Fock Space and Field Operators

A one-particle Hilbert space turns the abstract canonical algebra into a familiar many-particle representation. The construction is concrete but not canonical: it depends on the chosen one-particle structure. Its operator domains also matter. Bosonic creation, annihilation, and field operators are unbounded, so their commutation relations are first identities on a common dense invariant domain, not equations between everywhere-defined bounded operators.

Required background. Flat-Spacetime Quantum Field Theory supplies creation and annihilation operators, while Variational and Symplectic Geometry defines the classical symplectic pairing that the represented field must reproduce.

Helpful background. Partial Differential Equations and Green Functions provides the causal test-function quotient used to smear the field, and Causality and Global Structure establishes why causally disjoint smearings have vanishing symplectic pairing.

For immediate use, $[f]$ denotes a compact test function modulo the field equation, and its causal solution is $Ef$. This makes the represented field depend on the classical solution rather than on a redundant choice of source.

Symmetric and antisymmetric Fock spaces

Let $\mathcal H$ be a separable one-particle Hilbert space. The bosonic and fermionic Fock spaces are

$$\mathcal F_+(\mathcal H)=\bigoplus_{n=0}^{\infty}\mathcal H^{\otimes_s n}, \qquad \mathcal F_-(\mathcal H)=\bigoplus_{n=0}^{\infty}\mathcal H^{\otimes_a n},$$

where the zeroth sector is $\mathbb C\Omega$. The vector $\Omega=(1,0,0,\ldots)$ is the Fock vacuum for this representation. The algebraic direct sum

$$\mathcal D_{\mathrm{fin}} =\bigcup_{N=0}^{\infty}\bigoplus_{n=0}^{N}\mathcal H^{\otimes_{s/a}n}$$

is dense and is preserved by finite products of creation and annihilation operators. It is therefore the natural core on which to derive the canonical relations.

With $S_n$ and $A_n$ denoting orthogonal symmetrization and antisymmetrization, normalized creation is

$$a^*(f)\psi_n=\sqrt{n+1}\,S_{n+1}(f\otimes\psi_n)$$

for bosons, with $S$ replaced by $A$ for fermions; $a(f)$ is its adjoint on finite-particle vectors. We take the Hilbert product to be antilinear in the first argument, so $a(f)$ is antilinear in $f$ and $a^*(f)$ is linear. The normalization is chosen so that

$$[a(f),a^*(g)]=\langle f,g\rangle\mathbf1$$

on $\mathcal F_+$, whereas

$$\{a(f),a^*(g)\}=\langle f,g\rangle\mathbf1$$

on $\mathcal F_-$. The detailed symmetric construction and its common invariant core appear in Bär, Ginoux, and Pfäffle 2007, §§4.6–4.7, pp. 139–153.

Bosonic fields from a one-particle map

Let $K:\mathcal V\to\mathcal H$ be a real-linear one-particle map with dense complex span and

$$2\operatorname{Im}\langle Kv,Kw\rangle=\sigma(v,w).$$

For a field as an operator-valued distribution, also require the map $f\mapsto K[f]$ to be continuous from the compact-test-function topology to $\mathcal H$. The symplectic identity alone is an algebraic condition and does not supply this regularity. Continuity ensures, for example, that the vacuum two-point function $\langle K[f],K[g]\rangle$ is a bidistribution. Define the Segal field on $\mathcal D_{\mathrm{fin}}$ by

$$\Phi(v)=a(Kv)+a^*(Kv).$$

The derivation of its commutator uses only the creation-annihilation relations:

$$\begin{aligned} [\Phi(v),\Phi(w)] &=[a(Kv),a^*(Kw)]+[a^*(Kv),a(Kw)]\\ &=\bigl(\langle Kv,Kw\rangle-\langle Kw,Kv\rangle\bigr)\mathbf1\\ &=i\sigma(v,w)\mathbf1. \end{aligned}$$

For the scalar field $v=[f]$ and $\sigma([f],[g])=E(f,g)$, so this is exactly $[\Phi(f),\Phi(g)]=iE(f,g)\mathbf1$. The equation of motion follows because $[Pf]=0$ in the quotient, hence $K[Pf]=0$ and $\Phi(Pf)=0$. Causal support of $E$ then gives locality. This shows how a chosen one-particle structure represents the abstract algebra without changing its defining relations.

For each $v$, the Segal field is essentially self-adjoint on the finite-particle core in the standard Fock construction. Its closure generates the bounded Weyl unitary $W(v)=e^{i\overline{\Phi(v)}}$. Thus the polynomial field representation and the regular Weyl representation describe the same chosen Fock realization, while the latter avoids multiplying unbounded generators.

