Secular Equations and Secular Determinants

Introduction

The linear variation method reduces a quantum problem to a determinant that must equal zero. For two basis functions the determinant is only 2 × 2, yet its solution already explains why chemical bonds form, why antibonding orbitals exist and why atoms of very different electronegativity form polar bonds. The matrix elements in the determinant are given names — Coulomb, resonance and overlap integrals — that chemists use constantly, from simple Hückel theory to modern computational output.

Core explanation

The 2 × 2 determinant. For basis functions φ A and φ B (each normalised), the secular determinant has H AA − E and H BB − E on the diagonal and H AB − ES in both off-diagonal positions, using H BA = H AB and S AA = S BB = 1. Setting it to zero gives

(H AA − E)(H BB − E) − (H AB − ES)² = 0

The three integrals.

- Coulomb integral α A = H AA = ∫φ AĤφ A dτ. It is the energy an electron would have if confined to φ A in the full molecular potential. It is negative and roughly tracks the ionisation energy of the atomic orbital: more electronegative atoms have more negative α. - Resonance integral β = H AB = ∫φ AĤφ B dτ. It measures the energy of interaction between the two functions and has no classical counterpart. For two orbitals overlapping in phase it is negative; it becomes small when the orbitals are far apart. - Overlap integral S = ∫φ Aφ B dτ. It ranges from 0 (no overlap) towards 1 (identical functions) and depends strongly on distance and orientation.

The homonuclear case. If the two functions are equivalent, α A = α B = α. The determinant gives (α − E)² = (β − ES)², so α − E = ±(β − ES). Solving:

E₊ = (α + β)/(1 + S) and E₋ = (α − β)/(1 − S)

Because β is negative, E₊ lies below α: the in-phase, bonding combination. E₋ lies above α: the out-of-phase, antibonding combination. The coefficients follow from the secular equations: c A = c B for E₊ and c A = −c B for E₋.

Asymmetric splitting. Measured from α, the bonding level is lowered by (β − αS)/(1 + S) and the antibonding level raised by (β − αS)/(1 − S) in magnitude. The 1 − S denominator is smaller, so the antibonding orbital is destabilised more than the bonding orbital is stabilised. This is why filling both orbitals, as in He₂, gives net repulsion.

The Hückel simplification. Setting S = 0 gives E = α ± β, a symmetric splitting of 2 β . This is the starting point of Hückel theory for conjugated π systems.

The heteronuclear case. With α A ≠ α B and S = 0:

E = (α A + α B)/2 ± √[((α A − α B)/2)² + β²]

If α A − α B is large compared with β , the square root is close to α A − α B /2 and the levels barely shift: the orbitals hardly mix. The bonding orbital then resembles the lower-energy (more electronegative) atom's orbital, and the bond is polar.

Formulae

Homonuclear: E± = (α ± β)/(1 ± S)

Hückel (S = 0): E = α ± β

Heteronuclear (S = 0): E = ½(α A + α B) ± √[¼(α A − α B)² + β²]

Step-by-step reasoning

1. Identify α A, α B, β and S for the two basis functions. 2. Write the 2 × 2 determinant with E subtracted on the diagonal and ES off the diagonal. 3. Expand: (α A − E)(α B − E) − (β − ES)² = 0. 4. Solve the quadratic in E. 5. Substitute each root into one secular equation to find c A/c B. 6. Normalise to fix the absolute coefficients.

Visual explanation

Draw a correlation diagram: the energy level of φ A on the left, that of φ B on the right, and the two molecular levels in the centre. Tie lines join each atomic level to the molecular levels it contributes to. When the atomic levels are equal, the molecular levels split symmetrically about them; when they differ, the bonding level sits just below the lower atomic level.

Real-world analogy

Two identical pendulums joined by a weak spring have two shared modes: swinging together and swinging opposite, at slightly different frequencies. The spring plays the role of β. Join a heavy pendulum to a very light one and the modes barely mix, just like orbitals of very different energy.

Real-world example

In hydrogen fluoride, the fluorine 2p orbital lies roughly 5 eV below the hydrogen 1s orbital. The secular determinant predicts a bonding orbital largely on fluorine and an antibonding orbital largely on hydrogen, consistent with the strongly polar H–F bond and a partial negative charge on fluorine.

Why?

Why is β negative for in-phase overlap? In the overlap region an electron is attracted by both nuclei simultaneously, so the region between them is energetically favourable. The resonance integral captures this extra stabilisation of density shared between the centres.

Common misconception

"The resonance integral describes a molecule flipping between two structures." Despite the name, β does not describe any oscillation. It is simply the off-diagonal Hamiltonian matrix element coupling two basis functions.

Worked example

Question: With S = 0, α A = −10.0 eV, α B = −12.0 eV and β = −2.0 eV, find the two orbital energies.

Reasoning: Mean = (−10.0 − 12.0)/2 = −11.0 eV. Half-difference = 1.0 eV. Square root = √(1.0² + 2.0²) = √5.0 ≈ 2.24 eV.

Answer: E = −11.0 − 2.24 = −13.24 eV (bonding) and −11.0 + 2.24 = −8.76 eV (antibonding). The bonding level lies 1.24 eV below the lower atomic level.

Quick check

1. For two identical orbitals with overlap S, which is larger in magnitude: the stabilisation of the bonding level or the destabilisation of the antibonding level? Answer: The destabilisation of the antibonding level, because it is divided by 1 − S, which is smaller than 1 + S.

Exam focus

Examiners expect you to write the determinant correctly, with ES off the diagonal, and to solve it for identical atoms. Be ready to explain the signs of α and β, to set S = 0 for the Hückel result α ± β, and to explain qualitatively why mismatched atomic energies give weak mixing and polar bonds.

Advanced insight

In the heteronuclear case, when β is much smaller than the energy gap Δ = α A − α B , the bonding level is lowered by approximately β²/Δ. This is the same form as a second-order perturbation correction, showing that the variational and perturbational pictures agree in the weak-mixing limit.

Summary

For two basis functions the secular determinant contains Coulomb integrals α (energies of each function alone), the resonance integral β (their interaction, negative for bonding overlap) and the overlap S. Identical functions give E = (α ± β)/(1 ± S), with the antibonding level raised more than the bonding level is lowered. Different Coulomb integrals reduce mixing and produce polar bonds.

Practice questions

1. Write the 2 × 2 secular determinant for two identical normalised orbitals. Answer: α − E, β − ES; β − ES, α − E = 0. 2. With α = −10.0 eV, β = −4.0 eV and S = 0.25, calculate E₊ and E₋. Answer: E₊ = (−14.0)/1.25 = −11.2 eV and E₋ = (−6.0)/0.75 = −8.0 eV, so the bonding level falls by 1.2 eV while the antibonding level rises by 2.0 eV. 3. Explain why He₂ is not stable in simple molecular orbital theory. Answer: Four electrons fill both bonding and antibonding orbitals, and the antibonding orbital is raised more than the bonding one is lowered, so the net effect is repulsive. 4. What value of S is assumed in simple Hückel theory, and what energies result for two identical orbitals? Answer: S = 0, giving E = α + β and α − β.