Separation of Variables in Spherical Coordinates

Introduction

The hydrogen Schrödinger equation involves three coordinates at once, which looks daunting. The key idea that makes it solvable is separation of variables: guessing that the wavefunction is a product of a function of r and a function of the angles. The same technique split the two- and three-dimensional boxes into independent one-dimensional problems. Here it splits hydrogen into an angular problem we have already solved and a radial problem that is new.

Core explanation

The trial product. We start from the Hamiltonian in spherical coordinates,

Ĥ = −(ħ²/2μ)(1/r²) ∂/∂r (r² ∂/∂r) + L̂²/(2μr²) + V(r)

and try a wavefunction of the form ψ(r, θ, φ) = R(r) Y(θ, φ).

Why it works. The operator L̂² acts only on θ and φ, and every other term acts only on r. Substituting the product, the L̂² term acting on Y gives l(l + 1)ħ²Y, provided Y is a spherical harmonic. Every term then contains Y as a common factor, which cancels. What remains depends only on r:

−(ħ²/2μ)(1/r²) d/dr (r² dR/dr) + [l(l + 1)ħ²/(2μr²) + V(r)] R = E R

This is the radial equation. The quantity l(l + 1) links the two halves: it is the separation constant.

A further split. The spherical harmonic itself separates again: Y(θ, φ) = Θ(θ) Φ(φ), with Φ = e^(imₗφ) the particle-on-a-ring function. In total, one three-dimensional equation has become three one-dimensional ones, in φ, θ and r, and each supplies a quantum number: mₗ from the φ equation, l from the θ equation, and n from the radial equation.

The effective potential. The radial equation looks like a one-dimensional problem with an effective potential

V eff(r) = −e²/(4πε₀r) + l(l + 1)ħ²/(2μr²)

The second term is the centrifugal term. It is positive, grows steeply as 1/r² near the nucleus, and is zero for s orbitals. For l > 0 it pushes the electron away from the nucleus, which is why p, d and f wavefunctions vanish at r = 0 while s wavefunctions do not.

Which properties belong where. The angular part determines the shape of an orbital, its orientation and its angular nodes. The radial part determines its size, its energy and its radial nodes. In hydrogen the energy comes entirely from the radial equation; the angular part enters only through l in the centrifugal term.

Normalisation in three dimensions. The volume of a small region in spherical coordinates is dτ = r² sin θ dr dθ dφ. Because the wavefunction is a product, normalisation also separates: the integral of Y ² sin θ dθ dφ over all angles is 1 for a normalised spherical harmonic, and the integral of R²r² dr from 0 to infinity must separately equal 1. The factor r² in the radial integral is the origin of the radial distribution function met later.

Formulae

ψ = R(r) Y(l, mₗ)(θ, φ); Y = Θ(θ)Φ(φ); radial equation: −(ħ²/2μ)(1/r²) d/dr (r² dR/dr) + V eff R = ER; V eff = V(r) + l(l + 1)ħ²/(2μr²); dτ = r² sin θ dr dθ dφ.

Step-by-step reasoning

To separate a Schrödinger equation with a central potential:

1. Write the Hamiltonian in spherical coordinates with the angular part as L̂²/(2μr²). 2. Substitute ψ = R(r)Y(θ, φ). 3. Use L̂²Y = l(l + 1)ħ²Y and cancel Y throughout. 4. Collect the remaining terms into a radial equation with an effective potential. 5. Normalise the radial and angular parts separately, including r² in the radial integral.

Visual explanation

Plot V eff against r for l = 0, 1 and 2. For l = 0 the curve is the plain Coulomb funnel. For l = 1 a steep repulsive wall rises near r = 0, turning the funnel into a well with a minimum some distance out; for l = 2 the wall is higher and the minimum moves further away.

Real-world analogy

Describing a location on Earth, you can give the altitude and, separately, the latitude and longitude. How high you are and where on the globe you are are independent pieces of information. Separation of variables does the same for a wavefunction: how far from the nucleus, and in what direction.

Real-world example

Computational chemistry programs exploit exactly this separation. Atom-centred basis functions are built as a radial function multiplied by a spherical harmonic or its real Cartesian equivalent. Integrals over angles can then be evaluated analytically, which greatly speeds up calculations on molecules with hundreds of atoms.

Why?

Why does an electron with angular momentum avoid the nucleus? Conserving angular momentum as r shrinks requires the angular speed to rise, increasing rotational kinetic energy as 1/r². Near the nucleus this cost outweighs the Coulomb gain, which only grows as 1/r, so the electron is kept away.

Common misconception

"Separation of variables is an approximation." For a central potential it is exact: the true wavefunctions really are products of radial and angular functions. Only in many-electron atoms, where repulsions spoil central symmetry, does the product form become an approximation.

Worked example

Question: For a d electron (l = 2) in hydrogen, calculate the centrifugal term at r = 52.9 pm and compare it with the Coulomb term at the same distance.

Reasoning: Centrifugal term = l(l + 1)ħ²/(2μr²) = 6 × (1.055 × 10⁻³⁴)²/(2 × 9.104 × 10⁻³¹ × (5.29 × 10⁻¹¹)²) = 6 × 1.113 × 10⁻⁶⁸/(5.095 × 10⁻⁵¹) = 1.31 × 10⁻¹⁷ J. The Coulomb term at this distance is −4.36 × 10⁻¹⁸ J.

Answer: The centrifugal term is about +1.3 × 10⁻¹⁷ J, three times larger in size than the Coulomb attraction, so V eff is strongly repulsive there and a 3d electron is found much further out.

Quick check

1. What is the separation constant that connects the radial and angular equations for hydrogen? Answer: The separation constant is l(l + 1), the eigenvalue of L̂² divided by ħ², which also appears in the centrifugal term.

Exam focus

Show clearly why the product R(r)Y(θ, φ) separates the equation: L̂² acts only on the angles and the potential depends only on r. Be ready to write the effective potential and explain the centrifugal term. Do not forget r² in radial normalisation integrals.

Advanced insight

Substituting u(r) = rR(r) turns the radial equation into a form identical to a one-dimensional Schrödinger equation: −(ħ²/2μ) d²u/dr² + V eff u = Eu, with u(0) = 0. Hydrogen's radial problem is thus a one-dimensional particle in the effective potential, with a hard wall at r = 0, and many one-dimensional methods, including numerical integration, can be applied directly.

Summary

For any central potential, the Schrödinger equation separates exactly into ψ = R(r)Y(θ, φ). The angular part is a spherical harmonic with quantum numbers l and mₗ, fixing shape and orientation. The radial equation contains an effective potential equal to the Coulomb term plus a centrifugal term l(l + 1)ħ²/(2μr²), and it supplies the energy and the quantum number n. Normalisation uses dτ = r² sin θ dr dθ dφ.

Practice questions

1. Why can the angular part of the hydrogen wavefunction be taken directly from the particle on a sphere? Answer: The angular part of the Hamiltonian is L̂²/(2μr²), whose eigenfunctions are the spherical harmonics, and the potential does not depend on the angles. 2. For which orbitals is the centrifugal term zero, and what consequence does this have? Answer: For s orbitals (l = 0); with no repulsive wall the s wavefunction is non-zero at the nucleus. 3. Write the radial normalisation condition for hydrogen. Answer: The integral of R(r)² r² dr from r = 0 to infinity must equal 1. 4. Which quantum number arises from each of the three separated equations? Answer: mₗ from the φ equation, l from the θ equation and n from the radial equation.