The Radial Equation

Separation of variables left us with one equation for the radial function $R(r)$ . With a clever substitution it turns into something we already understand: a one-dimensional Schrödinger equation on the half-line $r \ge 0$ .

The substitution that simplifies everything

Start from the radial equation derived previously,

-\frac{\hbar^2}{2\mu}\frac{1}{r^2}\frac{d}{dr}\!\left(r^2\frac{dR}{dr}\right) + \left[V(r) + \frac{\hbar^2\,\ell(\ell+1)}{2\mu r^2}\right]R = E\,R.

Define the reduced radial function

u(r) = r\,R(r).

A short calculation using the product rule shows that

\frac{1}{r^2}\frac{d}{dr}\!\left(r^2\frac{dR}{dr}\right) = \frac{1}{r}\frac{d^2 u}{dr^2}.

Multiplying the radial equation through by $r$ then gives a remarkably clean result.

The effective one-dimensional problem

In terms of $u(r)$ the equation reads

-\frac{\hbar^2}{2\mu}\frac{d^2 u}{dr^2} + V_{\text{eff}}(r)\,u = E\,u,

which is identical in form to a 1D Schrödinger equation, with an effective potential

V_{\text{eff}}(r) = V(r) + \frac{\hbar^2\,\ell(\ell+1)}{2\mu r^2} = -\frac{1}{4\pi\epsilon_0}\frac{e^2}{r} + \frac{\hbar^2\,\ell(\ell+1)}{2\mu r^2}.

The first term is the attractive Coulomb well; the second is the repulsive centrifugal barrier, which grows as $1/r^2$ and pushes the electron away from the origin for $\ell > 0$ . Their competition sets the shape of the well in which $u(r)$ lives.

Behaviour at the boundaries

Two physical conditions pin down the allowed solutions:

At the origin ( $r \to 0$ ): because $R = u/r$ must stay finite, we need $u(0) = 0$ . For $\ell > 0$ the centrifugal barrier already forces $u$ to vanish there; for $\ell = 0$ the boundary condition does it directly.
At infinity ( $r \to \infty$ ): the potential goes to zero, so for a bound state ( $E < 0$ ) the equation becomes $u'' \approx (2\mu|E|/\hbar^2)\,u$ , whose normalizable solution decays exponentially, $u \sim e^{-\kappa r}$ with $\kappa = \sqrt{2\mu|E|}/\hbar$ .

Why this matters

The reduction to $u(r)$ means hydrogen is, at heart, a 1D bound-state problem on the half-line. Demanding that $u$ vanish at the origin and decay at infinity is exactly the kind of two-sided boundary condition that forces quantization: only discrete energies $E_n$ admit normalizable solutions. The next lesson carries out that quantization and recovers the famous $-13.6\,\text{eV}/n^2$ spectrum.

The takeaway

Substituting $u = rR$ turns the radial equation into a one-dimensional Schrödinger equation with effective potential $V_{\text{eff}} = -e^2/(4\pi\epsilon_0 r) + \hbar^2\ell(\ell+1)/(2\mu r^2)$ . The boundary conditions $u(0)=0$ and $u(\infty)\to 0$ select the normalizable bound states.