Method for calculating analytical solutions of the Schro¨dinger … · 2005-09-30 · Method for calculating analytical solutions of the Schro¨dinger equation: Anharmonic oscillators
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Method for calculating analytical solutions of the Schrodinger equation:Anharmonic�
oscillators and generalized Morse oscillators
L. Skala,1 J.�
Cızek,� 2 J.�
Dvora�´k,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
A generaldiscussionof the possibleforms of the potentialsandwavefunctionsthat arenecessaryto get theanalytical� solution is presented.In general,the analyticalsolutionsappearin multipletscorrespondingto thequantum numbern� of the harmonicoscillator. As an application,known solutionsfor the anharmonicoscilla-tors�
arecritically recalculatedanda few additionalresultsarefound.Analytical solutionsarealsofoundfor thegeneralized� Morseoscillators.
P�ACS number� s� : 03.65.Ge,31.15.� p�
I. INTRODUCTION
The�
solutionof theone-dimensionalSchrodinger�
equationrepresents� an importantproblemwith numerousapplicationsin many fields of physics. This equation can always besolved� numerically. Despitethis, analyticalsolutionsyield amore� detailedand exact descriptionof the physical realityand� arethereforeof considerableinterest.
The�
numberof potentialsV(�x� )�
for which the analyticso-lution�
of the one-dimensionalSchrodinger�
equation,
H�! #"
x�%$'& E(*),+
x�.- ,/ 0 11with2 the Hamiltonian
H�4365 d
7 28
dx7 2 9 V : x�%; < 2=?>
is known is ratherlimited. Exceptfor trivial cases,examplesof@ analyticallysolvableproblemsincludetheharmonicoscil-lator�
, some anharmonic oscillators A 1–9B ,/ the one-dimensional�
hydrogenatom, the Morse oscillator C 10D ,/ andsome� othersimplecasesE see,� e.g, F 11–14GIH .
wave functions have the samestructure.Thewave2 functions have the form of the exponentialor otherrelatedfunctions multiplied by a polynomial in a variablethatK
is a function of x� . In other words, the wave functionsLforM
all theseproblemscanbe written asa linear combina-tionK
of functions N mOQP fR mO gS ,/ wheref
R(�x� )�
andgS (�x� )�
aresuitablychosenT functionsandmU is
Van integer.
It is obviousthat thereis a chanceof finding ananalyticalsolution� if the Hamiltonian transformsthe set of the basisfunctionsM W
mO intoV
itself. Namely, if the result of H�!X
mO isV
afiniteY
linearcombinationof Z mO ,/ we canhopethat theresult-ingV
finite ordermatrix problemis analyticallysolvable.As-suming� thesepropertiesof the wave function and Hamil-tonian,K
we discussin this paperconditionsfor the functions
fR
and� gS and� the potentialV,/ which must be fulfilled to gettheK
analyticalsolutionof the Schrodinger�
equation.Using[
the approachindicatedabovewe first usethe basis\mO toK
transformtheSchrodinger�
equationto thematrix formwith2 a non-Hermitianmatrix ] Sec.
^II _ . Possibleforms of f
R,/
gS ,/ andV thatK
canyield analyticalsolutionsarediscussedinSec.^
III. In the next threesections,known analyticalresultsfor theanharmonicoscillatorsarecritically recalculated.Sec-tionK
IV is devotedto the problemof the quarticanharmonicoscillator@ . In Sec.V, a detailedanalysisof the sexticoscilla-torK
is performed and a few new analytical solutions arefound.Discussionof thehigher-orderanharmonicoscillatorsisV
presentedin Sec.VI. Another interestingproblemis thegeneralization` of theMorseoscillator. Thequadratic,quartic,sextic,� andhigher-ordergeneralizedMorseoscillatorsarein-vestigateda in Secs.VII –IX.
