Anharmonic Solution of Schrödinger Time-Independent Equation

PRAMANA c Indian Academy of Sciences Vol. 77, No. 2 — journal of August 2011 physics pp. 243–261

Anharmonic solution of Schrödinger time-independent equation

MOHAMMED ASHRAFUL ISLAM1,2,∗ and JAMAL NAZRUL ISLAM1 1Research Centre for Mathematical and Physical Sciences, University of Chittagong, Chittagong, Bangladesh 2Department of Mathematics, University of Chittagong, Chittagong, Bangladesh ∗Corresponding author. E-mail: [email protected]

MS received 3 January 2010; revised 2 January 2011; accepted 12 January 2011

Abstract. We present here a mathematical explanation of how the Schrödinger equation for a class of harmonic oscillators possesses exact solutions. Some of the extended potentials used here are not present in the literature.

Keywords. Anharmonic solutions; Yukawa potential.

PACS Nos 03.65.Ge; 33.10.Cs

1. Introduction

The problem of finding exact (classical and quantum mechanical) solutions for anharmonic oscillators is of methodological and practical interest in some model field theories (φ4, φ6 etc.), in the context of various fields of applied physics; for instance, in the mean field theory of first-order structural phase transitions [1], in many other cases in atomic and molecular physics or in charmonium systems which provide us with mathematical formu- lations of physical problems. Moreover, in other cases non-relativistic exact results can be used for extrapolation into the relativistic regime, for example in the context of the theory of S-matrix of strong interaction [2]. Keen interest was shown in the use of multi-term potential in various articles [3–8]. Most of them use doubly harmonic oscillators. Also, if the potential can be treated by perturbation methods [9], then the exact solutions provide an excellent check of the perturbation expansion for those states for which they exist. If the potential cannot be treated by a perturbation method, then the exact solution do at least give some information about the energy levels. Also, Yan [10] discussed solution for non-linear Schrödinger equation. For these reasons, various types of time-independent solutions and eigenvalues for the Schrödinger equation are important. So we present a class of new exact solutions and eigenvalues for the Schrödinger equation.

DOI: 10.1007/s12043-011-0101-8; ePublication: 9 July 2011 243 Mohammed Ashraful Islam and Jamal Nazrul Islam

2. One-dimensional oscillations

The Hamiltonian for one-dimensional oscillation is well known [4] to be 1 H = p2 + V(q) , (2.1) 2 where p is the momentum and q is the function of generalized coordinates. Let us now consider a more general potential ⎧ ⎫ ⎨ 2⎬ 1 N N V(q) = − 2n (2n − 1) a q2n−2 + 2na q2n−1 , (2.2) 2 ⎩ n n ⎭ n=2 n=1 where an are constants. The time-independent Schrödinger equation is 1 d2 − + V(q) ψ(q) = Eψ(q) . (2.3) 2 dq2 For the potential (2.2) the solution of eq. (2.3) is of the form N 2n ψ(q) = exp − anq , (2.4) n=1 where E = a1. This is a ground state solution of one-dimensional time-independent Schrödinger equation and the ground state energy level is positive or negative depending on whether a1 is positive or negative. For N = 2, in potential (2.2), the solution (2.4) can be reduced to the potential of Leach ([6], expression (1.3)), i.e. V(q) = aq2 + bq4 + cq6, (2.5) and his solution (expression (1.5)), i.e.

1 1 ψ(q) = exp − αq2 − βq4 , (2.6) 2 4 provided = 2 − , = , = 2 a 2E 6a2 b 8Ea2 c 8a2 and

α = 2a1 = 2E,β= 4a2. Also for N = 2 the potential (2.2) reduces to that of Flessas’s ([3–5], expression (1)) and of Khare ([7], expression (1)), i.e. 1 1 1 V(q) = ω2q2 + λq4 + ηq6,η>0, (2.7) 2 4 6 provided / 1 3 1 2 1 η 1/2 3λ2 E = a = λ , a = ,ω2 = − (3η)1/2 , 1 8 η 2 4 3 16η

