Separating Radial and Angular Parts

To solve the hydrogen Schrödinger equation we exploit its spherical symmetry. The strategy is separation of variables: we guess that the wavefunction factors into a part depending only on the radius and a part depending only on the angles, then check that the guess is consistent.

The Laplacian in spherical coordinates

In spherical coordinates the Laplacian is

\nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}\!\left(r^2\frac{\partial}{\partial r}\right) + \frac{1}{r^2}\!\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta} \!\left(\sin\theta\frac{\partial}{\partial\theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\right].

The bracketed angular operator is, up to a factor, the square of the orbital angular momentum operator:

\hat{L}^2 = -\hbar^2\!\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta} \!\left(\sin\theta\frac{\partial}{\partial\theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\right].

So the Laplacian becomes

\nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}\!\left(r^2\frac{\partial}{\partial r}\right) - \frac{\hat{L}^2}{\hbar^2 r^2}.

The separable ansatz

We try a product form

\psi(r,\theta,\phi) = R(r)\,Y(\theta,\phi).

Substituting into $\hat{H}\psi = E\psi$ and using the Laplacian above, the angular dependence appears only through $\hat{L}^2$ acting on $Y$ . The spherical harmonics $Y_\ell^m(\theta,\phi)$ are precisely the eigenfunctions of that operator:

\hat{L}^2\, Y_\ell^m = \hbar^2\,\ell(\ell+1)\, Y_\ell^m, \qquad \hat{L}_z\, Y_\ell^m = \hbar m\, Y_\ell^m,

with $\ell = 0, 1, 2, \ldots$ and $m = -\ell, \ldots, +\ell$ . Choosing $Y = Y_\ell^m$ replaces the operator $\hat{L}^2$ by the number $\hbar^2\ell(\ell+1)$ .

The radial equation falls out

With that substitution the angular part cancels on both sides, leaving an equation for $R(r)$ alone:

-\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.

The three-dimensional partial differential equation has become a one-dimensional ordinary differential equation. The angular momentum quantum number $\ell$ now appears as a parameter, and the term $\hbar^2\ell(\ell+1)/(2\mu r^2)$ is the centrifugal barrier — the same repulsive term that keeps orbits from collapsing in classical mechanics.

Counting the quantum numbers

The full wavefunction is therefore labelled by three quantum numbers:

\psi_{n\ell m}(r,\theta,\phi) = R_{n\ell}(r)\,Y_\ell^m(\theta,\phi).

The angular numbers $\ell$ and $m$ come from the spherical harmonics; the third number $n$ (the principal quantum number) will emerge when we impose that $R(r)$ be normalizable. Solving the radial equation is the subject of the next lessons.

The takeaway

Spherical symmetry lets the hydrogen wavefunction factor as $\psi = R(r)\,Y_\ell^m(\theta,\phi)$ . The angular part is universal — the spherical harmonics, eigenfunctions of $\hat{L}^2$ and $\hat{L}_z$ — leaving a single radial equation with a centrifugal term to solve for each $\ell$ .