II. TRANSFORMATION OF THE SCHRODINGERb
EQUATION INTO THE MATRIX FORM
Wc
e assumethe wavefunction d inV
the form
egfihmO cj mO,k mO ,/ l 3m?n
where2o
mOQprq fRts mO gS . u 4v?wThe standardapproachto the solutionof the Schrodinger
�equation� consistsin substitutingthe assumptionx 3m?y into Eq.z1{ . Introducingthe matrix elements
equation³1 we2 get anothermatrix formulationknown from the mo-
mentmethod µ 15–19¶·
n� h¢
mnO* M¸
n�º¹ EM(
mO . » 10¼Wc
e seethat thevectorof theoverlapsM¸
mO isV
the right eigen-vectora of the matrix h
¢. The advantageof Eq. ½ 10¾ is that, in
contrastT to Eq. ¿ 7�?À ,/ thereis no matrix S�
in this equation.Thematrix� h
¢isV
usuallysparse,which furthersimplifiestheprob-lem. On the otherhand,the matrix h
¢is non-Hermitian.The
equations�
HmnO}ÁiÂpà S�
mpO h¢
np� Ä 11Åand�
M mOQÆ¡Çn� S�
mnO cj n� È 12ÉfollowingM
from the assumptionsÊ 8¥?Ë and� Ì 3m?Í give` the relationof@ the quantitiesappearingin Eqs. Î 7�?Ï and� Ð 10Ñ . Applicationof@ theseequationsis usuallycomplicatedby theinfinite orderof@ the matrix S
�.
Thereis also anotherpossibility closeto the approachesgiven` above. If Eqs. Ò 3m?Ó and� Ô 8¥ÖÕ are� used directly in theSchro^
¨dinger�
equation× 1Ø theK
following result is obtained:
ÙmO ,n� cj mO h
¢mnOÛÚ n�ºÜ E
(ÞÝmO cj mO,ß mO . à 13á
Assuminglinear independenceof the functions â mO we2 get asimple� matrix problem,
ãmO cj mO h
¢mnO}ä Ec
(n� . å 14æ
The�
vectorof thecoefficientscj mO isV
theleft eigenvectorof thematrix h
¢.
The�
coefficientscj mO are� obtainedfrom Eq. ç 14è directly�
without2 thenecessityof usingthetransformationé 12ê as� theyare� in the momentmethod.
Anotherë
disadvantageof the momentmethodis that evenfor the analytically solvableproblemsthe overlapsM mO are�usuallyì different from zeroandsometimesevendiverge formU4íïîñð 9,15,16
ò. The problem ó 10ô of@ the infinite order is
difficult�
to solveanalyticallyandevenwhenit is solved,thetransformationK õ
12ö of@ the usually infinite ordermustbe ap-plied.� On the other hand, the left eigenvectorsof Eq. ÷ 14øwith2 a finite numberof nonzerocoefficientscj mO canT often befoundM
directly andtheanalyticalwavefunctioncanbefoundin theform of a finite linearcombinationof ù mO . For thesakeof@ simplicity, we discussin this paperone-dimensionalprob-lems�
only. We note,however, that the momentmethodhassuccessively� beenappliedto one-dimensionalaswell asmul-tidimensionalK
problemsú see,� e.g. û 16,17ü ,/ ý .The�
conditionthatonly a finite numberof thecoefficientscj mO is different from zero is known, for example,from thesolution� of the harmonicoscillatorwherecj mO are� the coeffi-cientsT of the Hermite polynomials.In the standardsolutionof@ theharmonicoscillatora simplerecurrencerelationfor thecoefficientsT cj mO of@ the Hermite polynomialsis obtained.Inour@ approach,sucha simple recurrencerelation is replacedbyJ
a generalmatrix equation þ 14ÿ and� can thereforelead toanalytical� solutionsthat havenot beenknown until now.
Aë
problemsimilar to Eq. � 14� isV
solvedalso in the Hilldeterminant�
method � see,� e.g. � 5,20,21,13,14� �
,/ � . As we showbelowJ
, our approachis more generalthan this method.WeconsiderT generalfunctions f
Rand� gS and� give a generaldis-
cussionT of Eq. � 14� . We are also interestedin a direct ana-lytical�
solutionof Eq. � 14 forM
a finite linear combinationin(�3) insteadof discussingthe infinite-orderproblem.
