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

The Closed-Shell Energy — Exercises

Exercises

Quantum Chemistry course › Unit 3 — Many-electron wavefunctions

The technique: reduce the energy of a Slater determinant to one- and two-electron integrals by expanding the expectation value and letting spin orthogonality kill terms. Every number in the answers below is from the converged H₂/STO-3G calculation, in the molecular-orbital basis: $h = - 1.252797$ , $J = 0.674594$ , $E_{nuc} = 0.714286$ (hartree throughout).

Worked example: E = 2h + J, from scratch

Problem. Two electrons occupy one spatial orbital φ with opposite spins: $Ψ = ∣ ϕ α ϕβ ⟩$ . Derive the energy $E = ⟨ Ψ∣ \hat{H} ∣Ψ ⟩$ in terms of $h = ⟨ ϕ ∣ \hat{h} ∣ ϕ ⟩$ and the Coulomb integral $J = (ϕϕ ∣ ϕϕ)$ .

Diagnosis. The Hamiltonian splits into one-electron parts ( $\hat{h}_{1} + \hat{h}_{2}$ ) and the repulsion $1/ r_{12}$ . Evaluate each against the determinant, tracking the spin inner products — they are what make terms survive or vanish.

Predict before reading on: Two candidate two-electron terms exist: a "direct" one and an "exchange" one with the electron labels crossed. Which spin factor will each acquire, and what is $⟨ α ∣ β ⟩$ ?

Solution. Write the determinant explicitly, $Ψ = \frac{1}{2} [ϕ (1) α (1) ϕ (2) β (2) - ϕ (1) β (1) ϕ (2) α (2)]$ .

One-electron part: each of $\hat{h}_{1}, \hat{h}_{2}$ acts on its own coordinate; the cross terms between the two determinant pieces carry $⟨ α ∣ β ⟩ = 0$ , and the diagonal pieces each give $h$ with weight $\frac{1}{2} \times 2$ . Total: $2 h$ — one h per electron.

Predict before reading on: Now the repulsion term. The direct piece pairs $ϕ (1)^{2}$ with $ϕ (2)^{2}$ — same-orbital densities. What spatial integral is that, in chemist notation?

Two-electron part: the direct terms give $(ϕϕ ∣ ϕϕ) = J$ with spin factors $⟨ α ∣ α ⟩ ⟨ β ∣ β ⟩ = 1$ ; the crossed (exchange) terms carry $⟨ α ∣ β ⟩^{2} = 0$ . Exchange dies by spin orthogonality, not by smallness:

E = 2 h + J

Verification. Plug in the real H₂ numbers: $2 (- 1.252797) + 0.674594 = - 1.831$ , and adding $E_{nuc} = 0.714286$ gives $- 1.116714$ — the converged SCF total energy to six digits.

Articulate: In one sentence: what single fact about spin made the closed-shell energy this simple?

Practice

P.1 real numbers — total energy

Using the H₂ values above, compute the electronic energy and the total energy. Which of the three ingredients (h, J, E_nuc) would you tune first if you wanted a lower total energy, and why is that question a trap?

Find the analogue: This is the worked example's verification step promoted to the main event — same plug-in, plus one interpretive twist.

show answer

E_{elec} = 2 h + J = - 1.831

E_{tot} = - 1.116714

. The trap: h, J are not independent knobs — they're functionals of the same orbital, and the variational principle has already balanced them. Lowering one raises the other.

P.2 orbital energies — the Fock eigenvalue

For this system the occupied orbital energy is

ε_{1} = h + J

. Compute it, and reconcile: why is it not

h + 2 J - K

with separate J and K?

Find the analogue: Track the same direct-vs-exchange bookkeeping as the worked example, but inside the Fock operator: which two integrals coincide when everything lives in one orbital?

show answer

It is

h + 2 J - K

— but with both electrons in φ,

K = J

, so

ε_{1} = h + J = - 0.578203

. (Koopmans preview: ionizing H₂ should cost about 0.578 hartree ≈ 15.7 eV.)

