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Fock Spaces in Rigorous Language

Fock space collects finite-particle sectors in a Hilbert direct sum. A vector may occupy infinitely many sectors in superposition, but its normalization does not guarantee a finite mean particle number. Here we relate particle-number probabilities to the domains of number operators; Fock Space Construction develops the completed construction, exponential-vector factorization, and second-quantized generator domains. The physical introductions are Bosonic Fock Space and Fermionic Fock Space.

Helpful background. Symmetric and Antisymmetric Tensor Products defines the fixed-particle sectors. Direct Sums versus Tensor Products distinguishes alternatives from simultaneous composition.

A Fock vector is a square-summable sequence of sectors

Section titled “A Fock vector is a square-summable sequence of sectors”

For a one-particle Hilbert space h\mathcal h, the bosonic and fermionic spaces have the form

F±(h)=⨁n=0∞h±(n),h+(n)=Sym⁡nh,h−(n)=⋀nh.\mathcal F_\pm(\mathcal h) =\bigoplus_{n=0}^\infty\mathcal h_\pm^{(n)}, \qquad \mathcal h_+^{(n)}=\operatorname{Sym}^n\mathcal h,\quad \mathcal h_-^{(n)}=\bigwedge\nolimits^n\mathcal h.

The zero-particle sector is C\mathbb C. A vector is a sequence Ψ=(ψ0,ψ1,…)\Psi=(\psi_0,\psi_1,\ldots) with ψn∈h±(n)\psi_n\in\mathcal h_\pm^{(n)} and

∥Ψ∥2=∑n∥ψn∥2<∞.\|\Psi\|^2=\sum_n\|\psi_n\|^2<\infty.

Thus different particle-number sectors are orthogonal. The normalized vacuum is Ω=(1,0,0,…)\Omega=(1,0,0,\ldots); it is not the zero vector.

For normalized Ψ\Psi, the number probabilities are pn=∥ψn∥2p_n=\|\psi_n\|^2. Their sum is one, even when infinitely many are nonzero. The probability that particle number exceeds a cutoff tends to zero as the cutoff tends to infinity. There is no extra sector labeled “infinitely many particles” in this direct sum.

Finite-particle vectors form a dense subspace

Section titled “Finite-particle vectors form a dense subspace”

The algebraic finite-particle subspace consists of sequences with only finitely many nonzero sectors. If P≤MP_{\leq M} projects onto the first M+1M+1 sectors, then

∥Ψ−P≤MΨ∥2=∑n>Mpn⟶0.\|\Psi-P_{\leq M}\Psi\|^2 =\sum_{n>M}p_n\longrightarrow0.

It follows that finite-particle vectors are dense. This does not say that a generic vector has a definite particle number, or that the sector projection is finite rank. For infinite-dimensional h\mathcal h, even the one-particle sector is infinite-dimensional.

“Finite particle number” is also used informally for finite expected number. Those are different properties: a vector can have infinitely many occupied sectors but finite mean, or it can be normalized with divergent mean.

The number operator multiplies the nn-particle component by nn. Its operator domain and quadratic-form domain are

D(N)={Ψ:∑nn2pn<∞},D(N1/2)={Ψ:∑nnpn<∞}.D(N)=\left\{\Psi:\sum_n n^2p_n<\infty\right\}, \qquad D(N^{1/2})=\left\{\Psi:\sum_n np_n<\infty\right\}.

The second condition is finite mean particle number. The first is finite second moment and permits NΨN\Psi as a Hilbert-space vector. These are direct applications of the spectral moment-domain distinction.

Finite-sector truncations converge in the graph norm of NN for vectors in D(N)D(N), because the omitted weighted square sum tends to zero. They therefore provide a core for this diagonal number operator. Hilbert-norm density alone would not have established that graph-norm statement.

For interacting Hamiltonians or creation operators, the correct domains require additional estimates. The word “Fock” does not make an unbounded formula defined on every normalized vector.

Normalized states with divergent number moments

Section titled “Normalized states with divergent number moments”

In one-mode bosonic Fock space, let ∣n⟩|n\rangle denote its normalized number basis. The vector

Ψ=C∑n=1∞1n∣n⟩\Psi=C\sum_{n=1}^\infty\frac1n|n\rangle

is normalizable because ∑n−2<∞\sum n^{-2}<\infty. Its mean number diverges, since ∑nnpn=C2∑nn−1\sum_nnp_n=C^2\sum_n n^{-1}. It lies outside D(N1/2)D(N^{1/2}).

Changing the amplitudes to C′n−3/2C'n^{-3/2} gives finite mean number but divergent second moment. That vector lies in D(N1/2)D(N^{1/2}) and outside D(N)D(N). In both examples, a number measurement still produces a finite integer with probability one.

The same sector-probability examples can be constructed in infinite-mode fermionic Fock space by choosing one unit vector in every nonzero exterior-power sector. They cannot be constructed with a fixed finite number of fermion modes, because those sectors vanish above that mode count.

Splitting modes is different from splitting particle labels

Section titled “Splitting modes is different from splitting particle labels”

A one-particle decomposition h=hA⊕hB\mathcal h=\mathcal h_A\oplus\mathcal h_B induces a bosonic Hilbert-space factorization

F+(hA⊕hB)≃F+(hA)⊗F+(hB).\mathcal F_+(\mathcal h_A\oplus\mathcal h_B) \simeq\mathcal F_+(\mathcal h_A)\otimes\mathcal F_+(\mathcal h_B).

This is a split into mode families, not a tensor product of distinguishable particle labels. A fixed total-number sector contains several allocations of particles between the mode families.

For fermions, a Hilbert-space factorization can also be chosen, but the field operators carry a grading. In an ordering with the AA modes first, an odd BB annihilation operator is represented with the AA parity factor (−1)NA(-1)^{N_A}. Treating all fermion operators as ordinary commuting operators on disjoint tensor factors would lose the canonical anticommutation signs.

Neither construction is simply a product of one-particle Hilbert spaces over infinitely many labeled particles. Infinite Tensor Products explains the different role of a chosen reference sequence.

  1. If dim⁡h=d<∞\dim\mathcal h=d<\infty, find the dimension of fermionic Fock space and compare with bosonic Fock space for d≥1d\geq1.
Solution

The fermionic dimension is ∑n=0d(dn)=2d\sum_{n=0}^d\binom dn=2^d. The bosonic space has a nonzero sector at every nn, so it is infinite-dimensional even for one mode. Both have a one-dimensional vacuum sector.

  1. Does finite mean particle number imply finite mean energy for every Hamiltonian on Fock space?
Solution

No. Even a state entirely in the one-particle sector has N=1N=1 with certainty, while its one-particle energy distribution can have divergent mean for an unbounded dispersion. Number and energy require their own moment conditions.

  • Dereziński, Jan, “Introduction to Representations of the Canonical Commutation and Anticommutation Relations,” in Large Coulomb Systems, Lecture Notes in Physics 695, Springer, 2006, pp. 63–143, Section 5 — Fock sectors, number operators, and factorization. Paper; preprint.
  • Dereziński, Jan, and Christian Gérard, Mathematics of Quantization and Quantum Fields, Cambridge University Press, 2013 — tensor algebras and bosonic/fermionic field operators. Publisher.