The wave functionsgiven in this paperare not normal-ized.V
III. CONDITIONS FOR f
AND g�In�
the previoussection,the validity of Eq. � 8¥� was2 as-sumed.� Now we deriveconditionsfor f
R,/ gS ,/ andV following
fromM
this assumption.Applyingë
theHamiltonian � 2=�� toK
thebasisfunction � 4v�� we2get`
H���
mO�� � mU�� mU�� 1 � fR�� 28fR 2 � mU 2
= fR �fR gS !
gS#" fR%$fR & gS('
gS*) V + mO .,15-
Here,.
fR /
denotes�
d f7
/0dx7
.In order to get H 1 mO as� a linear combinationof 2 n� the
Kexpression� in bracketsmust be a linear combinationof f
R n� .As differenttermsin Eq. 3 154 depend
�on mU in a differentway
any� of the terms fR 5 28 /0 fR 2
8,2(/ fR�6
/0fR
)(�
gS 7 /0 gS )�98
fR(:
/0fR
,/ and; gS(< /0 gS>= V must� be a linear combinationof fR n� . It follows
from the first and secondterms that fR ?
must be a linearcombinationT of f
R mO ,/
fR�@BADC
mO fR
mO>E fR(F mO ,/ G 16H
2010 53L. SKALA,I
J. CIZEK,J
J. DVORAK ´ K, AND V. SPIRKO
�
where2 fR
mO are� numbers.Analogously, the secondand thirdtermsK
leadto
gS LNMPO gS�QmO gS mO>R fR(S mO ,/ T 17U
where2 the minus sign on the right-handside is chosenforfurther convenience.Finally, the last term gives
V VDWmO VmO>X fR(Y mO . Z 18[
Wc
e seethat thepotentialsV consideredT in this papermusthavethe form givenby Eq. \ 18] . At thesametime, the func-tionK
fR
(�x� )�
appearingin this equationmust satisfy Eq. ^ 16_ .Thesetwo conditionsrestrictpossibleformsof thepotentialsfor which our methodis applicable.
Wc
e note that therearea numberof simple functionsful-filling Eq. ` 16a such� as x� ,/ exp(x� ),coth(
�x� ),�
and cot(x� ).�
How-ever� , therearealsomorecomplexfunctionssuchas the or-thogonalK
polynomialsthat canbe usedasthe function fR
.The�
coefficientsfR
mO ,/ gS mO ,/ andVmO are� arbitraryuntil now. IftheK
coefficientsfR
mO and� gS mO are� known, the functions fR
and�gS canT be obtainedby inverting
x��b fR(c9d 1emO fR
mO>f fR(g mO d f7 h
19i
and� calculating
gSkj x�ml9n exp� o pmO gS mO>q fR(r mO dx
7
s exp� t u mO gS mO>v fR(w mOxmO fR
mOky fR(z mO d f7
. { 20|
To get Eq. } 14~ ,/ the function gS cannotT be arbitraryandisgiven` by Eq. � 20� ,/ where gS mO are� parameters.The way todetermine�
the coefficientsgS mO isV
describedbelow.In�
the momentmethodand the Hill determinantmethodtheK
function � 20� is often replacedby a singleGaussianex-ponential.� Obviously, suchan approximateapproachcannotbeJ
usedif analyticsolutionsareto be found.As a resultof the integration,the function gS (
�x� )�
canhavea� rathercomplex form. It showsthat the assumptionabouttheK
polynomial form of the argument of the exponentialmadein the Hill determinantmethodis too restrictive � see�theK
sectionsdevotedto the generalizedMorsepotentials� .There�
is alsoanotherconclusionfollowing from Eq. � 20=��
in the form of a finite sum � 3m9� . Then, investigatingtheintegralV
in Eq. � 20=��
,/ it is easyto determinegS mO forM
whichgS (�x� )�
is finite. For example,let us assumethat fR
(�x� )�9�
x� ,/ gS mO�0�
for mU�� M¸
and� gS mO�� 0�
for mU�� 0�
and mU�� M¸
. It followsfrom Eq. � 20� that
KM must be odd, otherwisethe function
gS (�x� )�
divergesfor x� �#� or@ x� ����� . In fact, this is thereasonfor which theanalyticalsolutionsfor thequarticanharmonicoscillator@ with M � 2 cannothavethis form of gS (
methodof finding analyticalsolutionsof the Schro-dinger�
equationcanbe describedasfollows. First we deter-mine the function f
R(�x� )�
from the form of the potentialV(�x� )�ò
see� Eq. ó 18ôöõ . Then we try to find the coefficientsgS mO and�VmO for
Mwhich the left eigenvectorsof thematrix h
¢exist� with
a� finite numberof nonzerocomponents.This leadsto a so-lution�
of a systemof equationsfor gS mO and� VmO ,/ which isoften@ possibleto solve.If theanalyticalsolutionsof Eq. ÷ 14øare� found the wave functionsare determinedfrom Eqs. ù 3m�úand� û 20ü .
