Eigenvalues & Eigenvectors — Maths for Quantum Chemistry

Eigenvalues are the answer quantum chemistry keeps asking for

Special directions a matrix only stretches — and the energies of a molecule

Most vectors get rotated when a matrix acts on them. A few special ones — the eigenvectors — keep their direction and are simply scaled by a number, the eigenvalue $λ$ :

A v = λ v ⟹ (A - λ I) v = 0

For this to have a non-trivial solution the matrix $(A - λ I)$ must be singular — its determinant must be zero. That gives the characteristic (secular) equation:

det (A - λ I) = 0

For an $n \times n$ matrix this is a polynomial of degree $n$ in $λ$ . Its $n$ roots are the eigenvalues. Two useful checks: the trace (sum of diagonal entries) equals their sum, and the determinant of the full matrix equals their product. Symmetric matrices — all the matrices of real observables in quantum mechanics — always have real eigenvalues.

The Schrödinger equation $\hat{H} ψ = E ψ$ is an eigenvalue equation. The wavefunctions $ψ$ are eigenvectors of the Hamiltonian operator $\hat{H}$ , and the energies $E$ are its eigenvalues. Hückel theory makes this concrete: build a small matrix for a $π$ system, solve its secular determinant, and out come the molecular-orbital energies.

Key relationships at a glance

Eigenvalue equation

A v = λ v

Characteristic equation

det (A - λ I) = 0

Trace = sum of eigenvalues

tr (A) = λ_{1} + λ_{2} + \dots

det = product of eigenvalues

det (A) = λ_{1} λ_{2} \dots

Schrödinger / Hückel

\hat{H} ψ = E ψ; E = α + x β

Ethylene MO energies

E = α \pm β (x = \pm 1)

Eigenvectors as MO coefficients. Once you have an eigenvalue $λ_{i}$ , substituting it back into $(H - λ_{i} I) c = 0$ and solving gives the eigenvector $c_{i}$ — a column of coefficients telling you how much each atomic p-orbital contributes to that MO. Convention demands normalisation: the sum of squares of the coefficients must equal 1, $\sum_{r} c_{r i}^{2} = 1$ , so that the total probability of finding the electron in that MO is 1. For allyl, the non-bonding MO has coefficients $(1/ 2, 0, - 1/ 2)$ — no contribution from the central carbon, a node there, and equal but opposite lobes on C₁ and C₃.

Degeneracy occurs when two distinct eigenvectors share the same eigenvalue. Benzene has a degenerate pair at $E = α + β$ — both $E_{2}$ orbitals have the same energy, so any linear combination of them is also a valid MO. This degeneracy is protected by the six-fold symmetry; any perturbation that breaks the symmetry (e.g., distorting to an alternating bond pattern in cyclohexadiene) immediately splits the degenerate pair.

Allyl and butadiene at a glance

Allyl (3 atoms)

Eigenvalues:

2, 0, - 2

; non-bonding MO at

α

Butadiene (4 atoms)

Eigenvalues:

\pm 1.618, \pm 0.618

; deloc. energy

0.472 β

Normalisation check

Allyl bonding MO:

(1/2)^{2} + (1/ 2)^{2} + (1/2)^{2} = 1 ✓

Trace / det checks

Sum of eigenvalues = trace; product = det. Fast consistency check.

Common pitfalls

Forgetting to set the secular determinant to zero. You must solve $det (H - x I) = 0$ — not simply expand $det (H)$ . The eigenvalues are the values of $x$ (or $λ$ ) that make the whole determinant vanish, not the entries of $H$ itself.
Sign of β. $β$ is negative (typically $\approx - 75 kJ mo l^{- 1}$ ), so a more positive $x$ in $E = α + x β$ gives a lower, more stable orbital. The HOMO of butadiene is $x = + 0.618$ , not $x = + 1.618$ , because it is only the second-lowest.
Not normalising eigenvectors. The raw solution from $(H - λ I) c = 0$ gives a direction, not a magnitude. Divide every component by $\sum c_{r}^{2}$ before quoting MO coefficients or bond orders.
Confusing the Hückel $x$ with an energy. The secular determinant is written in terms of $x = (α - E) / β$ (so that all diagonal entries are $- x$ and off-diagonal entries are $1$ for adjacent atoms). The actual MO energy is recovered as $E = α + x β$ .

The quantum connection

The Schrödinger equation

\hat{H} ψ = E ψ

is an eigenvalue equation: the wavefunctions are eigenvectors and the energies are eigenvalues. Hückel theory makes this concrete — build a matrix for a π system, diagonalize it, and out come the molecular-orbital energies. The same mathematics gives normal vibrational modes (IR/Raman) and principal axes of inertia (rotational spectroscopy). Try the Hückel MO lab.

Going deeper

The Hückel secular determinant for a linear chain of $n$ atoms has the elegant closed-form eigenvalues $x_{k} = 2 cos (kπ / (n + 1)), k = 1, 2, \dots, n$ . For allyl (n = 3): $x_{1} = 2 cos (π /4) = 2, x_{2} = 2 cos (π /2) = 0, x_{3} = 2 cos (3 π /4) = - 2$ — reproducing the hand-calculated values instantly. For cyclic chains (benzene) the equivalent formula is $x_{k} = 2 cos (2 πk / n)$ .

