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Qm Identical Particles And Exclusion

1 · Prerequisites

2 · Summary

This page develops the framework's reading of the identical-particle sector assignment of P6 and its first consequences: the exchange operator, the occupation-number bases of the symmetric and antisymmetric sectors, the exclusion of repeated fermionic occupations, and the conditional indistinguishability of relabelled readouts. The readings are fixed in Declared species sectors and their preparations: the framework's reading of P6, which declares a species with its one-particle space, mass and spin parameters, statistics choice and adopted sector space $H_{n}^{σ} = ran P_{σ}^{(n)}$ , fixes the vocabulary of declared sector preparations and sector quantities, and records that a bounded quantity of the labelled space is read on the sector only when it preserves the sector. The physical scope is explicit: the sector assignment is an adopted model assumption, the spin-statistics connection is not derived, and no measurement is recorded.

The supporting mathematics precedes the physical reading. The two-particle exchange operator and the symmetric and antisymmetric combinations constructs the exchange operator $τ$ of a tensor square, proves that it is a self-adjoint involution, identifies the sector projections as $\frac{1}{2} (I \pm τ)$ , and computes the norms of the symmetric and antisymmetric combinations of two vectors. Slater and symmetric occupation-number bases of the finite-dimensional permutation sectors proves that the Slater vectors form an orthonormal basis of the antisymmetric sector and the symmetric sums an orthogonal basis of the symmetric sector, with the dimension formulas $(\binom{d}{n})$ and $(\binom{n + d - 1}{n})$ , and diagonalises the occupation operators $N_{j}$ on both bases. Finite-mode occupation configurations: fermionic exclusion counts, bosonic counts and truncated dimensions counts the occupation configurations, showing that the antisymmetric index set contains no occupation $\geq 2$ and that the truncated symmetric dimension is $(\binom{N + d}{d})$ . Regrouping two factors: the symmetry decomposition of a two-particle sector over an orbital and an internal factor proves the regrouping identity that splits a two-particle sector over an orbital and an internal factor into opposite-symmetry summands.

The physical results are conditional on P1, P2 and P6 in the page's readings. The declared finite-mode model and its occupation readouts declares the finite-mode occupation model, its readouts $N_{j}^{σ}$ and the joint occupation law. Permutation invariance of sector preparations and the indistinguishability of relabelled readouts proves that every declared sector preparation is invariant under the permutation unitaries and that the outcome law of any assigned sharp observable of the labelled model equals that of its relabelled conjugate, which is the precise content of indistinguishability inside the model. Pauli exclusion in the declared finite-mode occupation readout, with the bosonic contrast proves the Pauli exclusion statement of the declared readout: in the antisymmetric sector the probability of an occupation $\geq 2$ and the expectation of $N_{j} (N_{j} - 1)$ vanish and the joint law is supported on occupations in ${0, 1}$ , while in the symmetric sector every configuration of total occupation $n$ occurs. Scope and empirical status of the identical-particles page records the adopted assumptions, the conditional results and the empirical limits of the page.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

The two-particle exchange operator and the symmetric and antisymmetric combinations

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a complex Hilbert space with inner product linear in the first argument (Hilbert space, Real and complex inner-product spaces and their induced length), let $H \hat{\otimes} H$ be its completed tensor square with inclusion $κ$ (Hilbert tensor construction: inner product, completion and product bases), and for $n = 2$ let $(U_{π})_{π \in S_{2}}$ and the projections $P_{\pm}^{(2)}$ be those of Permutation sector projections on the n-fold tensor power. Write $τ : = U_{(0 1)}$ for the exchange operator, so that $τ (κ (u \otimes v)) = κ (v \otimes u)$ for $u, v \in H$ ; in the sequel the inclusion is suppressed and $u \otimes v$ denotes $κ (u \otimes v)$ . Then:

(a) $τ$ is self-adjoint and $τ^{2} = I$ ; consequently $τ$ is an isometry with $τ^{- 1} = τ = τ^{*}$ (The Hilbert-space adjoint of a bounded operator, A bounded linear operator between normed spaces);

(b) $P_{+}^{(2)} = \frac{1}{2} (I + τ)$ and $P_{-}^{(2)} = \frac{1}{2} (I - τ)$ , and $ran P_{+}^{(2)} = {x \in H \hat{\otimes} H : τ x = x}, ran P_{-}^{(2)} = {x \in H \hat{\otimes} H : τ x = - x};$

(c) for all $u, v \in H$ , $∥ u \otimes v + v \otimes u ∥^{2} = 2 (∥ u ∥^{2} ∥ v ∥^{2} + ∣ ⟨ u, v ⟩ ∣^{2}), ∥ u \otimes v - v \otimes u ∥^{2} = 2 (∥ u ∥^{2} ∥ v ∥^{2} - ∣ ⟨ u, v ⟩ ∣^{2});$ in particular, if $u, v$ are orthonormal, then $s : = \frac{1}{\sqrt{2}} (u \otimes v + v \otimes u)$ and $a : = \frac{1}{\sqrt{2}} (u \otimes v - v \otimes u)$ are orthonormal, $τ s = s$ and $τ a = - a$ , and both $ran P_{\pm}^{(2)}$ are nonzero;

(d) whenever $u, v \in H$ are orthonormal, $τ κ (u \otimes v) \neq κ (u \otimes v)$ . Consequently, if $H$ contains an orthonormal pair (equivalently, $\dim H \geq 2$ ), both exchange eigenspaces contain nonzero vectors and $τ \neq I$ . If $H = {0}$ or $\dim H = 1$ , then $τ = I$ , $P_{+}^{(2)} = I$ and $P_{-}^{(2)} = 0$ ; for $H = {0}$ the tensor square and both ranges are zero.

Facts & Assumptions

Given: The complex Hilbert space $H$ with inner product linear in the first argument, the completed tensor square $H \hat{\otimes} H$ with inclusion $κ$ , the permutation unitaries $U_{π}$ for $n = 2$ and the projections $P_{\pm}^{(2)}$ , and vectors $u, v \in H$ .

[F1]

Tensor square. $κ : H ⊙ H \to H \hat{\otimes} H$ is a linear isometry with dense image, $κ (u \otimes v)$ is bilinear in $(u, v)$ , $∥ κ (u \otimes v) ∥ = ∥ u ∥ ∥ v ∥$ and $⟨ κ (u \otimes v), κ (u^{'} \otimes v^{'}) ⟩ = ⟨ u, u^{'} ⟩ ⟨ v, v^{'} ⟩$ (Hilbert tensor construction: inner product, completion and product bases clauses (a)-(d), Real and complex inner-product spaces and their induced length).

[F2]

Permutation unitaries. $U_{(0 1)} \in B (H \hat{\otimes} H)$ is the unique bounded operator with $U_{(0 1)} κ (u \otimes v) = κ (v \otimes u)$ ; it is unitary with $U_{π} U_{σ} = U_{π σ}$ and $U_{π}^{*} = U_{π^{- 1}}$ . For every $π \in S_{2}$ , $U_{π} P_{+}^{(2)} = P_{+}^{(2)} U_{π} = P_{+}^{(2)}$ and $U_{π} P_{-}^{(2)} = P_{-}^{(2)} U_{π} = sgn (π) P_{-}^{(2)}$ . The orthogonal projections are $P_{+}^{(2)} = \frac{1}{2} \sum_{π \in S_{2}} U_{π}$ and $P_{-}^{(2)} = \frac{1}{2} \sum_{π \in S_{2}} sgn (π) U_{π}$ (Permutation sector projections on the n-fold tensor power clauses (a)-(c), The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F3]

Orthogonal projections. An orthogonal projection $P$ satisfies $P^{2} = P = P^{*}$ , and its range is a closed subspace (The Hilbert orthogonal projection onto a closed subspace, Hilbert projections are linear, self-adjoint and contractive).

[F4]

Adjoint and operator norm. For $T \in B (H \hat{\otimes} H)$ the Hilbert adjoint $T^{*}$ is the unique bounded operator with $⟨ T x, y ⟩ = ⟨ x, T^{*} y ⟩$ , and $T$ is an isometry exactly when $T^{*} T = I$ ; $∥ T x ∥ \leq ∥ T ∥ ∥ x ∥$ (The Hilbert-space adjoint of a bounded operator, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A bounded linear operator between normed spaces).