Wc
e note that the boundaryconditionsfor the wavefunc-tionK
havenot beentakeninto considerationuntil now. Thismeansthatthis methodcanbeusedfor thediscreteaswell ascontinuousT partof theenergy spectrum.It alsomeansthat toget` wave functionsfor the discreteenergies,only the solu-tionsK
In general,solution of Eq. ý 14þ leadsto two linearly in-dependent�
solutionsasit shouldbe for the differentialequa-tionK
of thesecondorder. For theboundstates,only oneof thesolutions� or their suitablelinear combinationmustbe taken.
Nowÿ
we searchfor the left eigenvectorof the matrix h¢
with2 a finite numberof nonzerocomponents.In this paper,we2 assumecj mO�� 0
�for mU�� 0
�and mU�� n� ,/ where n��� 0 i
�s an
integerV
. It meansthat we searchfor the wavefunction in theform
���mO� 0�
n�cj mO fR mO gS . � 22
=��
If necessary, thesummationin this equationcanbeextendedtoK
mU�� 0.�
The correspondingeigenvalueproblem � 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 equationsthan the number of unknown coefficientscj mO and� has in general only the trivial solution cj mO�& 0,
�mU�' 0
�,..., n� . To get nonzerocj mO ,/ the numberof equations
must� be reducedor they must be madelinearly dependent.
53 201(
1METHOD FOR CALCULATIONS ANALYTICAL SOLUTIONSOF . . .
Ourñ
aim is to reducethe problem ) 23* toK
a standardeigen-valuea problemwith a squarematrix.
General+
discussionof this problemis rathercomplex.InthisK
paper, we assumethe potentialin the form
V ,.-i / 1
2M
V i 0 fR21 i. 3 24=�4
If�
necessary, negativepowersi!#5
0�
canbealsoincluded.Thepotential� coefficients V1 ,/ ..., V2M appear� in h
¢mO ,mO�6 i ,/
i!87
1, . . . ,2M¸
. Assumingfurther gS mO�9 fR
mO�: 0�
for mU�; 0�
andmU�< M
¸,/ the matrix = h¢ i j > hasnonzeroelementsin the rows
i?8@
0�,..., nA and� columns j
¿2B0�,..., nA�C 2M . To reducethe
numberD of columns,we startwith thelastone j¿FE
nA�G 2=
M¸
and�determine�
gS M in sucha way that the only nonzeroelementh¢
n� ,n�IH 2M inV
this column becomes zero. This leads togS M
28KJ
V28
ML so� that gS M
LNMPOPQ V28
ML .
Let usassumefor a momentthat thepotentialis quadratic(�M¸�R
1). In this casewe calculategS 0S fromM
theconditionthattheK
remainingnonzeroelementh¢
n� ,n�IT 1 in the (nAVU 1)th rowequals� zero.As a result,the eigenvalueproblem W 23
=�Xwith2 a
square� matrix is obtained.We seethat the problem of thequadraticY oscillatorscanbe solvedeasily.
