Method for calculating analytical solutions of the Schrodinger equation:Anharmonic

oscillators and generalized Morse oscillators

L. Skala, 1 J. C zek, 2 J. Dvorak, 1 and V. Spirko 3

1Charles University, Faculty of Mathematics and Physics, Ke Karlovu 3, 12116 Prague 2, Czech Republic2University of Waterloo, Department of Applied Mathematics, Waterloo, Canada N2L 3G1

3 Czech Academy of Sciences, J. Heyrovsky Institute of Physical Chemistry, Dolejskova 3, 18000 Prague 8, Czech Republic

Received 12 October 1995

A method for calculating the analytical solutions of the one-dimensional Schrodinger equation is suggested.A general discussion of the possible forms of the potentials and wave functions that are necessary to get theanalytical solution is presented. In general, the analytical solutions appear in multiplets corresponding to thequantum number n of the harmonic oscillator. As an application, known solutions for the anharmonic oscilla-tors are critically recalculated and a few additional results are found. Analytical solutions are also found for thegeneralized Morse oscillators.

P ACS number s : 03.65.Ge,31.15. p

I. INTRODUCTION

The solution of the one-dimensional Schrodinger equationrepresents an important problem with numerous applicationsin many fields of physics. This equation can always besolved numerically. Despite this, analytical solutions yield amore detailed and exact description of the physical realityand are therefore of considerable interest.

The number of potentials V( x ) for which the analytic so-lution of the one-dimensional Schrodinger equation,

H! #" x%$'& E(*),+ x.- ,/ 0 11

with2 the Hamiltonian

H4365d7 28

dx7 2 9 V : x%; < 2=?>

is known is rather limited. Except for trivial cases, examplesof@ analytically solvable problems include the harmonic oscil-lator , some anharmonic oscillators A 19B ,/ the one-dimensional hydrogen atom, the Morse oscillator C 10D ,/ andsome other simple cases E see, e.g, F 1114GIH .

Analyzing these analytic solutions, we conclude that thebound-stateJ wave functions have the same structure. Thewave2 functions have the form of the exponential or otherrelated functions multiplied by a polynomial in a variablethatK is a function of x . In other words, the wave functionsL forM all these problems can be written as a linear combina-tionK of functions N mOQP f

R mO gS ,/ where fR ( x ) and gS ( x ) are suitablychosenT functions and mU isV an integer.

It is obvious that there is a chance of finding an analyticalsolution if the Hamiltonian transforms the set of the basisfunctionsM W mO into

V itself. Namely, if the result of H!X mO isV

afiniteY linear combination of Z mO ,/ we can hope that the result-ingV finite order matrix problem is analytically solvable. As-suming these properties of the wave function and Hamil-tonian,K we discuss in this paper conditions for the functions

fR and gS and the potential V ,/ which must be fulfilled to gettheK analytical solution of the Schrodinger equation.

Using[ the approach indicated above we first use the basis\

mO toK transform the Schrodinger equation to the matrix form

with2 a non-Hermitian matrix ] Sec.^ II_ . Possible forms of fR ,/gS ,/ and V thatK can yield analytical solutions are discussed inSec.^ III. In the next three sections, known analytical resultsfor the anharmonic oscillators are critically recalculated. Sec-tionK IV is devoted to the problem of the quartic anharmonicoscillator@ . In Sec. V, a detailed analysis of the sextic oscilla-torK is performed and a few new analytical solutions arefound. Discussion of the higher-order anharmonic oscillatorsisV presented in Sec. VI. Another interesting problem is thegeneralization` of the Morse oscillator. The quadratic, quartic,sextic, and higher-order generalized Morse oscillators are in-vestigateda in Secs. VIIIX.

II. TRANSFORMATION OF THE SCHRO DINGERb

EQUATION INTO THE MATRIX FORMWc e assume the wave function d inV the form

egfih

mOcj mO,k mO ,/ l 3

m?n

where2

o

mOQprq fRts mO gS . u 4v?w

The standard approach to the solution of the Schrodinger

equation consists in substituting the assumption x 3m?y into Eq.z

1{ . Introducing the matrix elements

HmnO}| ~ mO*H n dx7 5?

and

S mnO} mO* n dx7 6?

PHYSICAL REVIEW A APRIL 1996VOLUME 53, NUMBER 4

531050-2947/96/53 4 /2009 12 /$10.00 2009 1996 The American Physical Society

one@ gets the well-known eigenvalue problem

nH mnO cj n E

(*

nS mnO cj n . 7

?

In most cases, it appears impossible to calculate H mnO andS mnO and solve Eq. 7

?

analytically . However, if the matricesH and S are truncated, this method is suitable for calculatingapproximate solutions.

There is a chance of finding analytical solutions of theSchro^ dinger equation if the Hamiltonian H transformsK the setof@ the basis functions mO into

V itself. We assume thereforethatK the Hamiltonian H fulfills the relation

H! mOQn

h mnO} n ,/ 8?

where2 the coefficients h mnO are numbers. Let us introduce anoverlap@ between the basis function mO and the exact wavefunctionM

M mOQ mO* dx7

. 9?

Substituting^ Eqs. 8? and 9? intoV the Schrodinger equation

1 we2 get another matrix formulation known from the mo-ment method 1519

nh mnO* M

n EM(

mO . 10

Wc e see that the vector of the overlaps M mO isV

the right eigen-vectora of the matrix h . The advantage of Eq. 10 is that, incontrastT to Eq. 7? ,/ there is no matrix S in this equation. Thematrix h isV usually sparse, which further simplifies the prob-lem. On the other hand, the matrix h is non-Hermitian. Theequations

HmnO}ip

S mpO h

np 11

and

M mOQn

S mnO cj n 12

followingM from the assumptions 8? and 3m? give` the relationof@ the quantities appearing in Eqs. 7? and 10 . Applicationof@ these equations is usually complicated by the infinite orderof@ the matrix S .

