Variational Treatment of the Hydrogen Atom

Introduction

Hydrogen is one of the few atomic systems for which we know the exact Schrödinger solution. That makes it an ideal test of the variational method. Choose a simple spherically symmetric exponential wavefunction, let its decay rate vary, and calculate the expected kinetic and electrostatic energies. The minimum reproduces the ideal 1s result because the exact ground-state shape happens to be a member of the chosen family.

Core explanation

In the fixed-nucleus approximation, the electronic Hamiltonian is Ĥ = −ℏ²∇²/(2mₑ) − e²/(4πε₀r). Let ψα(r) = (α³/π)¹ᐟ² exp(−αr), with α > 0. The prefactor normalises the state over three-dimensional volume. The function has no angular dependence, stays finite at the origin and decays at infinity. Its parameter α, measured in inverse length, controls electron localisation: a larger α concentrates probability nearer the nucleus.

For this family, ⟨1/r⟩ = α. The Coulomb expectation is therefore V(α) = −e²α/(4πε₀). The kinetic expectation is T(α) = ℏ²α²/(2mₑ). Combining them gives E(α) = Aα² − Bα, with A = ℏ²/(2mₑ) and B = e²/(4πε₀). The quadratic kinetic cost grows faster than the linear attraction gained as α rises. That competition guarantees a finite minimum for positive α.

Differentiate: dE/dα = ℏ²α/mₑ − e²/(4πε₀). Setting it to zero gives α = mₑe²/(4πε₀ℏ²) = 1/a₀, where a₀ is the Bohr radius in the fixed-nucleus model. The second derivative ℏ²/mₑ is positive. At the minimum, E var = −mₑe⁴/[2(4πε₀)²ℏ²], approximately −13.6 eV. The optimised wavefunction becomes the exact ideal hydrogen 1s eigenfunction. If a different family, such as one restricted to Gaussian decays, were used, the best energy would normally remain above the exact ground energy.

For precise hydrogen energies, the reduced mass of the electron–proton system replaces mₑ in the relative-coordinate problem. The approximation here isolates the variational idea and matches the common fixed-nucleus Bohr-radius formula. An upper-bound claim always refers to the Hamiltonian actually used; one should not mix a fixed-nucleus trial result with a different Hamiltonian without saying so.

Step-by-step reasoning

Write the Hamiltonian and an admissible normalised trial function. Use the spherical volume element 4πr²dr when evaluating radial integrals, or cite verified expectation values for this exponential family. Calculate T and V separately, combine them into E(α), and minimise only over α > 0. Confirm the stationary point is a minimum. Finally compare the optimised function with the known 1s form to explain why equality with the exact fixed-nucleus result occurs.

Visual explanation

Plot T as an upward-curving parabola of α and V as a downward-sloping straight line. Their sum begins near zero as α approaches zero, dips below zero, and rises at large α. Mark the dip at 1/a₀. Sketch two electron-density clouds alongside: a diffuse cloud for small α and a compact cloud for large α.

Real-world analogy

A person choosing how close to sit beside a warm fire benefits from approaching it but eventually pays another cost, such as discomfort. The best position balances competing terms. In the atom, nuclear attraction favours concentration while quantum kinetic energy penalises excessive localisation. The analogy captures the tradeoff, not the mathematical origin of either energy.

Real-world example

Quantum-chemistry calculations often tune exponents of atom-centred basis functions to represent electronic density. Hydrogen's exponential calculation shows why a physically chosen shape matters: an exponential tail can capture a one-electron Coulomb ground state particularly well. In larger atoms or molecules, electron repulsion and multiple nuclei prevent this one-parameter calculation from being exact.

Why?

The exact 1s function is proportional to exp(−r/a₀), so the chosen family contains it at α = 1/a₀. The variational theorem says no allowed trial has an expectation below the exact ground energy. When the minimisation reaches the exact eigenfunction, its expectation equals that energy. This equality is a special success of the trial family, not a general promise of variational calculations.

Common misconception

One might think the attraction alone should drive α to infinity, concentrating the electron at the nucleus. That ignores the kinetic term proportional to α², which eventually dominates. Another error is to replace α by 1/a₀ before minimising; doing so hides what the method discovers. Finally, −13.6 eV is a negative bound-state energy relative to a free electron at infinity, not a negative photon energy.

Worked example

Use atomic units, in which ℏ = mₑ = e²/(4πε₀) = 1 for this model. Then E(α) = α²/2 − α in hartrees. Its derivative is α − 1, so α = 1 and E(1) = 1/2 − 1 = −1/2 hartree. Since one hartree is about 27.2 eV, this is −13.6 eV. A nonoptimal trial α = 2 gives E(2) = 2 − 2 = 0 hartree, which is higher than −1/2 and illustrates the upper-bound property.

Quick check

1. Why does the calculated energy rise again when α becomes very large? Answer: The positive kinetic contribution grows as α², faster than the magnitude of the negative Coulomb contribution, which grows only as α. Strong localisation eventually costs more kinetic energy than it gains in attraction.

Exam focus

Keep the three-dimensional normalisation and the α > 0 domain explicit. Differentiate the full expectation energy, then verify the second derivative. State that exact equality occurs here because the exponential trial family includes the exact 1s function for the specified fixed-nucleus Hamiltonian. Distinguish energy in hartrees from energy in electronvolts.

Advanced insight

At the optimum, the derivative condition gives Bα = 2Aα ², so V = −2T and E = −T for the Coulomb potential. This is consistent with the virial theorem for a bound Coulomb eigenstate. Away from the optimum, the same trial family generally does not satisfy that balance. This offers an independent check on the minimisation algebra.

Summary

The normalised hydrogen trial ψα = (α³/π)¹ᐟ²e⁻αʳ has E(α) = ℏ²α²/(2mₑ) − e²α/(4πε₀). Minimising yields α = 1/a₀ and E = −13.6 eV in the fixed-nucleus approximation. The result is exact because this trial family contains the ideal 1s eigenfunction; the general variational method only guarantees an upper bound.

Practice questions

1. In atomic units, calculate the trial energy at α = 0.5 and compare it with the optimum. Answer: E(0.5) = 0.5²/2 − 0.5 = 0.125 − 0.5 = −0.375 hartree. It lies above the optimum −0.5 hartree, so it is a weaker upper bound. 2. What change to the formula is needed if electron and proton motion are both included in the relative-coordinate hydrogen problem? Answer: Replace mₑ by the electron–proton reduced mass in the relevant kinetic term and associated optimum. This produces slightly different length and energy values while retaining the same variational procedure. 3. A student obtains −0.60 hartree for an admissible trial state using the same ideal hydrogen Hamiltonian. What should be checked? Answer: The exact ground energy is −0.50 hartree, so −0.60 violates the variational lower limit. Recheck normalisation, kinetic-energy signs, radial integration, units and whether the Hamiltonian actually matches the one claimed.