ForZ
quartic and higher-order potentials (M¸�[
2,3,=
. . . ),however, we get more nonzero elementsin the columnsj¿2\
nA�] 2=
M¸_^
1, . . . ,nA�` 1 than in the caseof the quadraticoscillators.@ In this case,gS M a 1 ,/ ..., gS 0
S must be determinedfromM
the condition that the columns j¿2b
nAVc 2=
M¸�d
1,nAe2=
M¸�f
2=,..., nA�g M
¸are� linearly dependenton the columns
j¿2h
0�,..., nA of@ the matrix h
¢�iE. To reducethe numberof
linearly�
independentcolumnsof h¢
,/ we mustcontinueto in-troduceK
some constraintson the potential coefficientsthatwere2 arbitrary until now. Considering the columnsj¿Fj
nA�k M¸_l
1, . . . ,nA�m 1 we cancalculateVMLon
1 ,/ ..., V1 as� afunctionof VM
L ,/ ..., V28
ML . Solvingthentheremainingprob-
lem p 23q with2 the squarematrix r h¢ i j s ,/ i?,/ j¿Ft 0
�,..., nA we2 can
find the solution in the form u 22v . We seethat the analyticsolution� in the form w 22x exists� for nonquadraticpotentialsonly@ if additionalconstraintson thepotentialcoefficientsareintroduced.
Wc
e note that, in general,the valuesof gS 0S ,/ ..., gS M
L and�V1 ,/ ..., VM y 1 depend
�on theenergy E and� nA . For nA�z 0,
�we
canT find only oneanalyticalsolutionwith the correspondingvaluesa of gS 0
S ,/ ..., gS M and� V1 ,/ ..., VM { 1 . Thenwe cangetanalytical� solutions for nA�| 1, etc. Thus, the solutions areobtained@ in certainmultipletscorrespondingto differentval-uesì of nA . Our nA correspondsT to thequantumnumbernA of@ theharmonic oscillator for which the matrix h
¢canT be easily
diagonalized�
and the energies En�~} (2�
nA�� 1)gS 1 � gS 0S2����� (
�nA�� 1/2) areobtained.
In general,the bestchanceto find the analyticalsolutionisV
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,�
j¿2�
2=
M¸
,/ ..., M¸
and� the potential constraintsfollow fromh¢
0S
j±�� 0,�
j¿F�
M � 1, . . . ,1. The ener� gy equalsE � h¢
00S and� the
correspondingT wave function is � (�x� )���
gS (�x� )�. With increas-
ing nA and� M ,/ theorderof theproblemandcomplexityof thepotential� constraintsincreaseandthe chanceto find explicitanalytic� expressionsfor the energies and wave functions islower. In generalcase,a numericalsolution of the problem�23� is necessary.
Let�
usdiscussnow thecaseof theanharmonicandMorseoscillators.@ For the anharmonicoscillatorswe put f
R(�x� )���
x� ,/fR
mO���� mO ,0 and� for the generalizedMorse oscillatorswe usefR
(�x� )���
1 � exp(� � x� ),�
fR
0SI� 1,f
R1 �P� 1 and f
RmO�� 0
�otherwise.
The�
potentialis assumedin the form � 24=�
. As follows fromour@ discussiongiven above,analyticalsolutionsfor the an-harmonic¡
oscillators exist only if M¸
isV
odd, i.e., if2M ¢ 4k
£2¤2, wherek
£is an integer. On the otherhand,ana-
lytical�
solutionsfor the generalizedMorse oscillatorsexistfor anyM . Theway to solvetheproblem ¥ 23¦ is thesameforbothJ
ing to gS ML28Nª V2M . After thatwe continuewith thesolutionof
theK
equationsh¢
n� ,n�I« i ¬ 0,�
i?#
2M ® 1, . . . ,M ,/ which yieldgS M ¯ 1 ,/ ..., gS 0
S as� a functionof VM ,/ ..., V2M . Consequently,all� thecoefficientsgS mO are� determinedandall columnsof thematrix h
¢,/ j¿F°
nA�± 2M ,/ ..., nA�² M are� equalto zero.ThenwecontinueT with the columns j
¿2³nAV´ M¸�µ
1, . . . ,nA�¶ 1 and de-termineK
the correspondingconstraintson the potentialcoef-ficientsY
VMLo·
1 , . . . ,/ V1 . The total numberof the nonzeroco-efficients� gS mO (
�M ¸ 1) plus the number of the potential
constraintsT (M¸_¹
1) equals2M¸
. If the potentialis even,thenumberof the constraintsreducesto one-half.
