11.3: Solutions independent of angular variables

Initially we shall just restrict ourselves to those cases where the wave function is independent of $\theta$ and $\varphi$, i.e.,

$$\phi(r, \theta, \varphi)=R(r) .$$

In that case the Schrödinger equation becomes (why?)

$$-\frac{\hbar^2}{2 m} \frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial}{\partial r} R(r)\right)+V(r) R(r)=E R(r)$$

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

$$-\frac{\hbar^2}{2 m} \frac{\partial^2}{\partial r^2} u(r)+V(r) u(r)=E u(r)$$

Of course we shall need to normalise solutions of this type. Even though the solution are independent of $\theta$ and $\varphi$, 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 $4 \pi r^2$. Thus the integration over $r, \theta, \varphi$ can be reduced to

$$\int_{\text {all space }} f(r) d x d y d z=\int_0^{\infty} f(r) 4 \pi r^2 d r .$$

Especially, the normalisation condition translates to

$$\int_0^{\infty}|R(r)|^2 4 \pi r^2 d r=\int_0^{\infty}|u(r)|^2 4 \pi d r=1$$