Oblomkov ldquoMonodromy free Schrodinger operators with quadratically increasing potentialsrdquo TheorMath Phys 121 1574ndash84 (1974)

[14] MM Nieto and VP Gutschick ldquoInequivalence of the classes of classical and quantum harmonic potentialsProof by examplerdquo Phys Rev D 23 922ndash26 (1981)

[15] J Dorignac ldquoOn the quantum spectrum of isochronous potentialsrdquo J Phys AMath Gen 38 6183ndash210(2005)

[16] R Jost and W Kohn ldquoEquivalent potentialsrdquo Phys Rev 88 382ndash385 (1952)[17] PB Abraham and HE Moses ldquoChanges in potentials due to changes in the point spectrum anharmonic

osillators with exact solutionsrdquo Phys Rev A 22 1333ndash1340 (1980)[18] AM Perelomov and YaB Zelrsquodovich Quantum Mechanics Selected Topics World Sci Singapore (1998)[19] JL Burchnall and TW Chaundy ldquoCommutative ordinary diifferential operatorsrdquo Proc London Math

Soc Ser 2 21 420ndash440 (1923) Proc Roy Soc London A 118 557ndash583 (1928)[20] H Airault HP McKean and J Moser ldquoRational and elliptic solutions of the Korteweg-de Vries equation

and a related many-body problemrdquo Commun Pure Appl Math 30 95ndash148 (1977)


13

[21] M Adler and J Moser ldquoOn a class of polynomials connected with the Korteweg de Vries equationrdquo CommunMath Phys 61 1ndash30 (1978)

[22] DJ Korteweg and G de Vries ldquoOn the change of form of long waves advancing in a rectangular canal andon a new type of long stationary wavesrdquo Phil Mag 39 422ndash443 (1895)

[23] AM Perelomov ldquoAlgebraic approach to the solution of the one-dimensional model of N interactingparticlesrdquo Theor Math Phys 6 263ndash282 (1971)

[24] AM Perelomov ldquoCompletely integrable classical systems connected with semi-simple Lie algebras IIrdquoPreprint ITEP-27 (1976) math-ph0111018

[25] MA Olshanetsky and AM Perelomov ldquoQuantum integrable systems related to Lie algebrasrdquo Phys Rep94 313ndash404 (1983)


14

Isochronous classical systems and quantum systems

with equally spaced spectra

Jose F Carinenaa) Askold M Perelomov 1b) and Manuel F Ranadac)

Departamento de Fısica Teorica Facultad de Ciencias Universidad de Zaragoza 50009Zaragoza Spain

E-mail ajfcunizares bperelomodftuzunizares cmfranunizares

Abstract

We study isoperiodic classical systems what allows us to find the classical isochronoussystems ie having a period independent of the energy The corresponding quantumanalog systems with an equally spaced spectrum are analysed by looking for possiblecreation-like differential operators The harmonic oscillator and the isotonic oscillatorare the two main essentially unique examples of such situation

Keywords Isochronous systems ladder operators equispaced spectra

1 Introduction

Inverse problems have been playing a very relevant role in the development of physical theoriesThe physical laws describing the different phenomena are constructed from experimental dataand their symmetries and regularities are used to discard possible laws or to pick up other lawsamong other mathematically possible formulations

In this paper we review some results on two important inverse problems and report somerecent progress on both The first one is the characterisation of possible convex potentials givingrise to isochronous motions ie systems for which the bounded motions have a period which doesnot depend on the energy With this aim we first review in Section 2 the very classical problemof the determination of the potential from the energy dependence of the period a problemwhich dates back to Abel [1] and therefore the characterisation of isoperiodic potentials thosegiving rise to the same dependence The harmonic oscillator is the prototype of an isochronoussystem and all other isochronous systems are isoperiodic with the harmonic oscillator After acharacterisation of such systems we report the recent approach used in [2] to recover a result byChalykh and Veselov [3] among rational potentials only the harmonic oscillator and the isotonicoscillator [4 5] produce isochronous motions There are however other isochronous systemsdescribed by non-rational potentials for instance potentials for which the second derivative hasa discontinuity

A related problem in quantum mechanics is the determination of those potentials with anequispaced energy spectrum Of course only in the best conditions of regularity the solution ofboth problems can be related and this is only possible to order ~ Our approach here is basedon the search for creation- and annihilation-like operators which are assumed to be differential

1 On leave of absence from Institute for Theoretical and Experimental Physics 117259 Moscow Russia


ccopy 2007 IOP Publishing Ltd 1


Ak =ksum

j=0

vkminusj(x)dj

dxj=

ksum

j=0





T (E) =radic

2m

int x+(E)

xminus

(E)

dxradicE minus U(x)

(1)



U(x) =



x =



T (E) = 2radic

2m

int E

0


dy = 2radic

2mE

int 1

0


dz


2


1

2radicE

int 1

0


dz +radicE

int 1

0


z dz = 0


int 1

0


dz = 0 =1radicE

int E

0

ζ(y)radicE minus y

dy forallE gt 0







πradic

2m

int U

0

T (E)radicU minus E

dE (2)


T (E) =radic

2m

int E

0


dy


(f1 lowast f2)(E) =

int E





intinfin






obtain

L[∆](s) =1radic

2π2m

radicπ

sL[T ](s) =

1radic2π2m

L[ψ](s)L[T ](s)



πradic

2

int U

0

τradicU minus E

dE =2 τ

πradic

2

radicU


3



radic2 τ

π

radicE =

2radic

2

ω

radicE


x+(E) =τradic2π

radicE =

radic2

ω

radicE (3)


