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Analytical solution of the Schrodinger equation with linear confinement potential

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1991 J. Phys. A: Math. Gen. 24 5267

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I Phyr A Math Gen. 24 (1991) 5267-5272 Pnnted :n the U K

Received 26 Apni 1941, in find form 15 July 1991

Abrhnct. It IS shann *.at the Schrodinger equation m ath xnear confineme nt potenoal hasan anaiytical sdut'or, The ex*stenceof rhe analyti~al olutmn require

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5268 H Tezuka2. Deduct ion of recurrence ~ Q Z E B U R ~ P ZSpherical symmetric Schradinger equation for a particle *,th mass m and orbitalangular momentum 1 is given by

( 2 . 1 )where R ( r ) is the radial part of the wavefunction MP);

*(~)=R(r)Y?(fl) . ( 2 . 2 )The energy eigenvalue E of the particir is determined by solving ( 2 1).

For the spherical symmetricpotential U( I ) , let us consider the confinement potentialwhich is made of a long range linear potential and a short range Coulomb typepotential [4]:

( 2 . 3bU ( r ) =ar-- PConstants Q and b are independent ofdistance r, and their values are determined byexperiments so as to reproduce proper physical values of the hadrons. Then equation(2.1) is rewritten as

At large r, equation (2.4) becomes

where

If W L require the term proportionai to r after double difierentiatton, u ( r ) must havethe form exp(-r3") and for the constant term, u ( r ) must have the form exp(-r). Sothe radiai part R(P) is assumed to have the form

R ( r ) = r .r"cxp(-er"'~-pr) (2.7)where a, and c, are constants independent of F and their values zre determined tosatisfy :he Schradinger equation ( 2 . 4 ) . The constant c s of course positive, becausethe radial wavefunction R ( r ) must become zero at r+w. If the divergence of the radial?an R ( r ) is prohibited at the origin r=O, n must be larger than zero.

Mter substitution of ( 2 7) into 2.4), the left-hand side of (2.4) becomes

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3. SOhXtbJ~Of the X&XWE'ellCe f5KmEhIf the iad ial wavefunction R ( r ) is required t o be normalizable, c must be zero at afinite number of n. If all c. are zera for n> /< the recurrence form ula (2 .11) gives thefallowing conditions:

3 @ = 02mP 2 + , E = 0n

i (2k+ =O (3.4)3.5)( l c + 1 ) p - - y b = Omh

andk ( k + 1)- I+ 1)= o (3.6)

Equations (3.1) and 3.5) fix (Y and p, bu t these a and p cannot satisfy (3.2) andwhere it is assumed that ck# 0.(3.4). So we intro duc e additxonai counter terms in the potentiai U ( r ) , hat is,

3.7)brU ?)=r - -+er ' /*+fr-" 'where canstants e and f are determined io satisfy the S chrodinger equaiion (2.1). Then(3.2) and (3.4) are rewritten into

(3.8)m3up 7 O?I

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5270 Tezuhoand

3 2 k + $ ) a + 7 f = Omh (3.9)

Now the simultaneous equations we have to solve are (3.1), (3.3), 13.5). (3 6) , (3 .8)and (3.9). From (3.1) we obtain a;~(3.10)

where a is positive. M s o . from ( 3 5 ) we obtain pb (3.11)=- k + l ) f i 2

m

It sezms as If p was dependent on k But k IS determined by (3.6) ask = i (3 12)Thus

bp =--I + I)+?IS constant. The condition (3 12) requires that all c should be zero for n < k if we useit in (2.10).The energy eigenvalue E is given by (3.3 ;

a 2 b 2 (3 14)?( +1) hwhich depends only OF the angular momentum I. Additional constants e and aredetermined by (3.8) and (3.9) ss

E = -

2nd(3.15)

(3.16)These are riependeni on In order rbst die SchrCdinger equa ibn with rhe concnementpotential given by (2 .3) nas an aaaiytical solution, the additional terms are necessary,2s shown in (3.7) , and these terms depend on the orbital angular momentum .4. C5bocksianIt has been shown that the spherical symmetnc Schriidlnger equation with the potential

(4.1)r j = m - - + -r 1 1has an analyticai solution. The iadial wavefunction with orbital angn?ar momentum 1can be written as

R i r f = c , r ' e x p - d - p i ) (4.2)

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Analytical solution of the Scb'odanger equation 5271where

and mp =- I + l)h2 IThe constant c is determmed by the normalization conditionThe energy eigenvalue E is given by

(4.3)

( 4 . 4 )

2nd depends only on The state is determmed only by the orbltal angular momentum1 and is degenerate (21+1)-fold.In the case where the potentizl does not inciude the Coulomb term proportlonalto l r , we may set b = O . From (3 5 ) or 4.41, p = O Then, from (4.5). the energyeigenvalue is always zero This potential is the sunpiest, but is not very interesting. Sowe propose the asymptotically linear confinement potential 4.1). The most popularphenomenological p~tentialhat is used to calculate hadronic properties is the inear+Coulomb type potential [4]. If necessary, we could add other terms, hut the potential41) is hp r i n q l ~ s t spp otica y linear po entia that w e can so ve snz y+cs ;z.

Calculations of hadron properties will be carried out with this potential.For the purpose of comparison, Eichten's potential [4]bU r )=a r - - 4.6)r

and the new proposed one ( 4 are shown in figure 1 with the same values of theparameters a and 6. It will be easy to determine the vdlues of the parameters a and6 to reproduce the physical quantities of hadrons.

Figure 1. Companson of he potential (4 1) wth 1 - 0 --I and E chten's potential [41I(".O, ,- - j wrih ihe same d u e s of a = o is: G e v , b = a 3 2 and ma s s in = i . @ Gcj'

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5272 H TezukaIf the quark state IS given by an analytical form, many physical properties of uarks

can be easily calculated and understood. It must be worthwhile to investigate quarksin the potentiai given by (4.1)

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