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11.3 Solutions independent of θ and φ

Initially we shall just restrict ourselves to those cases where the wave function is independent of θ and φ, i.e.,

ϕ(r,θ,φ) = R(r).
(11.8)

In that case the Schrödinger equation becomes (why?)

−{{ℏ}^{2}\over 2m} {1\over { r}^{2}} {∂\over ∂r}\left ({r}^{2} {∂\over ∂r}R(r)\right ) + V (r)R(r) = ER(r).
(11.9)

One often simplifies life even further by substituting u(r)∕r = R(r), and multiplying the equation by r at the same time,

−{{ℏ}^{2}\over 2m} {{∂}^{2}\over ∂{r}^{2}}u(r) + V (r)u(r) = Eu(r).
(11.10)

Of course we shall need to normalise solutions of this type. Even though the solution are independent of θ and φ, we shall have to integrate over these variables. Here a geometric picture comes in handy. For each value of r, the allowed values of x range over the surface of a sphere of radius r. The area of such a sphere is 4π{r}^{2}. Thus the integration over r,θ,φ can be reduced to

{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{\textrm{all space}}f(r)\kern 1.66702pt dxdydz ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{0}^{∞}f(r)\kern 1.66702pt 4π{r}^{2}dr.
(11.11)

Especially, the normalisation condition translates to

{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{0}^{∞}|R(r){|}^{2}\kern 1.66702pt 4π{r}^{2}dr ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{0}^{∞}|u(r){|}^{2}\kern 1.66702pt 4πdr = 1
(11.12)

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