244 Pramana – J. Phys., Vol. 77, No. 2, August 2011 Anharmonic solution of Schrödinger time-independent equation and the corresponding solution λ 3 1/2 1 η 2 ψ(q) = exp − q2 − q4 . (2.8) 8 η 4 3 Again for N = 3 the potential (2.2) reduces to the potential of Flessas ([4], expression (3)), i.e. 1 1 1 1 1 V(q) = ω2q2 + b q4 + b q6 + b q8 + b q10, b > 0, (2.9) 2 4 2 6 3 8 4 10 5 5 provided 2 2 E = a1,ω= 4 E − 3a2 , b2 = 4 (8Ea2 − 15a3) , = 2 + , = , = 2, b3 6 8a2 12Ea2 b4 192a2a3 b5 180a3 and the corresponding solution 3 2n ψ(q) = exp − anq . (2.10) n=1 Now, consider the potential 1 2 2 2 V = E − Q(q)e−2β q + Q(q)e−β q − 2βqQ(q)e−β q , (2.11) 2 where Q(q) is a function of q and Q(q) = dQ/dq,β= const. For this potential, Schrödinger equation (2.3) gives the solution q 2 ψ(q) = A exp Q q e−β q dq . (2.12)

For bounded support, one may consider the wave function [11,12] −α2 ψ(q) = A exp , if |q| < q , 2 − 2 0 (2.13) q0 q = 0, if |q| ≥ q0.

(Note that ψ(q) and all its derivatives vanish strictly for |q|≥q0, q0 >0.) Thus the Schrödinger equation (2.3) gives the potential α2q2 + 2α2 q2 − 1 q2 − 3α2q4 V = E − 0 0 . (2.14) 2 − 2 4 q0 q The time-independent one-dimensional Schrödinger equation (2.3) may also be written as [6] D2 + λ − X ψ = 0, (2.15) where D ≡ d/dq. λ is the eigenvalue and X(q) is the potential (to within a constant multiplier). This differential equation may be written in the operator form [6] as D2 + λ − X = (D − α)(D − β) (2.16)

Pramana – J. Phys., Vol. 77, No. 2, August 2011 245 Mohammed Ashraful Islam and Jamal Nazrul Islam from which it follows that

β =−α, α2 − α = X − λ. (2.17)

Equation (2.11) now reads [6] as

(D − α)(D + α) ψ = 0, (2.18) the general solution of (2.18) as follows: q q ψ(q) = A exp − α(u) du + B exp − α(u) du q u × exp 2 α(v) dv du. (2.19)

In the context of an oscillator potential, α(q) is a polynomial and either A or B will be zero according to the behaviour of α(q) as |q|→∞. The solution (2.19) is a ground-state solution. To obtain higher states factorization of the form [6] (D − α)(fD − β) = f D2 + λ − X , (2.20) is assumed, which gives β = f − α f , and

X − λ = α2 − α − 2α f − f /f. (2.21)

Then the solution changes to q q ψ(q) = Af(q) exp − α(u) du + Bf(q) exp − α(u) du q u × f −2 (u) exp 2 α(v) dv du, (2.22) having the same analysis as for (2.19). In (2.17) the choice of a polynomial for α(q) will give a polynomial X(q). For the wave function to be square integrable the degree of α(q) must be odd, i.e., 2N + 1, whence the degree of X(q) is 4N + 2. It is possible to obtain more than one explicit solution (corresponding to different energy levels) for a particular potential by assuming that ψ(q) has the form N 2n−1 ψ(q) = f (q) exp − αnq dq , (2.23) n=1 where αn are constants. For odd states, we need to consider

p N N−1 2n−1 2n−1 2n f (q) = f2n−1q ,α(q) = α2n−1q , A = a2nq . (2.24) n=1 n=1 n=0

246 Pramana – J. Phys., Vol. 77, No. 2, August 2011 Anharmonic solution of Schrödinger time-independent equation

Then (2.21) implies the choice 2α f − f = Af, (2.25) which gives N sets of relations for k = 1,2,..., where p is a positive integer, equating coefﬁcients of powers of q: p P 2k−1 q : 2 (2n − 1)α2k−2n +1 f2n−1 − 2k(2k + 1) f2k+1 = a2k−2n f2n−1, n=1 n=1 (2.26) from which we can obtain all unknowns a2i , i = 1,2,3,. . . . For even states,wehavetotake f as an even function, i.e. p 2n f (q) = f2nq . n=0 Then for this f , eq. (2.25) gives N number of relations p 2k q : 2 2nα2k−2n+1 f2n − 2(k + 1)(2k + 1) f2k+2 n=1 p = a2k−2n f2n, (2.27) n=0 where

a−n = f−n = 0,α2N+i = f2P+ j = a2N+k = 0. Using the above process we can obtain different states of solution of the form (2.23). To clarify this process we cite some examples.

3 5 Case 1. Consider α(q) = α1q + α3q + α5q , f = f1q. Then eq. (2.25), i.e., 2α f − f = 2 4 (a0 + a2q + a4q ) f ,gives

a0 = 2α1, a2 = 2α3, a4 = 2α5.

