Orbital Energies and Koopmans' Theorem

Introduction

Hartree–Fock produces orbital eigenvalues, and the highest occupied value is often compared with the energy needed to remove an electron. Koopmans' theorem gives a simple relation: the first vertical ionisation potential is approximately the negative of the Hartree–Fock HOMO energy when the remaining orbitals are frozen. The word approximately is essential. Orbital relaxation, electron correlation and geometry change separate a calculated orbital eigenvalue from a measured ionisation energy.

Core explanation

An ionisation energy is a total-energy difference between an N-electron neutral species and an (N−1)-electron cation. For vertical ionisation, both energies refer to the neutral molecule's nuclear geometry: IP vert=E {N−1}(R neutral)−E N(R neutral). An adiabatic ionisation energy allows the cation's nuclei to relax to its own equilibrium geometry, so it is a different quantity. A photoelectron experiment can probe vertical transitions with vibrational structure, while thermochemical values may be closer to adiabatic differences under specified conditions.

In Hartree–Fock, remove an electron from occupied orbital i but keep every remaining orbital shape fixed. Under this frozen-orbital approximation and a consistent single-determinant treatment, the resulting energy difference is −ε i. For the first ionisation, i is commonly the HOMO, giving IP≈−ε HOMO. A negative occupied eigenvalue then gives a positive removal energy. This relation is not an identity for the exact interacting-electron system and does not say every orbital eigenvalue is a directly measured electron energy.

After ionisation, the remaining electrons experience a different mean field and can rearrange. Allowing the cation's orbitals to relax usually lowers its variational energy relative to the frozen configuration and changes the energy difference. Correlation also differs between neutral and cation and is missing from basic Hartree–Fock. These effects can partly cancel numerically in some systems, making the Koopmans estimate useful, but such cancellation is not guaranteed.

The theorem is most naturally associated with vertical ionisation at fixed geometry. A molecule's geometry may relax after electron removal, lowering the cation's energy and changing the adiabatic IP. A single HOMO eigenvalue cannot include this nuclear relaxation. When comparing calculation with experiment, specify whether the reference is vertical or adiabatic and what electronic state of the ion is produced.

Using −ε LUMO as an electron affinity is more delicate. Hartree–Fock virtual orbitals belong to the neutral N-electron Fock field, which is not the optimised field of the N+1 electron anion. Relaxation and correlation can be especially important, and a finite basis may represent a weakly bound extra electron poorly. A total-energy difference between properly treated neutral and anion states is generally a safer approach than blindly applying a HOMO analogy to the LUMO.

The HOMO rule also has qualifications for open-shell and degenerate cases. Different spin orbitals or nearly degenerate occupied levels may lead to several ion states, and a simple one-determinant removal picture may not describe the observed band. Ionisation intensities and final-state correlations require more than the energy estimate. Koopmans' theorem is best presented as a model connection, not a universal photoelectron spectrum generator.

Hartree atomic units are often used in calculations: one hartree is about 27.2114 eV. If a HOMO energy is expressed in hartree, its negative magnitude must be converted before comparing with eV experimental values. The reference energy zero and method convention should also be checked; the sign alone without units is not a sufficient answer.

Step-by-step reasoning

Identify a converged Hartree–Fock HOMO and its units. Negate the eigenvalue for a frozen-orbital vertical IP estimate. Convert units if needed. State that the nuclei remain fixed and the remaining orbitals do not relax in this estimate. For a better comparison, compute neutral and ion total energies at the appropriate geometries and consider correlation and final-state identity.

Visual explanation

Draw a neutral energy well and a cation energy well along a nuclear coordinate. A vertical arrow from the neutral minimum to the cation curve at the same coordinate depicts vertical ionisation. A diagonal route toward the cation minimum indicates later nuclear relaxation. On a separate orbital ladder, show removal from HOMO and label the frozen estimate −ε HOMO.

Real-world analogy

Removing a person from a crowded room immediately changes occupancy while everyone else still stands where they were; afterward people can move and relax into new positions. Koopmans' frozen-orbital step resembles the immediate removal, while full ion calculation allows rearrangement. The analogy omits quantum correlation and geometry changes of nuclei.

Real-world example

Photoelectron spectroscopy measures energies associated with electron removal from molecules. Hartree–Fock HOMO values can provide rough first assignments, but discrepancies with band positions can reflect relaxation, correlation and vibrational structure. A measured sequence of bands should not be mapped one-to-one onto a raw orbital eigenvalue list without these considerations.

Why?

Why is the HOMO usually associated with first ionisation? Among occupied orbitals in a simple Aufbau-like ground state, its eigenvalue is highest, so its negative is the smallest frozen-orbital removal estimate. Electron rearrangement and state coupling can alter the precise ordering, but HOMO removal is the natural first approximation.

Common misconception

Koopmans' theorem is not the statement that HF total energy equals the sum of occupied orbital energies. It compares a frozen-orbital total-energy difference with one eigenvalue. Nor does it justify treating every virtual orbital as a measured anion energy or every calculated IP as adiabatic.

Worked example

Suppose a Hartree–Fock calculation reports ε HOMO=−0.400 hartree. The frozen-orbital estimate is IP≈0.400 hartree. Multiplying by 27.2114 eV per hartree gives about 10.9 eV. This is an illustrative vertical ionisation estimate; calculating relaxed cation and neutral total energies could shift the value, and cation geometry relaxation would change an adiabatic comparison further.

Quick check

1. What is the basic Koopmans estimate for a negative Hartree–Fock HOMO energy ε H? Answer: IP vert≈−ε H under a frozen-orbital approximation. 2. Does this estimate include cation orbital relaxation? Answer: No. Remaining orbitals are assumed unchanged after electron removal.

Exam focus

Define vertical versus adiabatic IP and state the frozen-orbital assumption. Convert hartree to eV carefully. Do not apply −ε LUMO as an equally reliable electron-affinity rule without discussing anion relaxation, correlation and basis effects.

Advanced insight

More sophisticated electron-propagator and many-body methods describe ionisation as removal into correlated cation states. Their spectral strengths can be distributed among multiple final states rather than one orbital line. Koopmans' one-orbital picture is therefore a useful starting point whose failure can reveal substantial final-state correlation.

Summary

Koopmans' theorem relates a frozen-orbital Hartree–Fock electron-removal energy to minus an occupied orbital eigenvalue, commonly the HOMO for first vertical ionisation. It omits orbital and nuclear relaxation and correlation changes. Total-energy differences and state-specific treatments are needed for more reliable ionisation and electron-affinity predictions.

Practice questions

1. A HF HOMO energy is −0.25 hartree. What is the Koopmans vertical IP estimate in eV? Answer: It is +0.25 hartree, or about 0.25×27.2114≈6.80 eV. The estimate retains frozen orbitals and fixed nuclei. 2. Why can an adiabatic IP differ from the vertical Koopmans estimate even if the electronic approximation were otherwise good? Answer: The cation can relax its nuclear geometry after electron removal, lowering its energy relative to the fixed neutral geometry used for vertical ionisation. 3. Why is −ε LUMO often a poor electron-affinity estimate in basic HF? Answer: The neutral virtual orbital is not an optimised orbital of the anion, and relaxation, correlation and diffuse-basis requirements can be large for the added electron.