A first look at second quantization

The last section ended with a hopeless tally: the antisymmetric wavefunction of 26 electrons, expanded in product states, has $26!\approx4\times10^{26}$ terms. Never mind solving it — merely writing it down exceeds the storage capacity of the universe.

But before despairing, ask how much information those terms actually carry.

The bookkeeping is exhausting because the ledger records what it shouldn’t

Take 3 fermions occupying orbitals $\phi_a,\phi_b,\phi_c$; the Slater determinant expands into 6 terms. How are these 6 terms related? They are 6 ways of saying the same sentence. Each one says “orbitals $a,b,c$ each house one particle”, merely permuting “which particle lives in which room” — and the very first lesson of section 8.1 was that particles have no labels, so “which particle” is not a legitimate question at all.

The $N!$ explosion is the translation tax we pay for insisting on describing a label-free world in a labelled language. The actual information content of the whole state is one sentence:

Orbital $a$: 1 particle; orbital $b$: 1 particle; orbital $c$: 1 particle; all others: 0.

So let the notation record only that sentence. Fix a set of single-particle orbitals $\{\phi_1,\phi_2,\dots\}$ and define the occupation-number representation:

$$\ket{n_1,n_2,n_3,\dots}\tag{8.5.1}$$

where $n_i$ is the number of particles in orbital $\phi_i$. The 6-term determinant above becomes $\ket{1,1,1,0,0,\dots}$; a ten-particle state of $10!=3\,628\,800$ terms is just $\ket{1,1,\dots,1,0,\dots}$. Every detail of symmetrization and antisymmetrization is built into the definition of the symbol (by convention it stands for that normalised determinant or symmetric sum) and never has to be written by hand again.

The space these states span is called Fock space — it merges the state spaces of “0 particles” (the vacuum $\ket{0}$), “1 particle”, “2 particles”, and so on into one. For the first time, particle number is a quantum number rather than a premise of the theory.

Moving particles between orbitals: creation and annihilation operators

New notation needs matching operators. Define the creation operator $\hat a_i^\dagger$: add one particle to orbital $\phi_i$; and the annihilation operator $\hat a_i$: remove one particle from orbital $\phi_i$ (returning zero if there is none to remove).

This algebra should feel familiar — in the harmonic oscillator of section 2.9, $\hat a^\dagger,\hat a$ climb up and down the ladder of levels. What read as “up one level” there reads here as “one more quantum in this mode”. The oscillator’s ladder is exactly the occupation number of a single mode; all we do now is give every orbital a ladder of its own.

For bosons, copy the oscillator’s commutation relations, one set per orbital, different orbitals independent:

$$[\hat a_i,\hat a_j^\dagger]=\delta_{ij},\qquad [\hat a_i,\hat a_j]=[\hat a_i^\dagger,\hat a_j^\dagger]=0\tag{8.5.2}$$

The matrix elements copy over too: $\hat a_i^\dagger\ket{\dots,n_i,\dots}=\sqrt{n_i+1}\,\ket{\dots,n_i+1,\dots}$, with $n_i$ unbounded — bosons love to bunch, and the $\sqrt{n+1}$ factor even rewards bunching (it is the root of stimulated emission beating spontaneous emission).

For fermions, the antisymmetry must be written into the algebra. The correct move is to replace commutators by anticommutators (writing $\{\hat A,\hat B\}=\hat A\hat B+\hat B\hat A$):

$$\{\hat c_i,\hat c_j^\dagger\}=\delta_{ij},\qquad \{\hat c_i,\hat c_j\}=\{\hat c_i^\dagger,\hat c_j^\dagger\}=0\tag{8.5.3}$$

Why anticommutators? Look at $i\ne j$: $\hat c_i^\dagger\hat c_j^\dagger=-\hat c_j^\dagger\hat c_i^\dagger$ — filling $j$ then $i$ differs from filling $i$ then $j$ by a minus sign. That is exactly the sign flip of exchanging two particles, now promoted to a grammatical rule of the operators: you could not violate it if you tried.

Pauli exclusion in two lines of algebrabasic~6 minexpand

Step 1: put two fermions in the same orbital. Set $i=j$ in the anticommutation relation:

$$\{\hat c_i^\dagger,\hat c_i^\dagger\}=2(\hat c_i^\dagger)^2=0 \quad\Longrightarrow\quad(\hat c_i^\dagger)^2=0\tag{8.5.4}$$

Creating two particles in the same orbital gives the zero vector — not some high-energy state, but no state at all. What section 8.3 said with determinants is one line of algebra here.

Step 2: occupation numbers can only be 0 or 1. Define the number operator $\hat n_i=\hat c_i^\dagger\hat c_i$ and square it:

$$\hat n_i^2=\hat c_i^\dagger\hat c_i\hat c_i^\dagger\hat c_i =\hat c_i^\dagger\bigl(1-\hat c_i^\dagger\hat c_i\bigr)\hat c_i =\hat c_i^\dagger\hat c_i-\hat c_i^\dagger\hat c_i^\dagger\hat c_i\hat c_i\tag{8.5.5}$$

The first step used $\hat c_i\hat c_i^\dagger=1-\hat c_i^\dagger\hat c_i$ (the $i=j$ component of the anticommutation relation). The last term contains $(\hat c_i^\dagger)^2=0$, so

$$\hat n_i^2=\hat n_i\quad\Longrightarrow\quad n_i(n_i-1)=0\quad\Longrightarrow\quad n_i\in\{0,1\}\tag{8.5.6}$$

The eigenvalue equation $n^2=n$ has only two roots. Pauli exclusion = the anticommutation relations of fermion operators, no more and no less. Compare bosons: $[\hat a_i,\hat a_i^\dagger]=1$ gives $n_i=0,1,2,\dots$ without bound.

