“Know how to solve every problem that has been solved.” “What I cannot create, I do not understand.” — Richard Feynman

The Slater-Condon Rules — Exercises

Exercises

Quantum Chemistry course › Unit 3 — Many-electron wavefunctions

The technique: evaluate $⟨ Ψ∣ \hat{H} ∣ Ψ^{'} ⟩$ between two determinants by counting how many spin-orbitals they differ in — zero, one, or two — and reading the answer off a short table instead of expanding N! terms. Every CI matrix, every MP2 coupling, every Fock element is one of these three cases. Numeric anchors throughout come from the converged H₂/STO-3G calculation (orbitals o = σ, v = σ*).

Worked example: the one-difference rule

Problem. Two determinants differ in exactly one spin-orbital: $Ψ = ∣ \dots m \dots ⟩$ and $Ψ^{'} = ∣ \dots p \dots ⟩$ (all other spin-orbitals identical and aligned). Show that

⟨ Ψ∣ \hat{H} ∣ Ψ^{'} ⟩ = ⟨ m ∣ \hat{h} ∣ p ⟩ + j \in occ \sum [(m p ∣ jj) - (mj ∣ j p)]

Diagnosis. Split $\hat{H}$ into one- and two-electron parts and ask, for each, which terms survive the orthonormality of the spectator orbitals.

Predict before reading on: The one-electron operator touches one coordinate at a time. If it acts on a spectator orbital (not m or p), what factor multiplies that term, and what is its value?

Solution. One-electron part: a term where $\hat{h}$ acts on a spectator leaves the mismatched pair ⟨m|p⟩ = 0 as a factor — dead. Only the term where $\hat{h}$ bridges the mismatch survives: $⟨ m ∣ \hat{h} ∣ p ⟩$ . Two-electron part: the operator must again bridge m→p with one of its two coordinates; the other coordinate pairs a spectator j with itself, giving the Coulomb-like $(m p ∣ jj)$ directly and the exchange-like $(mj ∣ j p)$ from the antisymmetrized partner, summed over every spectator.

Predict before reading on: Set m and p to the occupied and virtual MOs of a converged Hartree-Fock calculation. The right-hand side becomes a familiar matrix element — which one, and what is its value at convergence?

Verification. The right side is exactly the Fock matrix element $F_{m p}$ . For m = o (occupied), p = v (virtual) of converged H₂: the reference run prints $F_{o v} = 0.00 \times 1 0^{0}$ — Brillouin's theorem, met for the third time (variationally in derive-fock, numerically in project 6, structurally here).

Articulate: State the counting logic in one sentence: what role does spin-orbital orthonormality play in killing terms?

Practice

P.1 zero differences — the diagonal

Write the zero-difference rule (the energy of a single determinant) and use it to recover the closed-shell result

E = 2 h_{ϕ} + J_{ϕ}

for two electrons in one spatial orbital.

Find the analogue: The worked example bridged a mismatch; here there is none. Which sums over spectators survive when every orbital is its own partner?

show answer

E = \sum_{i} h_{ii} + \frac{1}{2} \sum_{ij} [(ii ∣ jj) - (ij ∣ ji)]

over spin-orbitals. For φα, φβ: the h-sum gives 2h; the pair sum gives (φφ|φφ) once (the αβ exchange dies by spin), so E = 2h + J — the closed-shell-energy set's result, now as a table lookup.

P.2 one difference — Brillouin, numerically

For converged H₂, the single excitation

Ψ_{o \to v}

couples to the ground determinant through which quantity? Evaluate it.

Find the analogue: This is the worked example's verification, posed as the question. What does SCF stationarity force?

show answer

⟨ Ψ_{0} ∣ \hat{H} ∣ Ψ_{o \to v} ⟩ = F_{o v} = 0

at convergence. Consequence with teeth: singles do not couple directly to the HF ground state, so CI-with- singles-only does nothing for the ground-state energy, and the first useful excitations in MP2/CI are doubles.

P.3 two differences — the number that runs the show

The doubly-excited determinant

∣ v \overset{v}{ˉ} ∣

differs from

∣ o \overset{o}{ˉ} ∣

in two spin-orbitals. Apply the two-difference rule and evaluate the coupling with the H₂ MO integrals.

