Hydrogen atom — solving in spherical coordinates

1. The starting equation (WHAT we solve)

Why reduced mass? The proton is not infinitely heavy, so both particles move about the common centre of mass. Replacing $m_e$ with $\mu = \dfrac{m_e m_p}{m_e+m_p}\approx m_e$ exactly converts the two-body problem into a one-body problem.

The Laplacian in spherical coordinates (this is a geometry fact, derived from the chain rule, not physics): $\nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}\!\left(r^2\frac{\partial}{\partial r}\right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\frac{\partial}{\partial\theta}\right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\phi^2}.$

2. Separation of variables (HOW it splits)

Substitute and multiply by $\dfrac{2\mu r^2}{\hbar^2 R Y}$. Everything containing $r$ goes on one side, everything angular on the other:

$\underbrace{\frac{1}{R}\frac{d}{dr}\!\left(r^2\frac{dR}{dr}\right) - \frac{2\mu r^2}{\hbar^2}\big(V-E\big)}_{\text{only }r} \;=\; \underbrace{-\frac{1}{Y}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\frac{\partial Y}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2 Y}{\partial\phi^2}\right]}_{\text{only }\theta,\phi}$

So we get two equations:

Angular: $\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\frac{\partial Y}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2 Y}{\partial\phi^2} = -\ell(\ell+1)\,Y$

Radial: $\frac{d}{dr}\!\left(r^2\frac{dR}{dr}\right) - \frac{2\mu r^2}{\hbar^2}\big(V-E\big)R = \ell(\ell+1)\,R$

3. The angular part → split again

The $\phi$ equation: $\frac{d^2\Phi}{d\phi^2} = -m_\ell^2\,\Phi \quad\Rightarrow\quad \Phi(\phi)=e^{im_\ell\phi}.$

The $\theta$ equation is the associated Legendre equation; its solutions are finite only when $\ell$ is a non-negative integer with $|m_\ell|\le\ell$. The full angular solutions are the spherical harmonics $Y_\ell^{m_\ell}(\theta,\phi)$.

4. The radial part → energy quantization

Define $u(r)=rR(r)$ to clean up the equation. It becomes a 1D Schrödinger equation: $-\frac{\hbar^2}{2\mu}\frac{d^2u}{dr^2} + \underbrace{\left[V(r)+\frac{\hbar^2\ell(\ell+1)}{2\mu r^2}\right]}_{V_{\text{eff}}}u = E\,u.$

Requiring $u$ to be normalizable (decay as $r\to\infty$) and finite at the origin forces another integer, the ==principal quantum number $n$==. Solving gives the famous energy levels.

5. Worked examples

Flashcards

Connections

• Bohr model of hydrogen — gives same $E_n$ but without wavefunctions
• Spherical harmonics — the $Y_\ell^{m_\ell}$ angular solutions
• Angular momentum operators — origin of $\ell,m_\ell$
• Separation of variables (PDEs)
• Reduced mass and two-body problem
• Zeeman effect — lifts the $m_\ell$ degeneracy with a magnetic field
• Quantum numbers and the periodic table

Concept Map

Hinglish (regional understanding)