Occupation-Number Representation and Fock Space

First quantization asks which labelled particle sits in which state, then forces you to antisymmetrize the answer by hand — Slater determinants for fermions, permanents for bosons — and to rebuild the whole wavefunction the moment a particle is created or destroyed. For a handful of electrons this is bookkeeping; for $10^{23}$ of them, or for a photon field where particle number is not even conserved, it is hopeless. The occupation-number representation throws the labels away. Since identical particles carry none, the only physical data are the tallies: how many particles $n_k$ sit in each single-particle mode $k$ . A state is written $|n_1,n_2,n_3,\dots\rangle$ , and that is the whole story. The arena for these tally-states is Fock space, the orthogonal direct sum $\mathcal F=\bigoplus_{N=0}^\infty\mathcal H^{(N)}$ of every particle-number sector at once — the vacuum $|0\rangle$ , the one-particle space, the symmetrized two-particle space, and so on up. A single space now holds states of any particle number, which is exactly what you need for creation and annihilation. This article owns the space, the occupation-number basis, and the tool that reads a particle count off any state: the number operator $\hat N=\sum_k a^\dagger_k a_k$ . The ladder operators $a^\dagger_k,a_k$ that move between sectors are introduced here as the moves, but their full algebra — the commutators and anticommutators — is handed to Creation and Annihilation Operators, its rightful owner, and the rewriting of Hamiltonians into $a^\dagger a$ form to Second Quantization. The owned payoff is a counting theorem you can check by hand. For $M$ single-particle modes, the fermionic Fock space has dimension exactly $2^M$ — each mode is either empty or full — and this splits sector by sector as $2^M=\sum_{N=0}^{M}\binom{M}{N}$ , one binomial coefficient per particle-number sector. Every fully-specified fermion tally $|1,1,\dots\rangle$ over $N$ occupied modes is a Slater determinant, so the count $\binom{M}{N}$ is simultaneously the number of $N$ -electron determinants you can build from $M$ orbitals. The whole determinant-and-permanent apparatus of the previous articles reappears here as the geometry of one unified space.

Key takeaways

The occupation-number representation drops particle labels: a many-body state of identical particles is specified entirely by the tallies $\{n_k\}$ — how many particles occupy each single-particle mode — written $|n_1,n_2,\dots\rangle$ .
The arena is Fock space $\mathcal F=\bigoplus_{N=0}^\infty\mathcal H^{(N)}$ , the orthogonal direct sum of all particle-number sectors, including the vacuum $|0\rangle$ — one space that holds any number of particles.
Creation/annihilation operators $a^\dagger_k,a_k$ move between adjacent sectors (add/remove a particle); their commutator/anticommutator algebra is owned by Creation and Annihilation Operators, stated here as the moves.
The number operator $\hat n_k=a^\dagger_k a_k$ counts particles in mode $k$ ; the total $\hat N=\sum_k a^\dagger_k a_k$ has eigenvalue $\sum_k n_k$ on $|n_1,n_2,\dots\rangle$ .
Occupation splits by statistics: bosons allow $n_k\in\{0,1,2,\dots\}$ (unbounded, each mode a harmonic oscillator); fermions allow $n_k\in\{0,1\}$ — Pauli exclusion, the operator form $(a^\dagger_k)^2=0$ owned by Pauli Exclusion Principle.
The owned counting theorem: the fermion Fock space over $M$ modes has dimension $2^M=\sum_{N=0}^M\binom{M}{N}$ ; each sector's $\binom{M}{N}$ is also the number of $N$ -electron Slater determinants over $M$ orbitals.
Key idea: forget the labels and keep the tally — a many-body state lives in Fock space $\bigoplus_N\mathcal H^{(N)}$ as $|n_1,n_2,\dots\rangle$ , the number operator reads off $\sum_k n_k$ , and the fermion space collapses to $2^M=\sum_N\binom{M}{N}$ because every occupied tally is one Slater determinant.

The core insight

Discard particle labels entirely: a many-body state of identical particles is the tally $|n_1,n_2,\dots\rangle$ of how many sit in each mode, living in Fock space $\mathcal F=\bigoplus_{N=0}^\infty\mathcal H^{(N)}$ — the direct sum of every particle-number sector. The number operator $\hat N=\sum_k a^\dagger_k a_k$ reads off the total; bosons stack ( $n_k$ unbounded), fermions cap ( $n_k\in\{0,1\}$ ), and the fermion space over $M$ modes has dimension $2^M=\sum_N\binom{M}{N}$ .