There is also another possibility close to the approachesgiven` above. If Eqs. 3m? and 8 are used directly in theSchro^ dinger equation 1 theK following result is obtained:

mO ,ncj mO h

mnO n E(

mOcj mO, mO . 13

Assuming linear independence of the functions mO we2 get asimple matrix problem,

mOcj mO h

mnO} Ec(

n . 14

The vector of the coefficients cj mO isV

the left eigenvector of thematrix h .

The coefficients cj mO are obtained from Eq. 14 directly

without2 the necessity of using the transformation 12 as theyare in the moment method.

Another disadvantage of the moment method is that evenfor the analytically solvable problems the overlaps M mO areusually different from zero and sometimes even diverge formU4 9,15,16 . The problem 10 of@ the infinite order isdifficult to solve analytically and even when it is solved, thetransformationK 12 of@ the usually infinite order must be ap-plied. On the other hand, the left eigenvectors of Eq. 14with2 a finite number of nonzero coefficients cj mO canT often befoundM directly and the analytical wave function can be foundin the form of a finite linear combination of mO . For the sakeof@ simplicity, we discuss in this paper one-dimensional prob-lems only. We note, however, that the moment method hassuccessively been applied to one-dimensional as well as mul-tidimensionalK problems see, e.g. 16,17 ,/ .

The condition that only a finite number of the coefficientscj mO is different from zero is known, for example, from thesolution of the harmonic oscillator where cj mO are the coeffi-cientsT of the Hermite polynomials. In the standard solutionof@ the harmonic oscillator a simple recurrence relation for thecoefficientsT cj mO of@ the Hermite polynomials is obtained. Inour@ approach, such a simple recurrence relation is replacedbyJ a general matrix equation 14 and can therefore lead toanalytical solutions that have not been known until now.

A problem similar to Eq. 14 isV solved also in the Hilldeterminant method see, e.g. 5,20,21,13,14 ,/ . As we showbelowJ , our approach is more general than this method. WeconsiderT general functions fR and gS and give a general dis-cussionT of Eq. 14 . We are also interested in a direct ana-lytical solution of Eq. 14 forM a finite linear combination in( 3) instead of discussing the infinite-order problem.

The wave functions given in this paper are not normal-ized.V

III. CONDITIONS FOR f AND g

In the previous section, the validity of Eq. 8 was2 as-sumed. Now we derive conditions for fR ,/ gS ,/ and V followingfromM this assumption.

Applying the Hamiltonian 2= toK the basis function 4v we2get`

H mO mU mU 1 fR 28

fR 2 mU 2=

fR fR

gS !

gS#"fR%$fR &

gS('

gS*) V + mO .

,

15-

Here,. fR / denotes d f7 /0 dx7 .In order to get H 1 mO as a linear combination of 2 n the

K

expression in brackets must be a linear combination of fR n .As different terms in Eq. 3 154 depend on mU in a different wayany of the terms fR 5 28 /0 fR 28 ,2(/ fR6 /0 fR )( gS 7 /0 gS )98 fR(: /0 fR ,/ and; gS(< /0 gS>= V must be a linear combination of fR n . It followsfrom the first and second terms that fR ? must be a linearcombinationT of fR mO ,/

fR@BADCmO

fR mO>E fR(F mO

,/ G 16H

2010 53L. SKA LA,I J. C IZ EK,J J. DVOR AK K, AND V. SPIRKO

where2 fR mO are numbers. Analogously, the second and thirdtermsK lead to

gS LNMPO gSQmO

gS mO>R fR(S mO

,/ T 17U

where2 the minus sign on the right-hand side is chosen forfurther convenience. Finally, the last term gives

V VDWmO

VmO>X fR(Y mO

. Z 18[

Wc e see that the potentials V consideredT in this paper musthave the form given by Eq. \ 18] . At the same time, the func-tionK fR ( x ) appearing in this equation must satisfy Eq. ^ 16_ .These two conditions restrict possible forms of the potentialsfor which our method is applicable.

Wc e note that there are a number of simple functions ful-filling Eq. ` 16a such as x ,/ exp(x ),coth( x ), and cot(x ). How-ever , there are also more complex functions such as the or-thogonalK polynomials that can be used as the function fR .

The coefficients fR mO ,/ gS mO ,/ and VmO are arbitrary until now. IftheK coefficients fR mO and gS mO are known, the functions f

R

andgS canT be obtained by inverting

xb fR(c9d 1emO fR

mO>f fR(g mO d f

7 h

19i

and calculating

gSkj xml9n exp o pmO

gS mO>q fR(r mO dx7

s exp t u mO gS mO>v f

R(w mO

x

mO fR

mOky fR(z mO d f

7

. { 20|

To get Eq. } 14~ ,/ the function gS cannotT be arbitrary and isgiven` by Eq. 20 ,/ where gS mO are parameters. The way todetermine the coefficients gS mO is

V described below.In the moment method and the Hill determinant method

theK function 20 is often replaced by a single Gaussian ex-ponential. Obviously, such an approximate approach cannotbeJ used if analytic solutions are to be found.

As a result of the integration, the function gS ( x ) can havea rather complex form. It shows that the assumption abouttheK polynomial form of the argument of the exponentialmade in the Hill determinant method is too restrictive seetheK sections devoted to the generalized Morse potentials .

There is also another conclusion following from Eq. 20= .Let us assume that we search for the bound-state wave func-tionK in the form of a finite sum 3m9 . Then, investigating theintegralV in Eq. 20= ,/ it is easy to determine gS mO for

M

whichgS ( x ) is finite. For example, let us assume that fR ( x )9 x ,/ gS mO 0 for mU M and gS mO 0

for mU 0 and mU M . It followsfrom Eq. 20 thatK M must be odd, otherwise the functiongS ( x ) diverges for x # or@ x . In fact, this is the reasonfor which the analytical solutions for the quartic anharmonicoscillator@ with M 2 cannot have this form of gS ( x ) see Sec.IV .

Substituting^ Eqs. 16 18 into Eq. 15 we2 get

H mOi

mU mU 1 j

fR j fR

i j 28 mU

j2= fR j gS i j 1

j f j fR

i j 289 jgS j gS

i j j f i j 1gS j9 V i mO i .