2×2 solver

Adjust the matrix entries and watch the characteristic equation update. The two roots are the eigenvalues — check them against the trace and determinant.

Set the matrix entries and read off the eigenvalues

[

]

λ^{2} - 4 λ + (3) = 0

trace = Σλ4

determinant = Πλ3

discriminant4

λ₁3.00

λ₂1.00

Real eigenvalues — the matrix has well-defined stretch directions. Symmetric matrices (b = c) like those in chemistry always give real eigenvalues.

Worked example 1Eigenvalues by hand: the Hückel matrix for ethylene

The Hückel matrix for ethylene, expressed in units where $α = 0, β = 1$ , is $H = (0110)$ .

Find its eigenvalues and hence the MO energies $E = α + x β$ .

Worked example 2Eigenvectors for the ethylene Hückel matrix

For each eigenvalue of $H = (0110)$ found above, find the eigenvector. Interpret these as the bonding and antibonding MO coefficients.

Hückel MO lab

Pick a conjugated π system. Its Hückel matrix eigenvalues (x values) give the MO energies E = α + xβ. Electrons fill from the bottom up.

Pick a π system — its eigenvalues are the MO energies

π electrons6

eigenvalues (x in α+xβ)2.00, 1.00, 1.00, -1.00, -1.00, -2.00

π energy6α + 8.00β

delocalization energy2.00β

Each energy level is an eigenvalue of the Hückel matrix — degenerate levels and big aromatic stabilization. Filling them with electrons (2 per orbital, lowest first) gives the total π energy. The delocalization energy is how much lower that is than isolated double bonds.

Worked example 3Allyl radical: the non-bonding orbital

The allyl radical has three carbon p-orbitals. Its Hückel matrix is $H = 010101010$ .

Show that the secular determinant gives eigenvalues $x = 0, \pm 2$ and calculate the delocalization energy.

Worked example 41,3-Butadiene: eigenvalues and delocalization

Butadiene has four π electrons and Hückel eigenvalues $x = \pm (1 + 5) /2 \approx \pm 1.618$ and $x = \pm (5 - 1) /2 \approx \pm 0.618$ . Calculate the total π energy and the delocalization energy (reference: two isolated C=C bonds each give $2 β$ ).

Worked examples

Characteristic equations, eigenvalue shortcuts, and connections to spectroscopy.

Worked example 5Eigenvalues of a symmetric 2×2 by the trace/det shortcut

Find the eigenvalues of $A = (5222)$ without expanding the full quadratic — use the trace/determinant relations as a check.

Worked example 6Normal vibrational modes: diagonalizing the force-constant matrix

A linear triatomic molecule (mass-weighted) has the approximate force-constant matrix (in units of the spring constant $k$ ):

F = (1 - 1 - 1 1)

Find its eigenvalues. What do they represent physically, and why is one of them zero?

Worked example 7Benzene aromaticity: the 2β delocalization energy

Benzene has six π electrons and Hückel eigenvalues $x = 2, 1, 1, - 1, - 1, - 2$ . Three isolated ethylene molecules would provide the reference energy. Confirm that the aromatic stabilization (delocalization energy) is exactly $2 β$ .

Worked example 8Allyl eigenvectors: normalised MO coefficients

The allyl Hückel matrix has eigenvalue $x = 2$ (bonding MO). The secular equations for the coefficients $(c_{1}, c_{2}, c_{3})$ are:

- 2 c_{1} + c_{2} = 0, c_{1} - 2 c_{2} + c_{3} = 0, c_{2} - 2 c_{3} = 0

Solve these and normalise the eigenvector. Then state the bond order between C₁–C₂ and C₂–C₃.

Worked example 9Trace and determinant as quick eigenvalue checks

Without expanding the full secular determinant, predict the sum and product of the eigenvalues of the allyl Hückel matrix $H = 010101010$ and confirm that the actual eigenvalues $2, 0, - 2$ are consistent.

ChallengeFull butadiene Hückel — energies, HOMO coefficients, and π-bond/delocalisation energy

The butadiene Hückel matrix is:

H = 0100101001010010

The eigenvalues are $x_{1} = 1.618, x_{2} = 0.618, x_{3} = - 0.618, x_{4} = - 1.618$ (using the golden ratio $ϕ = (1 + 5) /2 \approx 1.618$ ).

(a) Verify the trace and determinant checks. (b) Fill the 4 π electrons and compute the total π energy and delocalization energy. (c) The HOMO ( $ψ_{2}, x = 0.618$ ) has coefficients proportional to $(c_{1}, c_{2}, c_{3}, c_{4})$ ; using the secular equations show that $c_{2} / c_{1} = 1/0.618 = 1.618$ and normalise to get the exact HOMO vector.

Check yourself

Five quick questions on the eigenvalue equation, the secular determinant, and Hückel MOs.

Question 1 of 5 · Score 0

An eigenvector v of a matrix A satisfies which equation?

A v = λ v

Choose an answer.