A less generaldiscussionwas performedin º 8¥�» for theanharmonic� oscillatorswith the evenpotential.
The�
discussiongivenaboveshowsthatall theanalyticallysolvable� problemswith the wave function in the form of afinite linear combination ¼ 3m�½ havethe samealgebraicstruc-tureK
givenby thematrix ¾ 21=�¿
. If thefunction fR
isV
changedthegeneral` discussionregardingh
¢,/ gS ,/ gS mO ,/ andVmO remains� the
same.� Assuming that the potential coefficients VmO ,/mU�À M
¸,/ ..., 2M
¸remain� unchangedfor new f
Rwe2 get new
valuesa of gS mO and� potential constraints on VmO ,/mU�Á 1, . . . ,M
¸_Â1. However, becauseof theintegrationin Eq.Ã
20=�Ä
,/ the function gS and� the wave function Å canT changeconsiderablyT .
IV. QUARTIC ANHARMONIC OSCILLATOR
The�
potentialhasthe form
V Æ x�ÈÇ�É V1x�ËÊ V28 x� 2 Ì V3
� x� 3�~Í
V4Î x� 4,/ V4
Î~Ï 0�
correspondingT to M¸�Ð
2.=
AssuminggS mO�Ñ 0�
for mU�Ò 0,1,2�
andfR
(�x� )��Ó
x� theK
matrix h¢
equals�
h¢
mO ,mO�Ô i Õ×Ö mUÙØ mU�Ú 1 ÛÝÜ i, Þ 2 ß 2mgU 0S�à
i, á 1 âäã 2mgU 1 å gS 0S28~æ gS 1 çÝè i,0é�ê
2=
mgU 28~ë 2=
gS 1gS 0SIì 2=
gS 28~í V1 îÝï i,1ðäñ�ò 2
=gS 28 gS 0SIó gS 1
2 ô V28öõÝ÷
i,2øäù#ú 2=
gS 1gS 28~û V3
�öüÝýi,3þäÿ�� gS 2
2 � V4����
i,4 .
2012 53L. SKALA,I
J. CIZEK,J
J. DVORAK ´ K, AND V. SPIRKO
�
FirstZ
we discussthegroundstatecorrespondingto nA�� 0.�
Themost� simplewavefunctionwith no nodesis givenby thelefteigenvector� cj mO�� mO ,0 so� that (
�x� )���
gS (�x� )�. To find � it is
sufficient� to find gS mO and� the potentialconstrainton V1 forwhich2 h
¢0,S
i � 0,�
i?��
4v,..., 1. Two possible� solutionsof these
equations� areasfollows. The coefficientsgS 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
VV1 $ 2
=gS 1gS 0
S�% 2=
gS 28 .
The energy E equals�E('&
h¢
00S)( gS 1 * gS 0
S28 .
It caneasilybe verified that both functions
+-,x�/.10 gS32 x�54�6 exp�87�9 gS 0
S x�;: gS 1x� 2/2�=<
gS 28 x� 3�/3�?> @
25=BA
forM
gS 28�C�D�E V4
F satisfy� the Schrodinger�
equation G 1H . How-ever� , they diverge for x�;IKJ or@ x�;LNM=O ,/ as concludedin theprevious� section.