Therefore, the matrix h mnO appearing in Eq. 14 equals

h mO ,mO i P mU mU 1 jfR j fR

i j 2 mU

j2= fR j gS i j 1

j f j fR

i j 289 jgS j gS

i j j f i j 1gS j9 V i .

21

Our method of finding analytical solutions of the Schro-dinger equation can be described as follows. First we deter-mine the function fR ( x ) from the form of the potential V( x )

see Eq. 18 . Then we try to find the coefficients gS mO andVmO for

M

which the left eigenvectors of the matrix h exist witha finite number of nonzero components. This leads to a so-lution of a system of equations for gS mO and VmO ,/ which isoften@ possible to solve. If the analytical solutions of Eq. 14are found the wave functions are determined from Eqs. 3m

and 20 .Wc e note that the boundary conditions for the wave func-

tionK have not been taken into consideration until now. Thismeans that this method can be used for the discrete as well ascontinuousT part of the energy spectrum. It also means that toget` wave functions for the discrete energies, only the solu-tionsK satisfying the appropriate boundary conditions must betaken.K

In general, solution of Eq. 14 leads to two linearly in-dependent solutions as it should be for the differential equa-tionK of the second order. For the bound states, only one of thesolutions or their suitable linear combination must be taken.

Now we search for the left eigenvector of the matrix h

with2 a finite number of nonzero components. In this paper,we2 assume cj mO 0

for mU 0 and mU n ,/ where n 0 i s anintegerV . It means that we search for the wave function in theform

mO 0

n

cj mO fR mO gS . 22=

If necessary, the summation in this equation can be extendedtoK mU 0.

The corresponding eigenvalue problem 14 becomesJ

mO 0

n

cj mO h

mO ,mO i E mO ,mO i 0, 23

where2 i!#" . . . , $ 2, % 1,0,1,2, . . . . This formula representsmore equations than the number of unknown coefficientscj mO and has in general only the trivial solution cj mO& 0,

mU' 0 ,..., n . To get nonzero cj mO ,/ the number of equationsmust be reduced or they must be made linearly dependent.

53 201( 1METHOD FOR CALCULATIONS ANALYTICAL SOLUTIONS OF . . .

Our aim is to reduce the problem ) 23* toK a standard eigen-valuea problem with a square matrix.

General+ discussion of this problem is rather complex. InthisK paper, we assume the potential in the form

V ,.-i / 1

2M

V i 0 fR21 i

. 3 24=4

If necessary, negative powers i!#5 0 can be also included. Thepotential coefficients V1 ,/ ..., V2M appear in h

mO ,mO6 i ,/

i!87 1, . . . ,2M . Assuming further gS mO9 fR

mO: 0 for mU; 0 and

mU< M ,/ the matrix = h i j > has nonzero elements in the rowsi?8@ 0 ,..., nA and columns j2B 0 ,..., nAC 2M . To reduce thenumberD of columns, we start with the last one jFE nAG 2= M anddetermine gS M in such a way that the only nonzero elementh n ,nIH 2M in

V

this column becomes zero. This leads togS M

28KJ V28 ML so that gS MLNMPOPQ V28 ML .Let us assume for a moment that the potential is quadratic

( MR 1). In this case we calculate gS 0S fromM

the condition thattheK remaining nonzero element h n ,nIT 1 in the (nAVU 1)th rowequals zero. As a result, the eigenvalue problem W 23=X with2 asquare matrix is obtained. We see that the problem of thequadraticY oscillators can be solved easily.

ForZ quartic and higher-order potentials (M[ 2,3,= . . . ),however, we get more nonzero elements in the columnsj2\ nA] 2= M_^ 1, . . . ,nA` 1 than in the case of the quadraticoscillators.@ In this case, gS M a 1 ,/ ..., gS 0S must be determinedfromM the condition that the columns j2b nAVc 2= Md 1,nAe 2= Mf 2= ,..., nAg M are linearly dependent on the columnsj2h 0 ,..., nA of@ the matrix hi E . To reduce the number oflinearly independent columns of h ,/ we must continue to in-troduceK some constraints on the potential coefficients thatwere2 arbitrary until now. Considering the columnsjFj nAk M_l 1, . . . ,nAm 1 we can calculate VMLon 1 ,/ ..., V1 as afunction of VML ,/ ..., V28 ML . Solving then the remaining prob-lem p 23q with2 the square matrix r h i j s ,/ i

?

,/ jFt 0 ,..., nA we2 canfind the solution in the form u 22v . We see that the analyticsolution in the form w 22x exists for nonquadratic potentialsonly@ if additional constraints on the potential coefficients areintroduced.

Wc e note that, in general, the values of gS 0S ,/ ..., gS ML andV1 ,/ ..., VM y 1 depend

on the energy E and nA . For nAz 0, wecanT find only one analytical solution with the correspondingvaluesa of gS 0S ,/ ..., gS M and V1 ,/ ..., VM { 1 . Then we can getanalytical solutions for nA| 1, etc. Thus, the solutions areobtained@ in certain multiplets corresponding to different val-ues of nA . Our nA correspondsT to the quantum number nA of@ theharmonic oscillator for which the matrix h canT be easilydiagonalized and the energies En~} (2

nA 1)gS 1 gS 0S2 ( nA 1/2) are obtained.

