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

Operator Algebra & Normal Ordering — Exercises

Exercises

QC II — The Correlation Problem › Unit 2 — Second quantization

The technique: evaluate matrix elements and operator identities using only the two anticommutators — ${a_{p}, a_{q}^{†}} = δ_{pq}$ and ${a_{p}^{†}, a_{q}^{†}} = {a_{p}, a_{q}} = 0$ — by pushing annihilation operators rightward until they meet the vacuum and die. No determinants are expanded; the algebra does the antisymmetry. Throughout, $∣ ⟩$ is the vacuum, $a_{p} ∣ ⟩ = 0$ , and $⟨ ∣ ⟩ = 1$ .

Worked example: the number operator

Problem. For a single occupied orbital, the state is $∣Φ ⟩ = a_{p}^{†} ∣ ⟩$ . Evaluate $⟨ Φ∣ a_{p}^{†} a_{p} ∣Φ ⟩$ and confirm the number operator returns 1.

Diagnosis. Write everything out as a vacuum expectation value, $⟨ ∣ a_{p} (a_{p}^{†} a_{p}) a_{p}^{†} ∣ ⟩$ , then use ${a_{p}, a_{p}^{†}} = 1$ , i.e. $a_{p} a_{p}^{†} = 1 - a_{p}^{†} a_{p}$ , to move annihilation operators right.

Predict before reading on: The rightmost operator acting on the ket vacuum is $a_{p}^{†}$ , and the leftmost on the bra vacuum is $a_{p}$ (as $⟨ ∣ a_{p}$ ). Which one will you eliminate first, and what does $a_{p} ∣ ⟩$ equal?

Solution. Look at the innermost piece $a_{p} a_{p}^{†} ∣ ⟩$ . Replace $a_{p} a_{p}^{†} = 1 - a_{p}^{†} a_{p}$ : $a_{p} a_{p}^{†} ∣ ⟩ = ∣ ⟩ - a_{p}^{†} a_{p} ∣ ⟩ = ∣ ⟩ - 0 = ∣ ⟩$ , since $a_{p} ∣ ⟩ = 0$ . So $a_{p}^{†} a_{p} ∣Φ ⟩ = a_{p}^{†} a_{p} a_{p}^{†} ∣ ⟩ = a_{p}^{†} ∣ ⟩ = ∣Φ ⟩$ . The number operator returns the state unchanged: eigenvalue 1.

Therefore $⟨ Φ∣ a_{p}^{†} a_{p} ∣Φ ⟩ = ⟨ Φ∣Φ ⟩ = 1$ .

Articulate: State the one move that did all the work — what do you do, every single time, the instant you see $a_{p} a_{p}^{†}$ next to a ket?

Practice

P.1 the empty orbital

Evaluate

a_{p}^{†} a_{p} ∣ ⟩

(the number operator on the vacuum). What is the eigenvalue, and why does it differ from the worked example?

Find the analogue: Same operator as the worked example, but on the empty state. Which factor now annihilates the ket immediately?

show answer

a_{p}^{†} a_{p} ∣ ⟩ = a_{p}^{†} (a_{p} ∣ ⟩) = a_{p}^{†} \cdot 0 = 0

. Eigenvalue 0 — orbital p is empty. Occupation 1 vs 0 is the whole content of the number operator.

P.2 Pauli, algebraically

Using

{a_{p}^{†}, a_{q}^{†}} = 0

, show

(a_{p}^{†})^{2} = 0

, and interpret

a_{p}^{†} a_{q}^{†} = - a_{q}^{†} a_{p}^{†}

in words.

Find the analogue: No vacuum push needed — read the anticommutator at p = q, then at p ≠ q.

show answer

Set p = q:

a_{p}^{†} a_{p}^{†} + a_{p}^{†} a_{p}^{†} = 0 \Rightarrow 2 (a_{p}^{†})^{2} = 0

, so creating twice in one spin-orbital gives zero — Pauli. For p ≠ q, swapping two creations flips the sign: the antisymmetry of the determinant, now an operator identity.

P.3 a one-body matrix element

For

∣Φ ⟩ = a_{p}^{†} ∣ ⟩

evaluate

⟨ Φ∣ a_{q}^{†} a_{r} ∣Φ ⟩

. For which (q, r) is it nonzero, and what is the value?

Find the analogue: Push $a_{r}$ right to meet $a_{p}^{†} ∣ ⟩$ using ${a_{r}, a_{p}^{†}} = δ_{r p}$ , just like the worked example.

show answer

a_{r} a_{p}^{†} ∣ ⟩ = (δ_{r p} - a_{p}^{†} a_{r}) ∣ ⟩ = δ_{r p} ∣ ⟩

, so the element is

δ_{r p} ⟨ Φ∣ a_{q}^{†} ∣ ⟩ = δ_{r p} δ_{qp}

: nonzero only when q = r = p, value 1. The operator

a_{q}^{†} a_{r}

must annihilate what is there (p) and recreate it (p) — exactly the diagonal one-electron term.