Then for λ = 3α1, the potential (2.21) becomes ( ) = α2 − α 2 + (α α − α ) 4 + α2 + α α 6 X q 1 5 3 q 1 3 7 5 q 3 2 1 5 q + α α 8 + α2 10, 2 3 5q 5q (2.28) and the solution is α α α ψ(q) = ( f q) exp − 1 q2 − 3 q4 − 5 q6 . (2.29) 1 2 4 6

2 Case 2. Consider f = f0 + f2q . Then (2.27) gives for ( f2/ f0) = g, a0 = 2g, a2 = α + 2, = α − α 2 − 3 α = α − α 2 − 1 3, 4 1g 2g a4 4 3g 4 1g 2g , 5 3g 1g 2 g which indicates that g has three values.

Pramana – J. Phys., Vol. 77, No. 2, August 2011 247 Mohammed Ashraful Islam and Jamal Nazrul Islam

The potential (2.21) for λ = α1 + 2g gives ( ) = α2 − α − α − 2 2 + α α − α − α + α 2 + 3 4 X q 1 3 3 4 1g 2g q 2 1 3 5 5 4 3g 4 1g 2g q + α2 + α α 6 + α α 8 + α 10, 3 2 1 5 q 2 3 5q 5q (2.30) and the solution is α α α ψ(q) = f 1 + gq2 exp − 1 q2 − 3 q4 − 5 q6 . (2.31) 0 2 4 6 2 Now, if we take f = 1 + f2q and α5 = a4 = 0, then using (2.27) the above result after calculations, gives =−α ± α2 + α , f2 1 1 2 3 and for λ = α1 − 2 f2, the potential (2.21) becomes ( ) = α2 − α 2 + α α 4 + α2 6, X q 1 7 3 q 2 1 3q 3q (2.30a) and the corresponding solution is α2 α ψ = 1 + −α ± α2 + 2α q2 exp − 1 q2 − 3 q4 . (2.31a) 1 1 3 2 4 √ √ Now for α1 = β/2 γ,α3 = γ, eq. (2.30a) gives

β2 √ X(q) = − 7 γ q2 + βq4 + γ q6, (2.30b) 4γ √ 1 β β2 √ β γ ψ = 1 + − ± + 8 γ q2 exp − √ q2 − q4 , 2 γ γ 4 γ 4 (2.31b)

β β β2 √ E = λ = √ + 2 √ ± + 2 γ 2 γ 2 γ 4γ 3β β2 √ = √ ± + 8 γ (2.30c) 2 γ γ and β2 √ α = − 7 γ, (2.30d) 4γ which shows that our results (eqs (2.30b–d)) and (eq. (2.31b)) are exactly matching with the results of Atre et al ([13], expressions (13), (14)) and their potential in Hamiltonian (in = expression (6)) for n 2 except their polynomial P2± (in expression (15)) but our method √ + 1 − β ± β2 + γ 2 gives 1 2 γ γ 8 q . So eqs (2.30) and (2.31) are more general than that of [13].

248 Pramana – J. Phys., Vol. 77, No. 2, August 2011 Anharmonic solution of Schrödinger time-independent equation

α β = 1 + α 3, = + 2 β − = Again for q 3q f f0 f2q . Then using (2.25), i.e. 2 f f Af, gives f = 4α f /(1 − 2α ) and for λ = E =−(4α f / f ) gives the potential 2 3 0 1 3 0 2 2 α + α1 X(q) = 1 + (2α α − 7α ) q2 + α2q6, (2.30e) q2 1 3 3 3 and the solution is 4α f α 3 0 2 −α1 3 4 ψ = f0 ± q q exp − q . (2.31c) 1 − 2α1 4 Then the potential (2.30e) can be matched with the potential of the Hamiltonian (in expression (16)) of Atre et al [13] iff σ = α (α + ) ,α= α α − α ,γ= α2. 1 1 1 2 1 3 7 3 3 √ λ E If we choose f2 = γ = α3 = a ⇒ f0 =− =− that gives 4 4 √ 1 α = (2α − 7) γ =−4 (s − 2) a, for α =−2l =−2 s − , 1 1 4 and

1 3 σ =−2l (−2l + 1) = 4 s − s − , 4 4 this agrees with that in ref. [13] for μ =−5/2. Then the solution (2.31c) becomes E a ψ = − ± aq2 q2l exp − q4 , 4 4 which agrees with the solution of Atre et al [13] (p. 49).