Proof

1.1F2F4algebra

[The exchange operator is a self-adjoint involution] By [F2] with $π = σ = (0 1)$ one has $τ^{2} = U_{(0 1) (0 1)} = U_{e} = I$ , and $τ^{*} = U_{(0 1)^{- 1}} = U_{(0 1)} = τ$ ; hence $τ^{- 1} = τ = τ^{*}$ and $τ$ is an isometry, which is (a).

1.2F2algebra

[The projections in terms of the exchange operator] Since $S_{2} = {e, (0 1)}$ and $sgn (e) = 1$ , $sgn ((0 1)) = - 1$ , the formulas of [F2] give $P_{+}^{(2)} = \frac{1}{2} (U_{e} + U_{(0 1)}) = \frac{1}{2} (I + τ)$ and $P_{-}^{(2)} = \frac{1}{2} (U_{e} - U_{(0 1)}) = \frac{1}{2} (I - τ)$ , which is the first assertion of (b).

2.1step 1.1step 1.2F3algebra

[Eigenspaces as ranges] Let $x \in H \hat{\otimes} H$ . If $τ x = x$ , then $\frac{1}{2} (I + τ) x = x$ , so $x \in ran \frac{1}{2} (I + τ)$ ; conversely if $x = \frac{1}{2} (I + τ) y$ for some $y$ , then $τ x = \frac{1}{2} (τ y + τ^{2} y) = \frac{1}{2} (τ y + y) = x$ by step 1.1. Hence the fixed space of $τ$ is $ran \frac{1}{2} (I + τ) = ran P_{+}^{(2)}$ by step 1.2; the same substitution with $x = \frac{1}{2} (I - τ) y$ gives $τ x = - x$ , and conversely $τ x = - x$ implies $x = \frac{1}{2} (I - τ) x$ , so the $(- 1)$ -eigenspace is $ran P_{-}^{(2)}$ . This completes (b).

2.2F1F2step 1.1algebra

[Norm identities on products] By [F1] the elementary tensors satisfy $∥ u \otimes v ∥^{2} = ∥ u ∥^{2} ∥ v ∥^{2}$ and $⟨ u \otimes v, v \otimes u ⟩ = ⟨ u, v ⟩ ⟨ v, u ⟩ = ∣ ⟨ u, v ⟩ ∣^{2}$ ; expanding $∥ u \otimes v \pm v \otimes u ∥^{2} = ∥ u \otimes v ∥^{2} + ∥ v \otimes u ∥^{2} \pm 2 Re ⟨ u \otimes v, v \otimes u ⟩$ and using that the last number is the real number $∣ ⟨ u, v ⟩ ∣^{2}$ gives the two identities of (c). If $u, v$ are orthonormal then $∥ u ∥ = ∥ v ∥ = 1$ and $⟨ u, v ⟩ = 0$ , so $s$ and $a$ have norm one; moreover $τ s = \frac{1}{\sqrt{2}} (v \otimes u + u \otimes v) = s$ and $τ a = \frac{1}{\sqrt{2}} (v \otimes u - u \otimes v) = - a$ by [F2]; and $⟨ s, a ⟩ = \frac{1}{2} (∥ u \otimes v ∥^{2} - ∥ v \otimes u ∥^{2}) = 0$ , so $s, a$ are orthonormal and $s \neq 0 \neq a$ . This proves (c).

3.1F1F2step 2.2algebraconstruct

[Nonvanishing of the swapped product] Let $u, v$ be orthonormal. Then $u \otimes v$ and $v \otimes u$ are orthonormal in $H \hat{\otimes} H$ : both have norm one by [F1], and $⟨ u \otimes v, v \otimes u ⟩ = ⟨ u, v ⟩ ⟨ v, u ⟩ = 0$ . Hence $v \otimes u \neq u \otimes v$ , so $τ (u \otimes v) \neq u \otimes v$ by [F2]; taking $x : = u \otimes v$ and $x^{'} : = s$ and $x^{''} : = a$ from step 2.2 gives nonzero vectors with $τ x^{'} = x^{'}$ and $τ x^{''} = - x^{''}$ , so $τ \neq I$ . This proves (d).

4.1F1F2step 1.2step 3.1choosealgebra

[Zero and one-dimensional spaces] If $H = {0}$ , all elementary tensors are zero and their completion is zero, so $τ = I$ on that space. If $H$ is one-dimensional, fix a unit spanning vector $e$ . Every elementary tensor has the form $(α e) \otimes (β e) = α β (e \otimes e)$ and its swap is $β α (e \otimes e)$ , the same vector. Hence $τ = I$ on the dense algebraic tensor space and, by boundedness, on its completion. Step 1.2 then gives $P_{+}^{(2)} = I$ and $P_{-}^{(2)} = 0$ in both cases. If $\dim H \geq 2$ , normalize a nonzero vector $u$ and choose $w$ outside its span; the nonzero vector $w - ⟨ w, u ⟩ u$ normalizes to a unit $v$ orthogonal to $u$ , so step 3.1 applies. This proves all branches of (d).

5.1step 1.1step 1.2step 2.1step 2.2step 3.1step 4.1∎

[Assembly] Step 1.1 proves (a), steps 1.2 and 2.1 prove (b), step 2.2 proves (c) and steps 3.1 and 4.1 prove (d). Only the cited tensor and permutation mathematics and the Countable Choice hypothesis of [F2] were used; no physical reading of the exchange operator is asserted.

Remarks

Why the letter $τ$ . The operator is the $n = 2$ instance of the permutation unitaries; the notation $τ$ matches the exchange operator of the source treatments and is used on this pair's pages for the two-particle swap.
Scope. The lemma concerns a single Hilbert space and its tensor square; it asserts nothing about particles, spin, statistics or dynamics. The physical reading of $τ$ as the exchange of two labelled factors is fixed in Declared species sectors and their preparations: the framework's reading of P6 on the page for the identical-particle model.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

Slater and symmetric occupation-number bases of the finite-dimensional permutation sectors

Statement

Slot convention. For $ℓ \geq 1$ , $S_{ℓ}$ acts on ${0, \dots, ℓ - 1}$ as in Permutation sector projections on the n-fold tensor power. On one-based tuple positions use $\hat{π} (k) : = 1 + π (k - 1)$ for $1 \leq k \leq ℓ$ . Conjugation by $k \mapsto k - 1$ gives $\hat{π σ} = \hat{π} \circ \hat{σ}$ and $\hat{π^{- 1}} = {\hat{π}}^{- 1}$ ; thus the supplier's tensor action becomes $U_{π} (v_{1} \otimes \dots \otimes v_{ℓ}) = v_{{\hat{π}}^{- 1} (1)} \otimes \dots \otimes v_{{\hat{π}}^{- 1} (ℓ)}$ . The parameters of $U_{π}$ and $sgn (π)$ remain canonical zero-based permutations.

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a complex Hilbert space of finite dimension $d \geq 1$ with orthonormal basis $(e_{1}, \dots, e_{d})$ (Hilbert space, Orthonormal families, complete orthonormal systems and Hilbert bases). Let $n \geq 1$ , and let $H^{\otimes n}$ be the $n$ -fold completed tensor power with $H^{\otimes 0} : = C$ , $H^{\otimes 1} : = H$ and $H^{\otimes (k + 1)} : = H^{\otimes k} \hat{\otimes} H$ (Hilbert tensor construction: inner product, completion and product bases), and let $P_{\pm}^{(n)} \in B (H^{\otimes n})$ , $n \geq 1$ , be the permutation projections of Permutation sector projections on the n-fold tensor power. For a multi-index $i = (i_{1}, \dots, i_{n}) \in {1, \dots, d}^{n}$ write $e_{i} : = e_{i_{1}} \otimes \dots \otimes e_{i_{n}}, ∣ i ∣_{j} : = # {k : i_{k} = j} (j = 1, \dots, d) .$ For $J \subseteq {1, \dots, d}$ with $∣ J ∣ = n$ , written $J = {j_{1} < \dots < j_{n}}$ , put $e_{J}^{\land} : = \frac{1}{\sqrt{n!}} \sum_{σ \in S_{n}} sgn (σ) e_{j_{\hat{σ} (1)}} \otimes \dots \otimes e_{j_{\hat{σ} (n)}},$ and for $λ \in W (n, d) : = {λ \in N^{d} : \sum_{j} λ_{j} = n}$ put $orb (λ) : = {c \in {1, \dots, d}^{n} : ∣ c ∣ = λ}$ and $m_{λ} : = \sum_{c \in orb (λ)} e_{c}$ . Then:

(a) product basis: $(e_{i})_{i \in {1, \dots, d}^{n}}$ is an orthonormal basis of $H^{\otimes n}$ , and $\dim H^{\otimes n} = d^{n}$ ;