φ+(U) = φ(0)+ (U) + f(U) φminus(U) = φ

(0)minus

(U) + f(U)

where φ(0)+ and φ

(0)minus







φ(U) =1

2πradic

2m

int U

0

T (E)radicU minus E

dE




radicU

2+ f(U) φ+(U) =

T

π

radicU

2+ f(U)



radicU

2+ a φ+(U) =

T

π

radicU

2+ a


U(x) =ω2

2(xminus a)2 ω =

2π

T


f(U) = αT

π

radicU

2

then


π

radicU

2 φ+(U) = (1 + α)

T

π

radicU

2


4


U(x) =

12 mω2

1 x2 if x le 0

12mω2

2 x2 if x ge 0


ω1 =2π

(1 minus α)T ω2 =

2π

(1 + α)T


α =ω1 minus ω2

ω1 + ω2and ω0 =

2ω1 ω2

(ω1 + ω2)




πradic

2m

int U

0

T (E)radicV minus E

dE


U(x) =

(ax2 + bx+ c

x+ d

)2



Jg(x) =x

2minus 2α

x α ge 0



(x

2minus 2α

x

)




5




H =1


dx


a =1radic2

(1radicω

d

dx+

radicω x

) a+ =

1radic2

(minus 1radic

ω

d

dx+

radicω x

)


A =1radic2

(d

dx+ ω x

) A+ =

1radic2

(minus d

dx+ ω x

)









6


H =1

2

(p2 +

1

4ω2x2

)+g2

x2



U(x) =

18 ω

2 x2 for x gt 0

infin for x lt 0




Aminus

k =

ksum

j=0

vkminusj(x)dj

dxj=

ksum

j=0



k



H =1

2p2 + U(x) = minus1

2

d2

dx2+ U(x) = minus1

2part2 + U(x) (6)



7


H ψ = E ψ




k satisfying

[HAplusmn


k (7)


Aminus

k =dk

dxk+ v1(x)

dkminus1






1 k = 1 Here Aminus



1 ] = minusωAminus


[minus1

2part2 v1(x)

]= minus1



vprime1 = ω

12 v





U(x) =1

2ω2(xminus x0)

2 x0 = minusα1

ω2


Aminus

2 =d2

dx2+ v1(x)

d






2 ] = minusωAminus

2 we arrive to

[minus1







8


2 ] = minusωAminus


vprime1 = ω

12 v


12 v

primeprime2 + v1 U


(8)



12v




2 ω2 = ω v2 and



3U prime +1

ω(ω x+ α1)U



1

ω(ω x+ α1)w



w =C

(ω x+ α1)3


w1 =ω2

4x+

1

4ω α1


U prime(x) =C

(ω x+ α1)3+ω2

4x+

1


C1

(ω x+ α1)2+ω2

8x2 +

ωα1

4x


U(x) =g2

(x+ α)2+ω2

8(x+ α)2



2 ω is

v2 =ω

2minus 2 g2

(x+ α)2+ω2

4(x+ α)2

Therefore Aminus

2 is given by

Aminus

2 =

(d

dx+ω

2(x+ α)

)2

minus 2 g2

(x+ α)2


9


Aminus

3 =d3

dx3+ v1(x)

d2

dx2+ v2(x)

d


2 + v2(x) part + v3(x)


3 is given by

[H Aminus



1


]part2 +


1



]part

+[minus1



]


3 ] =minusωAminus


vprime1 = ω

vprime2 + 12 v


vprime3 + 12 v

primeprime2 + 2v1U


12 v

primeprime3 + v2U


(9)




U(x) sim 1



U(x) sim 1





radic2 τ

π

radicE =

2radic

2

ω

radicE


10


T (E) = τ(1 +

α1

E+α2


)

we obtain that


πradic

2

int U

0

T (E)radicU minus E

dE =2

πradic

2

radicU

(τ +

β1

U+β2


)


Ik =

int 1

0

2 dζ

(1 minus ζ2)k

and then


radic2

π

radicE

(τ +

β1

E+β2


) (10)


radic2 τπ)

radicE


k

(2

ω

)2n (E

2

)n

= E if U is regular

and

k 2n

(2

ω

)2n (E

2

)n

= E if U has a pole








11


U1(x) =lsum

j=0

mj(mj + 1)

(xminus xj)2(11)








U(x) sim

g2

x2 xrarr 0

18 ω

2 x2 +g21

x2 xrarr infin

where

g2 = m0(m0 + 1) g21 =

2r+1sum

j=0

mj(mj + 1)


x+(E) simradicE

ω+

g1radicE




ω+g1 minus gradic

E


U(x) =1

8ω2 x2 +

g2

x2


12



U(x) = αω2 x2 + U1(x) α =1

2or

1

8




Acknowledgements



















13







14


Ak =ksum

j=0

vkminusj(x)dj

dxj=

ksum

j=0





T (E) =radic

2m

int x+(E)

xminus

(E)

dxradicE minus U(x)

(1)



U(x) =



x =



T (E) = 2radic

2m

int E

0


dy = 2radic

2mE

int 1

0


dz


2


1

2radicE

int 1

0


dz +radicE

int 1

0


z dz = 0


int 1

0


dz = 0 =1radicE

int E

0

ζ(y)radicE minus y

dy forallE gt 0







πradic

2m

int U

0

T (E)radicU minus E

dE (2)


T (E) =radic

2m

int E

0


dy


(f1 lowast f2)(E) =

int E





intinfin






obtain

L[∆](s) =1radic

2π2m

radicπ

sL[T ](s) =

1radic2π2m

L[ψ](s)L[T ](s)



πradic

2

int U

0

τradicU minus E

dE =2 τ

πradic

2

radicU


3



radic2 τ

π

radicE =

2radic

2

ω

radicE


x+(E) =τradic2π

radicE =

radic2

ω

radicE (3)