3 Case 3. Consider f = f1q + f3q . Then eq. (2.25) gives for f3/ f1 = g, a0 = 2α1 − 2 2 3 2 3 6g, a2 = 2α3 + 4α1g + 6g , a4 = 6(α3g − α1g − 6g ), α5 = α3g − α1g − 6g , which indicates that g has three values. For λ = 6g − α1, eq. (2.21) reduces to ( ) = α2 − α − α − 2 2 + α α − α + α 2 + 3 4 X q 1 5 3 4 1g 6g q 2 1 3 6 3g 6 1g 36g q + α2 + α α 6 + α α 8 + α2 10, 3 2 1 5 q 2 3 5q 5q (2.32) and the solution is α α α ψ(q) = f q + gq3 exp − 1 q2 − 3 q4 − 5 q6 . (2.33) 1 2 4 6

= + 2 + 4 α = = Case 4. Consider f 1 f2q f4q ,for 5 a4 0. Then eq. (2.25) yields α f 2 α =− , = α , = 1 + 2 − 2 3 , 3 + α 2 + α2 − α − α α = , a0 2 f2 a2 8 3 f4 3 f2 6 3 f2 4 1 f2 4 1 4 3 f2 8 1 3 0 which indicates that f2 has three values. For λ = E = α1 + 2 f2, eq. (2.21) reduces to ( ) = α2 − α 2 + α α 4 + α2 6, X q 1 11 3 q 2 1 3q 3q

Pramana – J. Phys., Vol. 77, No. 2, August 2011 249 Mohammed Ashraful Islam and Jamal Nazrul Islam and the solution for f = P4(q) is α α ψ(q) = P (q) exp − 1 q2 − 3 q4 , 4 2 4 for the three values of f2, P4(q) gives three polynomials. Now consider the special case to match with the results of Atre et al [13] (p. 50). For α3 = 1, choose α1 = 0 yielding f2 = 0, ±4. Hence ⎧ ⎨ 2 − , for f = 0, = 2 f4 ⎩ 3 2, for f2 =±4. The potential now becomes 2 6 X(q) =−11q + q , q4 ψ(q) = P (q) exp − , 4 4 where ⎧ ⎨ 2 1 − q4, for f = 0 ( ) = 2 P4 q ⎩ 3 2 4 1 ± 4q + 2q , for f2 =±4 and E = λ = 0, ±8 which agree with the results of Atre et al [13] (p. 50). Similarly for higher states we can extend the result for X(q) which is a polynomial in q2. However, it is not necessary that X(q) be a polynomial in q2 but the polynomial of α(q) must be an odd degree. The solution ψ(q) admits solution of Magyari’s ([8], expression (3)).

3. Three-dimensional oscillators

The Schrödinger equation in cylindrical polar coordinate system when ψ is independent of θ and z (by replacing ρ by q) reduces to

1 d2 1 d − + ψ + (V − E)ψ = 0. (3.1) 2 dq2 q dq The potential, ⎧ ⎫ ⎨ 2 ⎬ 1 N N V = 2na q2n−1 − 4n2a q2n−2 , (3.2) 2 ⎩ n n ⎭ n=1 n=2 gives the solution of (3.1) for E = 3a1 as N 2n ψ(q) = A exp − anq . (3.3) n=1

250 Pramana – J. Phys., Vol. 77, No. 2, August 2011 Anharmonic solution of Schrödinger time-independent equation

Again, the Schrödinger equation in spherical polar coordinate system when ψ is independent of θ and ϕ (by replacing r by q) reduces to

1 d2 2 d − + ψ + (V − E) ψ = 0. (3.4) 2 dq2 q dq The potential ⎧ ⎫ ⎨ 2 ⎬ 1 N N V = 2na q2n−1 − 2n(2n + 1)a q2n−2 , (3.5) 2 ⎩ n n ⎭ n=1 n=2 yields the solution of (3.4) for E = 3a1 as N 2n ψ(q) = A exp − anq . (3.6) n=1 The time-independent Schrödinger equation for cylindrical polar coordinate system is

1 − ∇2 + V ψ = Eψ, (3.7) 2 where

1 ∂ ∂ 1 ∂2 ∂2 ∇2 = r + + . r ∂r ∂r r 2 ∂θ2 ∂z2 Consider ψ is independent of θ.

