-
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
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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
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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
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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 . . .
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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.
1 J. Killingbeck, M.N. Jones, and M.J. Thompson, J. Phys. A
18,793 1985 .
2 J. Killingbeck, Phys. Lett. 67A, 13 1978 .
3 D. Bessis, E.R. Vrscay, and C.R. Handy, J. Phys. A 20, 419
1987 .
4 E.R.M Vrscay, Int. J. Quantum Chem. 32 , 613 1987 .
5 V . Singh, S.N. Biswas, and K. Datta, Phys. Rev. D 18 ,
1901
1978 .
6 G.P . Flessas, Phys. Lett. 72A, 289 1979 .
7 G.P . Flessas and K.P. Das, Phys. Lett. 78A , 19 1980 .
8 E.M Magyari, Phys. Lett. 81A , 116 1981 .
9 M. Vicher, thesis, Charles University, Prague, 1994
unpub-lished .
10 PO .M. Morse, Phys. Rev. 34, 57 ! 1929" .#
11$ S.% Flugge,& Rechenmethoden' der Quantentheorie
dargestelt inAufgaben( und Losungen ) Springer% , Berlin, 1947*
.
+
12, I.I. Goldman and V.D. Krivcenko, Exercises in Quantum
Me-chanics- . Gosudarstvennoe Izdatelstvo
Techniko-TeoreticeskojLiteraturyL , Moscow, 1957/10 in2 Russian3
.
4
135 M. Znojil, review in Rigor' ous Results in Quantum
Mechanics,
edited by J. Dittrich and P. Exner 6 W7 orld Scientific,
Singapore,19918 .
9
14: M. Znojil, Phys. Lett. A 169 , 415 ; 1992< , and
referencestherein.
=
15> J.L. Richardson and R. Blankenbecler, Phys. Rev. D 19,
496?
1979@ .A
16B R. Blankenbecler, T. DeGrand, and R.L. Sugar, Phys. Rev.
D21, 1055 C 1980D .
E
17F E.R. Vrscay and C. R. Handy, J. Phys. A 22, 823 G 1989H
.I
18J C.R. Handy, D. Bessis, and T.D. Morley, Phys. Rev. A 37
,4557 K 1988L .
M
19N J. Killingbeck and M.N. Jones, J. Phys. A 19, 705 O 1986P
.Q
20RTS J. Killingbeck, J. Phys. A 18 ,U L1025 V 1985W .XY
21RTZ A. Hautot, Phys. Rev. D 33,U 437 [ 1986\ .]
22^ B. G. Wybourne, Classical_ Groups for Physicists ` John
Wiley& Sons, New York, 1988a .X
b
23RTc C.H. Townes and A.L. Schawlow, Micrd owave
Spectroscopye
McGraw-Hill, New York, 1955f .Xg
24h J. C zek and J. Paldus, Int. J. Quant. Chem. 12,U 875 i
1977j .X
2020R 53kL. SKA LA,l J. C IZ EK,m J. DVOR An K, AND V.
SPIRKOo