(b) Slater basis: for every $J$ with $∣ J ∣ = n$ one has $e_{J}^{\land} = \sqrt{n!} P_{-}^{(n)} (e_{j_{1}} \otimes \dots \otimes e_{j_{n}})$ and the family $(e_{J}^{\land})_{J \subseteq {1, \dots, d}, ∣ J ∣ = n}$ is an orthonormal family whose closed linear span is $ran P_{-}^{(n)}$ ; it is therefore an orthonormal basis of $ran P_{-}^{(n)}$ , $\dim ran P_{-}^{(n)} = (\binom{d}{n})$ , and $ran P_{-}^{(n)} = {0}$ when $n > d$ . Moreover for every multi-index $i$ one has $P_{-}^{(n)} (e_{i}) = 0$ when two entries of $i$ coincide, and $P_{-}^{(n)} (e_{i}) = sgn (ρ_{i}) \frac{e_{J}^{\land}}{\sqrt{n!}}$ when the entries of $i$ are pairwise distinct with underlying set $J$ and $ρ_{i} \in S_{n}$ is the unique permutation with $i_{k} = j_{\hat{ρ_{i}} (k)}$ ;

(c) symmetric basis: for every $λ \in W (n, d)$ the vector $m_{λ}$ is nonzero, the family $(m_{λ})_{λ \in W (n, d)}$ is orthogonal, $U_{π} m_{λ} = m_{λ}$ for every $π \in S_{n}$ , and for every multi-index $i$ there is a real number $γ_{i} > 0$ with $P_{+}^{(n)} (e_{i}) = γ_{i} m_{∣ i ∣}$ ; consequently $ran P_{+}^{(n)}$ is the linear span of $(m_{λ})$ , the family $(m_{λ})$ is an orthogonal basis of $ran P_{+}^{(n)}$ , and $\dim ran P_{+}^{(n)} = (\binom{n + d - 1}{n})$ ;

(d) occupation operators: for $j \in {1, \dots, d}$ the map $N_{j} : H^{\otimes n} \to H^{\otimes n}, N_{j} x : = \sum_{i \in {1, \dots, d}^{n}} ∣ i ∣_{j} ⟨ x, e_{i} ⟩ e_{i},$ is a self-adjoint bounded endomorphism, the $N_{j}$ commute pairwise, $\sum_{j = 1}^{d} N_{j} = n I$ , and $N_{j} e_{J}^{\land} = 1_{J} (j) e_{J}^{\land}, N_{j} m_{λ} = λ_{j} m_{λ} .$ Consequently both sectors are invariant under every $N_{j}$ , the joint eigenspaces of $(N_{1}, \dots, N_{d})$ inside $ran P_{-}^{(n)}$ are the lines $C e_{J}^{\land}$ (one for each $n$ -element subset $J$ , with eigenvalue vector $1_{J}$ ), and the joint eigenspaces inside $ran P_{+}^{(n)}$ are the lines $C m_{λ}$ (one for each $λ \in W (n, d)$ , with eigenvalue vector $λ$ ).

Facts & Assumptions

Given: The finite-dimensional complex Hilbert space $H$ of dimension $d \geq 1$ with orthonormal basis $(e_{1}, \dots, e_{d})$ , the tensor powers $H^{\otimes n}$ with permutation projections $P_{\pm}^{(n)}$ , multi-indices $i \in {1, \dots, d}^{n}$ , subsets $J$ with $∣ J ∣ = n$ , and $λ \in W (n, d)$ .

[F1]

Tensor powers. For all $n$ the tensor power carries bilinear elementary tensors $v_{1} \otimes \dots \otimes v_{n}$ with $∥ v_{1} \otimes \dots \otimes v_{n} ∥ = ∥ v_{1} ∥ \dots ∥ v_{n} ∥$ and $⟨ v_{1} \otimes \dots \otimes v_{n}, w_{1} \otimes \dots \otimes w_{n} ⟩ = \prod_{k} ⟨ v_{k}, w_{k} ⟩$ ; for orthonormal families $(f_{a})_{a \in A}$ in $H$ and $(g_{b})_{b \in B}$ in $K$ with dense spans, the products $f_{a} \otimes g_{b}$ form an orthonormal family in $H \hat{\otimes} K$ with dense span; the algebraic tensor product is dense in each completion (Hilbert tensor construction: inner product, completion and product bases clauses (b)-(d)).

[F2]

Permutation projections. $U_{π} (v_{1} \otimes \dots \otimes v_{n}) = v_{{\hat{π}}^{- 1} (1)} \otimes \dots \otimes v_{{\hat{π}}^{- 1} (n)}$ defines unitaries with $U_{π} U_{σ} = U_{π σ}$ and $U_{π}^{*} = U_{π^{- 1}}$ ; $P_{+}^{(n)} = \frac{1}{n!} \sum_{π \in S_{n}} U_{π}$ and $P_{-}^{(n)} = \frac{1}{n!} \sum_{π \in S_{n}} sgn (π) U_{π}$ are orthogonal projections with $U_{π} P_{-}^{(n)} = P_{-}^{(n)} U_{π} = sgn (π) P_{-}^{(n)}$ , and $P_{-}^{(n)} (v_{1} \otimes \dots \otimes v_{n}) = 0$ whenever two of the vectors $v_{k}$ coincide (Permutation sector projections on the n-fold tensor power clauses (a),(c),(d)).

[F3]

Determinant. For a square matrix $A$ of size $n$ over a commutative ring, write $a_{a b} : = A (a - 1, b - 1)$ for $1 \leq a, b \leq n$ . Relabeling the zero-based Leibniz formula then gives $\det (A) = \sum_{σ \in S_{n}} sgn (σ) \prod_{a = 1}^{n} a_{\hat{σ} (a), a}$ ; in particular $\det (I_{n}) = 1$ , and $\det (A) = 0$ if some row of $A$ is zero (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[F4]

Binomial coefficients as counts. $(\binom{d}{k}) = ∣ [d]^{k} ∣$ for $d, k \in N$ , so with $k = n$ the number of $n$ -element subsets of ${1, \dots, d}$ is $(\binom{d}{n})$ (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ).

[F5]

Weak compositions. For $m \geq 1$ and $n \in N$ , $∣ W (n, m) ∣ = (\binom{n + m - 1}{m - 1})$ (For $m \geq 1$ the number of weak compositions of $n$ into $m$ parts is $(\binom{n + m - 1}{m - 1})$ , and the number of compositions is $(\binom{n - 1}{m - 1})$ for $n \geq 1$ ); and $(\binom{n + d - 1}{d - 1}) = (\binom{n + d - 1}{n})$ ( $(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ ; hence $(\binom{n}{k}) k! = n^{\underline{k}}$ , the quotient $n! / (k! (n - k)!)$ is a natural number, and $(\binom{n}{k}) = (\binom{n}{n - k})$ clause 3, with $N = n + d - 1$ and $k = d - 1$ ).

[F6]

Orthonormal expansions. For an orthonormal family $(f_{a})_{a \in A}$ in a Hilbert space whose closed linear span is the whole space, every $x$ equals the limit of the finite partial sums $\sum_{a \in F} ⟨ x, f_{a} ⟩ f_{a}$ ; for a finite complete orthonormal family this is the finite equality $x = \sum_{a} ⟨ x, f_{a} ⟩ f_{a}$ (Parseval equivalences for an orthonormal family clauses 1 and 4, Orthonormal families, complete orthonormal systems and Hilbert bases).

[F7]

Self-adjointness and boundedness. A bounded endomorphism $T$ of $H^{\otimes n}$ is self-adjoint when $⟨ T x, y ⟩ = ⟨ x, T y ⟩$ for all $x, y$ , equivalently $T^{*} = T$ for the Hilbert adjoint of The Hilbert-space adjoint of a bounded operator (A bounded linear operator between normed spaces).

Proof

technique · direct; build the product basis by induction from the tensor construction, determine the two permutation projections on basis vectors by explicit reindexing sums, evaluate the resulting orthonormality as a determinant, read the dimension counts off the index sets, and compute the occupation operators on the two resulting bases

Given: The data of the statement and [F1]-[F7].