In general, the best chance to find the analytical solutionisV for nA 0 when the matrix h reduces to one row. The co-efficients gS mO are then given by equations h

0S j# 0,

j2 2= M ,/ ..., M and the potential constraints follow fromh 0S j 0,

jF M 1, . . . ,1. The ener gy equals E h 00S and thecorrespondingT wave function is ( x ) gS ( x ) . With increas-ing nA and M ,/ the order of the problem and complexity of thepotential constraints increase and the chance to find explicitanalytic expressions for the energies and wave functions islower. In general case, a numerical solution of the problem

23 is necessary.Let us discuss now the case of the anharmonic and Morse

oscillators.@ For the anharmonic oscillators we put fR ( x ) x ,/fR mO mO ,0 and for the generalized Morse oscillators we usefR ( x ) 1 exp( x ), fR 0SI 1,f

R

1 P 1 and fR

mO 0

otherwise.The potential is assumed in the form 24= . As follows fromour@ discussion given above, analytical solutions for the an-harmonic oscillators exist only if M isV odd, i.e., if2M 4k2 2, where k is an integer. On the other hand, ana-lytical solutions for the generalized Morse oscillators existfor any M . The way to solve the problem 23 is the same forbothJ types of oscillators. First, we choose nA fromM the rangenA 0,1, . . . . Then we solve the equation h n ,n~ 2M 0

lead-ing to gS ML

28N V2M . After that we continue with the solution oftheK equations h n ,nI i 0,

i?# 2M 1, . . . ,M ,/ which yieldgS M 1 ,/ ..., gS 0S as

a function of VM ,/ ..., V2M . Consequently,all the coefficients gS mO are determined and all columns of thematrix h ,/ jF nA 2M ,/ ..., nA M are equal to zero. Then wecontinueT with the columns j2 nAV M 1, . . . ,nA 1 and de-termineK the corresponding constraints on the potential coef-ficientsY VMLo 1 , . . . ,/ V1 . The total number of the nonzero co-efficients gS mO (

M 1) plus the number of the potentialconstraintsT (M_ 1) equals 2M . If the potential is even, thenumber of the constraints reduces to one-half.

A less general discussion was performed in 8 for theanharmonic oscillators with the even potential.

The discussion given above shows that all the analyticallysolvable problems with the wave function in the form of afinite linear combination 3m have the same algebraic struc-tureK given by the matrix 21= . If the function fR isV changed thegeneral` discussion regarding h ,/ gS ,/ gS mO ,/ and VmO remains thesame. Assuming that the potential coefficients VmO ,/mU M ,/ ..., 2M remain unchanged for new fR we2 get newvaluesa of gS mO and potential constraints on VmO ,/mU 1, . . . ,M_ 1. However, because of the integration in Eq.

20= ,/ the function gS and the wave function canT changeconsiderablyT .

IV. QUARTIC ANHARMONIC OSCILLATORThe potential has the form

V x V1x V28 x 2 V3 x 3~

V4 x 4,/ V4~ 0

correspondingT to M 2.= Assuming gS mO 0 for mU 0,1,2 and

fR ( x ) x theK matrix h equals

h mO ,mO i mU mU 1 i , 2 2mgU 0S i , 1 2mgU 1 gS 0S28~ gS 1 i ,0

2= mgU 28~ 2=

gS 1gS

0SI 2=

gS 28~ V1 i ,1 2=

gS 28 gS

0SI gS

12 V28 i ,2 # 2

=

gS 1gS

28~ V3 i ,3 gS 22 V4 i ,4 .

2012 53L. SKA LA,I J. C IZ EK,J J. DVOR AK K, AND V. SPIRKO

FirstZ we discuss the ground state corresponding to nA 0. Themost simple wave function with no nodes is given by the lefteigenvector cj mO mO ,0 so that (

x ) gS ( x ) . To find it issufficient to find gS mO and the potential constraint on V1 forwhich2 h 0,S i 0,

i? 4v ,..., 1. Two possible solutions of theseequations are as follows. The coefficients gS mO are given by

gS 2

V4,/ gS 1 V3 / 2gS 2 ,/ gS 0S V2 gS 1

2 ! /" 2gS 2 #

and the potential constraint giving V1 as a function ofV2 ,/ ..., V4 is

V

V1 $ 2=

gS 1gS

0S% 2=

gS 28 .

The energy E equals

E('& h 00S)( gS 1 * gS 0S28

.

It can easily be verified that both functions

+-,

x/.10 gS32 x546 exp879 gS 0S x;: gS 1x 2/2= @ 25=BA

forM gS 28CDE V4F satisfy the Schrodinger

equation G 1H . How-ever , they diverge for x;IKJ or@ x;LNM=O ,/ as concluded in theprevious section.

For the higher multiplets nAQP 0 the situation is analogous.Wc e see therefore that the wave functions of the quartic an-harmonic oscillator cannot have the form R 22S with2 gS ( x )given` by Eq. T 25U .

VV . SEXTIC ANHARMONIC OSCILLATOR

The potential is assumed in the form

V W x/XY V1x-Z\[[[] V6^ x 6^

,/ V6^_ 0.

Assuming further gS mO` 0,

mUba 0 ,..., 3 the matrix h becomesJ

h mO ,mOc i dfe mUhg mUji 1 kl i , m 2 n 2mgU 0Spo i , q 1 rs 2mgU 1 t gS 0S2 u gS 1 vw i ,0

xy 2= mgU 28z 2=

gS 1gS

0S|{ 2=

gS 28} V1 ~ i ,1 2=

mgU 3 2=

gS 28 gS

0S| gS

128 3m gS 3| V28 i ,2

2gS 3 gS 0S| 2gS 1gS 2 V3 i ,3 2gS 1gS 3| gS 282 V4 i ,4 2gS 2gS 3 V5 i ,5 gS 3

2 V6^ i ,6 .

A. nQ 0

The values of gS mO and the potential constraints are foundbyJ solving successively h 0S j 0,

j 6 ,..., 1.The coefficients gS mO equal

gS 3 V6^ ,/ gS 2 V5 / 2gS 3 ,/ gS 1 V4 gS 28

28 / 2gS 3 ,/

gS 0S V3| 2gS 1gS 28 / 2gS 3 ,/ 26

where2 the sign before V6^ follows from the boundaryconditionsT at x;N= . These equations for gS mO are also validforM all the higher multiplets nAQ 1,2, . . . .

The coefficients V3 ,/ ..., V6^ canT be arbitrary. Two re-maining coefficients are given by the potential constraints

V1 2=

gS 1gS

0S| 2=

gS 28 ,/ V28| gS 128| 2= gS 28 gS 0S| 3

m

gS 3 .

The corresponding energy and wave function equal

E(' h 00S) gS 1 gS 0S28

and

-

x/ exp8 gS 0S x; gS 1x 28

/2 gS 2x 3

/3 gS 3 x 4F

/4? .