Find the analogue: Now both coordinates of the two-electron operator must bridge mismatches — no spectator sums at all. Which single integral remains?

show answer

⟨ o \overset{o}{ˉ} ∣ \hat{H} ∣ v \overset{v}{ˉ} ⟩ = (o v ∣ o v) = 0.1813

hartree. You have met this number twice: it is the MP2 coupling of project 7, and it will be the off-diagonal element of the FCI matrix in QC II. One integral, three theories.

P.4 three or more — the cutoff

Two determinants differ in three spin-orbitals. What is their Hamiltonian matrix element, and what property of

\hat{H}

decides it?

Find the analogue: Count bridges: the worked example needed one operator coordinate per mismatch. How many can H supply?

show answer

Exactly zero. H contains at most two-body operators, so it can bridge at most two mismatches; a third leaves an untouched ⟨spectator|mismatch⟩ = 0 factor. This cutoff is why CI matrices are sparse and why coupled-cluster's disconnected triples come from products of doubles rather than direct couplings.

P.5 bookkeeping — the alignment sign

The rules assume the two determinants are written with their common orbitals in identical positions ("maximum coincidence"). The determinant

∣ \overset{o}{ˉ} v ∣

is the same state as

\pm ∣ v \overset{o}{ˉ} ∣

— which sign, and why does the bookkeeping matter?

Find the analogue: Recall the column-swap property from the determinants lesson — what does one transposition do?

show answer

One column swap flips the sign:

∣ \overset{o}{ˉ} v ∣ = - ∣ v \overset{o}{ˉ} ∣

. Skipping the alignment step silently flips matrix-element signs — in a CI matrix that error doesn't crash anything, it just converges to a wrong, plausible-looking energy. Production codes carry explicit permutation parities for exactly this reason.

P.6 selection rules — spin kills a coupling

The determinants

∣ o \overset{o}{ˉ} ∣

and

∣ o \overset{v}{ˉ} ∣

differ in one spin-orbital (β: o→v), and the one-difference rule applies as usual. Now consider instead the spin-flip excitation:

∣ o \overset{o}{ˉ} ∣

∣ o v ∣

(both electrons α, so Ms jumped from 0 to 1). What is the coupling, before computing any integral?

Find the analogue: The worked example's surviving term was ⟨m|ĥ|p⟩ plus exchange-like sums. What spin factor rides along when m and p carry different spins?

show answer

Zero. Every surviving integral carries the spin inner product ⟨α|β⟩ = 0 — the Hamiltonian is spin-free, so it cannot connect determinants of different Ms. Block structure for free: CI matrices factor by spin sector before any integral is computed.

Check problems

Check 1 derivation

Derive the one-difference rule honestly for the smallest nontrivial case: N = 2,

Ψ = \frac{1}{2} det [m, k]

Ψ^{'} = \frac{1}{2} det [p, k]

. Expand both determinants, evaluate

⟨ Ψ∣ \hat{h}_{1} + \hat{h}_{2} + r_{12}^{- 1} ∣ Ψ^{'} ⟩

, and confirm the general formula's prediction

⟨ m ∣ \hat{h} ∣ p ⟩ + (m p ∣ kk) - (mk ∣ k p)

show solution sketch

Each determinant expands to two products; the four cross terms pair off. The ⟨k|k⟩ = 1 spectator factors keep the ⟨m|ĥ|p⟩ and direct/exchange two-electron terms; every term containing ⟨m|k⟩, ⟨k|p⟩, or ⟨m|p⟩ as a bare overlap dies by orthonormality. The ½ normalization cancels against the two identical surviving copies. Matching the general rule term-for-term is the point: the N-electron proof is this calculation plus induction on spectators.

Check 2 transfer

Use the rules to assemble the complete Hamiltonian matrix of H₂ in the Ms = 0 singlet space spanned by

∣ o \overset{o}{ˉ} ⟩

and

∣ v \overset{v}{ˉ} ⟩

. Express every entry in named quantities (E_HF, the doubly-excited diagonal, and the coupling), and insert the numbers you know.

show solution sketch

H = (E_{HF} K_{o v} K_{o v} E_{vv})

with

E_{HF} = - 1.1167

K_{o v} = (o v ∣ o v) = 0.1813

, and

E_{vv} = 2 h_{vv} + J_{vv} + E_{nuc}

by the zero-difference rule applied to the excited determinant. Diagonalizing this 2×2 is Full CI for H₂ — which is precisely the first project of the correlation course. You have already built its every ingredient.

Three rules, every matrix element in the course. The two-difference number you just computed runs both MP2 and the FCI matrix waiting in QC II.