φ+(U) = φ(0)+ (U) + f(U) φminus(U) = φ

(0)minus

(U) + f(U)

where φ(0)+ and φ

(0)minus







φ(U) =1

2πradic

2m

int U

0

T (E)radicU minus E

dE




radicU

2+ f(U) φ+(U) =

T

π

radicU

2+ f(U)



radicU

2+ a φ+(U) =

T

π

radicU

2+ a


U(x) =ω2

2(xminus a)2 ω =

2π

T


f(U) = αT

π

radicU

2

then


π

radicU

2 φ+(U) = (1 + α)

T

π

radicU

2


4


U(x) =

12 mω2

1 x2 if x le 0

12mω2

2 x2 if x ge 0


ω1 =2π

(1 minus α)T ω2 =

2π

(1 + α)T


α =ω1 minus ω2

ω1 + ω2and ω0 =

2ω1 ω2

(ω1 + ω2)




πradic

2m

int U

0

T (E)radicV minus E

dE


U(x) =

(ax2 + bx+ c

x+ d

)2



Jg(x) =x

2minus 2α

x α ge 0



(x

2minus 2α

x

)




5




H =1


dx


a =1radic2

(1radicω

d

dx+

radicω x

) a+ =

1radic2

(minus 1radic

ω

d

dx+

radicω x

)


A =1radic2

(d

dx+ ω x

) A+ =

1radic2

(minus d

dx+ ω x

)









6


H =1

2

(p2 +

1

4ω2x2

)+g2

x2



U(x) =

18 ω

2 x2 for x gt 0

infin for x lt 0




Aminus

k =

ksum

j=0

vkminusj(x)dj

dxj=

ksum

j=0



k



H =1

2p2 + U(x) = minus1

2

d2

dx2+ U(x) = minus1

2part2 + U(x) (6)



7


H ψ = E ψ




k satisfying

[HAplusmn


k (7)


Aminus

k =dk

dxk+ v1(x)

dkminus1






1 k = 1 Here Aminus



1 ] = minusωAminus


[minus1

2part2 v1(x)

]= minus1



vprime1 = ω

12 v





U(x) =1

2ω2(xminus x0)

2 x0 = minusα1

ω2


Aminus

2 =d2

dx2+ v1(x)

d






2 ] = minusωAminus

2 we arrive to

[minus1







8


2 ] = minusωAminus


vprime1 = ω

12 v


12 v

primeprime2 + v1 U


(8)



12v




2 ω2 = ω v2 and



3U prime +1

ω(ω x+ α1)U



1

ω(ω x+ α1)w



w =C

(ω x+ α1)3


w1 =ω2

4x+

1

4ω α1


U prime(x) =C

(ω x+ α1)3+ω2

4x+

1


C1

(ω x+ α1)2+ω2

8x2 +

ωα1

4x


U(x) =g2

(x+ α)2+ω2

8(x+ α)2



2 ω is

v2 =ω

2minus 2 g2

(x+ α)2+ω2

4(x+ α)2

Therefore Aminus

2 is given by

Aminus

2 =

(d

dx+ω

2(x+ α)

)2

minus 2 g2

(x+ α)2


9


Aminus

3 =d3

dx3+ v1(x)

d2

dx2+ v2(x)

d


2 + v2(x) part + v3(x)


3 is given by

[H Aminus



1


]part2 +


1



]part

+[minus1



]


3 ] =minusωAminus


vprime1 = ω

vprime2 + 12 v


vprime3 + 12 v

primeprime2 + 2v1U


12 v

primeprime3 + v2U


(9)




U(x) sim 1



U(x) sim 1





radic2 τ

π

radicE =

2radic

2

ω

radicE


10


T (E) = τ(1 +

α1

E+α2


)

we obtain that


πradic

2

int U

0

T (E)radicU minus E

dE =2

πradic

2

radicU

(τ +

β1

U+β2


)


Ik =

int 1

0

2 dζ

(1 minus ζ2)k

and then


radic2

π

radicE

(τ +

β1

E+β2


) (10)


radic2 τπ)

radicE


k

(2

ω

)2n (E

2

)n

= E if U is regular

and

k 2n

(2

ω

)2n (E

2

)n

= E if U has a pole








11


U1(x) =lsum

j=0

mj(mj + 1)

(xminus xj)2(11)








U(x) sim

g2

x2 xrarr 0

18 ω

2 x2 +g21

x2 xrarr infin

where

g2 = m0(m0 + 1) g21 =

2r+1sum

j=0

mj(mj + 1)


x+(E) simradicE

ω+

g1radicE




ω+g1 minus gradic

E


U(x) =1

8ω2 x2 +

g2

x2


12



U(x) = αω2 x2 + U1(x) α =1

2or

1

8




Acknowledgements



















13







14


1

2radicE

int 1

0


dz +radicE

int 1

0


z dz = 0


int 1

0


dz = 0 =1radicE

int E

0

ζ(y)radicE minus y

dy forallE gt 0







πradic

2m

int U

0

T (E)radicU minus E

dE (2)


T (E) =radic

2m

int E

0


dy


(f1 lowast f2)(E) =

int E





intinfin






obtain

L[∆](s) =1radic

2π2m

radicπ

sL[T ](s) =

1radic2π2m

L[ψ](s)L[T ](s)



πradic

2

int U

0

τradicU minus E

dE =2 τ

πradic

2

radicU


3



radic2 τ

π

radicE =

2radic

2

ω

radicE


x+(E) =τradic2π

radicE =

radic2

ω

radicE (3)


φ+(U) = φ(0)+ (U) + f(U) φminus(U) = φ

(0)minus

(U) + f(U)

where φ(0)+ and φ

(0)minus







φ(U) =1

2πradic

2m

int U

0

T (E)radicU minus E

dE




radicU

2+ f(U) φ+(U) =

T

π

radicU

2+ f(U)



radicU

2+ a φ+(U) =

T

π

radicU

2+ a


U(x) =ω2

2(xminus a)2 ω =

2π

T


f(U) = αT

π

radicU

2

then


π

radicU

2 φ+(U) = (1 + α)