This function has no nodes and is therefore the ground-statewave2 function. We see that the analytic solution exists for theasymmetric potential with general potential coefficientsV3 ,/ ..., V6^ . In 8

p

,/ the solutions were found for the evenpotential only.

In a special case of the even potential,

V x5 V2x 28|

V4x 4F|

V6^ x 6^

,/

much more simple formulas are obtained,

V28 V4F28 / 4V6^1 3

m

V6^ ,/

E(' V4F / 2=Q V6^ ,/

and

;

x5 exp V4x 28

/ 4vQ V6^ V6^ x 4

/4 .

This result has one more parameter than the example giveninV 1 . These equations give the ground state of the sexticdouble-well potential. If V4 0,

the energy E lies below themaximum of the potential at x; 0 and the wave function hastwoK maxima at x; V4 /(2

V6 ).

B. n 1

In this case, we solve successively the equations

m 0S

1

c m h

m j E m j 0 27

for j 7 ,..., 0. First we solve these equations forj 7 ,..., 4. This leads to Eqs. 26 . Then, Eq. 27 forj 3 gives

V2!#" g$ 12!&% 2g$ 2! g$ 0S#' 5

(

g$ 3) .

Assuming* for simplicity c 1 + 1 we get from Eq. , 27.- for

j/ 2

c 0S&021 h3

12 / h3 02S5476 V1 8 2g$ 1g$ 0S&9 4g$ 2 : /

= .

53 2013METHOD FOR CALCULATIONS ANALYTICAL SOLUTIONS OF . . .

Then we solve two equations following from Eq. ? 27@ forjA 0 and jB 1 and get the cubic equation for V1V1

3)#C7D 10g$ 2 E 6F

g$ 1g$

0S>G V12 H7I 32 g$ 2!

2 J 4g$ 1g$ 3)#K 12g$ 12g$ 0S

2

L 40g$ 1g$ 0S g$ 2!>M V1 N 32

g$ 2!3)#O 8P g$ 3)

2! g$ 0S#Q 8P

g$ 13) g$ 0S

3)&R 64F g$ 1g$ 0S g$ 2!2!

S 8P g$ 12g$ 0S g$ 3)#T 16g$ 1g$ 2g$ 3)#U 40

V

g$ 12g$ 0S

2g$ 2 W 0.

Thus, depending on the values of V3) ,X ..., V6 ,X we can get uptoY three real values of V1 for

which the analytical solution oftheY SchrodingerZ equation exists. The corresponding energyobtained[ from Eq. \ 27.] for j^ 0,1 equals

E_a`7b V12 c7d 6F g$ 2!&e 4

V

g$ 1g$

0S>f V1 g 4V

g$ 12g$ 0S

2 h 12g$ 1g$ 0S g$ 2!#i 8P

g$ 22

j 6F g$ 1g$ 3)#k 2g$ 0S2! g$ 3)ml /

o

andp the wave function isqsr

xtvuw7x c 0S#y c

1xtvz exp{}|m~ g$ 0S xts g$ 1xt 2/2

g$ 2! xt 3) /3 g$ 3) xt 4/4

.

This function has one node and represents the first excited-state wave function.

In a special case c 0S& 0

a more simple result with threepotential constraints instead of two is obtained. The potentialconstraints are

V1 2 4V

g$ 2! ,X V2!&2 5(

g$ 3)# g$

12

,X V3)& 2

g$ 1g$

2! .

The last constraint leads to g$ 0S& 0.

The energy and wavefunction with one node corresponding to this potential equal

E_ 3 g$ 1

andp

s

xtv xt exp{ g$ 1xt 2!

/2 g$ 2xt 3)

/3 g$ 3) xt 4

/4 . 28

In 8P ,X a special analytic solution corresponding to Eq.

28. for the even potential was given. In this paper, we havefound solutions for a more general asymmetric potential.

C. n 2

General+ discussion leads to rather complicated expres-sions that will not be given here. We discuss only the specialcase c 0S# 0,

c 1 0,

c 2 0. Analyzing the equations

m 0S

2

c m h3

m j E_

m j 0, j 0 ,..., 8 29.

we get conditions g$ 0S# g$ 2!& 0.

It follows from these equationsthatY the potential V( xt ) must be even,

V V2xt 2!#

V4xt 4&

V6 xt 6

.

The same form of the potential also will be assumed for thehigher -order multiplets. Because of the symmetry of the po-tentialY the number of potential constraints reduces to one,

V2 g$ 12!# 7 g$ 3) .

There are two energies,

E 3 g$ 1 2 g$ 12!# 2g$ 3)

andp wave functions

xtv7 1 7 g$ 1 E_

xt 2/2 exp{}m g$ 1xt 2/2

g$ 3) xt 4/4

solving the SchrodingerZ equation in this case. The signdenotesZ the ground state theY wave function has nonodes . The sign denotes the second excited state theY

wave function has two nodes . We also see that E_ E .

D. n 3

Now we search for a special solution of Eq. 14 corre-sponding to c m 0

for m 1,3 theY solution with the oddparity andp the even potential. We get the following result:

V2!& g$ 12!# 9 g$ 3) ,X

E_ 5( g$ 1 2

g$ 12 6F g$ 3)

andp

xt xt 1 3 g$ 1 E_

xt 2/6 exp{ g$ 1xt 2/2

g$ 3) xt 4/4 .

The sign denotes the first excited state theY correspondingwave function has one node . The sign denotes the thirdexcited{ state theY wave function has three nodes .

E. Higher-order multiplets for n even

The energies and wave functions corresponding to theeven{ potential are given by the constraint

V2! g$ 12! 2 nA 3 "! g$ 3)

andp eigenvalues and eigenvectors of the matrix

g$ 1 # 2ngA 3) 0 0 0 0

$ 2 5 g$ 1 % & 2

nA ' 4V( g$ 3) 0 0 0

0 ) 12 9g$ 1 *+ 2nA , 8P-

g$ 3) 0 0

././. ././. ././. ././. ././. ../.