Where this lives today

Fock space is the working language of nearly all of many-body and field theory. In condensed-matter physics, model Hamiltonians — the Hubbard model, the BCS pairing Hamiltonian, spin chains via the Jordan-Wigner transformation — are written and solved entirely in the occupation-number representation, where a lattice of sites becomes a set of modes and the ground state is a specific pattern of tallies. In quantum optics and cold-atom physics, the photon number states $|n\rangle$ and the atom-number states of a Bose-Einstein condensate are literal occupation-number states, and experiments now prepare and measure them one quantum at a time. In relativistic quantum field theory, the Fock vacuum is the starting point for perturbation theory, and particle creation — Hawking radiation, the Unruh effect, pair production in strong fields — is the statement that a Hamiltonian fails to conserve $\hat N$ , so the state climbs the tower. And in quantum computing, the Jordan-Wigner and Bravyi-Kitaev mappings encode fermionic Fock states onto qubits, so that simulating molecular electronic structure on a quantum processor is, underneath, arithmetic in exactly the $2^M$ -dimensional space this article counts.

The physical picture

Picture the single-particle modes — orbitals, momentum states, lattice sites — as a row of numbered boxes. A Fock state is nothing more than a tally sheet: how many balls sit in each box. Because the balls are identical and unlabelled, the tally is the complete description; asking "which ball is in box 3" is meaningless, and no information is lost by refusing to answer. For bosons a box holds any number of balls; for fermions each box holds at most one. The vacuum $|0\rangle$ is the tally sheet with every entry zero — not "nothing," but a genuine, normalized state that is the foundation of the whole construction. What makes this more than notation is that it unifies particle numbers. In first quantization, a two-electron wavefunction and a three-electron wavefunction live in different Hilbert spaces, and you cannot even write an operator that turns one into the other. Fock space stacks them all — vacuum, one-particle, two-particle, $\dots$ — into one direct sum, so a single operator can carry a state up or down the tower. That is precisely what you need to describe emission and absorption of photons, the creation of electron-hole pairs, or any process where particle number is not conserved. The ladder operators $a^\dagger_k$ and $a_k$ are the "add a ball / remove a ball" moves between adjacent rungs.

Figure

The Fock-space tower: one space stacks every particle-number sector, and the fermion count collapses to 2^M

The figure is the whole article in one image. The fermionic Fock space over three modes is eight-dimensional, and that eight splits, rung by rung, into $1+3+3+1$ — the binomial coefficients $\binom{3}{N}$ . Read it as a headcount of Slater determinants: there is one way to leave all three modes empty, three ways to fill exactly one, three ways to fill exactly two, one way to fill all three. This is the counting theorem this article owns, and it is exact.

Occupation-number states and the vacuum

A basis state of Fock space is a list of occupation numbers, one per mode, with the total particle number read off by summing them.

|n_1,n_2,n_3,\dots\rangle,\qquad N=\sum_k n_k,\qquad \mathcal F=\bigoplus_{N=0}^\infty\mathcal H^{(N)},\quad \mathcal H^{(0)}=\mathrm{span}\{|0\rangle\} \qquad (1)

where: $n_k$ — occupation of single-particle mode $k$ ; $N$ — total particle number; $\mathcal H^{(N)}$ — the (anti)symmetrized $N$ -particle subspace; $|0\rangle$ — the vacuum, the unique normalized state with every $n_k=0$ . The physical read: a state is a tally, the tally lives in a definite particle-number sector, and Fock space is the direct sum of all such sectors. Different sectors are orthogonal — a two-particle state has zero overlap with a three-particle state — so the tower does not mix particle numbers unless an operator deliberately moves between rungs. The vacuum deserves emphasis because it is easy to misread as "the absence of a state." It is a full-fledged, normalized vector $\langle 0|0\rangle=1$ , annihilated by every lowering operator, $a_k|0\rangle=0$ for all $k$ (there is nothing to remove), and it carries $\hat N|0\rangle=0$ . Every other Fock state is built by acting with creation operators on it, so the vacuum is the anchor of the entire construction — the empty tally sheet from which all filled ones are generated.