Case (a). Consider the potential ⎧ ⎫ ⎨ N N 2⎬ 1 α − α − V (r) = − 4α2m r 2 i 2 + 2α m r 2 i 1 . (3.8) 2 ⎩ i i i i ⎭ i=2 i=1

1 2 For E = k = 2m1 and α1 = 1, the solution of (3.7) for the above potential is 2 N kz −kz 2αi ψ(rz) = ae + be exp mi r . (3.9) i=1 Case (b). Let us assume the potential ⎧ ⎫ 2 ⎨ 2 N N ⎬ 1 m β − β − V (rz) = + β α z i 1 − β (β − 1)α z i 2 . (3.10) 2 ⎩ r 2 i i i i i ⎭ i=1 i=1 = 1 2 For E 2 k , the required solution of (3.7) for the above potential is N βi ψ = {C1 Jm (kr) + C2 J−m (kr)} exp − αi z , for m-fraction, (3.11) i=1 N βi ψ = {C1 Jm (kr) + C2Ym (kr)} exp − αi z , for m-integer, (3.12) i=1

Pramana – J. Phys., Vol. 77, No. 2, August 2011 251 Mohammed Ashraful Islam and Jamal Nazrul Islam where Jm , Ym are respectively Bessel’s functions of the ﬁrst kind and Neumann’s Bessel’s function of the second kind, of order n.

Case (c). Again, consider the potential

V (r, z) = V (r) + V (z) n n n 2 − − − = β 2qi 1 − β 2 2qi 2 + α 2mi 1 2 i qi r 2 i qi r 2 mi i z i=1 i=2 i=1 n 2m−2 + mi (2mi − 1)z . (3.13) i=1

For q1 = 1, E = 2β1, the corresponding solution of (3.7) for the above potential is n 2mi 2qi ψ = exp αi z − βi r . (3.14) i=1 The time-independent Schrödinger equation for the spherical coordinate system is

1 − ∇2 + V ψ = Eψ, (3.15) 2 where

1 ∂ ∂ 1 ∂ ∂ 1 ∂2 ∇2 ≡ r 2 + sin θ + . r 2 ∂r ∂r r 2 sin θ ∂θ ∂θ r 2 sin2 θ ∂ϕ2 Let V = V (r) is the function of r only. Then for ψ = R, where R = R(r), = (θ), = (ϕ), the above equation reduces to Morse and Freshbach [9],

d dR r 2 + 2(E − V (r))r 2 − l(l + 1) R = 0, (3.16a) dr dr

d2 + k2 = 0, (3.16b) dϕ2

1 d d k2 sin θ + l(l + 1) − = 0. (3.16c) sin θ dθ dϕ sin2 θ The solution of eqs (3.16b) and (3.16c) are respectively

= a1 cos kϕ + b1 sin kϕ, (3.17a)

( θ) = k ( θ)+ k ( θ), cos aPl cos bQl cos (3.17b)

252 Pramana – J. Phys., Vol. 77, No. 2, August 2011 Anharmonic solution of Schrödinger time-independent equation

k ( ) k ( ) where Pl x and Ql x are the associated Legendre polynomials or functions of the ﬁrst and second kind respectively, and n ψ = ( ϕ + ϕ) k ( θ) + k ( θ) − γ 2mi a1 cos k b1 sin k aPl cos bQl cos exp ai i=1

dk Q (cos θ) Qk (cos θ) = (−1)l sink θ l . l d(cos θ)k

Case (i). Consider n(n + 1) − l(l + 1) V (r) = , 2r 2 then eq. (3.16a) gives

d dR r 2 + 2Er2 − n(n + 1) R = 0. (3.18) dr dr Choosing 2E = k2 and then using the transformation x = kr, eq. (3.18) reduces to d2 R dR x2 + 2x + x2 − n (n + 1) R = 0. dx2 dx √ Again, by choosing Z = xR, the above equation becomes

2 d Z dZ 2 x2 + 2x + x2 − n + 1 Z = 0, dx2 dx 2 + 1 which is a Bessel’s equation of order n 2 and so the solution of eq. (3.16a) will be 1 2 n+1 R = AJ 1 (kr) + (−1) BJ 1 (kr) . (3.19) n+ −n− kr π 2 2 The required solution of (3.15) is 1 2 n+1 ψ = AJ 1 (kr) + (−1) BJ 1 (kr) n+ −n− kr π 2 2 × k ( θ)+ k ( θ) { ϕ + ϕ} . aPl cos bQl cos a1 cos k b1 sin k (3.20)

Case (ii). Now, consider the potential n 2 n 2mi −1 2mi −2 2 V (r) = 2 mi ai r − mi (2mi +1)ai r −l(l +1)/2r . (3.21) i=1 i=2

For E = 3a1 and m1 = 1, the solution of eq. (3.16a) for the potential (eq. (3.21)) is n 2mi R = exp − ai r . (3.22) i=1