0 0/0/0 0 12 nA 3 2465 nA 7 3 "8:9 2 nA ; 3 "< g$ 1 = 4V

g$ 3)

0 >/>> 0 0 ? nAA@ nA B 1 C E D2 nA F 1 G g$ 1

.

The left eigenvectors of this matrix with the components c 0S ,X c 2! ,X ..., c nH giveI nA /2KJ 1 even wave functions

2014 53L. SKA LA,L J. C IZ EK,M J. DVOR AN K, AND V. SPIRKOO

PRQ

xtTSVUXW

mZY 0[

nH /2\

c 2! m xt 2m exp{]^ g$ 1xt 2/2

_

g$ 3) xt 4/4` .

F. Higher-order multiplets for n odd

Again,* we assume the even potential. The energies and wave functions corresponding to the potential constraint

V2!a g$ 12 bc 2 nd e 3 "f g$ 3)

arep given by the eigenvalues and eigenvectors of the matrix

3 g$ 1 gh 2nd i 2 j g$ 3) 0 0 0 0

k 6 7F g$ 1 l n m 2nd o 6Fp

g$ 3) 0 0 0

0 q 20 11g$ 1 rns 2

nd t 10 u g$ 3) 0 0

vv/v v/v/v v/v/v v/v/v vv/v v/v/v

0 w/w/w 0 xy ndAz 2 {}| nd ~ 3 2nd 3 g$ 1 4g$ 3)

0 // 0 0 nd nd 1 2 nd 1 g$ 1

.

The left eigenvectors of this matrix with the components c 1 ,X c 3) ,X ..., c nH giveI (nd 1)/2 odd wave functions:

xtV

mZ 0[

nH 1 /2\

c 2! mZ 1xt 2mZ 1exp{ g$ 1xt 2/2

K

g$ 3) xt 4/4 .

VI. HIGHER-ORDER ANHARMONIC OSCILLATORS

As shown in Sec. III, analytically solvable anharmonic oscillators are only those with the highest-order term xt 4 k6 2!

,X wherek is an integer.

The solution of the problem of the higher-order oscillators is analogous to that for the sextic oscillator. As an example weconsider the decadic oscillator with

V V1xtRn V10xt 10,X V10 0.

Assuming* g$ mZ 0 for m 0 ,..., 5 the matrix h3 has the form

h3 m ,mZ i m m 1 } i , 2 2

mg 0[ i , 1 2

mg 1 g$ 0[2 g$ 1 } i ,0

2mg 2! 2g$ 1g$ 0[ 2g$ 2! V1 } i ,1 2mg 3) 2g$ 2! g$ 0[ g$ 12! 3 g$ 3) V2! i ,2

2 mg 4 2

g$ 3) g$ 0[ 2

g$ 1g$ 2 4V

g$ 4 V3)} i ,3 2

mg 5 2

g$ 1g$ 3) 2

g$ 4g$ 0[ g$ 22 5( g$ 5 V4 } i ,4

n 2g$ 1g$ 4 2g$ 2g$ 3) 2g$ 5 g$ 0[ V5} i ,5 n 2g$ 1g$ 5 2g$ 2g$ 4 g$ 3)2! V6 i ,6

n 2 g$ 3) g$ 4 2

g$ 2! g$ 5 V7} i ,7 n 2

g$ 3) g$ 5 g$ 42 V8

i ,8 2

g$ 4 g$ 5 V9 i ,9 g$ 52 V10 i ,10 .

Solving equations h3 n ,n i 0 for i 10, . . . ,5 the following

values of g$ m arep obtained:

g$ 5 ! V10,X g$ 4 " V9 /#%$ 2g$ 5& ,X g$ 3)'( V8 ) g$ 4

2 * /#,+ 2g$ 5- ,X

g$ 2!./ V70 2

g$ 3) g$

41 /#,2 2 g$ 53 ,X g$ 1 45 V66 g$ 3)

2 7 2 g$ 2! g$ 48 /#,9 2 g$ 5:

andp

g$ 0[;< V5= 2g$ 2g$ 3)> 2g$ 1g$ 4 ? /#%@ 2g$ 5A .

Similarly to the sextic anharmonic oscillator we takeg$ 5B C V10. These values are the same for all the multiplets.Let us consider for example ndED 0. Then, the potential con-straints following from h3 0,[ i F 0,

i,G 4V ,..., 1 are

V1 H 2g$ 1g$ 0[I 2g$ 2 ,X V2 J 2g$ 2g$ 0[K 3

g$ 3)L g$

12

,X

V3)M 2g$ 3) g$ 0[N 2g$ 1g$ 2 O 4g$ 4 ,X

V4P 2g$ 1g$ 3)Q 2g$ 4 g$ 0[R 5(

g$ 5S g$

2!2!

.

Again, the analytic solution exists for the asymmetric poten-tial.Y The ground-state wave function of the singlet ndET 0 i sgivenI by

UWV

xtYX[Z exp{ \^]m`_ 1

6

g$ m`a 1xt m

/# mb .

The corresponding energy equals

53 2015METHOD FOR CALCULATIONS ANALYTICAL SOLUTIONS OF . . .

E_dc g$ 1 e g$ 0[2

.

Resultsf for the higher-order multiplets are analogous tothoseY for the sextic oscillator and will not be given here.

VII.g QUADRATIC MORSE OSCILLATORThe Morse oscillator h 10i with the potential

V j rk[l[m Dnpo 1 q exp{srtu rkwv rk 0[x /y

az|{~} 2!

30

is of considerable interest in molecular physics. In this paper,we use the variable xt ( rk rk 0[ )/

az andp discuss generalizedMorse potentials in the form

V xtYi 1

2! M

V i 1 exp{ xtY% i. 31

Such potentials are more general than the original Morsepotential and can describe, for example, potentials with reso-nances when the barrier higher than the value of the potentialatp xt exists.{ As we pointed out in Sec. III, in case of thegeneralizedI Morse oscillators we are not limited by the2 M^ 4V k 2 rule valid for the anharmonic oscillators andM can be an arbitrary positive integer.

W e take now f ( xt )[ 1 exp({ xt ) so that f 0[ 1,f

1 1,andp f m` 0 otherwise.