T

π

radicU

2


4


U(x) =

12 mω2

1 x2 if x le 0

12mω2

2 x2 if x ge 0


ω1 =2π

(1 minus α)T ω2 =

2π

(1 + α)T


α =ω1 minus ω2

ω1 + ω2and ω0 =

2ω1 ω2

(ω1 + ω2)




πradic

2m

int U

0

T (E)radicV minus E

dE


U(x) =

(ax2 + bx+ c

x+ d

)2



Jg(x) =x

2minus 2α

x α ge 0



(x

2minus 2α

x

)




5




H =1


dx


a =1radic2

(1radicω

d

dx+

radicω x

) a+ =

1radic2

(minus 1radic

ω

d

dx+

radicω x

)


A =1radic2

(d

dx+ ω x

) A+ =

1radic2

(minus d

dx+ ω x

)









6


H =1

2

(p2 +

1

4ω2x2

)+g2

x2



U(x) =

18 ω

2 x2 for x gt 0

infin for x lt 0




Aminus

k =

ksum

j=0

vkminusj(x)dj

dxj=

ksum

j=0



k



H =1

2p2 + U(x) = minus1

2

d2

dx2+ U(x) = minus1

2part2 + U(x) (6)



7


H ψ = E ψ




k satisfying

[HAplusmn


k (7)


Aminus

k =dk

dxk+ v1(x)

dkminus1






1 k = 1 Here Aminus



1 ] = minusωAminus


[minus1

2part2 v1(x)

]= minus1



vprime1 = ω

12 v





U(x) =1

2ω2(xminus x0)

2 x0 = minusα1

ω2


Aminus

2 =d2

dx2+ v1(x)

d






2 ] = minusωAminus

2 we arrive to

[minus1







8


2 ] = minusωAminus


vprime1 = ω

12 v


12 v

primeprime2 + v1 U


(8)



12v




2 ω2 = ω v2 and



3U prime +1

ω(ω x+ α1)U



1

ω(ω x+ α1)w



w =C

(ω x+ α1)3


w1 =ω2

4x+

1

4ω α1


U prime(x) =C

(ω x+ α1)3+ω2

4x+

1


C1

(ω x+ α1)2+ω2

8x2 +

ωα1

4x


U(x) =g2

(x+ α)2+ω2

8(x+ α)2



2 ω is

v2 =ω

2minus 2 g2

(x+ α)2+ω2

4(x+ α)2

Therefore Aminus

2 is given by

Aminus

2 =

(d

dx+ω

2(x+ α)

)2

minus 2 g2

(x+ α)2


9


Aminus

3 =d3

dx3+ v1(x)

d2

dx2+ v2(x)

d


2 + v2(x) part + v3(x)


3 is given by

[H Aminus



1


]part2 +


1



]part

+[minus1



]


3 ] =minusωAminus


vprime1 = ω

vprime2 + 12 v


vprime3 + 12 v

primeprime2 + 2v1U


12 v

primeprime3 + v2U


(9)




U(x) sim 1



U(x) sim 1





radic2 τ

π

radicE =

2radic

2

ω

radicE


10


T (E) = τ(1 +

α1

E+α2


)

we obtain that


πradic

2

int U

0

T (E)radicU minus E

dE =2

πradic

2

radicU

(τ +

β1

U+β2


)


Ik =

int 1

0

2 dζ

(1 minus ζ2)k

and then


radic2

π

radicE

(τ +

β1

E+β2


) (10)


radic2 τπ)

radicE


k

(2

ω

)2n (E

2

)n

= E if U is regular

and

k 2n

(2

ω

)2n (E

2

)n

= E if U has a pole








11


U1(x) =lsum

j=0

mj(mj + 1)

(xminus xj)2(11)








U(x) sim

g2

x2 xrarr 0

18 ω

2 x2 +g21

x2 xrarr infin

where

g2 = m0(m0 + 1) g21 =

2r+1sum

j=0

mj(mj + 1)


x+(E) simradicE

ω+

g1radicE




ω+g1 minus gradic

E


U(x) =1

8ω2 x2 +

g2

x2


12



U(x) = αω2 x2 + U1(x) α =1

2or

1

8




Acknowledgements



















13







14



radic2 τ

π

radicE =

2radic

2

ω

radicE


x+(E) =τradic2π

radicE =

radic2

ω

radicE (3)


φ+(U) = φ(0)+ (U) + f(U) φminus(U) = φ

(0)minus

(U) + f(U)

where φ(0)+ and φ

(0)minus







φ(U) =1

2πradic

2m

int U

0

T (E)radicU minus E

dE




radicU

2+ f(U) φ+(U) =

T

π

radicU

2+ f(U)



radicU

2+ a φ+(U) =

T

π

radicU

2+ a


U(x) =ω2

2(xminus a)2 ω =

2π

T


f(U) = αT

π

radicU

2

then


π

radicU

2 φ+(U) = (1 + α)

T

π

radicU

2


4


U(x) =

12 mω2

1 x2 if x le 0

12mω2

2 x2 if x ge 0


ω1 =2π

(1 minus α)T ω2 =

2π

(1 + α)T


α =ω1 minus ω2

ω1 + ω2and ω0 =

2ω1 ω2

(ω1 + ω2)




πradic

2m

int U

0

T (E)radicV minus E

dE


U(x) =

(ax2 + bx+ c

x+ d

)2



Jg(x) =x

2minus 2α

x α ge 0



(x

2minus 2α

x

)




5




H =1


dx


a =1radic2

(1radicω

d

dx+

radicω x

) a+ =

1radic2

(minus 1radic

ω

d

dx+

radicω x

)


A =1radic2

(d

dx+ ω x

) A+ =

1radic2

(minus d

dx+ ω x

)









