ISSN: 0973-4945; CODEN ECJHAO E- Chemisry hp://www.e-journals.ne Vol. 5, No.3, pp. 659-66, July 008 NOTE Schrodinger Equaion for he Hydrogen Aom - A Simplified Treamen L.R.GANESAN and M.BALAJI* Former Professor of Chemisry, Madura College, Madurai. *Deparmen of Chemisry, Ramakrishna Mission Vivekananda College, Chennai. balajimanisharma@gmail.com Received 4 November 007; Acceped January 008 Absrac: A simple mehod is presened here for solving he wave mechanical problem of he hydrogen aom. The normal mehod of convering he Caresian coordinaes ino polar coordinaes is edious and also requires an undersanding of he Legendre and Lagurre polynomials. In his paper we are using an alernaive mehod, which requires only minimal familiariy wih mahemaical conceps and echniques. Keywords: Schrödinger equaion, Euler s heorem on homogeneous funcions, Laplace equaion. The equaion µ ψ + E + ψ = 0 is usually solved by expressing in spherical polar coordinaes -4. The following is an alernaive reamen of he same problem. Implicily i is equivalen o he usual exbook reamen bu explicily i is easier o follow The following assumpions are made. Their jusificaion is a poseriori, since hey yield he correc soluion. Assumpion Since he poenial energy conains r explicily, we chooseψ, a produc funcion i.e., ψ = ψ ψ where ψ depends on r alone, while no resricion of a similar kind is placed on ψ ; we have hus, a cerain freedom regarding he choice of ψ. We exercise his freedom in selecing a convenien form for ψ. Thus µ ( ψψ ) + E + ( ψψ ) = 0 ()
660 M.BALAJI e al. µ ψ ( ψ ) + ψ ( ψ ) + (grad ψ) (grad ψ ) + E ( + ψψ ) = 0 Noe: grad ψ = gradien vecor of ψ r = ψ r ψ r ψ i + j + k x y z (grad ψ). (grad ψ ) is he do produc of he gradien vecors i.e., ψ ψ ψ ψ ψ ψ + + x x y y z z Assumpion We now choose Ψ o be (i) a homogeneous funcion of degree l in (x, y, z), of such a naure ha (ii) i is a soluion of he equaion Ψ=0 (Laplace equaion 5 ), such a funcion is called a spherical harmonic. The advanages of his choice are wo. Firs, he erm Ψ drops ou in he equaion () above. Furher since ψ is a funcion of r alone, ψ ψ r x = = ψ x r x r r wih similar expressions for y and z. Therefore, he erm (gradψ ).(gradψ )= dψ ψ ψ ψ x + y + z r dr x y z and his, by Euler s heorem 6, becomes l ψ dψ r dr Afer cancellaion of ψ equaion () may now be wrien as I is now easy o show ha l dψ µ ψ + + E + ψ = 0 r dr ψ = e is a soluion of equaion (3a) viz d ψ ( l + ) dψ µ + + E + ψ = 0 dr r dr Subsiuion of ψ = e and simplificaion gives in his case ( l + ) µ r K K + E + = 0 Equaing separaely o zero, erms involving r h µ E = K and = ( + ) K l µ h 4 π µ E = ( l + ) h and erms independen of r, one ges () (3a) (3b)
Schrodinger Equaion for he Hydrogen aom 66 I is furher possible o selec a funcion ψ for each value of l. l ψ 0 Consan x, y, z xy, zx, yz, (x y ).. (x z ), (y z ) Noe The six funcions chosen for ψ for l = are no, all linearly independen. Thus one chooses xy, xz, yz, (x y ) and a linear combinaion of linearly independen funcions. Assumpion 3 The choice e (x z ) and ψ = is oo resricive. One may herefore ry ψ = is a polynomial in r of degree. Equaion (3) now gives. ( l + ) µ (D K) f + (D K)f + E + f = 0 r Where D = d dr. fe (y z ) as five, where f The coefficien of r in his is µ E K + a. h Equaing his o zero, we have recovered he relaionship shown earlier beween E and K. e Furher he coefficien of r is µ K ( l + )K + a. h Equaed o zero, his gives µ K = ( ) l + + h = p= 0 4 Thus π µ E = ( l + + ) h We may now idenify ( l + + ) wih he principal quanum number, n and l wih he (so called) azimuhal quanum number. The choice l = 0 gives he s orbials, for which = (n ) is he degree of he polynomial f, for any given value of n. For n =, (l = 0, =0) is he only possible choice of l and, viz, he s orbial. For n =, (l = 0, = ) gives he s orbial, which is accidenally degenerae wih (l =, =0) viz. he hree p orbials (see able ). Their funcional forms are (xe ), (ye ) and (ze ). Wih n = 3, l = 0, = ), (l =, = ) and (l =, = 0) yield he 3s, 3p and 3d orbials.the funcional forms of he orbials presen no difficuly. An examinaion of equaion (4) shows ha he coefficien of r s equaed o zero gives ( s )K as+ = a (5) s+ (s + )(l + s + 3) a r p p (4)
66 M.BALAJI e al. The larges and smalles values of (s + ) are and 0. Thus s varies from o ( ). The choice s = gives he resul ha a 0 is indeerminae, since, in equaion (5), a S+ and (S+) are boh zero fors =. For he oher values of s, equaion (5) is he recurrence relaion for he coefficiens of powers of r in f. e.g., s orbial, l =0, =, SxK as a. Pu S = - K = + a S ( S + )( S + 3) + = a 0. Therefore f has he form a 0( kr) and ψ has he form, (consan) (I kr)e e.g., By a similar reasoning, we ge he following resuls for n = 3 (l =0,=): ψ = (consan) kr + k r e 3 ( l =,=): ψ= (consan) kr e ( x or y or z), hree funcions. kr (l =, = 0):ψ= (consan) e (any of he five funcions, xz, xy, (c.f Table). The consans in he above forms may be deermined by normalizaion. Thus, he mehod oulined above yields he real forms of he orbials of he hydrogen like aom, in a sraighforward manner. Is meri is ha here is no need here o sruggle wih he Legendre and Lagurre differenial equaions. The mehod gives he real forms of orbials direcly. This has he advanage of visualizing he shapes of orbials, including heir nodes, and overlaps in chemical bonding. Furher one may easily consruc heir linear combinaions as eigen funcions of L z operaor, if and when needed. For sudens and wih no familiariy wih he special funcions of mahemaical physics, he above mehod is mahemaically ransparen and demands no furher knowledge han elemenary calculus. References. Schrödinger E, Ann.Physik, 96, 79, 36.. Schrödinger E, Ann.Physik, 96, 80, 437. 3. Epsein P S, Phys.Rev. 96, 8, 695. 4. Waler I, Z.Physik, 96, 38, 635. 5. Séroul R, Programming for Mahemaicians, Springer-Verlag, 000, p.5. 6. Shanks D, Solved and Unsolved Problems in Number Theory, 4 h Ed., 993, p., 3-5.
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