First we discuss briefly the quadratic Morse oscillatorwith the potential

V V1 1 exp{ xt, V2! 1 exp{ xt% 2,X V2! 0, 32

which is equivalent to the original Morse potential 30 .For the quadratic Morse oscillator (M^ 1) there are no

potential constraints so that all the multiplets ndE 0,1, . . .belong to the same potential.

Assuming g$ m` 0 for m 1 the matrix h3 becomes

h3 m ,m` i m m 1 i , 2 m 2m 1 2g$ 0[ i , 1

m m 2g$ 0[ 2g$ 1 g$ 1 g$ 0[2!

i ,0 g$ 1 2

m 2 g$ 0[ 1 [ V1 i ,1 V2! g$ 12

i ,2 .

T aking into account the expression for g$ ( xt ),

g$ xt[ exp{ g$ 0[ g$ 1 f

xt, dx exp{ g$ 0[ g$ 1 xt

g$ 1exp{ xt

theY value

g$ 1

V2!

must be taken. Similarly, to get h3 n ,n 1 0 for a given nd ,X the

value

g$ 0[ V1 /y 2g$ 1 1/2 nd

must be used. In contrast to the anharmonic oscillators, g$ 0[ isap function of nd . In order to get bound states the wave func-tionY must be finite for xt . It follows from

!#"

xt$%'&

m)( 0[

n

c m)* 1 + exp{,.- xt0/21 m

g$43 xt05

thatY to fulfill these boundary conditions the relation

g$ 0[6 g$ 1 7 0

must be valid. Taking into account the form of g$ 0[8 g$ 0[ (

nd ) w esee that there is a maximum value of nd:9 nd max; for which theboundary conditions are obeyed. We get

nd max;= V1 /y2@ 2 g$ 1 AB 1/2 C g$ 1 D ,X

where EGF denotesZ the integer part. Therefore, only a finitenumber of bound states for ndIH 0 ,..., nd max; exists.{ There areno bound states for V1 /(2

y

g$ 1)J 1/2 K g$ 1 L 0.

T o get the eigenvalues we assume h3 n ,nM 1 N h3

n ,nO 2!P 0 for

ap given nd andp make use of the summation rule

Q

ih3 m ,m)R i ST 2

ndIU 1 V g$ 1 W g$ 0[2

.

This equation shows that the energies

E_ nXY 2

ndIZ 1 [ g$ 1 \ g$ 0[2 ]^ 2 ndI_ 1 ` g$ 1 ab V1 /

yc 2 g$ 1 def ndIg 1/2hi 2j

33 lk

arep the eigenvalues of the matrix m h3 i j n ,X io

,X jprq 0 ,..., nd sincetheY columns of the matrix s h3 i j tvu E

_

,X io ,X jpxw 0 ,..., nd arep lin-early{ dependent. It is worth noting that, except for the ex-pression for g$ 0[ ,X Eq. y 33

{z is the same as that for the energy oftheY harmonic oscillator with the potential V | V1xt#} V2xt 2

!

.

T o get corresponding c m we solve Eq. ~ 23{

,X leading to thefollowing system of recurrence equations:

c n 1,

c n 1h3

n 1,n c

n h3

n ,n E_ 0,

c n 2! h3

n 2,! n 1 c

n 1 h3

n 1,n 1 E_

c n h3

n ,n 1 0,

34 {c i 1h

3

i 1,i c

i h3

i ,i E_

c i 1h3

i 1,i c

i 2! h3

i 2,! i 0,

io2 ndI 2, . . . ,1,

c 0[ h3

0,0[ E c 1h3

10 c

2h3

20 0.

It can be shown that the results of this section agree withknown results for the standard Morse oscillator with the po-tentialY 30 l .

VIII. QUARTIC MORSE OSCILLATORNow we discuss the quartic oscillator with the potential

31 { for M 2. For the quartic and higher-order Morse oscil-lators, we write the function g$ asp

g$: xt exp{ 'm) 0[

M

g$ m G

mI xt0

,X

where

2016 53L. SKA LA, J. C IZ EK, J. DVOR A K, AND V. SPIRKO

G mI xt 1 f:

xt m

dx .

These functions equal for the quartic oscillator

G 0[ xt0 xt ,X G

1 xt xt exp{ xt0 ,X

G 2! xt xt 2

exp . xt0 exp{. 2 xt0 /2.y

The matrix h3 for the quartic oscillator is given by theformula

h3 m ,m) i ? m m 1 i , 2 2m 2!

m 2mg 0[ i , 1 m 2!

2mg 0[ 2mg 1 g$ 0[2! g$ 1 i ,0

2 mg 1 2

mg 2! 2

g$ 1g$

0[ 2

g$ 2! g$

1 V1 i ,1 2

mg 2! 2

g$ 2! g$

0[ g$

12 2 g$ 2! V2! i ,2

2g$ 1g$ 2 V3) i ,3 !" g$ 2!2 # V4 $% i ,4 .

Solving successively h3 n ,n& i ' 0,

io)( 4V ,..., 2 for a givenI nd wegetI the coefficients g$ m :

g$ 2!*,+,- V4. ,X g$ 1 / V3) /y)0 2g$ 2!1 ,X

g$ 0[23 V2!4 g$ 12!5 /y76 2 g$ 2!8:9 nd

To get bound states, we take

g$ 2 @,A,B V4,X

where V4 C 0.

A further obvious condition for the existenceof[ the bound states is

g$ 0[D g$

1 E g$

2 F 0.

The expression for the maximum nd givingI the bound states isasp follows:

nd max GH)I V2 J g$ 12!K /y7L 2g$ 2 M:N 1 O g$ 1 P g$ 2 Q .

If the argument of the integer part is less than or equal tozero, there are no bound states.

Assuming that nd is given, the summation rule for thequarticR Morse oscillator equals

S

ih3 m ,mUT i V 2

XW

nd

C. ng

If the argument of the integer part is less than or equal tozero,P there are no bound states.