1.1F1algebrainduction

[Product basis] We show by induction on $n \geq 1$ that $(e_{i})_{i \in {1, \dots, d}^{n}}$ is an orthonormal basis of $H^{\otimes n}$ . For $n = 1$ this is the hypothesis. If it holds for $n$ , then by [F1] applied to the orthonormal basis $(e_{i})_{i \in {1, \dots, d}^{n}}$ of $H^{\otimes n}$ and $(e_{j})_{j \in {1, \dots, d}}$ of $H$ , the products $e_{i} \otimes e_{j}$ form an orthonormal family with dense span in $H^{\otimes (n + 1)} = H^{\otimes n} \hat{\otimes} H$ . This family is finite, indexed by $Q : = {1, \dots, d}^{n} \times {1, \dots, d}$ . Its span is closed: if $z_{r} = \sum_{q \in Q} c_{r, q} g_{q} \to z$ , where $g_{q}$ denotes the corresponding product vector, orthonormality gives $∥ z_{r} - z_{s} ∥^{2} = \sum_{q \in Q} ∣ c_{r, q} - c_{s, q} ∣^{2}$ , so each coefficient sequence is Cauchy. Let $c_{q}$ be its limit in $C$ ; the same finite norm identity gives $z_{r} \to \sum_{q \in Q} c_{q} g_{q}$ , hence $z = \sum_{q \in Q} c_{q} g_{q}$ lies in the span. Since the span is also dense, it equals the whole tensor product. Thus the family is an orthonormal basis, its cardinality is $d \cdot d^{n} = d^{n + 1}$ and $\dim H^{\otimes (n + 1)} = d^{n + 1}$ . This proves the induction step and (a).

1.2F2algebra

[The alternating projection on basis vectors] Let $i \in {1, \dots, d}^{n}$ . By the action formula of [F2], $P_{-}^{(n)} (e_{i}) = \frac{1}{n!} \sum_{π \in S_{n}} sgn (π) e_{i_{{\hat{π}}^{- 1} (1)}} \otimes \dots \otimes e_{i_{{\hat{π}}^{- 1} (n)}} = \frac{1}{n!} \sum_{τ \in S_{n}} sgn (τ) e_{i_{\hat{τ} (1)}} \otimes \dots \otimes e_{i_{\hat{τ} (n)}}$ after the substitution $τ = π^{- 1}$ . If two entries of $i$ coincide, say $i_{k} = i_{l}$ with $k \neq l$ , then the transposition $υ = (k - 1 l - 1)$ satisfies $e_{i_{\hat{υ τ} (1)}} \otimes \dots = e_{i_{\hat{τ} (1)}} \otimes \dots$ for every $τ$ and $sgn (υ τ) = - sgn (τ)$ , so the sum pairs to zero and $P_{-}^{(n)} (e_{i}) = 0$ by [F2]. If the entries are pairwise distinct with underlying set $J = {j_{1} < \dots < j_{n}}$ and $i_{k} = j_{\hat{ρ_{i}} (k)}$ , then $e_{i_{\hat{τ} (1)}} \otimes \dots = e_{j_{\hat{ρ_{i} τ} (1)}} \otimes \dots$ and $sgn (τ) = sgn (ρ_{i}) sgn (ρ_{i} τ)$ ; substituting $ω = ρ_{i} τ$ gives $P_{-}^{(n)} (e_{i}) = \frac{sgn (ρ_{i})}{n!} \sum_{ω \in S_{n}} sgn (ω) e_{j_{\hat{ω} (1)}} \otimes \dots \otimes e_{j_{\hat{ω} (n)}} = sgn (ρ_{i}) \frac{e_{J}^{\land}}{\sqrt{n!}}$ , which is the second formula of (b).

1.3F1F2F3algebra

[Orthonormality of the Slater family] For subsets $J, K$ with $∣ J ∣ = ∣ K ∣ = n$ let $A$ be the canonical matrix with $A (a - 1, b - 1) = ⟨ e_{j_{a}}, e_{k_{b}} ⟩ = δ_{j_{a}, k_{b}}$ , and abbreviate this entry as $A_{a b}$ for $1 \leq a, b \leq n$ . By [F1], $⟨ e_{J}^{\land}, e_{K}^{\land} ⟩ = \frac{1}{n!} \sum_{σ, π \in S_{n}} sgn (σ) sgn (π) \prod_{l = 1}^{n} A_{\hat{σ} (l), \hat{π} (l)}$ . For each fixed $π$ substitute $σ = ω π$ and reindex the product by $t = \hat{π} (l)$ . Then $\prod_{l} A_{\hat{σ} (l), \hat{π} (l)} = \prod_{t} A_{\hat{ω} (t), t}$ and $sgn (σ) sgn (π) = sgn (ω)$ , so the sum over $π$ contributes $n!$ and the remaining sum is $\det (A)$ by [F3]. For $J = K$ , $A = I_{n}$ and this is $1$ ; for $J \neq K$ , some $j_{a}$ is absent from $K$ , giving a zero row and determinant $0$ . Thus the Slater family is orthonormal.

2.1F6F7step 1.1algebra

[Occupation operators] For $j \in {1, \dots, d}$ the formula of (d) defines a linear endomorphism of the finite-dimensional space $H^{\otimes n}$ ; by [F6] applied to the complete orthonormal family $(e_{i})$ we have $x = \sum_{i} ⟨ x, e_{i} ⟩ e_{i}$ and $N_{j} x = \sum_{i} ∣ i ∣_{j} ⟨ x, e_{i} ⟩ e_{i}$ for every $x$ . Self-adjointness: for $x, y \in H^{\otimes n}$ the expansions give $⟨ N_{j} x, y ⟩ = \sum_{i} ∣ i ∣_{j} ⟨ x, e_{i} ⟩ \overline{⟨ y, e_{i} ⟩} = ⟨ x, N_{j} y ⟩$ because the numbers $∣ i ∣_{j}$ are real, so $N_{j}$ is self-adjoint by [F7]; it is bounded because its range is spanned by the finitely many $e_{i}$ with eigenvalue at most $n$ , and $∥ N_{j} x ∥ \leq n ∥ x ∥$ . The same computation on the basis vectors $e_{i}$ , on which $N_{j}$ is diagonal with the real eigenvalues $∣ i ∣_{j}$ , gives $N_{j} N_{k} e_{i} = ∣ i ∣_{k} ∣ i ∣_{j} e_{i} = N_{k} N_{j} e_{i}$ and $\sum_{j} N_{j} e_{i} = (\sum_{j} ∣ i ∣_{j}) e_{i} = n e_{i}$ ; since the $e_{i}$ form a basis by step 1.1 and both sides of each identity are linear, $N_{j} N_{k} = N_{k} N_{j}$ and $\sum_{j = 1}^{d} N_{j} = n I$ . This proves the operator statements of (d) except the two eigenvector formulas.

2.2F4step 1.1step 1.2step 1.3algebra

[Slater basis and its dimension] By step 1.2 each $e_{J}^{\land} = \sqrt{n!} P_{-}^{(n)} (e_{j_{1}} \otimes \dots \otimes e_{j_{n}})$ lies in $ran P_{-}^{(n)}$ , and each $P_{-}^{(n)} (e_{i})$ is either zero or a multiple of some $e_{J}^{\land}$ ; since the $e_{i}$ form a basis of $H^{\otimes n}$ by step 1.1 and $P_{-}^{(n)}$ is linear, the closed linear span of $(e_{J}^{\land})$ is exactly $ran P_{-}^{(n)}$ (a finite-dimensional, hence closed, subspace). By step 1.3 the family is orthonormal, so it is an orthonormal basis of $ran P_{-}^{(n)}$ , indexed by the $n$ -element subsets $J$ of ${1, \dots, d}$ ; by [F4] there are $(\binom{d}{n})$ of them, so $\dim ran P_{-}^{(n)} = (\binom{d}{n})$ . If $n > d$ , then two of the $n$ entries of any $i$ coincide, so $P_{-}^{(n)} (e_{i}) = 0$ for every basis vector $e_{i}$ by step 1.2, whence $P_{-}^{(n)} = 0$ and $ran P_{-}^{(n)} = {0}$ . In particular $e_{J}^{\land} \neq 0$ whenever $∣ J ∣ = n \leq d$ . This completes (b).