Pramana – J. Phys., Vol. 77, No. 2, August 2011 253 Mohammed Ashraful Islam and Jamal Nazrul Islam

So the required solution of (3.15) is ψ = ( ϕ + ϕ) k ( θ) + k ( θ) a1 cos k b1 sin k aPl cos bQl cos n 2mi × exp − ai r . (3.23) i=1

Case (iii). For the extended Yukawa potential, which may be considered as a hadron potential, V (r) = E − a + br + cr 2 e−μr . (3.24) For the above potential, eq. (3.16a) gives ∝ l m R = r exp − amr , (3.25) m=1 provided that they satisfy the following recurrence relations:

∝ − + 2 2 + ( − ) − ( + ) 4lma2m m am 2 p 2m p apa(2m−p) 2m 2m 1 a2m p=1

2aμ2m−2 2bμ2m−3 2cμ2m−4 + − + = 0, m = p (3.26a) (2m − 2)! (2m − 3)! (2m − 4)!

∝ −2l(2m − 1)a2m−1 + 2 p(2m − p − 1)apa(2m−p−1) − 2m(2m − 1)a2m−1 p=1

2aμ2m−3 2bμ2m−4 2cμ2m−5 − + − = 0, 2m = 2p + 1 (3.26b) (2m − 3)! (2m − 4)! (2m − 5)! which gives a (b − aμ) a = , a = , 2 (2l + 3) 3 (3l + 6) 1 4a2 a = aμ2 − 2bμ + 2c − , 4 4(2l + 5) (2l + 3)2 1 aμ3 6a(b − aμ) a = bμ2 − 2cμ − + , 5 (10l + 3) 3 (2l + 3)(3l + 6) and

ar = fr (a, b, c, l,μ), r ≥ 4.

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4. Higher states

To obtain wave functions and eigenvalues for higher energy states we make a modiﬁcation of the factorization as was done in (2.16). We assume that the energy states occur for the radial functions only.

Case (α). For the cylindrical coordinate system, we write the radial equation as d2 R 1 dR + + λ − X(r) − k2 R = 0. dr 2 r dr So, for higher states we assume

1 (D − α)( fD− β) ≡ f D2 + D + λ − X(r) − k2 , (4.1) r which yields f β = f − α f − , r

α 1 1 f X − λ + k2 = α2 + − α + + f − 2α f − . (4.2) r r 2 f r The solution of R(r) is given by A r B r R = f (r) exp − α(u)du + f (r) exp − α(u)du r r r u u × exp 2α(v)dv du. (4.3) f 2(u) For X to represent an oscillator potential, the constant A will be zero if α(r) →−∞ as r →−∞and B will be zero if α(r) →∞as r →∞. For higher state solutions, we choose n+1 n 2i−1 m 2i b α = α − r + , A = a r − . (4.4) 2i 1 r 2i r 2 i=1 i=1

! = Q 2i−1 − α − f = Case (i). For odd states: Consider f i=1 f2i−1r . Then using (4.4) f 2 f r Af gives the following recurrence relations, for m = 0, b = 1: n−1 2 (2i + 1) + 1 f2i+1 − 2 (2i − 2 j − 1)α2i+1 f2(i− j)−1 j=0 n = a2 j f2(i− j)−1, (4.5) j=0 where i = 1, 2, 3,... ,n − 1 and Q is a positive integer, which determines the unknown values of a2i uniquely.

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2 The potential is, to within a constant multiplier, for λ = k − a0; n+1 2 n+1 n 2i−1 2i−2 2i X(r) = α2i−1r − 2 (i − 1)α2i−1r + a2i r , i=1 i=2 i=1 and the solution is Q n+1 1 2i−1 1 2i R = f − r exp − α − r r 2i 1 2i 2i 1 i=1 i=1 ⎡ ⎤ r n+1 ⎢ u 1 2i ⎥ × ⎣ + α − ⎦ . A B ! 2 exp 2 2i 1u du Q − 2i 2i 1 i=1 i=1 f2i−1r The solution of eq. (3.7) is Q n+1 1 2i−1 1 2i ψ = f − r exp − α − r r 2i 1 2i 2i 1 i=1 i=1 ⎡ ⎤ r n+1 ⎢ u 1 2i ⎥ × ⎣ + α − ⎦ A B ! 2 exp 2 2i 1u du Q − 2i 2i 1 i=1 i=2i−1 f2i−1r × aekz + beikz .

