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LESSON NOTES · 06

6. Second quantization and determinant matrix elements

Position in the course: Lesson 6 of 10. Complete the preceding derivation and use the explained exercises to check understanding.

Prerequisites: Slater determinants and operator algebra

1. Change bookkeeping, not particle physics

Second quantization describes occupation changes of a chosen orthonormal spin-orbital basis. The number of electrons need not change in the physical problem. Label basis orbitals \(p,q,r,s\) and define an ordered occupation state. A creation operator \(a_p^\dagger\) inserts orbital \(p\) with the fermionic sign needed by that ordering; an annihilation operator \(a_p\) removes it. The anticommutation relations are

\[ \{a_p,a_q^\dagger\}=\delta_{pq},\qquad\{a_p,a_q\}=\{a_p^\dagger,a_q^\dagger\}=0. \]

Consequently \((a_p^\dagger)^2=0\): one spin orbital cannot be occupied twice. The number operator \(n_p=a_p^\dagger a_p\) has eigenvalues zero or one, and \(N=\sum_p n_p\). Opposite-spin spatial partners are distinct spin orbitals, so both may be occupied.

2. Write the electronic Hamiltonian consistently

Use physicists' spin-orbital integrals \(\langle pq|rs\rangle=\iint\varphi_p^*(1)\varphi_q^*(2)r_{12}^{-1}\varphi_r(1)\varphi_s(2)d1d2\) and \(\langle pq\Vert rs\rangle=\langle pq|rs\rangle-\langle pq|sr\rangle\). Then

\[ H=\sum_{pq}h_{pq}a_p^\dagger a_q+ \frac14\sum_{pqrs}\langle pq\Vert rs\rangle a_p^\dagger a_q^\dagger a_s a_r+V_{NN}. \]

The factor one-quarter accompanies antisymmetrized integrals and unrestricted index sums. With ordinary two-electron integrals it becomes one-half. Mixing conventions is a common factor-of-two error. Nuclear repulsion is a scalar at a fixed geometry and adds to every electronic diagonal energy.

3. Derive why only singles and doubles couple directly

A one-body term annihilates one occupied orbital and creates one replacement, so two determinants differing by more than one occupation cannot be connected by it. A two-body term changes at most two occupations. Therefore Hamiltonian matrix elements between determinants differing by three or more substitutions vanish. This is an algebraic consequence of the operator rank, not an assumption that triple excitations have no physical effect: they can couple through intermediate determinants.

For the diagonal determinant element the surviving contractions give \(\sum_i h_{ii}+\tfrac12\sum_{ij}\langle ij\Vert ij\rangle+V_{NN}\). For a single substitution \(i\to a\) the matrix element is the corresponding Fock element, up to a consistent determinant sign convention. A stationary canonical HF reference makes that element zero, giving Brillouin's theorem.

4. Worked fermionic sign

Choose \(|12\rangle=a_1^\dagger a_2^\dagger|vac\rangle\). Remove the second orbital:

\[ a_2|12\rangle=a_2a_1^\dagger a_2^\dagger|vac\rangle =-a_1^\dagger(1-a_2^\dagger a_2)|vac\rangle=-|1\rangle. \]

Removing the first instead gives \(|2\rangle\). These signs are essential in CI matrix assembly. A different declared occupation ordering changes intermediate signs consistently but leaves eigenvalues and expectation values unchanged.

5. Exercises with explained answers

Show \(n_p^2=n_p\). Use \(a_pa_p^\dagger=1-a_p^\dagger a_p\): \(n_p^2=a_p^\dagger(1-a_p^\dagger a_p)a_p=n_p\), because squared creation and annihilation operators vanish.

Can a two-electron Hamiltonian directly connect a reference to a triple substitution? No; its two annihilations cannot remove three reference occupations. Higher-order correlation can still generate triple amplitudes through repeated coupling.

6. Use and limitations

The notation compresses many-electron algebra and makes method truncations precise. It does not choose an accurate orbital basis or solve the Hamiltonian by itself. Nonorthogonal orbital bases need a different algebraic treatment. The following MP2, CI and coupled-cluster lessons use this orthonormal convention throughout; verify whether software documentation instead prints spatial-orbital formulas before translating factors and index ranges.

Further conceptual check

An occupation basis conserves total particle number when each Hamiltonian term contains the same number of creations and annihilations. In the present Hamiltonian every term has that property, so \([H,N]=0\). This explains why a fixed-electron-number determinant sector can be solved independently even though creation operators are used in its notation.

7. References and study connections

The derivations and toy arithmetic are original teaching constructions. No molecular simulation is reported here.

Quantum mechanics · Density functional theory · Quantum Monte Carlo · Gaussian


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