6


H =1

2

(p2 +

1

4ω2x2

)+g2

x2



U(x) =

18 ω

2 x2 for x gt 0

infin for x lt 0




Aminus

k =

ksum

j=0

vkminusj(x)dj

dxj=

ksum

j=0



k



H =1

2p2 + U(x) = minus1

2

d2

dx2+ U(x) = minus1

2part2 + U(x) (6)



7


H ψ = E ψ




k satisfying

[HAplusmn


k (7)


Aminus

k =dk

dxk+ v1(x)

dkminus1






1 k = 1 Here Aminus



1 ] = minusωAminus


[minus1

2part2 v1(x)

]= minus1



vprime1 = ω

12 v





U(x) =1

2ω2(xminus x0)

2 x0 = minusα1

ω2


Aminus

2 =d2

dx2+ v1(x)

d






2 ] = minusωAminus

2 we arrive to

[minus1







8


2 ] = minusωAminus


vprime1 = ω

12 v


12 v

primeprime2 + v1 U


(8)



12v




2 ω2 = ω v2 and



3U prime +1

ω(ω x+ α1)U



1

ω(ω x+ α1)w



w =C

(ω x+ α1)3


w1 =ω2

4x+

1

4ω α1


U prime(x) =C

(ω x+ α1)3+ω2

4x+

1


C1

(ω x+ α1)2+ω2

8x2 +

ωα1

4x


U(x) =g2

(x+ α)2+ω2

8(x+ α)2



2 ω is

v2 =ω

2minus 2 g2

(x+ α)2+ω2

4(x+ α)2

Therefore Aminus

2 is given by

Aminus

2 =

(d

dx+ω

2(x+ α)

)2

minus 2 g2

(x+ α)2


9


Aminus

3 =d3

dx3+ v1(x)

d2

dx2+ v2(x)

d


2 + v2(x) part + v3(x)


3 is given by

[H Aminus



1


]part2 +


1



]part

+[minus1



]


3 ] =minusωAminus


vprime1 = ω

vprime2 + 12 v


vprime3 + 12 v

primeprime2 + 2v1U


12 v

primeprime3 + v2U


(9)




U(x) sim 1



U(x) sim 1





radic2 τ

π

radicE =

2radic

2

ω

radicE


10


T (E) = τ(1 +

α1

E+α2


)

we obtain that


πradic

2

int U

0

T (E)radicU minus E

dE =2

πradic

2

radicU

(τ +

β1

U+β2


)


Ik =

int 1

0

2 dζ

(1 minus ζ2)k

and then


radic2

π

radicE

(τ +

β1

E+β2


) (10)


radic2 τπ)

radicE


k

(2

ω

)2n (E

2

)n

= E if U is regular

and

k 2n

(2

ω

)2n (E

2

)n

= E if U has a pole








11


U1(x) =lsum

j=0

mj(mj + 1)

(xminus xj)2(11)








U(x) sim

g2

x2 xrarr 0

18 ω

2 x2 +g21

x2 xrarr infin

where

g2 = m0(m0 + 1) g21 =

2r+1sum

j=0

mj(mj + 1)


x+(E) simradicE

ω+

g1radicE




ω+g1 minus gradic

E


U(x) =1

8ω2 x2 +

g2

x2


12



U(x) = αω2 x2 + U1(x) α =1

2or

1

8




Acknowledgements



















13







14


U(x) =

12 mω2

1 x2 if x le 0

12mω2

2 x2 if x ge 0


ω1 =2π

(1 minus α)T ω2 =

2π

(1 + α)T


α =ω1 minus ω2

ω1 + ω2and ω0 =

2ω1 ω2

(ω1 + ω2)




πradic

2m

int U

0

T (E)radicV minus E

dE


U(x) =

(ax2 + bx+ c

x+ d

)2



Jg(x) =x

2minus 2α

x α ge 0



(x

2minus 2α

x

)




5




H =1


dx


a =1radic2

(1radicω

d

dx+

radicω x

) a+ =

1radic2

(minus 1radic

ω

d

dx+

radicω x

)


A =1radic2

(d

dx+ ω x

) A+ =

1radic2

(minus d

dx+ ω x

)









6


H =1

2

(p2 +

1

4ω2x2

)+g2

x2



U(x) =

18 ω

2 x2 for x gt 0

infin for x lt 0




Aminus

k =

ksum

j=0

vkminusj(x)dj

dxj=

ksum

j=0



k



H =1

2p2 + U(x) = minus1

2

d2

dx2+ U(x) = minus1

2part2 + U(x) (6)



7


H ψ = E ψ




k satisfying

[HAplusmn


k (7)


Aminus

k =dk

dxk+ v1(x)

dkminus1






1 k = 1 Here Aminus



1 ] = minusωAminus


[minus1

2part2 v1(x)

]= minus1



vprime1 = ω

12 v





U(x) =1

2ω2(xminus x0)

2 x0 = minusα1

ω2


Aminus

2 =d2

dx2+ v1(x)

d






2 ] = minusωAminus

2 we arrive to

[minus1







8


2 ] = minusωAminus


vprime1 = ω

12 v


12 v

primeprime2 + v1 U


(8)



12v




2 ω2 = ω v2 and



3U prime +1

ω(ω x+ α1)U



1

ω(ω x+ α1)w



w =C

(ω x+ α1)3


w1 =ω2

4x+

1

4ω α1


U prime(x) =C

(ω x+ α1)3+ω2

4x+

1


C1

(ω x+ α1)2+ω2

8x2 +

ωα1

4x


U(x) =g2

(x+ α)2+ω2

8(x+ α)2



2 ω is

v2 =ω

2minus 2 g2

(x+ α)2+ω2

4(x+ α)2

Therefore Aminus

2 is given by

Aminus

2 =

(d

dx+ω

2(x+ α)