! = G 2i Case (ii). For even states: Consider f i=0 f2i r , where G is a positive integer. Then − α − f = =−, using (4.4) f 2 f r Af gives the following recurrence relation, for m 1 b = 0, a0 = 0: i n 2 4 (i + 1) f2i+2 − 2 2(i − j)α2 j+1 f2(i− j) = a2 j f2(i− j), j=0 j=1 where i = 1, 2, 3,... ,n − 1, which determines the unknown values of a2i uniquely. 2 The potential is, to within a constant multiplier, for λ = k − a0 + 2α1, n+1 2 n+1 n 2i−1 2i−2 2i X(r) = α2i−1r − 2iα2i−1r + a2i r , i=1 i=2 i=1 and the solution of R(r) is G n+1 2i 1 2i R = f r exp − α − r 2i 2i 2i 1 i=1 i=1 ⎡ ⎤ r n+1 ⎢ 1 1 2i ⎥ × ⎣ + α − ⎦ . A B ! 2 exp 2 2i 1u du G 2i 2i i= u i=0 f2i r 1

256 Pramana – J. Phys., Vol. 77, No. 2, August 2011 Anharmonic solution of Schrödinger time-independent equation

So the solution of eq. (3.7) is G n+1 2i 1 2i ψ = f r exp − α − r 2i 2i 2i 1 ⎡i=1 i=1 ⎤ r n+1 ⎢ 2 1 2i ⎥ kz ikz ×⎣ + α − ⎦ + . A B ! 2 exp 2 2i 1u du ae be G 2i 2i i= i=0 f2i r 1 We now cite two examples.

= = − = − α − f = Case (i). Let f f1r, p n 1 and b 1. Then f 2 f r Af gives the recurrence relation

2i(2i + 1)α2i+1 − 2 f1α2i−1 = f1a2i−2, (4.5a) for i = 1,...,n + 1, the above relation determines a2i uniquely. 2 For λ = k − a0, the potential (4.2) becomes n+1 2 n+1 n−1 2i−1 2i−2 2i X(r) = α2i−1r − 2 (i − 1)α2i−1r + a2i r , (4.6) i=1 i=2 i=1 which gives n+1 n+1 2 1 2i 2 1 2i R = A f r exp − α − r + B f r exp − α − r 1 2i 2i 1 1 2i 2i 1 i=1 i=1 n+1 1 1 2i × exp 2 α − r dr. f r 2 2i 2i 1 1 i=1 Hence the solution of (3.7) is n+1 n+1 2 1 2i 2 1 2i ψ = A f r exp − α − r + B( f r ) exp − α − r 1 2i 2i 1 1 2i 2i 1 i=1 i=1 n+1 1 1 2i kz −kz × exp 2 α − r dr ae + be (4.7) ( f r)2r 2i 2i 1 1 i=1 which is the second state solution.

2 Case (ii). Let f = f0 + f2r , p = n−2, b = 0. Then using (4.4) f −2α f −( f /r) = Af gives the recurrence relation

4 f2α2i−1 + f0a2i + f2a2i−2 = 0, for i = 1, 2,...,n − 1. (4.8) 2 Relation (4.8) determines a2i uniquely, which gives a0 = 0. For λ = k , we get the potential n+1 2 n+1 n−2 2i−1 2i−2 2i 1 X(r) = α − r + 2 (i − 1)α − r + a r + (4.9) 2i 1 2i 1 2i r 2 i=1 i=2 i=1

Pramana – J. Phys., Vol. 77, No. 2, August 2011 257 Mohammed Ashraful Islam and Jamal Nazrul Islam and n+1 2 1 2i R = f + f r r exp − α − r 0 2 2i 2i 1 i=1 n+1 1 1 2i × A + B exp 2 α2i−1r dr . (4.10) + 2 2 2i f0 f1r r i=1

Hence the solution of (3.7) is n+1 2 1 2i 2 ψ = A f + f r r exp − α − r + B f + f r r 0 2 2i 2i 1 0 2 i=1 n+1 n+1 1 2i 1 1 2i × exp − α2i−1r exp 2 α2i−1r dr 2i + 2 2 2i i=1 f0 f1r r i=1 × aekz + be−kz which is a third state solution. Similarly we can extend for higher states for different values of G.

Case (β). For the spherical polar coordinate systems the radial part of the equation for higher state is assumed to be

2 (D − α)( fD− β) ≡ f D2 + D + λ − X − l(l + 1)/r (4.11) r which gives

2 f β = f − α f − (4.12) r

α 2 1 f X − λ + l(l + 1)/r 2 = α2 − α + 2 + + f − 2α f − 2 . (4.13) r r f r

Then the function R is given by A r B r R = f (r) exp − α(u)du + f (r) exp − α(u)du r 2 r 2 r u × u2 f −2(u) exp 2α(v)dv du. (4.14)

Let us consider

n+1 p 2i−1 m 2i p α = α − r + , A = a r + . 2i 1 r 2i r 2 i=1 i=0

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! = Q 2i−1 − α − f = Case (1). f i=0 f2i−1r , for odd states. Then f 2 f 2 r Af gives the recurrence relation by equating the coefﬁcients of r 2i−1, which are

j≤i 2i(2i + 1) f2i+1 − 2 (2i − 2 j − 1)α2 j+1 f2i−2 j−1 j=0

j≤i = a2 j f2i−2 j−1 + pf2i+1 j=0 which give the values of unknown a2i , i = 0,1,2, . . . . The potential, to within a constant multiplier, for m = l, p =−2 (l + 1) and λ = −a0 − (2l + 1) α1,is n+1 2 n+1 n 2i−1 2i−2 2i X = α2i−1r + (2l + 3 − 2i) α2i−1r + a2i r i=1 i=2 i=1 and the solution of R(r) is Q n+1 2i−1 l−2 1 2i R = f − r r exp − α − r 2i 1 2i 2i 1 i=1 i=1 ⎧ ⎛ ⎛ ⎞⎞ ⎫ ⎪ ⎪ ⎨ r l+2 n+1 ⎬ ⎜ u ⎝ 1 2i ⎠⎟ × + ⎝ α − ⎠ . A B ! 2 exp 2 2i 1u du ⎩⎪ Q − 2i ⎭⎪ 2i 1 j=1 i=1 f2i−1u

! = Q 2i = − α − f = Case 2. Consider f i=0 f2i r , for even stages. Then for p 0, f 2 f 2 r Af. This gives the recurrence relation by equating the coefﬁcient of r 2i as

j≤i 2(i + 1)(2i + 1) f2i+2 − 4 (i − j)α2 j+1 f2(i− j) − 2(i + 1) f2i+2 j=0

j≤i = a2 j f2(i− j) j=1 which gives the values of unknown a2i uniquely for i = 0,1,2,3, . . . . The potential, to within a constant multiplier for m = l −1, and λ =−a0 −(2l − 1) α1 is n+1 2 n+1 n 2i−1 2i−2 2i X = α2i−1r + (2l + 1 − 2i) α2i−1r + a2i r i=1 i=2 i=1

Pramana – J. Phys., Vol. 77, No. 2, August 2011 259 Mohammed Ashraful Islam and Jamal Nazrul Islam and the solution for R(r) is Q n+1 2i−1 l−3 1 2i R = f − r r exp − α − r 2i 1 2i 2i 1 i=1 i=1 ⎧ ⎛ ⎛ ⎞⎞ ⎫ ⎪ ⎪ ⎨ r l+1 n+1 ⎬ ⎜ u ⎝ 1 2i ⎠⎟ × + ⎝ α − ⎠ . A B ! 2 exp 2 2i 1u du ⎩⎪ Q − 2i ⎭⎪ 2i 1 j=1 i=1 f2i−1u Similarly, we can extend for higher states for different values of m.

5. Conclusion

In this article, we have discussed various types of new exact solutions of Schrödinger time- independent equations. To obtain exact solutions of different states we used factorization method for isotropic anharmonic oscillator. For different values of Q or G we get different states of exact solutions. Various theories have been proposed for the interaction of hadrons such as quantum chromodynamics, and certain modiﬁcations and approxima- tions, e.g. the non-linear sigma model (Feynman [14], Islam[15–18]). When one considers non-relativistic limits of these and related theories, one arrives at different forms of non- linear Schrödinger equations. Some of the solutions considered in this article may be useful in these contexts. In general, this article (with possible extensions) can be considered as a ‘Laboratory’ for discussing solutions of various and diverse forms of the Schrödinger and other nonlinear equations in different classical and quantum mechanical contexts of physical interest. The present paper may also be useful in connection with the fairly exten- sive and interesting on quasi-exactly solvable models in quantum mechanics (Refs. 4–7 of Atre and Panigrahi [13]). There is a scope for examining precisely the extent to which quasi-exact solutions (as opposed to exact solution) are relevant and meaningful in various physical contexts. We hope to examine these and related questions (almost philosophical: see e.g. David Bohm, Wholeness and the implicate order, ARK Edition, 1983, London WCIE 7DD, England), in a future paper.

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