The Hydrogen Atom Hamiltonian

Introduction

The hydrogen atom is the only real atom whose Schrödinger equation can be solved exactly, and its solutions underpin all of chemistry's orbital language. Before we can solve anything we must write down the correct Hamiltonian: the operator for the total energy. This page assembles it term by term, explains why the electron mass is replaced by a reduced mass, and shows why spherical polar coordinates are the natural choice.

Core explanation

Two particles, one problem. A hydrogen atom contains a proton and an electron. The full Hamiltonian contains the kinetic energy of each particle plus their Coulomb interaction. As in classical mechanics, the motion can be separated into the motion of the centre of mass, which is just a free particle carrying the whole atom through space, and the relative motion of the electron with respect to the nucleus. Only the relative motion determines the internal energy levels.

Reduced mass. The relative motion behaves like that of a single particle with the reduced mass

μ = m e m p/(m e + m p)

Because the proton is 1836 times heavier than the electron, μ = 0.99946 m e, only about 0.05% smaller than the electron mass. The nucleus is not perfectly stationary; it wobbles slightly about the common centre of mass, and the reduced mass accounts for this exactly.

Kinetic energy. The kinetic-energy operator for the relative motion is −(ħ²/2μ)∇², where ∇² = ∂²/∂x² + ∂²/∂y² + ∂²/∂z².

Potential energy. The electron and proton attract with the Coulomb potential energy

V(r) = −e²/(4πε₀r)

For a hydrogen-like ion with nuclear charge Ze, this becomes −Ze²/(4πε₀r). The potential is negative because the particles attract, tends to zero at infinite separation and depends only on r: it is a central potential.

The complete Hamiltonian.

Ĥ = −(ħ²/2μ)∇² − e²/(4πε₀r)

Choosing coordinates. Since V depends only on r, Cartesian coordinates would mix everything together. In spherical polar coordinates (r, θ, φ), the Laplacian becomes

∇² = (1/r²) ∂/∂r (r² ∂/∂r) + (1/r²) Λ²

where Λ² is the angular operator met for the particle on a sphere, related to angular momentum by L̂² = −ħ²Λ². The Hamiltonian can then be written

Ĥ = −(ħ²/2μ)(1/r²) ∂/∂r (r² ∂/∂r) + L̂²/(2μr²) − e²/(4πε₀r)

The middle term is the rotational kinetic energy: the particle-on-a-sphere Hamiltonian with moment of inertia μr². This form makes it clear that the angular problem is already solved, and that the new work lies entirely in the radial direction.

Scale of the energy. At a distance equal to the Bohr radius, 52.9 pm, the Coulomb potential energy is about −4.36 × 10⁻¹⁸ J, or −27.2 eV. Atomic energies are therefore measured in electronvolts, and the natural unit of 27.2 eV, called the hartree, appears again when we meet atomic units.

Formulae

Ĥ = −(ħ²/2μ)∇² − Ze²/(4πε₀r); μ = m e m N/(m e + m N); ∇² = (1/r²)∂/∂r(r²∂/∂r) + Λ²/r²; L̂² = −ħ²Λ².

Step-by-step reasoning

To build a Hamiltonian for any one-electron atom:

1. Separate off the centre-of-mass motion, which does not affect the energy levels. 2. Replace the electron mass by the reduced mass μ. 3. Write the kinetic energy as −(ħ²/2μ)∇². 4. Add the Coulomb potential −Ze²/(4πε₀r). 5. Rewrite ∇² in spherical coordinates, splitting it into radial and angular parts.

Visual explanation

Plot V(r) against r: a curve that plunges towards minus infinity at r = 0 and rises smoothly towards zero at large r. Unlike the flat-bottomed, steep-walled box, this well is funnel-shaped. Energy levels drawn as horizontal lines inside it will crowd together near the top, a first hint of the 1/n² pattern.

Real-world analogy

Two skaters holding hands and spinning both circle their common centre, the heavier one moving less. If one skater is enormously heavier, the lighter one does almost all the moving. The reduced mass captures the small but real motion of the heavier partner, the proton.

Real-world example

In 1931 Harold Urey identified deuterium by spotting faint lines beside the normal hydrogen lines. Because the deuteron is heavier than the proton, the reduced mass is slightly larger and the levels slightly deeper. The red Balmer line shifts from 656.28 nm for hydrogen to 656.10 nm for deuterium, a difference of about 0.18 nm.

Why?

Why does a heavier nucleus make the energy levels slightly more negative? The energy levels are proportional to μ. A heavier nucleus moves less, so μ approaches the free-electron mass more closely and increases slightly, making every level a little deeper and every transition a little more energetic.

Common misconception

"The Hamiltonian must include the electron's orbit radius as a fixed number." Unlike the Bohr model, the quantum Hamiltonian contains r as a variable. The electron has no fixed orbit; the solutions give a probability distribution over all values of r.

Worked example

Question: Calculate the reduced mass of a hydrogen atom and the percentage by which it differs from the electron mass. (m e = 9.109 × 10⁻³¹ kg; m p = 1.673 × 10⁻²⁷ kg)

Reasoning: μ = m e m p/(m e + m p) = (9.109 × 10⁻³¹ × 1.673 × 10⁻²⁷)/(1.673 × 10⁻²⁷ + 0.000911 × 10⁻²⁷) = (1.5239 × 10⁻⁵⁷)/(1.6739 × 10⁻²⁷) = 9.104 × 10⁻³¹ kg. The ratio μ/m e = 0.99946.

Answer: μ = 9.104 × 10⁻³¹ kg, about 0.054% smaller than the electron mass.

Quick check

1. Which term of the hydrogen Hamiltonian is responsible for binding the electron, and what sign does it have? Answer: The Coulomb potential energy term, −e²/(4πε₀r), which is negative because the proton and electron attract.

Exam focus

Be able to write the hydrogen Hamiltonian in full, name each term and justify the reduced mass. Examiners frequently ask why spherical coordinates are used: the answer is that the potential depends only on r. Remember to include Z for hydrogen-like ions.

Advanced insight

The Hamiltonian written here is non-relativistic and ignores electron spin. Adding relativistic corrections and spin–orbit coupling splits the levels into fine structure, of order 10⁻⁴ eV for the n = 2 level. Even smaller effects, such as the Lamb shift arising from quantum electrodynamics, were measured precisely in hydrogen and became landmark tests of modern physics.

Summary

The hydrogen Hamiltonian for relative motion is Ĥ = −(ħ²/2μ)∇² − e²/(4πε₀r). The reduced mass μ = m e m p/(m e + m p) corrects for the small motion of the nucleus. The Coulomb potential depends only on r, so spherical coordinates separate the Laplacian into a radial part and an angular part that equals L̂²/(2μr²), the particle-on-a-sphere operator already solved.

Practice questions

1. Write the Hamiltonian for the He⁺ ion. Answer: Ĥ = −(ħ²/2μ)∇² − 2e²/(4πε₀r), with μ the reduced mass of the electron and the helium nucleus. 2. Why is the centre-of-mass motion ignored when finding the energy levels? Answer: It is simply the free translation of the whole atom and does not change the internal energy of the electron relative to the nucleus. 3. Would the reduced mass of positronium (an electron bound to a positron) be close to the electron mass? Explain. Answer: No; the two masses are equal, so μ = m e/2, exactly half the electron mass. 4. Identify the rotational kinetic energy term in the hydrogen Hamiltonian written in spherical coordinates. Answer: It is L̂²/(2μr²), the angular momentum squared divided by twice the moment of inertia μr².