)2

minus 2 g2

(x+ α)2


9


Aminus

3 =d3

dx3+ v1(x)

d2

dx2+ v2(x)

d


2 + v2(x) part + v3(x)


3 is given by

[H Aminus



1


]part2 +


1



]part

+[minus1



]


3 ] =minusωAminus


vprime1 = ω

vprime2 + 12 v


vprime3 + 12 v

primeprime2 + 2v1U


12 v

primeprime3 + v2U


(9)




U(x) sim 1



U(x) sim 1





radic2 τ

π

radicE =

2radic

2

ω

radicE


10


T (E) = τ(1 +

α1

E+α2


)

we obtain that


πradic

2

int U

0

T (E)radicU minus E

dE =2

πradic

2

radicU

(τ +

β1

U+β2


)


Ik =

int 1

0

2 dζ

(1 minus ζ2)k

and then


radic2

π

radicE

(τ +

β1

E+β2


) (10)


radic2 τπ)

radicE


k

(2

ω

)2n (E

2

)n

= E if U is regular

and

k 2n

(2

ω

)2n (E

2

)n

= E if U has a pole








11


U1(x) =lsum

j=0

mj(mj + 1)

(xminus xj)2(11)








U(x) sim

g2

x2 xrarr 0

18 ω

2 x2 +g21

x2 xrarr infin

where

g2 = m0(m0 + 1) g21 =

2r+1sum

j=0

mj(mj + 1)


x+(E) simradicE

ω+

g1radicE




ω+g1 minus gradic

E


U(x) =1

8ω2 x2 +

g2

x2


12



U(x) = αω2 x2 + U1(x) α =1

2or

1

8




Acknowledgements



















13







14




H =1


dx


a =1radic2

(1radicω

d

dx+

radicω x

) a+ =

1radic2

(minus 1radic

ω

d

dx+

radicω x

)


A =1radic2

(d

dx+ ω x

) A+ =

1radic2

(minus d

dx+ ω x

)









6


H =1

2

(p2 +

1

4ω2x2

)+g2

x2



U(x) =

18 ω

2 x2 for x gt 0

infin for x lt 0




Aminus

k =

ksum

j=0

vkminusj(x)dj

dxj=

ksum

j=0



k



H =1

2p2 + U(x) = minus1

2

d2

dx2+ U(x) = minus1

2part2 + U(x) (6)



7


H ψ = E ψ




k satisfying

[HAplusmn


k (7)


Aminus

k =dk

dxk+ v1(x)

dkminus1






1 k = 1 Here Aminus



1 ] = minusωAminus


[minus1

2part2 v1(x)

]= minus1



vprime1 = ω

12 v





U(x) =1

2ω2(xminus x0)

2 x0 = minusα1

ω2


Aminus

2 =d2

dx2+ v1(x)

d






2 ] = minusωAminus

2 we arrive to

[minus1







8


2 ] = minusωAminus


vprime1 = ω

12 v


12 v

primeprime2 + v1 U


(8)



12v




2 ω2 = ω v2 and



3U prime +1

ω(ω x+ α1)U



1

ω(ω x+ α1)w



w =C

(ω x+ α1)3


w1 =ω2

4x+

1

4ω α1


U prime(x) =C

(ω x+ α1)3+ω2

4x+

1


C1

(ω x+ α1)2+ω2

8x2 +

ωα1

4x


U(x) =g2

(x+ α)2+ω2

8(x+ α)2



2 ω is

v2 =ω

2minus 2 g2

(x+ α)2+ω2

4(x+ α)2

Therefore Aminus

2 is given by

Aminus

2 =

(d

dx+ω

2(x+ α)

)2

minus 2 g2

(x+ α)2


9


Aminus

3 =d3

dx3+ v1(x)

d2

dx2+ v2(x)

d


2 + v2(x) part + v3(x)


3 is given by

[H Aminus



1


]part2 +


1



]part

+[minus1



]


3 ] =minusωAminus


vprime1 = ω

vprime2 + 12 v


vprime3 + 12 v

primeprime2 + 2v1U


12 v

primeprime3 + v2U


(9)




U(x) sim 1



U(x) sim 1





radic2 τ

π

radicE =

2radic

2

ω

radicE


10


T (E) = τ(1 +

α1

E+α2


)

we obtain that


πradic

2

int U

0

T (E)radicU minus E

dE =2

πradic

2

radicU

(τ +

β1

U+β2


)


Ik =

int 1

0

2 dζ

(1 minus ζ2)k

and then


radic2

π

radicE

(τ +

β1

E+β2


) (10)


radic2 τπ)

radicE


k

(2

ω

)2n (E

2

)n

= E if U is regular

and

k 2n

(2

ω

)2n (E

2

)n

= E if U has a pole








11


U1(x) =lsum

j=0

mj(mj + 1)

(xminus xj)2(11)








U(x) sim

g2

x2 xrarr 0

18 ω

2 x2 +g21

x2 xrarr infin

where

g2 = m0(m0 + 1) g21 =

2r+1sum

j=0

mj(mj + 1)


x+(E) simradicE

ω+

g1radicE




ω+g1 minus gradic

E


U(x) =1

8ω2 x2 +

g2

x2


12



U(x) = αω2 x2 + U1(x) α =1

2or

1

8




Acknowledgements



















13







14


H =1

2

(p2 +

1

4ω2x2

)+g2

x2



U(x) =

18 ω

2 x2 for x gt 0

infin for x lt 0




Aminus

k =

ksum

j=0

vkminusj(x)dj

dxj=

ksum

j=0



k



H =1

2p2 + U(x) = minus1

2

d2

dx2+ U(x) = minus1

2part2 + U(x) (6)



7


H ψ = E ψ




k satisfying

[HAplusmn


k (7)


Aminus

k =dk

dxk+ v1(x)

dkminus1






1 k = 1 Here Aminus



1 ] = minusωAminus


[minus1

2part2 v1(x)