P.3 spin — the triplet configuration

Now put the two electrons in different orbitals with the same spin:

Ψ = ∣ ϕ_{1} α ϕ_{2} α ⟩

. Write E in terms of

h_{1}, h_{2}, J_{12}, K_{12}

Find the analogue: Rerun the worked example's spin bookkeeping. Which inner products are now $⟨ α ∣ α ⟩ = 1$ that used to be zero?

show answer

E = h_{1} + h_{2} + J_{12} - K_{12}

. With parallel spins the exchange term survives — and it lowers the energy, the quantitative root of Hund's first rule.

P.4 scaling up — four electrons

A closed-shell system has four electrons in two orbitals (

ϕ_{1}^{2} ϕ_{2}^{2}

). Assemble E from

h_{1}, h_{2}, J_{11}, J_{22}, J_{12}, K_{12}

Find the analogue: The worked example handled one doubly-occupied orbital. Apply its result to each orbital, then add the between-orbital pairs with the worked example's spin-counting.

show answer

E = 2 h_{1} + 2 h_{2} + J_{11} + J_{22} + 4 J_{12} - 2 K_{12}

. The 4 and 2: four between-orbital electron pairs, of which two are same-spin and feel exchange. This is the general closed-shell rule

E = 2 \sum_{i} h_{i} + \sum_{ij} (2 J_{ij} - K_{ij})

at its smallest nontrivial size.

P.5 ionization — Koopmans by subtraction

Remove one electron from H₂ without changing φ (frozen orbital): the ion's energy is just

h

. Compute the ionization potential

IP = E^{+} - E

and compare it with

- ε_{1}

from P.2.

Find the analogue: Both energies come from the worked example's formula — one with two electrons, one with one. The subtraction is the whole derivation.

show answer

IP = h - (2 h + J) = - (h + J) = 0.578203

hartree — exactly

- ε_{1}

. You have just derived Koopmans' theorem for the two-electron case, no approximation beyond the frozen orbital.

P.6 bookkeeping — which terms die

For the determinant

∣ ϕ_{1} α ϕ_{2} β ⟩

(different orbitals, opposite spins), which of

J_{12}

and

K_{12}

appears in E? Write the energy expression.

Find the analogue: The worked example's exchange term died from $⟨ α ∣ β ⟩ = 0$ . Does that spin factor reappear here even though the orbitals differ?

show answer

E = h_{1} + h_{2} + J_{12}

— exchange needs same-spin electrons regardless of orbitals, so

K_{12}

is absent. (This "open-shell singlet-ish" determinant is not a spin eigenfunction, which is its own story — see the correlation lesson.)

Check problems

Check 1 diagnosis

A student computes the H₂ electronic energy as "the sum of orbital energies of both electrons":

E_{elec} = 2 ε_{1} = - 1.156

. The correct value is −1.831. Identify precisely what was double-counted, and produce the corrected formula in terms of

ε_{1}

and J — verify it numerically.

show solution sketch

Each

ε_{1} = h + J

already contains the full interaction with the other electron; summing two of them counts the single electron-electron repulsion twice. Corrected:

E = 2 ε_{1} - J = - 1.156406 - 0.674594 = - 1.831

✓. This is the n = 2 case of

E = \sum ε_{i} - \frac{1}{2} V_{ee}

— and the reason the SCF energy formula carries its ½.

Check 2 derivation

Generalize the worked example: derive the closed-shell energy for 2N electrons in N doubly-occupied orbitals,

E = 2 \sum_{i} h_{i} + \sum_{ij} (2 J_{ij} - K_{ij})

, by counting electron pairs and their spin cases. (P.4 is your N = 2 sanity check; note the diagonal trick

K_{ii} = J_{ii}

that lets the double sum run unrestricted.)

show solution sketch

One-electron: two electrons per orbital gives

2 \sum h_{i}

. Pairs between orbitals i and j: 2×2 = 4 spin combinations, all four feel Coulomb (

4 J_{ij}

per ordered pair counted once →

2 J_{ij}

in the unrestricted double sum), and the two same-spin combinations feel exchange (

- K_{ij}

likewise). Within one orbital: a single αβ pair contributes

J_{ii}

, which is exactly what the unrestricted sum's diagonal

2 J_{ii} - K_{ii} = J_{ii}

delivers. Hence the formula — and the diagonal identity is why quantum chemistry programs never special-case it.

These expressions are what your project-6 SCF loop evaluates numerically, every iteration.