The number operator

The tool this article owns for reading a particle count off any state is the number operator. For each mode it is the product $\hat n_k=a^\dagger_k a_k$ , and the total counts all particles at once.

\hat n_k=a^\dagger_k a_k,\qquad \hat N=\sum_k a^\dagger_k a_k,\qquad \hat N|n_1,n_2,\dots\rangle=\Big(\sum_k n_k\Big)\,|n_1,n_2,\dots\rangle \qquad (2)

where: $\hat n_k$ — counts particles in mode $k$ (eigenvalue $n_k$ ); $\hat N$ — total number operator; $a^\dagger_k,a_k$ — creation and annihilation operators for mode $k$ . The physical read: occupation-number states are the simultaneous eigenstates of all the $\hat n_k$ , which is why the tally is a good quantum number. The eigenvalue of $\hat N$ is just the total occupation — a number you can literally count off the tally sheet. That $\hat N$ commutes with a particle-conserving Hamiltonian is what makes particle number a conserved quantity; when it does not commute (a photon-emitting Hamiltonian, say), particle number changes, and only Fock space can hold the before-and-after states. The ladder operators $a^\dagger_k,a_k$ that build $\hat n_k$ are introduced here as the moves between sectors; their defining algebra — bosonic commutators $[a_k,a^\dagger_l]=\delta_{kl}$ , fermionic anticommutators $\{a_k,a^\dagger_l\}=\delta_{kl}$ , and the resulting ladder normalizations — is owned by Creation and Annihilation Operators. What we use here is only that $\hat n_k=a^\dagger_k a_k$ is diagonal in the occupation basis with eigenvalue $n_k$ , which is enough to fix every counting statement below.

Derivation AThe occupation states are the eigenstates of the number operator

Establish that occupation-number states $|n_1,n_2,\dots\rangle$ are the simultaneous eigenstates of every mode's number operator $\hat n_k=a^\dagger_k a_k$ , with eigenvalue $n_k$ , so that the total $\hat N=\sum_k\hat n_k$ has eigenvalue $\sum_k n_k$ . The goal is to show that the tally sheet is not just suggestive notation but a genuine list of eigenvalues — that "occupation number $n_k$ " means literally " $\hat n_k$ eigenvalue $n_k$ " — which is what makes particle number a good quantum number and lets $\hat N$ read a count off any state by inspection.

Step 1. Single-mode eigenvalue. For each mode $k$ , the operator $\hat n_k=a^\dagger_k a_k$ is Hermitian and, by the ladder algebra owned by Creation and Annihilation Operators, satisfies $[\hat n_k,a^\dagger_k]=+a^\dagger_k$ and $[\hat n_k,a_k]=-a_k$ . Its eigenstates are the single-mode number states $|n_k\rangle$ with $\hat n_k|n_k\rangle=n_k|n_k\rangle$ — for bosons $n_k=0,1,2,\dots$ , for fermions $n_k=0,1$ . This is the harmonic-oscillator spectrum (bosons) or the two-level spectrum (fermions).

Step 2. Modes are independent. Operators for distinct modes commute (bosons) or anticommute in pairs that leave $\hat n_k$ commuting (fermions): $[\hat n_k,\hat n_l]=0$ for all $k,l$ . Hence the $\hat n_k$ form a mutually commuting set, and their simultaneous eigenstates are the tensor-product occupation states $|n_1,n_2,\dots\rangle=\bigotimes_k|n_k\rangle$ , with $\hat n_k|n_1,n_2,\dots\rangle=n_k|n_1,n_2,\dots\rangle$ .

Step 3. Total number operator. Summing, $\hat N=\sum_k\hat n_k$ acts as $\hat N|n_1,n_2,\dots\rangle=\big(\sum_k n_k\big)|n_1,n_2,\dots\rangle$ . The eigenvalue is the total occupation — the sum of the tally. Because the occupation state is an eigenstate of $\hat N$ , it lies in a definite particle-number sector $\mathcal H^{(N)}$ with $N=\sum_k n_k$ , confirming that the direct-sum structure of Eq. (1) is diagonalized by exactly this basis. Machine check: building $\hat N$ as a diagonal matrix over the $M=3$ fermion Fock basis gives eigenvalues $\{0,1,1,1,2,2,2,3\}$ — one per basis state, each equal to its total occupation; $\hat N|1,0,1\rangle=2|1,0,1\rangle$ and $\hat N|0,0,0\rangle=0$ explicitly. Verified: `verify_aq-qm-occupation-number-fock-space.py` — $\hat N$ eigenvalues equal the total occupation for all eight $M=3$ states, including the vacuum ( $0$ ) and the fully-filled state ( $3$ ).

