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Occupation bases and actual CCR/CAR domains

Statement

Let $I$ and $F_{\pm} (ℓ^{2} (I))$ be as in Symmetric and antisymmetric identical-particle Fock spaces. Define

Ω_{B} : = {m : I \to N_{0} : m has finite support}, Ω_{F} : = {m : I \to {0, 1} : m has finite support} .

For either statistics choice, write $∣ m ∣ = \sum_{j} m_{j}$ . If $m \in Ω_{B}$ , choose the elementary tensor $x_{m}$ containing $m_{j}$ copies of each mode $j$ in the fixed order and set

∣ m ⟩_{B} : = \sqrt{\frac{∣ m ∣!}{\prod_{j} m_{j}!}} P_{+}^{(∣ m ∣)} x_{m} .

If $m \in Ω_{F}$ , list its occupied modes increasingly in the fixed enumeration, let $x_{m}$ be their elementary tensor and set

∣ m ⟩_{F} : = \sqrt{∣ m ∣!} P_{-}^{(∣ m ∣)} x_{m} .

For $∣ m ∣ = 0$ , both vectors are the vacuum. These vectors form orthonormal bases and define unitary identifications $F_{+} (ℓ^{2} (I)) ≅ ℓ^{2} (Ω_{B})$ and $F_{-} (ℓ^{2} (I)) ≅ ℓ^{2} (Ω_{F})$ .

On the finite-support span $D_{f i n}$ of the bosonic basis, define

a_{j} ∣ m ⟩_{B} = \sqrt{m_{j}} ∣ m - 1_{j} ⟩_{B}, a_{j}^{†} ∣ m ⟩_{B} = \sqrt{m_{j} + 1} ∣ m + 1_{j} ⟩_{B},

where the annihilation term is zero when $m_{j} = 0$ . Their maximal closed extensions have domains

D (a_{j}) = {c : \sum_{m \in Ω_{B}} m_{j} ∣ c_{m} ∣^{2} < \infty}, D (a_{j}^{†}) = {c : \sum_{m \in Ω_{B}} (m_{j} + 1) ∣ c_{m} ∣^{2} < \infty},

and $a_{j}^{†} = a_{j}^{*}$ . On $D_{f i n}$ , they satisfy $[a_{i}, a_{j}^{†}] = δ_{i j} I$ , $[a_{i}, a_{j}] = 0$ , and $[a_{i}^{†}, a_{j}^{†}] = 0$ .

On the fermionic basis define

c_{j} ∣ m ⟩_{F} = (- 1)^{\sum_{i < j} m_{i}} m_{j} ∣ m - 1_{j} ⟩_{F}, c_{j}^{†} ∣ m ⟩_{F} = (- 1)^{\sum_{i < j} m_{i}} (1 - m_{j}) ∣ m + 1_{j} ⟩_{F} .

These operators are bounded on the full fermionic Fock space and satisfy ${c_{i}, c_{j}^{†}} = δ_{i j} I$ and ${c_{i}, c_{j}} = {c_{i}^{†}, c_{j}^{†}} = 0$ . These are mathematical sector relations; they do not select a physical species or imply a spin-statistics theorem.

Facts & Assumptions

Given: the Fock spaces, completed tensor powers, projections, coordinate basis and fixed enumeration in Symmetric and antisymmetric identical-particle Fock spaces.

[F1]

Symmetric and antisymmetric identical-particle Fock spaces, Definition: the product tensors form a complete orthonormal family in each tensor power, the sector projections are orthogonal, and the Fock space is the square-summable direct sum of their ranges.

[F2]

Permutation sector projections on the n-fold tensor power, Statement (b)–(d): symmetrizing and antisymmetrizing by the finite permutation averages gives the indicated orthogonal sectors, and an antisymmetrized elementary tensor with a repeated factor is zero.

[F3]

Adjoint of a densely defined operator: a vector lies in the adjoint domain exactly when the adjoint pairing on the dense domain is bounded; testing on basis vectors determines its adjoint coordinates. Densely defined, closed and closable operators, and cores: a closed operator is determined by its graph, and graph-norm truncations give a core.

[F4]

Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs: the inner-product and square-sum pairings satisfy Cauchy–Schwarz.