]= minus1



vprime1 = ω

12 v





U(x) =1

2ω2(xminus x0)

2 x0 = minusα1

ω2


Aminus

2 =d2

dx2+ v1(x)

d






2 ] = minusωAminus

2 we arrive to

[minus1







8


2 ] = minusωAminus


vprime1 = ω

12 v


12 v

primeprime2 + v1 U


(8)



12v




2 ω2 = ω v2 and



3U prime +1

ω(ω x+ α1)U



1

ω(ω x+ α1)w



w =C

(ω x+ α1)3


w1 =ω2

4x+

1

4ω α1


U prime(x) =C

(ω x+ α1)3+ω2

4x+

1


C1

(ω x+ α1)2+ω2

8x2 +

ωα1

4x


U(x) =g2

(x+ α)2+ω2

8(x+ α)2



2 ω is

v2 =ω

2minus 2 g2

(x+ α)2+ω2

4(x+ α)2

Therefore Aminus

2 is given by

Aminus

2 =

(d

dx+ω

2(x+ α)

)2

minus 2 g2

(x+ α)2


9


Aminus

3 =d3

dx3+ v1(x)

d2

dx2+ v2(x)

d


2 + v2(x) part + v3(x)


3 is given by

[H Aminus



1


]part2 +


1



]part

+[minus1



]


3 ] =minusωAminus


vprime1 = ω

vprime2 + 12 v


vprime3 + 12 v

primeprime2 + 2v1U


12 v

primeprime3 + v2U


(9)




U(x) sim 1



U(x) sim 1





radic2 τ

π

radicE =

2radic

2

ω

radicE


10


T (E) = τ(1 +

α1

E+α2


)

we obtain that


πradic

2

int U

0

T (E)radicU minus E

dE =2

πradic

2

radicU

(τ +

β1

U+β2


)


Ik =

int 1

0

2 dζ

(1 minus ζ2)k

and then


radic2

π

radicE

(τ +

β1

E+β2


) (10)


radic2 τπ)

radicE


k

(2

ω

)2n (E

2

)n

= E if U is regular

and

k 2n

(2

ω

)2n (E

2

)n

= E if U has a pole








11


U1(x) =lsum

j=0

mj(mj + 1)

(xminus xj)2(11)








U(x) sim

g2

x2 xrarr 0

18 ω

2 x2 +g21

x2 xrarr infin

where

g2 = m0(m0 + 1) g21 =

2r+1sum

j=0

mj(mj + 1)


x+(E) simradicE

ω+

g1radicE




ω+g1 minus gradic

E


U(x) =1

8ω2 x2 +

g2

x2


12



U(x) = αω2 x2 + U1(x) α =1

2or

1

8




Acknowledgements



















13







14


H ψ = E ψ




k satisfying

[HAplusmn


k (7)


Aminus

k =dk

dxk+ v1(x)

dkminus1






1 k = 1 Here Aminus



1 ] = minusωAminus


[minus1

2part2 v1(x)

]= minus1



vprime1 = ω

12 v





U(x) =1

2ω2(xminus x0)

2 x0 = minusα1

ω2


Aminus

2 =d2

dx2+ v1(x)

d






2 ] = minusωAminus

2 we arrive to

[minus1







8


2 ] = minusωAminus


vprime1 = ω

12 v


12 v

primeprime2 + v1 U


(8)



12v




2 ω2 = ω v2 and



3U prime +1

ω(ω x+ α1)U



1

ω(ω x+ α1)w



w =C

(ω x+ α1)3


w1 =ω2

4x+

1

4ω α1


U prime(x) =C

(ω x+ α1)3+ω2

4x+

1


C1

(ω x+ α1)2+ω2

8x2 +

ωα1

4x


U(x) =g2

(x+ α)2+ω2

8(x+ α)2



2 ω is

v2 =ω

2minus 2 g2

(x+ α)2+ω2

4(x+ α)2

Therefore Aminus

2 is given by

Aminus

2 =

(d

dx+ω

2(x+ α)

)2

minus 2 g2

(x+ α)2


9


Aminus

3 =d3

dx3+ v1(x)

d2

dx2+ v2(x)

d


2 + v2(x) part + v3(x)


3 is given by

[H Aminus



1


]part2 +


1



]part

+[minus1



]


3 ] =minusωAminus


vprime1 = ω

vprime2 + 12 v


vprime3 + 12 v

primeprime2 + 2v1U


12 v

primeprime3 + v2U


(9)




U(x) sim 1



U(x) sim 1





radic2 τ

π

radicE =

2radic

2

ω

radicE


10


T (E) = τ(1 +

α1

E+α2


)

we obtain that


πradic

2

int U

0

T (E)radicU minus E

dE =2

πradic

2

radicU

(τ +

β1

U+β2


)


Ik =

int 1

0

2 dζ

(1 minus ζ2)k

and then


radic2

π

radicE

(τ +

β1

E+β2


) (10)


radic2 τπ)

radicE


k

(2

ω

)2n (E

2

)n

= E if U is regular

and

k 2n

(2

ω

)2n (E

2

)n

= E if U has a pole








11


U1(x) =lsum

j=0

mj(mj + 1)

(xminus xj)2(11)








U(x) sim

g2

x2 xrarr 0

18 ω

2 x2 +g21

x2 xrarr infin

where

g2 = m0(m0 + 1) g21 =

2r+1sum

j=0

mj(mj + 1)


x+(E) simradicE

ω+

g1radicE




ω+g1 minus gradic

E


U(x) =1

8ω2 x2 +

g2

x2


12



U(x) = αω2 x2 + U1(x) α =1

2or

1

8




Acknowledgements



















13







14


2 ] = minusωAminus


vprime1 = ω

12 v


12 v

primeprime2 + v1 U


(8)