Go deeper · Derivation B · 2 minThe fermion Fock-space dimension and its binomial split

Derivation BThe fermion Fock-space dimension and its binomial split

Establish the owned counting theorem: for $M$ single-particle modes, the fermionic Fock space has dimension exactly $2^M$ , and this decomposes sector by sector as $2^M=\sum_{N=0}^M\binom{M}{N}$ , with $\binom{M}{N}$ the number of Slater determinants of $N$ electrons in $M$ orbitals. The goal is to turn an abstract direct sum into an honest headcount you can verify by listing states, and to show that the whole Slater-determinant apparatus of the previous article is nothing but the sector-by-sector geometry of one space. We prove the two claims — total dimension and its split — and confirm both numerically.

Step 1. Each fermion mode is a two-state system. By Pauli exclusion (owned by Pauli Exclusion Principle, whose operator form is $(a^\dagger_k)^2=0$ ), a fermionic mode has occupation $n_k\in\{0,1\}$ — it is either empty or full. There are exactly two choices per mode, and the modes are independent, so a fermion Fock state over $M$ modes is a binary string of length $M$ .

Step 2. Total dimension. The number of length- $M$ binary strings is $2^M$ . Hence the fermionic Fock space over $M$ modes has dimension $\dim\mathcal F_{\text{ferm}}=2^M$ — one basis state per subset of occupied modes.

Step 3. Sector split. Grouping the strings by their particle number $N=\sum_k n_k$ , the number with exactly $N$ ones is the number of ways to choose which $N$ of the $M$ modes are occupied, namely $\binom{M}{N}$ . Summing over all sectors reproduces the total by the binomial theorem, $\sum_{N=0}^M\binom{M}{N}=(1+1)^M=2^M$ . So the fermion Fock space splits as $2^M=\sum_{N=0}^M\binom{M}{N}$ .

Step 4. Each occupied tally is a Slater determinant. A fermion Fock state with $N$ occupied modes, $|\dots1\dots1\dots\rangle$ , is the totally antisymmetric $N$ -particle state built from those $N$ orbitals — precisely a Slater determinant (owned by Slater Determinants). The number of such determinants from $M$ orbitals is the number of ways to choose $N$ occupied modes, $\binom{M}{N}$ — identical to the sector dimension. The Fock-space bookkeeping and the determinant count are the same count. Numerical confirmation: for $M=3$ the dimension is $2^3=8=1+3+3+1$ ; for $M=4$ it is $16=1+4+6+4+1$ ; for $M=5$ it is $32=1+5+10+10+5+1$ — all verified, and each sector's $\binom{M}{N}$ matched against the number of occupied-mode choices $\binom{M}{N}$ . Verified: `verify_aq-qm-occupation-number-fock-space.py` — $2^M=\sum_N\binom{M}{N}$ for $M=3,4,5,6$ ; sector splits $[1,3,3,1]$ etc.; and the identity $\#\{\text{occupied-mode choices}\}=\binom{M}{N}$ for $(M,N)=(4,2),(5,3),(6,2)$ .

Boson versus fermion occupation

The single framework of tallies specializes by the statistics, and the split is exactly the occupancy split of Bosons and Fermions, now read at the level of the whole space.

\text{fermions: } n_k\in\{0,1\}\ \ ((a^\dagger_k)^2=0);\qquad \text{bosons: } n_k\in\{0,1,2,\dots\}\ \ (\text{each mode a harmonic oscillator}) \qquad (3)

where: the fermion cap comes from antisymmetry (two fermions cannot share a mode), the boson freedom from symmetry (any number may). The physical read: a bosonic mode behaves exactly like a harmonic oscillator — its states are the ladder $|0\rangle_k,|1\rangle_k,|2\rangle_k,\dots$ and its Fock space is a tensor product of one oscillator per mode. A fermionic mode is a two-level switch. The dimension counting therefore differs dramatically: with $M$ modes and $N$ particles, the fermion sector has dimension $\binom{M}{N}$ (choose which modes are filled), while the boson sector has dimension $\binom{N+M-1}{M-1}$ (stars and bars — distribute $N$ identical balls among $M$ boxes with no cap). For $M=3,N=2$ these are $3$ and $6$ ; the boson space is larger precisely because it permits double occupancy.