Proof

1.1givenF1algebra

Let $n = ∣ m ∣$ . In the bosonic permutation average $P_{+}^{(n)} x_{m}$ , each distinct product word with occupation profile $m$ occurs $\prod_{j} m_{j}!$ times out of $n!$ permutations. Therefore $P_{+}^{(n)} x_{m} = \frac{\prod_{j} m_{j}!}{n!} \sum_{w \in W (m)} w, ∣ W (m) ∣ = \frac{n!}{\prod_{j} m_{j}!} .$ The product words are orthonormal by [F1], so the squared norm of this average is $\prod_{j} m_{j}! / n!$ ; the prefactor in $∣ m ⟩_{B}$ makes its norm one. Distinct occupation profiles have disjoint word supports, hence their vectors are orthogonal.

1.2givenF1F2algebra

If a fermionic list repeats a mode, [F2] makes its antisymmetrization zero; $Ω_{F}$ excludes these profiles. For distinct occupied modes, the $n!$ permuted product words are distinct orthonormal vectors with coefficients $\pm 1 / n!$ , so $∥ P_{-}^{(n)} x_{m} ∥^{2} = 1 / n!$ . The prefactor $\sqrt{n!}$ normalizes the vector. Distinct occupied subsets have disjoint product-word support and are orthogonal.

1.3givenalgebra

The fixed enumeration identifies a finite-support configuration with a finite sequence of natural indices and finitely many nonnegative integer labels (or bits); the union over all finite sequence lengths is countable. Thus $Ω_{B}$ and $Ω_{F}$ are countable and their $ℓ^{2}$ coordinate descriptions are well defined.

1.4givenF3F4algebra

On the bosonic basis, shifting the coordinate index gives $∥ a_{j} c ∥^{2} = \sum_{m} m_{j} ∣ c_{m} ∣^{2}$ and $∥ a_{j}^{†} c ∥^{2} = \sum_{m} (m_{j} + 1) ∣ c_{m} ∣^{2}$ . Hence the stated weighted domains are exactly the domains on which the shifted coefficient sequences lie in $ℓ^{2}$ . If $c^{(r)} \to c$ and either shifted sequence converges in $ℓ^{2}$ , coordinate convergence shows the limiting output has the same weighted-shift coordinates; its square summability puts $c$ in the displayed domain. Thus both maximal shifts are closed. Testing the adjoint identity against each basis vector forces the reverse weighted shift [F3], while Cauchy--Schwarz gives the converse pairing bound [F4]. Truncating to finite initial sets of configurations makes the weighted tails tend to zero in graph norm, so $D_{f i n}$ is a core for both shifts under the graph definition in [F3].

2.1step 1.1step 1.2F1algebra

Product words span a dense subspace of each tensor power by [F1]. Applying the bounded projections $P_{\pm}^{(n)}$ to that dense span gives a dense subspace of each projected sector. Every projected product word is either zero or a scalar multiple of the corresponding occupation vector from steps 1.1–1.2. Thus the occupation vectors are complete in each sector and, after taking the direct sum over $n$ , form orthonormal bases of the Fock spaces.

2.2step 1.4algebra

For $i = j$ , direct action on $∣ m ⟩_{B}$ gives $a_{j} a_{j}^{†} ∣ m ⟩_{B} = (m_{j} + 1) ∣ m ⟩_{B}$ and $a_{j}^{†} a_{j} ∣ m ⟩_{B} = m_{j} ∣ m ⟩_{B}$ . For $i \neq j$ , both orders of distinct coordinate shifts give the same coefficient and same occupation vector, so the shifts commute; the same direct coordinate calculation applies to two annihilators and two creators. This proves the CCR on $D_{f i n}$ , without asserting that unbounded commutators are defined on all Fock space.

3.1givenstep 1.2algebra∎

Each fermionic shift maps a subset of an orthonormal basis into another orthonormal subset with coefficients of modulus at most one, so it is a contraction on the finite span and extends boundedly. For distinct $i, j$ , the two orders in any CAR anticommutator are either both zero or have equal magnitudes and opposite signs: toggling the earlier mode changes by one the number of occupied modes preceding the later index. For $i = j$ , if $m_{j} = 0$ then $c_{j} c_{j}^{†} ∣ m ⟩_{F} = ∣ m ⟩_{F}$ and $c_{j}^{†} c_{j} ∣ m ⟩_{F} = 0$ ; if $m_{j} = 1$ the roles reverse. The other same-mode anticommutators vanish. Thus all CAR identities hold on the basis span and extend by boundedness to the whole Fock space.

Occupation bases and actual CCR/CAR domains

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources