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 equa- tion 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.