Vacuum correlations and number normalization

Because $a(f)\Omega=0$ and $a^*(f)\Omega=f$ in the one-particle sector,

$$\langle\Omega,\Phi(v)\Phi(w)\Omega\rangle =\langle Kv,Kw\rangle.$$

Its antisymmetric part is fixed by the CCR; its symmetric part records the choice of $K$, and therefore the choice of Fock vacuum. This is already the simplest example of a quasifree two-point function.

The number operator is defined sectorwise by

$$N\psi_n=n\psi_n.$$

On $\mathcal D_{\mathrm{fin}}$ it satisfies $[N,a^*(f)]=a^*(f)$ and $[N,a(f)]=-a(f)$. For a normalized $f$, the vectors

$$\lvert n_f\rangle=\frac{(a^*(f))^n}{\sqrt{n!}}\Omega$$

have unit norm and number $n$. These checks fix the square-root normalization in the creation formula. They do not define a representation-independent particle observable: a different one-particle splitting generally gives a different $N$.

The separation of the abstract algebra from this vacuum representation, including the role of cyclic states, is developed in Wald 1994, Chapter 3, pp. 23–52.

Domains and the boson–fermion contrast

For nonzero $f$, bosonic creation and annihilation operators are unbounded. Set $\widehat f=f/\lVert f\rVert$. The normalized number states of that mode obey

$$\lVert a^*(f)\lvert n_{\widehat f}\rangle\rVert =\sqrt{n+1}\,\lVert f\rVert,$$

which grows without bound. A generic Fock vector is an infinite sequence of particle sectors and need not lie in the domain of $a(f)$, $a^*(f)$, $N$, or their products. Writing a commutator without naming a common domain hides a genuine mathematical condition. The finite-particle space is invariant and sufficient for the algebraic derivations; closures are used for spectral statements.

A finite norm is not enough to guarantee finite expected particle number. For one normalized mode, the vector $\psi=c\sum_{n\geq1}n^{-1}\lvert n_f\rangle$ is normalizable because $\sum n^{-2}$ converges, but $\langle N\rangle$ diverges because $\sum n\,n^{-2}$ is harmonic. More precisely, it is outside the quadratic-form domain of $N$. This is why a proof of Bogoliubov nonimplementability cannot merely argue that an arbitrary normalizable vector must have finite particle number.

Fermionic operators behave differently. Antisymmetry and the CAR imply

$$\lVert a(f)\rVert=\lVert a^*(f)\rVert=\lVert f\rVert,$$

so they extend to bounded operators on all of $\mathcal F_-$. A free Dirac field normally uses particle and antiparticle one-particle subspaces, or equivalently a self-dual CAR representation. Its smeared fields remain bounded after the one-particle norm is completed. This analytic advantage does not make a fermionic polarization canonical, and it does not turn the bosonic Shale theorem into the fermionic Shale–Stinespring theorem.

Representation is a choice, not a new algebra

The vacuum vector defines a state $\omega_0(A)=\langle\Omega,\pi_K(A)\Omega\rangle$ on the abstract CCR or CAR algebra. The triple $(\mathcal F_\pm,\pi_K,\Omega)$ is its cyclic representation. Changing $K$ changes the state and may change the unitary-equivalence class of the representation, but it leaves the abstract causal commutator or anticommutator intact.

It is therefore unsafe to identify “the field algebra” with operators on one preferred Fock space. In curved spacetime there may be several useful Fock representations and no universal vacuum. Local observables can still be compared algebraically, and suitable physically admissible representations can often be locally equivalent even when no global unitary identifies their particle bases.

Exercise: check the creation normalization

Let $f$ be normalized and suppose $a(f)\lvert n_f\rangle=\sqrt n\lvert n-1_f\rangle$. Use the CCR to derive the norm and number of $a^*(f)\lvert n_f\rangle$.

Solution

The CCR gives $a(f)a^*(f)=a^*(f)a(f)+\mathbf1$. Taking its expectation in the normalized number state yields $\lVert a^*(f)\lvert n_f\rangle\rVert^2=n+1$. Hence the normalized result is $a^*(f)\lvert n_f\rangle=\sqrt{n+1}\lvert n+1_f\rangle$, up to the fixed phase convention. The commutator with $N$ then gives $N a^*(f)\lvert n_f\rangle=(n+1)a^*(f)\lvert n_f\rangle$. This calculation separates two facts: the canonical relation fixes the norm increase, while the number-operator commutator fixes the sector reached. Both checks would fail if the square-root factor were omitted, and the error would grow with particle number rather than remaining a harmless overall convention.

References

• Bär, Christian, Nicolas Ginoux, and Frank Pfäffle. Wave Equations on Lorentzian Manifolds and Quantization. ESI Lectures in Mathematics and Physics. EMS Press, 2007. DOI 10.4171/037.
• Wald, Robert M. Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics. Chicago Lectures in Physics. University of Chicago Press, 1994. Publisher record.