12v




2 ω2 = ω v2 and



3U prime +1

ω(ω x+ α1)U



1

ω(ω x+ α1)w



w =C

(ω x+ α1)3


w1 =ω2

4x+

1

4ω α1


U prime(x) =C

(ω x+ α1)3+ω2

4x+

1


C1

(ω x+ α1)2+ω2

8x2 +

ωα1

4x


U(x) =g2

(x+ α)2+ω2

8(x+ α)2



2 ω is

v2 =ω

2minus 2 g2

(x+ α)2+ω2

4(x+ α)2

Therefore Aminus

2 is given by

Aminus

2 =

(d

dx+ω

2(x+ α)

)2

minus 2 g2

(x+ α)2


9


Aminus

3 =d3

dx3+ v1(x)

d2

dx2+ v2(x)

d


2 + v2(x) part + v3(x)


3 is given by

[H Aminus



1


]part2 +


1



]part

+[minus1



]


3 ] =minusωAminus


vprime1 = ω

vprime2 + 12 v


vprime3 + 12 v

primeprime2 + 2v1U


12 v

primeprime3 + v2U


(9)




U(x) sim 1



U(x) sim 1





radic2 τ

π

radicE =

2radic

2

ω

radicE


10


T (E) = τ(1 +

α1

E+α2


)

we obtain that


πradic

2

int U

0

T (E)radicU minus E

dE =2

πradic

2

radicU

(τ +

β1

U+β2


)


Ik =

int 1

0

2 dζ

(1 minus ζ2)k

and then


radic2

π

radicE

(τ +

β1

E+β2


) (10)


radic2 τπ)

radicE


k

(2

ω

)2n (E

2

)n

= E if U is regular

and

k 2n

(2

ω

)2n (E

2

)n

= E if U has a pole








11


U1(x) =lsum

j=0

mj(mj + 1)

(xminus xj)2(11)








U(x) sim

g2

x2 xrarr 0

18 ω

2 x2 +g21

x2 xrarr infin

where

g2 = m0(m0 + 1) g21 =

2r+1sum

j=0

mj(mj + 1)


x+(E) simradicE

ω+

g1radicE




ω+g1 minus gradic

E


U(x) =1

8ω2 x2 +

g2

x2


12



U(x) = αω2 x2 + U1(x) α =1

2or

1

8




Acknowledgements



















13







14


Aminus

3 =d3

dx3+ v1(x)

d2

dx2+ v2(x)

d


2 + v2(x) part + v3(x)


3 is given by

[H Aminus



1


]part2 +


1



]part

+[minus1



]


3 ] =minusωAminus


vprime1 = ω

vprime2 + 12 v


vprime3 + 12 v

primeprime2 + 2v1U


12 v

primeprime3 + v2U


(9)




U(x) sim 1



U(x) sim 1





radic2 τ

π

radicE =

2radic

2

ω

radicE


10


T (E) = τ(1 +

α1

E+α2


)

we obtain that


πradic

2

int U

0

T (E)radicU minus E

dE =2

πradic

2

radicU

(τ +

β1

U+β2


)


Ik =

int 1

0

2 dζ

(1 minus ζ2)k

and then


radic2

π

radicE

(τ +

β1

E+β2


) (10)


radic2 τπ)

radicE


k

(2

ω

)2n (E

2

)n

= E if U is regular

and

k 2n

(2

ω

)2n (E

2

)n

= E if U has a pole








11


U1(x) =lsum

j=0

mj(mj + 1)

(xminus xj)2(11)








U(x) sim

g2

x2 xrarr 0

18 ω

2 x2 +g21

x2 xrarr infin

where

g2 = m0(m0 + 1) g21 =

2r+1sum

j=0

mj(mj + 1)


x+(E) simradicE

ω+

g1radicE




ω+g1 minus gradic

E


U(x) =1

8ω2 x2 +

g2

x2


12



U(x) = αω2 x2 + U1(x) α =1

2or

1

8




Acknowledgements



















13







14


T (E) = τ(1 +

α1

E+α2


)

we obtain that


πradic

2

int U

0

T (E)radicU minus E

dE =2

πradic

2

radicU

(τ +

β1

U+β2


)


Ik =

int 1

0

2 dζ

(1 minus ζ2)k

and then


radic2

π

radicE

(τ +

β1

E+β2


) (10)


radic2 τπ)

radicE


k

(2

ω

)2n (E

2

)n

= E if U is regular

and

k 2n

(2

ω

)2n (E

2

)n

= E if U has a pole








11


U1(x) =lsum

j=0

mj(mj + 1)

(xminus xj)2(11)








U(x) sim

g2

x2 xrarr 0

18 ω

2 x2 +g21

x2 xrarr infin

where

g2 = m0(m0 + 1) g21 =

2r+1sum

j=0

mj(mj + 1)


x+(E) simradicE

ω+

g1radicE




ω+g1 minus gradic

E


U(x) =1

8ω2 x2 +

g2

x2


12



U(x) = αω2 x2 + U1(x) α =1

2or

1

8




Acknowledgements



















13







14


U1(x) =lsum

j=0

mj(mj + 1)

(xminus xj)2(11)








U(x) sim

g2

x2 xrarr 0

18 ω

2 x2 +g21

x2 xrarr infin

where

g2 = m0(m0 + 1) g21 =

2r+1sum

j=0

mj(mj + 1)


x+(E) simradicE

ω+

g1radicE




ω+g1 minus gradic

E


U(x) =1

8ω2 x2 +

g2

x2


12



U(x) = αω2 x2 + U1(x) α =1

2or

1

8




Acknowledgements



















13







14



U(x) = αω2 x2 + U1(x) α =1

2or

1

8




Acknowledgements



















13







14







14

Isochronous classical systems and quantum systems with equally

Documents

Transcript of Isochronous classical systems and quantum systems with equally