Page 71 - Electrical Properties of Materials

54 The hydrogen atom and the periodic table

Thus, the differential equation to be solved is
2 2 e 2
∇ ψ + + E ψ = 0. (4.3)
2m 4π 0 r

It would be hard to imagine a physical configuration much simpler than that
of a proton and an electron, and yet it is difficult to solve the corresponding
differential equation. It is difficult because the 1/r term does not lend itself
readily to analytical solutions. Thanks to the arduous efforts of nineteenth-
century mathematicians, the general solution is known, but it would probably
mean very little to you. Unless you have a certain familiarity with the prop-
erties of associated Legendre functions, it will not make you much happier if
you learn that associated Legendre functions happen to be involved. So I will
not quote the general solution because that would be meaningless, nor shall I
derive it because that would be boring. But just to give an idea of the mathem-
atical operations needed, I shall show the derivation for the simplest possible
case, when the solution is spherically symmetric, and even then only for the
lowest energy.
z The potential energy of the electron depends only on the distance, r; it there-
fore seems advantageous to solve eqn (4.3) in the spherical coordinates r, θ, φ
(Fig. 4.1). If we restrict our attention to the spherically symmetrical case, when
r ψ depends neither on φ nor on θ but only on r, then we can transform eqn (4.3)
θ
without too much trouble. We shall need the following partial derivatives
y
φ ∂ψ ∂r ∂ψ
= (4.4)
x
∂x ∂x ∂r
Fig. 4.1 and
Coordinate system used to transform
2
eqn (4.3) to spherical coordinates. ∂ ψ ∂ ∂ψ ∂r
= . (4.5)
∂x 2 ∂x ∂r ∂x
When we differentiate eqn (4.5), we have to remember that ∂ψ/∂r is a function
of r and ∂r/∂x is still a function of x; therefore,
2
∂ ∂ψ ∂r ∂r ∂ ∂ψ ∂r ∂ψ ∂ r
= +
∂x ∂r ∂x ∂x ∂r ∂r ∂x ∂r ∂x 2
2
2
∂ ψ ∂r 2 ∂ψ ∂ r
= + . (4.6)
∂r 2 ∂x ∂r ∂x 2
Obtaining the derivatives with respect to y and z in an analogous manner,
we finally get
2
2
2
∂ ψ ∂ ψ ∂ ψ
2
∇ ψ = + +
∂x 2 ∂y 2 ∂z 2
2 2 2
2
∂ ψ ∂r ∂r ∂r
= + +
∂r 2 ∂x ∂y ∂z
2 2 2
∂ψ ∂ r ∂ r ∂ r
+ + + . (4.7)
∂r ∂x 2 ∂y 2 ∂z 2