For the highermultipletsnAQP 0�
the situationis analogous.Wc
e seethereforethat the wave functionsof the quartic an-harmonic oscillator cannot have the form R 22S with2 gS (
�x� )�
given` by Eq. T 25U .VV
. SEXTIC ANHARMONIC OSCILLATOR
The potentialis assumedin the form
V W x�/X�Y V1x�-Z\[�[�[�] V6^ x� 6^,/ V6
^�_ 0.�
Assumingë
further gS mO` 0,�
mUba 0�,..., 3 the matrix� h
¢becomesJ
h¢
mO ,mOc i dfe mUhg mUji 1 k�l i, m 2 n 2mgU 0Spo
i, q 1 r�s 2mgU 1 t gS 0S2 u gS 1 v�w i,0x�y
2=
mgU 28�z 2=
gS 1gS 0S|{ 2=
gS 28�} V1 ~�� i,1��� 2= mgU 3
��� 2=
gS 28 gS 0S|� gS 1
28��
3m
gS 3�|� V2
8����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. n£Q¤ 0¥
The valuesof gS mO and� the potentialconstraintsare foundbyJ
The ground-statewave function of the singlet ndET 0 i�
sgivenI by
UWVxtYX[Z exp{ \^]
m�`_ 1
6þ
g$ m�`a 1xt m� /#mb .
The�
correspondingenergy equals
53 2015METHOD FOR CALCULATIONS ANALYTICAL SOLUTIONSOF . . .
E_dc
g$ 1 e g$ 0[2 .
Resultsf
for the higher-order multiplets are analogoustothoseY
for the sexticoscillatorandwill not be given here.
VII.g
QUADRATIC MORSE OSCILLATOR
The Morseoscillator h 10i with¶ the potential
V j rk[l[m Dnpo
1 q exp{sr�t�u rkwv rk 0[ x /y az|{~} 2! �
30 ��
is of considerableinterestin molecularphysics.In this paper,we¶ use the variable xt�� (
�rk�� rk 0
[ )/� az andp discussgeneralizedMorsepotentialsin the form
V � xtY�����i � 1
2!
M�
V i � 1 � exp{���� xtY�%� i. � 31 ��
Such�
potentialsare more generalthan the original Morsepotential� andcandescribe,for example,potentialswith reso-nanceswhenthebarrierhigherthanthevalueof thepotentialatp xt���� exists.{ As we pointedout in Sec.III, in caseof thegeneralizedI Morse oscillators we are not limited by the2�
M�^�
4V
k� �
2�
rule valid for the anharmonicoscillators andM can� be an arbitrarypositive integer.
is the eigenvalue.The correspondingwavefunction equals
µ·¶xt¹¸:º¼» c� 0
[�½ c� 1 ¾ 1 ¿ exp{ÁÀ� xtzÃ7ÄÆÅ exp{ Ç~Èm�UÉ 0[
2!
g$ m� G¾
m�<Ê xtzË .
Becauseof Eq. Ì 39 ?Í
,X we can get two wave functions.Onefunction�
hasno nodesandthe otherhasonenode.There is also a special solution correspondingto c� m�ÎÐÏm� 1 . This assumptionleadsto theadditionalpotentialcon-
straint� 2g$ 0[�Ñ 1 Ò 0 o
�r
V2!�Ó V3
)2! /y)Ô 4V4.�Õ×Ö 3
XØV4. .
The energy and wave function with one nodeequal in thiscase�
E_ÚÙ
3
g$ 1 Û g$ 0[2
andpÜxÝ
xtzÞ:ß�à 1 á exp{Áâ�ã xt¹ä7å exp{ æèçm�Ué 0[
2
g$ m� Gê
m�Xë xt¹ì .
53 2017METHOD FOR CALCULATIONS ANALYTICAL SOLUTIONSOF . . .
C. ng<í 2
Similarly�
to case ndXî 1, we solve Eq. ï 29�?ð
for�
jñóò
6F,..., 0. The equations{ for j
ñóô6,5,4F
aresatisfiedbecauseof[ Eq. õ 35
?ö. Assuming c� 2 ÷ 1 we first solve Eq. ø 27ù for
jñóú
3
andthenfor jñ�û
2.�
This leadsto expressionsfor c� 1 andpc� 0[ . Substitutingtheseexpressionsto Eq. ü 23ý for j
ñ�þ1, we
getI the cubic equationfor V1 . The resultingexpressionsforV1 ,X c� 1 ,X andc� 0
[ arep complexandwill not be given here.Theener{ gy is given by Eq. ÿ 36
��andp the wavefunction equals���
xt���� c� 0[�� c� 1 1 � exp{���� xt������ c� 2
!�� 1 � exp{���� xt���� 2�
exp{ !#"m�%$ 0[
2
g$ m� Gê
m�'& xt�( .