2.3F1F2step 1.1algebra

[The symmetric sums] Fix $λ \in W (n, d)$ . The word $c$ that lists $1$ with multiplicity $λ_{1}$ , then $2$ with multiplicity $λ_{2}$ , and so on through $d$ has $∣ c ∣ = λ$ , so $orb (λ) \neq \emptyset$ and $m_{λ} = \sum_{c \in orb (λ)} e_{c}$ is a nonempty sum of orthonormal vectors of $H^{\otimes n}$ , hence nonzero, with $∥ m_{λ} ∥^{2} = ∣ orb (λ) ∣$ by the orthonormality of the product basis in step 1.1 [F1]. If $λ \neq μ$ then a word has a unique multiplicity vector, so $orb (λ) \cap orb (μ) = \emptyset$ and the inner product of $m_{λ}$ and $m_{μ}$ is the empty sum $0$ . For $π \in S_{n}$ the map $c \mapsto π \cdot c$ , $(π \cdot c)_{k} : = c_{{\hat{π}}^{- 1} (k)}$ , permutes $orb (λ)$ and $U_{π} e_{c} = e_{π \cdot c}$ by [F2], so $U_{π} m_{λ} = m_{λ}$ . Finally, $P_{+}^{(n)} e_{i} = \frac{1}{n!} \sum_{π \in S_{n}} e_{π \cdot i}$ has coefficients supported on $orb (∣ i ∣)$ ; for $c \in orb (∣ i ∣)$ the set ${π : π \cdot i = c}$ is the coset $π_{0} Stab (i)$ for any $π_{0}$ with $π_{0} \cdot i = c$ , hence has cardinality $∣ Stab (i) ∣ \geq 1$ , and it is empty for $c \notin orb (∣ i ∣)$ ; therefore $P_{+}^{(n)} e_{i} = γ_{i} m_{∣ i ∣}$ with $γ_{i} : = ∣ Stab (i) ∣ / n! > 0$ . This proves all the assertions of (c) except the dimension.

3.1F5step 2.3algebra

[Symmetric basis and its dimension] By step 2.3 the $m_{λ}$ are nonzero, pairwise orthogonal and lie in $ran P_{+}^{(n)}$ (indeed $m_{λ} = γ_{i}^{- 1} P_{+}^{(n)} e_{i}$ for any multi-index $i$ with $∣ i ∣ = λ$ , and such $i$ exists because $λ \in W (n, d)$ ); conversely every $P_{+}^{(n)} e_{i}$ is a multiple of some $m_{λ}$ , so by linearity and the basis property of the $e_{i}$ the span of $(m_{λ})_{λ \in W (n, d)}$ is $ran P_{+}^{(n)}$ . A pairwise orthogonal family of nonzero vectors spanning a finite-dimensional space is a basis, so $\dim ran P_{+}^{(n)} = ∣ W (n, d) ∣ = (\binom{n + d - 1}{d - 1}) = (\binom{n + d - 1}{n})$ by [F5]. This completes (c).

4.1step 2.1step 2.2step 3.1algebra

[Occupation numbers on the two bases] Every summand $e_{c}$ of $e_{J}^{\land}$ has exactly one entry equal to $j$ when $j \in J$ and none otherwise, so $∣ c ∣_{j} = 1_{J} (j)$ for every such $c$ ; and every $c \in orb (λ)$ has $∣ c ∣_{j} = λ_{j}$ . Since the $N_{j}$ are linear and act diagonally on the $e_{c}$ by the formula of step 2.1, this gives $N_{j} e_{J}^{\land} = 1_{J} (j) e_{J}^{\land}$ and $N_{j} m_{λ} = λ_{j} m_{λ}$ , the remaining assertions of (d). Both basis families displayed in (b) and (c) therefore consist of joint eigenvectors of $(N_{1}, \dots, N_{d})$ , and the two maps $J \mapsto 1_{J}$ and $λ \mapsto λ$ are injective; hence the joint eigenspaces inside the two sectors are the lines spanned by the corresponding basis vectors, as asserted in (d), and each sector is invariant under every $N_{j}$ .

5.1step 1.1step 1.2step 1.3step 2.2step 2.3step 3.1step 2.1step 4.1∎

[Assembly] Step 1.1 proves (a), steps 1.2, 1.3 and 2.2 prove (b), steps 2.3 and 3.1 prove (c), and steps 2.1 and 4.1 prove (d). The proof uses only the completed tensor powers, the permutation projections, the orthonormal expansions and the stated counting facts, under the Countable Choice hypothesis of [F1] and [F2]; no physical reading of the sectors or of the occupation operators is asserted.

Remarks

Reading of the operators. $N_{j}$ counts the particles occupying the declared basis vector $e_{j}$ ; the theorem proves the exact statements that the two permutation sectors diagonalise the family $(N_{1}, \dots, N_{d})$ and that the antisymmetric sector has only the occupation values $0$ and $1$ . The physical reading of this diagonalisation as an occupation readout of the framework belongs to The declared finite-mode model and its occupation readouts and Pauli exclusion in the declared finite-mode occupation readout, with the bosonic contrast.
Finite dimension. The dimension of $H$ is finite and is part of the hypotheses; no infinite-dimensional Slater basis or occupation operator is constructed here. The open case is treated by different machinery on the Fock-space pair.

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Finite-mode occupation configurations: fermionic exclusion counts, bosonic counts and truncated dimensions

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a complex Hilbert space of finite dimension $d \geq 1$ with orthonormal basis $(e_{1}, \dots, e_{d})$ (Hilbert space), let $H^{\otimes n}$ and $P_{\pm}^{(n)}$ be as in Slater and symmetric occupation-number bases of the finite-dimensional permutation sectors (with $P_{\pm}^{(0)} = {i d}_{C}$ ), and let $W (n, d) = {λ \in N^{d} : \sum_{j} λ_{j} = n}$ and $m_{λ}$ be the symmetric vectors of that theorem. Then:

(a) fermionic configurations: the map $J \mapsto 1_{J}$ is a bijection from the set of $n$ -element subsets of ${1, \dots, d}$ onto the set ${λ \in {0, 1}^{d} : \sum_{j} λ_{j} = n}$ ; consequently $\dim ran P_{-}^{(n)} = # {λ \in {0, 1}^{d} : \sum_{j} λ_{j} = n}, \sum_{n = 0}^{d} \dim ran P_{-}^{(n)} = 2^{d};$

(b) bosonic configurations: the assignment $λ \mapsto m_{λ}$ induces a bijection from $W (n, d)$ onto the set of lines spanned by the orthogonal basis $(m_{λ})$ , so the number of bosonic configurations of total occupation exactly $n$ is $(\binom{n + d - 1}{n}) = (\binom{n + d - 1}{d - 1})$ , and for every $N \geq 0$ the number of bosonic configurations of total occupation at most $N$ is $\sum_{n = 0}^{N} \dim ran P_{+}^{(n)} = \sum_{n = 0}^{N} (\binom{n + d - 1}{d - 1}) = (\binom{N + d}{d});$

(c) exclusion counts: no fermionic configuration has an entry $\geq 2$ , so the antisymmetric Slater index set contains no $λ$ with $λ_{j} \geq 2$ for some $j$ ; for $n > d$ there is no fermionic configuration at all and $ran P_{-}^{(n)} = {0}$ , while for every $n$ and every $j$ the bosonic configuration with $λ_{j} = n$ and all other entries $0$ occurs, so no occupation bound holds in the symmetric sector;

(d) consistency with the total occupation: for every $λ \in W (n, d)$ one has $\sum_{j} λ_{j} = n$ , and for every fermionic configuration $λ \in {0, 1}^{d} \cap {\sum_{j} λ_{j} = n}$ likewise; in particular the total occupation equals the particle number $n$ in both sectors.

Facts & Assumptions

Given: The finite-dimensional complex Hilbert space $H$ of dimension $d \geq 1$ with orthonormal basis $(e_{1}, \dots, e_{d})$ , the sectors $ran P_{\pm}^{(n)}$ of the cited theorem, and $n, N \geq 0$ .

[F0]

Empty configuration, directly. At $n = 0$ , the Statement's conventions give $H^{\otimes 0} = C$ and $P_{\pm}^{(0)} = {i d}_{C}$ . The empty tensor is the scalar $1$ , so the empty symmetric vector is $m_{0} = 1$ , where $0$ is the unique element of $W (0, d)$ . Thus both sector ranges are one-dimensional and have this single normalized basis vector; the unique fermionic configuration is the indicator of the empty subset and is also $0$ . These are direct calculations in $C$ , not consequences of the cited theorem.

[F1]

Occupation basis for positive particle number. For $n \geq 1$ , the Slater vectors $e_{J}^{\land}$ , $J \subseteq {1, \dots, d}$ with $∣ J ∣ = n$ , form an orthonormal basis of $ran P_{-}^{(n)}$ , and $\dim ran P_{-}^{(n)} = (\binom{d}{n})$ with $ran P_{-}^{(n)} = {0}$ for $n > d$ ; the symmetric vectors $(m_{λ})_{λ \in W (n, d)}$ form an orthogonal basis of $ran P_{+}^{(n)}$ with $\dim ran P_{+}^{(n)} = (\binom{n + d - 1}{n})$ ; each $e_{J}^{\land}$ has occupation vector $1_{J}$ and each $m_{λ}$ has occupation vector $λ$ (Slater and symmetric occupation-number bases of the finite-dimensional permutation sectors clauses (b),(c),(d)).

