The Virial Theorem for Hydrogen

Introduction

The total energy of a hydrogen atom in its ground state is −13.6 eV. But this single number hides two competing contributions: a positive kinetic energy from the electron's motion and a negative potential energy from its attraction to the nucleus. How is the total shared between them? For any potential that varies as a power of distance, the virial theorem gives an exact and remarkably simple answer. For Coulomb forces, the kinetic energy is always minus one half of the potential energy.

Core explanation

The general statement. For a stationary state of a particle in a potential of the form V(r) = k rˢ, the expectation values of kinetic energy T and potential energy V satisfy

2⟨T⟩ = s⟨V⟩.

The theorem is exact for eigenstates of the Hamiltonian. It has a classical counterpart for time-averaged energies of bound orbits, first used in the nineteenth century for gases and later for planets and star clusters.

Coulomb systems. For the electron–nucleus attraction V = −Ze²/(4πε₀r), the power is s = −1, so

2⟨T⟩ = −⟨V⟩, or ⟨T⟩ = −½⟨V⟩.

Because E = ⟨T⟩ + ⟨V⟩, two neat results follow:

E = −⟨T⟩ and E = ½⟨V⟩.

The total energy is negative and equal in size to the kinetic energy, while the potential energy is twice as negative as the total.

Applying it to hydrogen. In atomic units the energy of level n for a hydrogen-like atom is E n = −Z²/(2n²). Therefore

⟨T⟩ = +Z²/(2n²) and ⟨V⟩ = −Z²/n².

For the hydrogen ground state (Z = 1, n = 1): ⟨T⟩ = +½ Eₕ = +13.6 eV and ⟨V⟩ = −1 Eₕ = −27.2 eV, giving E = −13.6 eV. The potential-energy result can be checked independently: ⟨1/r⟩ for a hydrogen-like orbital equals Z/(n²a₀), so ⟨V⟩ = −Z × Z/n² in atomic units, exactly as the theorem predicts.

Other potentials. For a harmonic oscillator, V ∝ r², s = 2, and the theorem gives ⟨T⟩ = ⟨V⟩: the energy is shared equally. For a particle in a box, where V is zero inside, all the energy is kinetic. The Coulomb case is special because the attraction and the kinetic energy scale with size in opposite ways.

Molecules. The theorem also holds, with Coulomb interactions between all nuclei and electrons, for molecules at their equilibrium geometry (and for exact wavefunctions within the Born–Oppenheimer approximation). A molecule therefore has E = −⟨T⟩ at equilibrium, so forming a stable bond, which lowers E, must overall raise the average kinetic energy of the electrons while lowering the potential energy by twice as much.

Step-by-step reasoning

To partition the energy of any hydrogen-like level:

1. Find the total energy, E n = −13.6 × Z²/n² eV. 2. The kinetic energy is ⟨T⟩ = −E n, a positive number. 3. The potential energy is ⟨V⟩ = 2E n, a negative number. 4. Check that ⟨T⟩ + ⟨V⟩ = E n.

Visual explanation

Draw a vertical energy axis. Place a bar rising from zero to +13.6 eV for kinetic energy and a bar falling from zero to −27.2 eV for potential energy. The total, −13.6 eV, sits exactly halfway down the potential bar: the kinetic energy cancels precisely half of the attraction.

Real-world analogy

Think of a satellite in a circular orbit. If it loses energy through drag, it drops to a lower orbit — and speeds up. Its kinetic energy rises while its total energy falls, because the potential energy drops by twice as much. An electron pulled closer to a nucleus behaves in the same counter-intuitive way.

Real-world example

In quantum chemistry software, the virial ratio −⟨V⟩/⟨T⟩ is often printed alongside the total energy of a molecule. For a well-converged calculation at an optimised geometry with a flexible basis set, it is close to 2.000; a significant departure warns that the basis set is poorly balanced or the geometry is not at a stationary point.

Why?

