Separation in Spherical Coordinates

When an electron orbits a proton, the interaction energy depends only on the distance between them — not on any particular direction in space. Exploiting that symmetry starts with rewriting the Schrödinger equation in spherical coordinates $(r, \theta, \phi)$ , then splitting one three-dimensional problem into three one-dimensional problems via separation of variables.

From Cartesian to Spherical

The time-independent Schrödinger equation for a particle of mass $m$ in a potential $V(r)$ is

-\frac{\hbar^2}{2m}\nabla^2 \psi + V(r)\,\psi = E\,\psi.

In Cartesian coordinates the Laplacian $\nabla^2$ mixes all three variables uncomfortably. In spherical coordinates it becomes

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

The key observation is that the last two terms — those involving $\theta$ and $\phi$ — form a well-known differential operator: $-\hat{L}^2 / (\hbar^2 r^2)$ , where $\hat{L}^2$ is the squared orbital angular-momentum operator. The equation therefore separates naturally into a radial part and an angular part.

The Separation Ansatz

We assume a product form

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

substitute into the Schrödinger equation, and divide through by $RY$ . Because the potential $V(r)$ depends only on $r$ , every term involving only $\theta$ and $\phi$ can be collected on one side and every term involving only $r$ on the other. Both sides must equal the same constant, which — for reasons that become clear in the angular problem — we call $\ell(\ell+1)$ where $\ell$ is a non-negative integer.

This yields two independent ordinary differential equations:

\frac{1}{Y}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\frac{\partial Y}{\partial\theta}\right) + \frac{1}{\sin^2\!\theta}\frac{\partial^2 Y}{\partial\phi^2}\right] = -\ell(\ell+1)

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

The Angular Equation and Spherical Harmonics

The angular equation separates once more. Writing $Y(\theta,\phi) = \Theta(\theta)\,\Phi(\phi)$ and introducing a second separation constant $m_\ell^2$ gives a simple equation in $\phi$ :

\frac{d^2\Phi}{d\phi^2} = -m_\ell^2\,\Phi \;\;\Longrightarrow\;\; \Phi(\phi) = e^{i m_\ell \phi},

with the single-valuedness condition $\Phi(\phi + 2\pi) = \Phi(\phi)$ forcing $m_\ell$ to be an integer. The equation for $\Theta$ is then the associated Legendre equation, whose well-behaved solutions exist only when $\ell = 0, 1, 2, \ldots$ and $m_\ell \in \{-\ell, -\ell+1, \ldots, \ell\}$ . The joint solutions

Y_\ell^{m_\ell}(\theta,\phi) = \mathcal{N}\,P_\ell^{|m_\ell|}(\cos\theta)\,e^{i m_\ell\phi}

are the spherical harmonics, where $\mathcal{N}$ is a normalization constant and $P_\ell^{|m_\ell|}$ is an associated Legendre polynomial. These functions are the same for every central potential; only the radial equation knows about the specific form of $V(r)$ .

The Radial Equation

With the angular part solved, the radial equation becomes

-\frac{\hbar^2}{2m}\frac{d^2 u}{dr^2} + \left[V(r) + \frac{\hbar^2}{2m}\frac{\ell(\ell+1)}{r^2}\right]u = E\,u,

where the substitution $u(r) = r\,R(r)$ turns it into an equation that looks like the familiar one-dimensional Schrödinger equation but with an effective potential

V_{\mathrm{eff}}(r) = V(r) + \underbrace{\frac{\hbar^2\,\ell(\ell+1)}{2m\,r^2}}_{\text{centrifugal barrier}}.

The second term — the centrifugal barrier — pushes the electron away from the nucleus for $\ell \geq 1$ and grows as the electron tries to come closer. For $\ell = 0$ (s-states) the barrier vanishes and the wave function can be non-zero at the origin.

Why This Matters for the Hydrogen Atom

For hydrogen the potential is the Coulomb attraction

V(r) = -\frac{e^2}{4\pi\varepsilon_0}\frac{1}{r}.

The radial equation with this $V(r)$ can be solved analytically. The requirement that $R(r) \to 0$ as $r \to \infty$ (so the state is normalizable) imposes a quantization condition on the energy, which depends only on the principal quantum number $n = 1, 2, 3, \ldots$ . The angular quantum numbers $\ell$ and $m_\ell$ label the angular shape of the state but — in the pure Coulomb problem — do not affect its energy. This accidental degeneracy in $\ell$ is special to $V \propto 1/r$ and disappears in multi-electron atoms where the full Coulomb symmetry is broken.

The full solution is a product $\psi_{n\ell m_\ell}(r,\theta,\phi) = R_{n\ell}(r)\,Y_\ell^{m_\ell}(\theta,\phi)$ , which is the starting point for all of atomic and molecular quantum mechanics.