Figure

Two occupation ladders: the fermion switch (0 or 1) versus the boson oscillator (0, 1, 2, …)

Limiting cases

Case	Result	Meaning
Single mode, fermion	dimension 2, states $\{\|0\rangle,\|1\rangle\}$	a two-level switch; $(a^\dagger)^2=0$ caps it (owned by Pauli Exclusion Principle)
Single mode, boson	infinite-dimensional oscillator ladder	$\|0\rangle,\|1\rangle,\|2\rangle,\dots$ ; a bosonic mode is a harmonic oscillator
$M$ modes, fermion	$\dim=2^M=\sum_N\binom{M}{N}$	one determinant per occupied tally; the owned counting theorem
Fixed sector $N$	$\mathcal H^{(N)}$ of dim $\binom{M}{N}$ (ferm) / $\binom{N+M-1}{M-1}$ (bos)	project Fock space onto one particle number — recovers first quantization
Vacuum $\|0\rangle$	annihilated by all $a_k$ , $\hat N\|0\rangle=0$	a normalized state, the anchor; not "nothing"
Particle-number conserved	$[\hat N,\hat H]=0$	dynamics stays within one sector; the tower is a formality
Number not conserved	$[\hat N,\hat H]\neq0$	emission/absorption; the full tower is essential (photons, QFT)

Common trapThe vacuum is a state, not nothing — and the fermion sign lives in the operators, not the tally

Two traps sit here. First, $|0\rangle$ is a genuine normalized vector with $\langle0|0\rangle=1$ , annihilated by every $a_k$ and carrying zero particles; it is the foundation of the construction, not an absence to be ignored — writing $a_k|0\rangle=0$ is a physical statement (nothing to remove), not $|0\rangle=0$ . Second, the occupation-number tally $|n_1,n_2,\dots\rangle$ records magnitudes only; for fermions the crucial minus signs live in the ordering of the creation operators, since $a^\dagger_k a^\dagger_l=-a^\dagger_l a^\dagger_k$ , so a fixed mode ordering must be chosen and an annihilation operator acting deep in the list picks up a Jordan-Wigner sign $(-1)^{\sum_{j<k}n_j}$ . The full algebra of these signs is owned by Creation and Annihilation Operators; the trap is to read the bare tally and forget that the sign convention is part of the state's definition.

Worked examples

Example 1 — the three-mode fermion Fock space. Enumerate the fermionic Fock space over $M=3$ modes, verify its dimension and sector split, and read the number-operator eigenvalue off a specific state.

Step 1. Enumerate. With three modes and $n_k\in\{0,1\}$ , the basis states are the eight binary strings $|000\rangle,|001\rangle,|010\rangle,|100\rangle,|011\rangle,|101\rangle,|110\rangle,|111\rangle$ . The dimension is $2^3=\mathbf{8}$ .

Step 2. Split by particle number. Grouping by $N=\sum_k n_k$ : one state with $N=0$ (the vacuum), three with $N=1$ , three with $N=2$ , one with $N=3$ . This is $1+3+3+1=\mathbf{8}$ , matching $\binom{3}{0}+\binom{3}{1}+\binom{3}{2}+\binom{3}{3}$ . Each of the three $N=2$ states is a two-electron Slater determinant over the chosen pair of orbitals.

Step 3. Apply the number operator. Take the specific state $|1,0,1\rangle$ : modes 1 and 3 are occupied, mode 2 empty. Then $\hat N|1,0,1\rangle=(1+0+1)|1,0,1\rangle=\mathbf{2}\,|1,0,1\rangle$ — an eigenstate with eigenvalue equal to the total occupation. The tally sheet is literally the list of $\hat n_k$ eigenvalues, and $\hat N$ just adds them up.

Example 2 — boson versus fermion sector sizes. Count the two-particle sector for $M=3$ modes in both statistics and read off the physical difference.

Step 1. Fermions. With $n_k\in\{0,1\}$ and $N=2$ , choose which two of the three modes are filled: $\binom{3}{2}=\mathbf{3}$ states ( $|110\rangle,|101\rangle,|011\rangle$ ). No mode is doubly occupied — that is the Pauli cap.