Why must the kinetic energy rise as an electron is confined more tightly? Squeezing a wavefunction into a smaller region makes it curve more sharply, and kinetic energy measures curvature. The Coulomb attraction grows as 1/r while the kinetic energy grows as 1/r², so there is an optimal size where the balance gives exactly ⟨T⟩ = −½⟨V⟩.

Common misconception

"A lower-energy state has less kinetic energy." For Coulomb systems the opposite is true. The hydrogen 1s electron has more kinetic energy (13.6 eV) than a 2s electron (3.4 eV), even though 1s is the more stable state, because E = −⟨T⟩.

Worked example

Question: Find ⟨T⟩ and ⟨V⟩ for an electron in the n = 2 level of He⁺ (Z = 2).

Reasoning: E₂ = −13.6 × Z²/n² = −13.6 × 4/4 = −13.6 eV. By the virial theorem, ⟨T⟩ = −E = +13.6 eV and ⟨V⟩ = 2E = −27.2 eV. Check: 13.6 + (−27.2) = −13.6 eV.

Answer: ⟨T⟩ = +13.6 eV and ⟨V⟩ = −27.2 eV, the same as hydrogen 1s, because Z²/n² = 1 in both cases.

Quick check

1. If the total energy of a hydrogen-like state is −3.4 eV, what are its average kinetic and potential energies? Answer: ⟨T⟩ = +3.4 eV and ⟨V⟩ = −6.8 eV, since ⟨T⟩ = −E and ⟨V⟩ = 2E.

Exam focus

Quote the theorem as 2⟨T⟩ = s⟨V⟩ and specialise it to s = −1. Examiners often ask for ⟨T⟩ and ⟨V⟩ given E, or for the sign and size of each term; remember that ⟨T⟩ is always positive and that E = −⟨T⟩ = ½⟨V⟩ for Coulomb systems.

Advanced insight

The theorem can be proved by scaling. Replace ψ(r) by λ^(3/2)ψ(λr): kinetic energy scales as λ² and Coulomb potential energy as λ. For an exact eigenfunction the energy must be stationary with respect to λ at λ = 1, so 2⟨T⟩ + ⟨V⟩ = 0. The same argument shows that any variational wavefunction optimised with respect to an overall scale factor automatically satisfies the virial theorem.

Summary

For a potential V ∝ rˢ, the virial theorem states 2⟨T⟩ = s⟨V⟩. For Coulomb attraction s = −1, so ⟨T⟩ = −½⟨V⟩ and E = −⟨T⟩ = ½⟨V⟩. The hydrogen ground state has ⟨T⟩ = +13.6 eV and ⟨V⟩ = −27.2 eV. The virial ratio −⟨V⟩/⟨T⟩ = 2 provides a useful test of approximate wavefunctions.

Practice questions

1. State the virial theorem for a harmonic oscillator and describe how its energy is shared. Answer: With V ∝ r², s = 2, so ⟨T⟩ = ⟨V⟩; the energy is shared equally between kinetic and potential. 2. Calculate ⟨T⟩ and ⟨V⟩ for the hydrogen n = 3 level. Answer: E₃ = −13.6/9 = −1.51 eV, so ⟨T⟩ = +1.51 eV and ⟨V⟩ = −3.02 eV. 3. In atomic units, what is ⟨V⟩ for the ground state of Li²⁺? Answer: E₁ = −Z²/2 = −4.5 Eₕ, so ⟨V⟩ = 2E₁ = −9 Eₕ. 4. When two hydrogen atoms form H₂ at its equilibrium bond length, does the total electronic kinetic energy rise or fall? Explain. Answer: It rises. The virial theorem gives E = −⟨T⟩ at equilibrium, and bond formation lowers E, so ⟨T⟩ must increase. 5. An approximate calculation on an atom gives ⟨T⟩ = 1.00 Eₕ and ⟨V⟩ = −1.90 Eₕ. What does this suggest? Answer: The virial ratio is 1.90 rather than 2, so the wavefunction is not optimised with respect to scaling and is not the exact eigenfunction.