P.4 ordering and signs

Write the two-orbital state

a_{q}^{†} a_{p}^{†} ∣ ⟩

in the standard order

a_{p}^{†} a_{q}^{†} ∣ ⟩

(with p < q). What sign appears?

Find the analogue: Use the p ≠ q part of P.2 — one transposition.

show answer

a_{q}^{†} a_{p}^{†} ∣ ⟩ = - a_{p}^{†} a_{q}^{†} ∣ ⟩

. The order you create electrons in is a convention; choosing a canonical order (ascending) and tracking the sign on every reordering is precisely the determinant's permutation parity, now mechanical.

P.5 normal order

An operator string is normal-ordered when all creation operators stand left of all annihilation operators. Rewrite

a_{p} a_{q}^{†}

in normal order, naming the leftover.

Find the analogue: This is the single move from the worked example, isolated and named.

show answer

a_{p} a_{q}^{†} = δ_{pq} - a_{q}^{†} a_{p}

. The normal-ordered part is

- a_{q}^{†} a_{p}

; the leftover

δ_{pq}

is the contraction. Wick's theorem generalizes exactly this: a string equals the sum over all ways of contracting pairs, with the fully-contracted terms giving the vacuum expectation value.

P.6 a contraction count

By Wick's theorem,

⟨ ∣ a_{a} a_{b} a_{c}^{†} a_{d}^{†} ∣ ⟩

equals the sum of full pairwise contractions. Write it in terms of Kronecker deltas (mind the sign on the crossed contraction).

Find the analogue: Each contraction is a ${a, a^{†}} = δ$ from P.5; a crossing of contraction lines costs a minus sign (P.4's transposition).

show answer

δ_{a d} δ_{b c} - δ_{a c} δ_{b d}

. The first pairs the nested lines (a–d, b–c), the second the crossed lines (a–c, b–d) with its parity sign. This antisymmetrized pair is exactly the

(J - K)

structure of a two-electron matrix element — Slater-Condon's two-difference rule, generated by the algebra.

Check problems

Check 1 derivation

Derive the one-difference Slater-Condon rule in this language. With

\hat{H}_{1} = \sum_{pq} h_{pq} a_{p}^{†} a_{q}

and two determinants differing in one spin-orbital (

m \to n

, common orbitals collected in a set K), show

⟨ Φ∣ \hat{H}_{1} ∣ Φ^{'} ⟩ = h_{mn}

show solution sketch

Write

∣Φ ⟩ = a_{m}^{†} \prod_{k \in K} a_{k}^{†} ∣ ⟩

and

∣ Φ^{'} ⟩ = a_{n}^{†} \prod_{k \in K} a_{k}^{†} ∣ ⟩

. The term

a_{p}^{†} a_{q}

must annihilate an orbital of

Φ^{'}

and create one of

Φ

so the two states coincide: it must remove n and add m, i.e.

q = n, p = m

, contributing

h_{mn}

. Any term acting only on the shared set K leaves the

m \neq = n

mismatch unpaired, and the bra-ket vanishes by orthogonality. Result:

h_{mn}

— the rule, with no determinant expansion and no memorized sign table.

Check 2 articulation

In 100–300 words, explain why second quantization is not just notation — what does it make possible that hand-tracking determinants does not? Use coupled cluster as your example.

show solution sketch

First quantization forces you to carry explicit N-electron determinants and their permutation signs through every manipulation; for anything past a few excitations the bookkeeping is unmanageable, and the antisymmetry is a constraint you must keep imposing. Second quantization absorbs the antisymmetry, the spin-orbital labels, and the particle number into the operator algebra, so a state is just a string of creation operators and a matrix element is a vacuum expectation value evaluated by contractions. The payoff is compositionality: the coupled-cluster ansatz

e^{\hat{T}}

is a sensible object only because

\hat{T}

is an operator that can be exponentiated and its powers organized by Wick's theorem into connected and disconnected pieces — the very distinction that makes CC size-extensive. You cannot write, let alone differentiate, an exponential of "expand-the-determinant-and-track-signs." The algebra turns a combinatorial nightmare into a diagram you can manipulate — which is why every post-HF method beyond toy size is derived this way.

The contraction structure you just practiced is the skeleton of every CI and coupled-cluster equation. Next you put it to work: catch CISD failing size consistency, the defect coupled cluster's exponential was built to cure.