Schrodinger Equation for the Hydrogen Atom-A Simplified Treatment

A simple method is presented here for solving the wave mechanical problem of the hydrogen atom. The normal method of converting the Cartesian coordinates into polar coordinates is tedious and also requires an understanding of the Legendre and Lagurre polynomials. In this paper we are using an alternative method, which requires only minimal familiarity with mathematical 
concepts and techniques.

The equation 1 2 e E 0 r is usually solved by expressing '∇ 2' in spherical polar coordinates 2-4 .The following is an alternative treatment of the same problem.Implicitly it is equivalent to the usual textbook treatment but explicitly it is easier to follow The following assumptions are made.Their justification is a posteriori, since they yield the correct solution.

Assumption 1
Since the potential energy contains r explicitly, we chooseψ, a product function i.e., ψ = ψ 1 ψ 2 where ψ 1 depends on r alone, while no restriction of a similar kind is placed on ψ 2 ; we have thus, a certain freedom regarding the choice of ψ 2 .We exercise this freedom in selecting a convenient form for ψ 2 .

Assumption 2
We now choose Ψ 2 to be (i) a homogeneous function of degree l in (x, y, z), of such a nature that (ii) it is a solution of the equation "∇ 2 Ψ=0" (Laplace equation 5 ), such a function is called a spherical harmonic.The advantages of this choice are two.First, the term∇ 2 Ψ 2 drops out in the equation ( 2) above.Further since 1 ψ is a function of r alone, with similar expressions for y and z.Therefore, the term and this, by Euler's theorem 6 , becomes After cancellation of 2 ψ equation ( 2) may now be written as It is now easy to show that 2( 1) Equating separately to zero, terms involving 1 r and terms independent of r, one gets Where Equating this to zero, we have recovered the relationship shown earlier between E and K.
Further the coefficient of We may now identify ( t 1) + + l with the principal quantum number, n and l with the (so called) azimuthal quantum number.The choice " l = 0" gives the 's' orbitals, for which t = (n -1) is the degree of the polynomial 'f', for any given value of n.For n = 1, ( l = 0, t =0) is the only possible choice of l and t, viz, the 1s orbital.For n = 2, ( l = 0, t = 1) gives the '2s' orbital, which is accidentally degenerate with ( l = 1, t =0) viz. the three p orbitals (see table 1). .With n = 3, l = 0, t = 2), ( l = 1, t = 1) and ( l = 2, t = 0) yield the 3s, 3p and 3d orbitals.The functional forms of the orbitals present no difficulty.An examination of equation (4) shows that the coefficient of r s equated to zero gives s 2 s 1

(
)( ) .  ).The constants in the above forms may be determined by normalization.Thus, the method outlined above yields the real forms of the orbitals of the hydrogen like atom, in a straightforward manner.Its merit is that there is no need here to struggle with the Legendre and Lagurre differential equations.
The method gives the real forms of orbitals directly.This has the advantage of visualizing the shapes of orbitals, including their nodes, and overlaps in chemical bonding.Further one may easily construct their linear combinations as eigen functions of L z operator, if and when needed.For students and with no familiarity with the special functions of mathematical physics, the above method is mathematically transparent and demands no further knowledge than elementary calculus.

It is further possible to select a function 2 ψThe six functions chosen for 2 ψ 2 (
for each value of l .l for l = 2 are not, all linearly independent.Thus one chooses xy, xz, yz, 2 in r of degree t.Equation (3) now gives.