Step 2. Bosons. With $n_k$ unbounded and $N=2$ , distribute two identical particles among three modes (stars and bars): $\binom{2+3-1}{3-1}=\binom{4}{2}=\mathbf{6}$ states. The extra three are the double-occupancy tallies $|200\rangle,|020\rangle,|002\rangle$ , forbidden to fermions.

Step 3. The physical difference. The boson sector is twice as large because it permits double occupancy — the same symmetric-versus-antisymmetric split that gives lasers and condensates on the boson side and the exclusion principle on the fermion side. The Fock-space dimension is where the occupancy rule of Bosons and Fermions becomes arithmetic.

Origins · 1 minfock space and how this was discovered

fock space

introduced 1932 · second-quantization formalism

The occupation-number arena is named for Vladimir Fock, who in 1932 gave the systematic construction of the space of variable particle number — the orthogonal direct sum of all $N$ -particle sectors — that underlies quantized field theory. The idea grew out of the 1927–28 work of Dirac, who quantized the electromagnetic field by treating each mode as a harmonic oscillator, and of Jordan, Klein, and Wigner, who introduced the creation and annihilation operators (with anticommutators for fermions) that move between sectors. Fock's contribution was to package these into a single coherent state space with a distinguished vacuum, turning the ad-hoc oscillator quantization of the radiation field into a general framework for many-body and field-theoretic problems. Fock space is now the standard setting for everything from condensed-matter many-body theory to relativistic quantum field theory; its defining feature — that particle number is a dynamical quantity rather than a fixed label — is exactly what first quantization cannot express, and is why the formalism became indispensable once physics needed to create and destroy particles.

History

1927

— Dirac quantizes the electromagnetic field, treating each mode as a harmonic oscillator (the seed of the occupation picture)

1928

— Jordan and Wigner introduce fermionic creation/annihilation operators with anticommutators, encoding exclusion

1928

— Jordan and Klein formulate the bosonic many-body operators

1932

— Fock constructs the space of variable particle number, the direct sum of all N-particle sectors, with a distinguished vacuum

1934

— Pauli and Weisskopf quantize the scalar field, cementing second quantization as the language of field theory

1957

— Bardeen, Cooper, and Schrieffer build the superconducting ground state directly in Fock space, a landmark many-body application

Going further

References. Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., Ch. 7 · Fetter & Walecka, Quantum Theory of Many-Particle Systems, Ch. 1 · Negele & Orland, Quantum Many-Particle Systems, Ch. 1 · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. II, Ch. XV · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 7.

Provenance. The Fock-space construction, occupation-number basis, and number operator are standard across Sakurai Ch. 7, Fetter & Walecka Ch. 1, and Negele & Orland Ch. 1. A verifier should: (1) confirm the fermion Fock dimension $2^M=\sum_N\binom{M}{N}$ by direct enumeration; (2) check the number-operator eigenvalue $\hat N|n_1,\dots\rangle=(\sum_k n_k)|\dots\rangle$ is the total occupation; (3) verify the boson stars-and-bars sector dimension $\binom{N+M-1}{M-1}$ against the fermion $\binom{M}{N}$ ; (4) confirm that each occupied fermion tally corresponds to exactly one Slater determinant, so the sector count and the determinant count agree. Every combinatorial claim in the article — the $2^M$ dimension, the binomial splits, the $M=3$ enumeration, the $3$ -versus- $6$ two-particle sector sizes — is recomputed in the committed verification script. No base physics errors were found: the ladder normalizations, $(a^\dagger)^2=0$ , and the direct-minus-exchange two-body structure of the base are all standard and correct, and are stated here as pointers to their rightful owners rather than re-derived.

Builds on: Indistinguishability of Identical Particles · Symmetric and Antisymmetric States · Bosons and Fermions · Slater Determinants Leads to: Creation and Annihilation Operators · Second Quantization · Field Operators and Many-Body Wavefunctions · Pauli Exclusion Principle

What exactly is the difference between the occupation-number representation and Fock space?

They are two sides of the same idea. Fock space is the arena — the direct sum $\bigoplus_N\mathcal H^{(N)}$ of all particle-number sectors, one big Hilbert space that holds states of any particle number. The occupation-number representation is the basis you use inside it — labelling each basis vector by the tally $|n_1,n_2,\dots\rangle$ instead of by a symmetrized wavefunction of labelled coordinates. You could in principle use a different basis for Fock space, but the occupation-number one is natural because identical particles carry no labels, so the tally is the complete physical data.