[F2]

Binomial counts and symmetry. $(\binom{d}{n})$ is the number of $n$ -element subsets of ${1, \dots, d}$ , $(\binom{n}{0}) = (\binom{n}{n}) = 1$ , and $(\binom{n}{k}) = 0$ for $k > n$ (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ); also $(\binom{n + d - 1}{d - 1}) = (\binom{n + d - 1}{n})$ ( $(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ ; hence $(\binom{n}{k}) k! = n^{\underline{k}}$ , the quotient $n! / (k! (n - k)!)$ is a natural number, and $(\binom{n}{k}) = (\binom{n}{n - k})$ clause 3, with $N = n + d - 1$ and $k = d - 1$ ).

[F3]

[F4]

Pascal rule and hockey stick. For all $n, k \in N$ , $(\binom{n + 1}{k + 1}) = (\binom{n}{k}) + (\binom{n}{k + 1})$ and $\sum_{i < n + 1} (\binom{i}{k}) = (\binom{n + 1}{k + 1})$ as an $N$ -valued finite sum (Pascal's rule $(\binom{n + 1}{k + 1}) = (\binom{n}{k}) + (\binom{n}{k + 1})$ , and the hockey-stick identity $\sum_{i \leq n} (\binom{i}{k}) = (\binom{n + 1}{k + 1})$ ); the sector formula of the theorem is stated for the finite sums there defined.

[F5]

Row sum. $\sum_{k < n + 1} (\binom{n}{k}) = 2^{n}$ in $N$ ( $\sum_{k < n + 1} (\binom{n}{k}) = 2^{n}$ , and $\sum_{k < n + 1} (- 1)^{k} ι ⁣ (\binom{n}{k}) = 0$ for $n \geq 1$ clause 1).

Proof

technique · direct; compare the two index sets of the occupation basis with the configuration sets, count them with the cited finite-counting facts, and read the vanishing statement off the antisymmetric sector

Given: The data of the statement and [F0]-[F5].

1.1F0F1F2algebraconstruct

[Fermionic configurations] Let $J \subseteq {1, \dots, d}$ have $∣ J ∣ = n$ ; its indicator $1_{J}$ lies in ${0, 1}^{d}$ with $\sum_{j} 1_{J} (j) = n$ . Conversely, if $λ \in {0, 1}^{d}$ has $\sum_{j} λ_{j} = n$ , then $J_{λ} : = {j : λ_{j} = 1}$ satisfies $∣ J_{λ} ∣ = n$ and $1_{J_{λ}} = λ$ . The two assignments are inverse bijections, so the number of fermionic configurations of total occupation $n$ equals the number of $n$ -element subsets, which is $(\binom{d}{n})$ by [F2]. For $n \geq 1$ , this is $\dim ran P_{-}^{(n)}$ by [F1]; for $n = 0$ , the dimension is $1 = (\binom{d}{0})$ by [F0] and the boundary value in [F2]. This proves (a)'s dimension assertion in all cases.

1.2F0F1F2F3algebra

[Bosonic configurations] If $n = 0$ , then $W (0, d) = {0}$ and $m_{0} = 1$ by [F0], so its assignment is a bijection onto the single basis line of $ran P_{+}^{(0)} = C$ . If $n \geq 1$ , the assignment $λ \mapsto m_{λ}$ is injective by the linear independence of the orthogonal family [F1], and its image spans $ran P_{+}^{(n)}$ , so it is a bijection onto the set of basis lines. In either case, the number of its elements is $∣ W (n, d) ∣ = (\binom{n + d - 1}{d - 1}) = (\binom{n + d - 1}{n})$ by [F3] and [F2], the first assertion of (b).

1.3F0F1F2F4algebra

[Truncated bosonic dimension] For $n \geq 1$ the dimension $\dim ran P_{+}^{(n)} = (\binom{n + d - 1}{n}) = (\binom{n + d - 1}{d - 1})$ is given by [F1] and [F2]; at $n = 0$ it is $1 = (\binom{d - 1}{d - 1})$ by [F0] and the boundary value of the binomial coefficient. Thus $\sum_{n = 0}^{N} \dim ran P_{+}^{(n)} = \sum_{n = 0}^{N} (\binom{n + d - 1}{d - 1})$ . Substituting $i : = n + d - 1$ , the terms with $i < d - 1$ vanish by the boundary values of the binomial coefficient, so the hockey stick [F4] with $k = d - 1$ and upper index $N + d - 1$ gives $\sum_{i = 0}^{N + d - 1} (\binom{i}{d - 1}) = (\binom{N + d}{d})$ ; this completes (b).

2.1F0F1F5step 1.1algebra

[Fermionic row sum] Summing the dimensions over all particle numbers, using [F0] at $n = 0$ and [F1] for $1 \leq n \leq d$ and the vanishing for $n > d$ , gives $\sum_{n = 0}^{d} \dim ran P_{-}^{(n)} = \sum_{n = 0}^{d} (\binom{d}{n}) = 2^{d}$ by the row sum [F5], which completes (a).

2.2F1step 1.1step 1.2algebra

[Exclusion counts] By construction every fermionic configuration lies in ${0, 1}^{d}$ , so no entry is $\geq 2$ ; for $n > d$ the set ${0, 1}^{d} \cap {\sum_{j} λ_{j} = n}$ is empty because at most $d$ entries can be equal to $1$ , and then $ran P_{-}^{(n)} = {0}$ by [F1], while for the bosonic sector the configuration $λ^{(j)}$ with $λ_{j}^{(j)} = n$ and all other entries $0$ lies in $W (n, d)$ for every $j$ , so no occupation bound holds there; this is (c).

3.1step 1.1step 1.2step 2.1step 1.3step 2.2algebra∎

[Consistency and assembly] Each $λ \in W (n, d)$ satisfies $\sum_{j} λ_{j} = n$ by definition, and each fermionic configuration in ${0, 1}^{d} \cap {\sum_{j} λ_{j} = n}$ does so as well, which is (d); step 1.1 and 2.1 prove (a), steps 1.2 and 1.3 prove (b), and step 2.2 proves (c). Only the cited spectral basis and finite-counting facts were used.

Remarks

Reading. The theorem is the finite-mode bookkeeping behind the occupation-number description: for a $d$ -mode one-particle space there are $(\binom{d}{n})$ antisymmetric $n$ -particle sector dimensions and $(\binom{n + d - 1}{n})$ symmetric ones, and the antisymmetric configurations have all occupations in ${0, 1}$ while the symmetric ones do not. The physical reading of the occupations as a readout of the framework is fixed on The declared finite-mode model and its occupation readouts and Pauli exclusion in the declared finite-mode occupation readout, with the bosonic contrast.
Truncation. The sums over $n \leq N$ are finite sums of finite dimensions; no infinite Fock space and no convergence statement is made here.

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Regrouping two factors: the symmetry decomposition of a two-particle sector over an orbital and an internal factor

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H, V$ be nonzero finite-dimensional complex Hilbert spaces (Hilbert space), $\dim H = m$ , $\dim V = k$ . All tensor products below are the completed tensor products of Hilbert tensor construction: inner product, completion and product bases, with elementary tensors $u \otimes v$ , and all permutation projections $P_{\pm}^{(2)}$ and exchange operators $τ_{W}$ are those of Permutation sector projections on the n-fold tensor power and The two-particle exchange operator and the symmetric and antisymmetric combinations for the displayed Hilbert space $W$ . Then:

(a) identification: the inclusion $κ : H ⊙ V \to H \hat{\otimes} V$ of the algebraic tensor product is a surjective linear isometry, so elementary tensors and finite sums of them may be treated as elements of the completed product without ambiguity; the same holds for $(H \otimes V) \hat{\otimes} (H \otimes V)$ and $(H \otimes H) \hat{\otimes} (V \otimes V)$ ;