Step 3 (a bookkeeping detail): the sign convention. Fermion operators act with a tracking sign:

$$\hat c_i^\dagger\ket{n_1,\dots,n_i,\dots}=(-1)^{\sum_{j<i}n_j}\,(1-n_i)\ket{n_1,\dots,n_i+1,\dots}\tag{8.5.7}$$

The factor $(-1)^{\sum_{j<i}n_j}$ counts the occupied orbitals listed before $i$ — it guarantees that repeated operations agree term by term with the determinant expansion. The good news: in practice one almost never writes it by hand; the anticommutation relations take care of everything automatically.

Translating the Hamiltonian

For the new language to earn its keep, it must express Hamiltonians. The rules are stated without proof (to verify: sandwich both sides between occupation-number states and compare matrix elements).

One-body terms (kinetic energy, external potentials — operators that “touch” one particle at a time):

$$\hat H_1=\sum_{ij}h_{ij}\,\hat a_i^\dagger\hat a_j, \qquad h_{ij}=\int\phi_i^*(\vec{r})\Bigl[-\frac{\hbar^2}{2m}\nabla^2+V(\vec{r})\Bigr]\phi_j(\vec{r})\,\dd^3r\tag{8.5.8}$$

Two-body terms (interparticle interactions):

$$\hat H_2=\frac12\sum_{ijkl}V_{ijkl}\,\hat a_i^\dagger\hat a_j^\dagger\hat a_l\hat a_k, \qquad V_{ijkl}=\int\phi_i^*(\vec{r})\phi_j^*(\vec{r}\,')\,U(\vec{r}-\vec{r}\,')\,\phi_k(\vec{r})\phi_l(\vec{r}\,')\,\dd^3r\,\dd^3r'\tag{8.5.9}$$

∎Key formulas

• Occupation-number representation

  $$\ket{n_1,n_2,\dots}\ \in\ \text{Fock space}$$

  Records only "how many particles per orbital"; (anti)symmetrization is built into the definition

• Boson algebra

  $$[\hat a_i,\hat a_j^\dagger]=\delta_{ij},\quad \hat a_i^\dagger\ket{n_i}=\sqrt{n_i+1}\,\ket{n_i+1}$$

  n unbounded; the √(n+1) factor is the root of stimulated emission

• Fermion algebra

  $$\{\hat c_i,\hat c_j^\dagger\}=\delta_{ij},\quad(\hat c_i^\dagger)^2=0$$

  Anticommutation = sign flip on exchange; square zero = Pauli exclusion

• Hamiltonian

  $$\hat H=\sum_{ij}h_{ij}\hat a_i^\dagger\hat a_j+\frac12\sum_{ijkl}V_{ijkl}\,\hat a_i^\dagger\hat a_j^\dagger\hat a_l\hat a_k$$

  One-body = hops; two-body = two-particle scattering

?Self-check4 questions

1. 1.

   Why can the occupation-number representation compress an N!-term antisymmetric wavefunction into a single symbol without losing information?

   • Because those N! terms are just permutations of "which label goes with which orbital", labels carry no physical meaning, and the real information is only the occupation of each orbital
   • Because it makes an approximation and drops the small terms
   • Because it applies only to non-interacting systems
   • Because the phase of the wavefunction does not matter

   Submit
2. 2.

   The direct algebraic reason fermion occupation numbers can only be 0 or 1 is:

   • The anticommutation relations give (ĉ†)² = 0, hence n̂² = n̂, whose eigenvalues satisfy n(n−1) = 0
   • Fermions are too massive for two to fit in one orbital
   • The normalisation factor √(n+1) diverges at n = 1
   • It is an extra imposed assumption unrelated to the operator algebra

   Submit
3. 3.

   10 fermions occupy 10 specified orbitals. Fully expanding this state's Slater determinant into product states gives how many terms? (In the occupation-number representation it is just the symbol |1,1,…,1,0,…⟩.)

   项0% relative tolerance

   Submit
4. 4.

   Which statements about second quantization are correct? (Select all that apply.)

   Select all that apply

   • It introduces no new physics; it is only a change of representation and notation
   • The operator â†ᵢâⱼ describes a particle hopping from orbital j to orbital i
   • It naturally describes processes where particle number changes, such as an atom emitting a photon
   • The boson-fermion difference appears as the difference between commutation and anticommutation relations

   Submit

What comes next

The language is ready; back to the unfinished practical question: how does one actually compute a many-electron atom? Helium already pushed perturbation theory and the variational method (chapter 7) to their limits; carbon has 6 electrons, iron 26.

The next section introduces an idea that has ruled computational physics for nearly a century: instead of making each electron face all the others, let it face one averaged “electron cloud” — then let the cloud and the orbitals feed each other, iterating to self-consistency. The Hartree–Fock method: the many-body problem’s first genuinely computable answer.