Why bother with the whole tower of particle numbers if my problem has a fixed number of electrons?

If particle number really is fixed and conserved, the dynamics stays within one sector $\mathcal H^{(N)}$ and Fock space is a convenient formality — the occupation-number bookkeeping is still cleaner than hand-antisymmetrized wavefunctions, but you never leave one rung. The tower becomes essential the moment particle number changes: photon emission and absorption, electron-hole pair creation, particle production in strong fields, or any process where $\hat N$ does not commute with the Hamiltonian. Then the initial and final states live on different rungs, and only a space that holds both — Fock space — can describe the transition. Second quantization is written in this language precisely so that both fixed- and variable-number problems use one formalism.

How is the number operator different from a Hamiltonian?

The number operator $\hat N=\sum_k a^\dagger_k a_k$ just counts particles; its eigenvalue is the total occupation, independent of any energies. A Hamiltonian assigns energy, and in the occupation-number language it is written in terms of the same $a^\dagger,a$ operators — a one-body kinetic/potential term plus interaction terms. When $[\hat N,\hat H]=0$ the Hamiltonian conserves particle number and its eigenstates can be labelled by a definite $N$ ; when they fail to commute (a field-theoretic Hamiltonian), energy eigenstates are superpositions across sectors. The rewriting of Hamiltonians into $a^\dagger a$ form is owned by Second Quantization; this article owns only the counting operator $\hat N$ .

Where do the fermion minus signs go in the occupation-number tally?

Into the operators, not the tally. The state $|n_1,n_2,\dots\rangle$ records only how many particles sit in each mode — the magnitudes. The antisymmetry shows up when you act with creation and annihilation operators: because $a^\dagger_k a^\dagger_l=-a^\dagger_l a^\dagger_k$ , you must fix a convention for the order in which modes are filled, and an operator acting on a mode buried in the list picks up a Jordan-Wigner sign $(-1)^{\sum_{j<k}n_j}$ from anticommuting past the occupied modes before it. Forgetting these signs gives correct magnitudes but wrong relative phases, and hence wrong matrix elements. The full sign algebra is owned by Creation and Annihilation Operators.

Is a bosonic mode really the same as a harmonic oscillator?

Yes, exactly. A single bosonic mode has states $|0\rangle,|1\rangle,|2\rangle,\dots$ with the creation operator raising the occupation and the annihilation operator lowering it, and the algebra $[a,a^\dagger]=1$ is identical to the ladder algebra of the quantum harmonic oscillator (owned by Ladder Operators and the Algebraic Oscillator). The bosonic Fock space over many modes is therefore a tensor product of independent oscillators, one per mode — which is why the quantized electromagnetic field, a collection of independent modes, is a collection of oscillators, and why "a photon in mode $k$ " means "mode $k$ 's oscillator is in its first excited state."

How does the fermion Fock state connect to the Slater determinants I already know?

Directly: a fermion Fock state $|\dots1\dots1\dots\rangle$ with $N$ occupied modes is the totally antisymmetric $N$ -electron state built from exactly those $N$ orbitals — that is, one Slater determinant (owned by Slater Determinants). So the $\binom{M}{N}$ states in the $N$ -particle sector of Fock space are in one-to-one correspondence with the $\binom{M}{N}$ Slater determinants you can form from $M$ orbitals. The occupation-number representation is just the compact way of naming those determinants by which orbitals are filled, and the sum $2^M=\sum_N\binom{M}{N}$ is the total count of all determinants over all particle numbers.

Why do bosons and fermions give different sector dimensions for the same M and N?

Because the occupation rules differ. For $N$ particles in $M$ modes, fermions must choose which $N$ modes are singly occupied — $\binom{M}{N}$ ways, since no mode can hold two. Bosons may pile multiple particles in a mode, so the count is the number of ways to distribute $N$ identical balls among $M$ boxes with no cap, $\binom{N+M-1}{M-1}$ by stars and bars. For $M=3,N=2$ these are $3$ and $6$ — the boson space is larger, and the extra states are exactly the double-occupancy tallies forbidden to fermions. The dimension gap is the exclusion principle made quantitative.