(b) regrouping unitary: there is a unique bounded linear operator $Θ : (H \otimes V) \hat{\otimes} (H \otimes V) \to (H \otimes H) \hat{\otimes} (V \otimes V)$ with $Θ ((u \otimes χ) \otimes (v \otimes χ^{'})) = (u \otimes v) \otimes (χ \otimes χ^{'})$ for all $u, v \in H$ , $χ, χ^{'} \in V$ ; $Θ$ is unitary and its inverse is the unique bounded operator with $Θ^{- 1} ((u \otimes v) \otimes (χ \otimes χ^{'})) = (u \otimes χ) \otimes (v \otimes χ^{'})$ ;

(c) exchange intertwining: $Θ τ_{H \otimes V} = (τ_{H} \otimes τ_{V}) Θ$ , where $τ_{H} \otimes τ_{V}$ is the bounded tensor product of the two exchange operators;

(d) symmetriser identities: $Θ P_{-}^{(H \otimes V)} = (P_{-}^{(H)} \otimes P_{+}^{(V)} + P_{+}^{(H)} \otimes P_{-}^{(V)}) Θ$ and $Θ P_{+}^{(H \otimes V)} = (P_{+}^{(H)} \otimes P_{+}^{(V)} + P_{-}^{(H)} \otimes P_{-}^{(V)}) Θ$ ; consequently $Θ$ maps $ran P_{-}^{(H \otimes V)}$ onto the orthogonal direct sum $(ran P_{-}^{(H)} \otimes ran P_{+}^{(V)}) + (ran P_{+}^{(H)} \otimes ran P_{-}^{(V)})$ and $ran P_{+}^{(H \otimes V)}$ onto $(ran P_{+}^{(H)} \otimes ran P_{+}^{(V)}) + (ran P_{-}^{(H)} \otimes ran P_{-}^{(V)})$ ;

(e) vector identities: for all $u, v \in H$ and $χ, χ^{'} \in V$ , $Θ (P_{-}^{(H \otimes V)} ((u \otimes χ) \otimes (v \otimes χ^{'}))) = P_{+}^{(H)} (u \otimes v) \otimes P_{-}^{(V)} (χ \otimes χ^{'}) + P_{-}^{(H)} (u \otimes v) \otimes P_{+}^{(V)} (χ \otimes χ^{'}),$ $Θ (P_{+}^{(H \otimes V)} ((u \otimes χ) \otimes (v \otimes χ^{'}))) = P_{+}^{(H)} (u \otimes v) \otimes P_{+}^{(V)} (χ \otimes χ^{'}) + P_{-}^{(H)} (u \otimes v) \otimes P_{-}^{(V)} (χ \otimes χ^{'});$

(f) dimensions: $\dim ran P_{-}^{(H \otimes V)} = (\binom{m}{2}) (\binom{k + 1}{2}) + (\binom{m + 1}{2}) (\binom{k}{2}) = (\binom{m k}{2}),$ with the two summands of (d) having dimensions $(\binom{m}{2}) (\binom{k + 1}{2})$ and $(\binom{m + 1}{2}) (\binom{k}{2})$ .

Facts & Assumptions

Given: The finite-dimensional nonzero complex Hilbert spaces $H, V$ of dimensions $m, k$ , their completed tensor products, the elementary tensors and the permutation projections and exchange operators of the cited items.

[F1]

Tensor construction. For complex Hilbert spaces $X, Y$ the inclusion $κ : X ⊙ Y \to X \hat{\otimes} Y$ is a linear isometry with dense image; $κ (u \otimes v)$ is bilinear, $∥ κ (u \otimes v) ∥ = ∥ u ∥ ∥ v ∥$ and $⟨ κ (u \otimes v), κ (u^{'} \otimes v^{'}) ⟩ = ⟨ u, u^{'} ⟩ ⟨ v, v^{'} ⟩$ ; every element of $X ⊙ Y$ is a finite sum of elementary tensors, and if $(e_{a})$ and $(f_{b})$ are orthonormal bases of $X$ and $Y$ then $(κ (e_{a} \otimes f_{b}))$ is an orthonormal basis of $X \hat{\otimes} Y$ (Hilbert tensor construction: inner product, completion and product bases clauses (a)-(d)).

[F2]

Multilinear factorisation. For a finite list of modules $M_{1}, \dots, M_{k}$ over a commutative ring, every parenthesised tensor product represents multilinear maps from $M_{1} \times \dots \times M_{k}$ , and different parenthesizations are connected by the unique isomorphism preserving pure tensors (Finite iterated tensor products represent multilinear maps independently of parenthesization).

[F3]

Finite-dimensional subspaces are closed. A subspace of a normed space that admits a finite basis is closed (A finite-dimensional normed subspace is closed).

[F4]

Exchange operators and projections. For a Hilbert space $W$ , the exchange operator $τ_{W} : = U_{(0 1)}$ satisfies $τ_{W}^{2} = I$ and $τ_{W}^{*} = τ_{W}$ and $τ_{W} (w \otimes w^{'}) = w^{'} \otimes w$ , and $P_{\pm}^{(W)} = \frac{1}{2} (I \pm τ_{W})$ are the orthogonal projections with the corresponding ranges (The two-particle exchange operator and the symmetric and antisymmetric combinations clauses (a),(b), Permutation sector projections on the n-fold tensor power clauses (a),(c)).

[F5]

Bounded tensor maps. For bounded operators $S, S^{'}$ on $X$ and $T, T^{'}$ on $Y$ there is a unique bounded $S \otimes T$ with $(S \otimes T) κ (u \otimes v) = κ^{'} (S u \otimes T v)$ , $∥ S \otimes T ∥ \leq ∥ S ∥ ∥ T ∥$ , $(S \otimes T) (S^{'} \otimes T^{'}) = (S S^{'}) \otimes (T T^{'})$ , $(S \otimes T)^{*} = S^{*} \otimes T^{*}$ , and unitarity of $S, T$ gives unitarity of $S \otimes T$ (Bounded tensor maps: extension, norm estimate, adjoint and commutation clauses (a)-(d), A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[F6]

Dimensions. For finite-dimensional vector spaces $X, Y$ over a field one has $\dim (X \otimes Y) = \dim X \cdot \dim Y$ ( $R^{m} \otimes_{R} R^{n} ≅ R^{m n}$ with the product basis, and $\dim_{F} (V \otimes_{F} W) = \dim_{F} V \dim_{F} W$ ); and $\dim ran P_{-}^{(n)} = (\binom{d}{n})$ , $\dim ran P_{+}^{(n)} = (\binom{n + d - 1}{n})$ for a $d$ -dimensional space (Slater and symmetric occupation-number bases of the finite-dimensional permutation sectors clauses (b),(c)).

[F7]

Orthogonal projections. If $P, Q$ are orthogonal projections with $P Q = Q P = 0$ , then $P + Q$ is an orthogonal projection with $ran (P + Q) = ran P + ran Q$ a direct sum of orthogonal subspaces; a product $P \otimes Q$ of orthogonal projections is an orthogonal projection (The Hilbert orthogonal projection onto a closed subspace, Permutation sector projections on the n-fold tensor power clause (b)).

Proof

technique · direct; identify the finite-dimensional algebraic and completed tensor products, build the regrouping map from the multilinear factorisation of the four-variable assignment, verify inversion and isometry on elementary tensors, transfer the exchange and symmetriser identities from elementary tensors by linearity, and read the decompositions and dimensions off the resulting projection identities

Given: The data of the statement and [F1]-[F7].

1.1F1F3algebra

[Identification] By [F1] the inclusion $κ : H ⊙ V \to H \hat{\otimes} V$ is a linear isometry with dense image; if $(e_{1}, \dots, e_{m})$ and $(f_{1}, \dots, f_{k})$ are orthonormal bases of $H$ and $V$ , then the $m k$ elements $e_{a} \otimes f_{b}$ span the algebraic tensor product by bilinearity, so $κ [H ⊙ V]$ is a finite-dimensional subspace of the Hilbert space $H \hat{\otimes} V$ , hence closed by [F3], and a dense closed subspace is the whole space; thus $κ$ is onto and identifies the two spaces, and the same argument applies to the algebraic and completed tensor products of $H \otimes V$ with itself and of $H \otimes H$ with $V \otimes V$ , whose algebraic parts are spanned by finitely many elementary tensors. This proves (a).