The summation rule for the sextic Morse oscillator,

Q

ih3 m ,mSR i TU g$ 0[

2!7V g$ 12!W 2g$ 1g$ 0[7X 2g$ 2g$ 0[7Y 2g$ 1g$ 2 Z 2g$ 3) g$ 0[

[ V1 \ V2 ] V3) ,X

leads to the energies

E_ n^_a` g$ 0[2 b g$ 1

2 c 2 g$ 1g$ 0[d 2

g$ 2! g$

0[e 2

g$ 1g$

2!7f 2

g$ 3) g$

0[7g V1 h V2!i V3) ,X

where g$ 0[7j g$ 0[ (

nd ) and constraints on V1 ,X V2! ,X V3) dependZ

alsoon[ nd .

Forr example, for ndlk 0 we get

V1 m 2g$ 0[ g$ 1 n 2g$ 2 o g$ 1 ,X V2 p g$ 12 q 2g$ 2g$ 0[7r 2g$ 2 s 3

g$ 3) ,X

V3)7t 2g$ 1g$ 2 u 2g$ 3) g$ 0[v 3

g$ 3) ,X

E w g$ 1 x g$ 0[2!

andp

yz

xt|{3} exp{ ~mS 0[

3)

g$ m G

ml xt .

The other calculations for the sextic and higher-orderMorse oscillators are analogous to that for the quadraticMorse oscillator. They will not be given here.

X. CONCLUSIONS

In this paper, a method for calculating the analytic solu-tionsY of the SchrodingerZ equation similar to the momentmethod and the Hill determinant method has been suggested.

First, the potential is assumed in the form V( xt )

m Vm f m

,X where fl f ( xt ) is a function that must satisfycertain conditions described below. In general, the summa-tionY can also run over the negative values of m . Then, thewave function is assumed to be a finite linear combination oftheY functions mS f

m g$ ,X where g$l g$ ( xt ) is a convenient func-tion.Y To get analytical solutions, it is assumed that the Hamil-tonianY transforms this basis set into itself. From the last as-sumption, we conclude that the derivative of f must be afinite linear combination of f m with the coefficients f m . Thesame condition must be valid for the logarithmic derivativeof[ g$ ,X i.e., g$ /y g$ . For a given function f ,X the function g$ caneasily{ be calculated from the equation g$ ( xt ) exp({ m g$ m f

m dx ), where g$ m arep constants. If the last ex-pression and the expression for f arep used in the Schro-dingerZ equation, a simple eigenvalue problem 14 with thematrix 21 is obtained. To get the analytic solution, the con-stants g$ m must

be determined in such a way that the analyticeigenvalues{ and left eigenvectors of this matrix exist. In gen-eral,{ some constraints on the potential coefficients also mustbe introduced. It appears that the solutions exist in multiplets

corresponding to different values of the quantum number ndof[ the harmonic oscillator. In general, different solutions cor-respond to different potentials.

Let us assume now that the potential has the formV m

2! M Vm f m

,X V2M 0.

It has been shown that the condi-tionsY for g$ m necessary for the existence of bound states fol-low from the form of the function g$ ( xt ) . For f ( xt )3 xt ,X ana-lytic solutions exist only for 2M 4V k 2, where k is aninteger .

This method is a generalization of the approaches knownfrom the moment method and the Hill determinant methodandp its main advantages are 1 known properties of f ( xt ) forwhich the analytical solution exist, 2 ap formula for g$ ( xt )with parameters g$ m that

Y

can be found from the solution oftheY eigenvalue problem 14 ,X 3 ap straightforward discussionof[ the conditions for the existence of the bound states, 4V apunique approach to all analytically solvable problems of thiskind leading to the matrix 21 in which only f m andp g$ mappearp . In this way, a common algebraic representation forallp these problems has been found.

As the first application of our method, known results fortheY anharmonic oscillators have been critically recalculatedandp some new results have been obtained. It has been shownthatY the analytic solution is possible only if 2M 4k 2,where k is an integer. For the sextic (k 1) and decadic( k 2 ) oscillators a few new solutions for the asymmetricpotential V have been given.

Another* interesting problem is the generalized Morse os-cillator , which is of interest in molecular physics. In contrasttoY the anharmonic oscillators, the analytic solutions exist foranyp 2M . We have discussed analytic solutions for the qua-dratic,Z quartic, sextic, and higher-order oscillators. New re-sults have been found for the quartic and higher-order gen-eralized{ Morse oscillators. For the quartic oscillator, analyticsolutions for the multiplets nd5 0,1 and ndl 2 have been dis-cussed. The transition from the quartic Morse oscillator totheY quartic anharmonic oscillator has also been made, con-firming our previous conclusions. For the sextic oscillator,generalI formulas for g$ m andp the multiplet nd5 0

have beeninvestigated.

Our method is applicable to any problem with the poten-tialY V andp function f satisfying assumptions given above.Generalization to more dimensions is also possible.

ACKNOWLEDGMENTS

This work was supported by a Natural Sciences and En-gineeringI Research Council Grant in Aid of Research J.C . ,Xwhich is hereby acknowledged. Further, we would like toexpress{ our gratitude to Professor R. LeRoy, Department ofChemistry , University of Waterloo, for his assistance withtheY use of the Silicon Graphics computer where the majorityof[ our results was obtained.

APPENDIX

It is interesting to notice that the case of the quadraticMorse potential, Eq. 30 ,X can be treated using the algebraic

53 2019METHOD FOR CALCULATIONS ANALYTICAL SOLUTIONS OF . . .

methods. This was recognized by many authors and it isdescribed,Z for example, in 22 . However, it is convenient forexperimental{ purposes see, e.g., 23 , pX . 8 toY consider thepotential in the form 32 . For the algebraic approaches, we

refer to the paper 25 ,X namely, to Eq. 45V . If we put ndl 1,B V2 ,X D V1 2V2 ,X b

0[ 1/2 V2 V1 E andp

V1 V2! E_ 1/4 into Eqs. 42V andp 52( of[ 24 ,X the for-

mula 33 is obtained.

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2020R 53kL. SKA LA,l J. C IZ EK,m J. DVOR An K, AND V. SPIRKOo