In
this case,up to threeanalyticalsolutionscanbeobtained.Thesesolutionshaveone,two, andthreenodes.
Similarly�
to case nd*) 1 a special solution with c� 0[�+ 0,�
c� 1 , 0,�
and c� 2!.- 0�
correspondingto two additionalpotentialconstraints�
g$ 1 / g$ 2! , 2X g$ 0
[�0 3 *1
0�
exists.{ Detaileddiscussionwill not be given here.
D. Higher-order multiplets
Thesolutionof Eq. 2 233 for highernd can� beobtainedin asimilar� way as describedabove.However, the results arecomplex� andin generalthenumericalsolutionof Eq. 4 23
�65is¢
necessaryÝ .
E. Transition to anharmonic oscillator
The�
transition from the quartic Morse potential to thequarticR anharmonicpotentialcanbe madeif the function
same� conditionmustbe valid for the logarithmicderivativeof[ g$ ,X i.e., g$�� /y g$ . For a given function f
�,X the function g$ can�
easily{ be calculated from the equation g$ (�xt )�
� exp({ �����m� g$ m� f� m� dx�
),�
whereg$ m� arep constants.If the last ex-pression� and the expressionfor f
���arep used in the Schro-
dingerZ
equation,a simple eigenvalueproblem � 14� with¶ thematrix � 21� is obtained.To get theanalyticsolution,thecon-stants� g$ m� must bedeterminedin sucha way that theanalyticeigenvalues{ andleft eigenvectorsof this matrix exist.In gen-eral,{ someconstraintson the potentialcoefficientsalsomustbeÂ
introduced.It appearsthat thesolutionsexist in multiplets
corresponding� to different valuesof the quantumnumberndof[ theharmonicoscillator. In general,differentsolutionscor-respondto differentpotentials.
Let us assumenow that the potential has the formV ��� m�2! M
Vm� f� m� ,X V2M ¡ 0.
�It has beenshown that the condi-
tionsY
for g$ m� necessaryfor the existenceof boundstatesfol-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 methodis a generalizationof the approachesknown
forwhich¶ the analytical solution exist, « 2�¬ ap formula for g$ (
�xt )�
with¶ parametersg$ m� thatY
can be found from the solution oftheY
eigenvalueproblem ® 14 ,X±° 3 ² ap straightforwarddiscussionof[ the conditionsfor the existenceof the boundstates,³ 4V´ apunique� approachto all analyticallysolvableproblemsof thiskind leading to the matrix µ 21¶ in which only f
�m� andp g$ m�
appearp . In this way, a commonalgebraicrepresentationforallp theseproblemshasbeenfound.
As the first applicationof our method,known resultsfortheY
the analytic solution is possibleonly if 2M · 4k¦¨¸
2,where¶ k
¦is an integer. For the sextic (k
¦±¹1) and decadic
(�k¦±º
2�
) oscillatorsa few new solutionsfor the asymmetricpotential� V havebeengiven.
Another*
interestingproblemis the generalizedMorseos-cillator� , which is of interestin molecularphysics.In contrasttoY
the anharmonicoscillators,the analyticsolutionsexist foranyp 2M
£. We havediscussedanalytic solutionsfor the qua-
dratic,Z
quartic, sextic,and higher-order oscillators.New re-sults� havebeenfound for the quartic and higher-order gen-eralized{ Morseoscillators.For thequarticoscillator, analyticsolutions� for the multipletsnd5» 0,1
�andndl¼ 2 havebeendis-
cussed.� The transition from the quartic Morse oscillator totheY