2.1F1F2step 1.1algebraconstruct

[The regrouping map and its inverse] Define $Φ (u, v, χ, χ^{'}) : = (u \otimes v) \otimes (χ \otimes χ^{'})$ for $u, v \in H$ , $χ, χ^{'} \in V$ ; it is multilinear in its four arguments, so by [F2] it corresponds to a unique linear map on the four-fold algebraic tensor product and, by the parenthesization independence of [F2] together with the identification of step 1.1, to a unique linear map $Θ$ on $(H \otimes V) \hat{\otimes} (H \otimes V)$ with the displayed values on elementary tensors; uniqueness holds because those elementary tensors span the space. Likewise $Ψ ((u \otimes v) \otimes (χ \otimes χ^{'})) : = (u \otimes χ) \otimes (v \otimes χ^{'})$ is multilinear and defines a linear map $Ψ$ on $(H \otimes H) \hat{\otimes} (V \otimes V)$ . On elementary tensors $Ψ Θ = i d$ and $Θ Ψ = i d$ , and two linear maps agreeing on a spanning set agree everywhere; hence $Θ$ is a linear isomorphism with inverse $Ψ$ , which is the existence and inverse statement of (b).

3.1F1step 2.1algebra

[Unitarity] For elementary tensors $x = (u \otimes χ) \otimes (v \otimes χ^{'})$ and $y = (u^{'} \otimes χ^{''}) \otimes (v^{'} \otimes χ^{'''})$ in the domain the inner-product identities of [F1] give $⟨ Θ x, Θ y ⟩ = ⟨ u \otimes v, u^{'} \otimes v^{'} ⟩ ⟨ χ \otimes χ^{'}, χ^{''} \otimes χ^{'''} ⟩ = ⟨ u, u^{'} ⟩ ⟨ v, v^{'} ⟩ ⟨ χ, χ^{''} ⟩ ⟨ χ^{'}, χ^{'''} ⟩ = ⟨ x, y ⟩$ ; the difference of the two sesquilinear forms vanishes on all pairs of elementary tensors, and since every vector is a finite sum of elementary tensors by [F1] it vanishes identically, so $Θ$ is an isometry; being invertible by step 2.1 it is unitary, which completes (b).

3.2F4F5step 2.1algebra

[Intertwining with the exchange operators] For $u, v \in H$ , $χ, χ^{'} \in V$ one computes with [F4] and [F5] that $Θ τ_{H \otimes V} ((u \otimes χ) \otimes (v \otimes χ^{'})) = Θ ((v \otimes χ^{'}) \otimes (u \otimes χ)) = (v \otimes u) \otimes (χ^{'} \otimes χ)$ and $(τ_{H} \otimes τ_{V}) Θ ((u \otimes χ) \otimes (v \otimes χ^{'})) = (τ_{H} \otimes τ_{V}) ((u \otimes v) \otimes (χ \otimes χ^{'})) = (v \otimes u) \otimes (χ^{'} \otimes χ)$ ; both sides are bounded linear maps agreeing on the spanning set of elementary tensors, so they agree everywhere, which is (c).

4.1F4F5step 3.2algebra

[Symmetriser identities] By [F4], $P_{\pm}^{(H \otimes V)} = \frac{1}{2} (I \pm τ_{H \otimes V})$ , $P_{\pm}^{(H)} = \frac{1}{2} (I \pm τ_{H})$ and $P_{\pm}^{(V)} = \frac{1}{2} (I \pm τ_{V})$ ; using step 3.2 to move $Θ$ past $τ_{H \otimes V}$ and the linearity and multiplicativity of the tensor product of operators [F5], one obtains on the whole space $Θ P_{-}^{(H \otimes V)} = \frac{1}{2} Θ - \frac{1}{2} (τ_{H} \otimes τ_{V}) Θ = (\frac{1}{2} - \frac{1}{2} τ_{H} \otimes τ_{V}) Θ = (P_{-}^{(H)} \otimes P_{+}^{(V)} + P_{+}^{(H)} \otimes P_{-}^{(V)}) Θ$ , because $\frac{1}{2} (I - τ_{H} \otimes τ_{V}) = \frac{1}{2} (I - τ_{H}) \otimes \frac{1}{2} (I + τ_{V}) + \frac{1}{2} (I + τ_{H}) \otimes \frac{1}{2} (I - τ_{V})$ and its sign-reversed analogue are identities of bounded operators; this proves the first assertion of (d), and the computation with both signs reversed gives the second.

5.1F7step 3.1step 4.1algebra

[Direct-sum decompositions] By [F7] the operators $P_{-}^{(H)} \otimes P_{+}^{(V)}$ and $P_{+}^{(H)} \otimes P_{-}^{(V)}$ are orthogonal projections whose product is $(P_{-}^{(H)} P_{+}^{(H)}) \otimes (P_{+}^{(V)} P_{-}^{(V)}) = 0$ , so their sum is an orthogonal projection with range the orthogonal direct sum of the two ranges; by step 4.1 and the unitarity of step 3.1, the image of $ran P_{-}^{(H \otimes V)}$ under $Θ$ is that direct sum, and the same argument with the second identity of step 4.1 gives the symmetric-sector decomposition, completing (d).

5.2step 4.1algebra

[Vector identities] Applying the first identity of step 4.1 to the elementary tensor $(u \otimes χ) \otimes (v \otimes χ^{'})$ and using the linearity of $Θ$ and of tensor products of operators gives $Θ P_{-}^{(H \otimes V)} ((u \otimes χ) \otimes (v \otimes χ^{'})) = (P_{-}^{(H)} \otimes P_{+}^{(V)} + P_{+}^{(H)} \otimes P_{-}^{(V)}) ((u \otimes v) \otimes (χ \otimes χ^{'})) = P_{+}^{(H)} (u \otimes v) \otimes P_{-}^{(V)} (χ \otimes χ^{'}) + P_{-}^{(H)} (u \otimes v) \otimes P_{+}^{(V)} (χ \otimes χ^{'})$ , which is the first identity of (e); the second follows in the same way from the second identity of step 4.1.

6.1F6step 3.1step 5.1algebra

[Dimensions] By step 5.1 and the unitarity of step 3.1, $\dim ran P_{-}^{(H \otimes V)} = \dim (ran P_{-}^{(H)} \otimes ran P_{+}^{(V)}) + \dim (ran P_{+}^{(H)} \otimes ran P_{-}^{(V)})$ ; by [F6] applied to the $m$ - and $k$ -dimensional spaces this is $(\binom{m}{2}) (\binom{k + 1}{2}) + (\binom{m + 1}{2}) (\binom{k}{2})$ , which are the displayed summand dimensions, and expanding the two products gives $\frac{m (m - 1) k (k + 1) + m (m + 1) k (k - 1)}{4} = \frac{m k (m k - 1)}{2} = (\binom{m k}{2})$ , which is (f).

7.1step 1.1step 2.1step 3.1step 3.2step 4.1step 5.1step 5.2step 6.1∎

[Assembly] Step 1.1 proves (a), steps 2.1 and 3.1 prove (b), step 3.2 proves (c), steps 4.1 and 5.1 prove (d), step 5.2 proves (e) and step 6.1 proves (f). Only the cited linear algebra, tensor products and permutation projections were used, under the Countable Choice hypotheses of [F1], [F2] and [F5]; no physical reading of the factors is asserted.

Remarks

Reading of the decomposition. With $H$ an orbital factor and $V$ an internal (spin) factor, the first summand of the antisymmetric sector pairs antisymmetric orbital two-vectors with symmetric internal two-vectors and the second pairs symmetric orbital two-vectors with antisymmetric internal two-vectors, so the internal permutation symmetry is opposite to the orbital one. The physical use of this identity for two spin-half fermions is developed on the identical-particles examples page.
Finite dimension. Both factors are finite dimensional, so the algebraic and completed tensor products are identified in (a); no infinite-dimensional regrouping is asserted, and the identities themselves are transferred from elementary tensors by linearity.

DefinitionDefinition: AI-adaptedProof: Not applicableOpen item page →

Declared species sectors and their preparations: the framework's reading of P6

Physical scope

The framework's vocabulary for identical-particle sectors: a declared species with its one-particle space, spin label and empirical parameters, the adopted symmetric or antisymmetric n-particle sector, its declared sector preparations, and the rule that a bounded quantity of the labelled tensor product is read on the sector only when it preserves the sector. The sector assignment is an adopted model assumption of P6; the definition derives nothing about nature, asserts no spin-statistics theorem, no dynamics, no measurement law beyond P2 and no empirical fact.

Definition

(a) Adopted readings. Adopt the state and composition postulate State and composition postulate P1: Hilbert spaces, preparations and composite systems (P1) and the identical-particle and spin postulate Identical-particle and spin assignments P6: permutation sectors and finite-spin representations (P6), read with the state space vocabulary of The state space of the framework: preparations, pure rays and density operators and the quantities and units of Dimensions, SI unit conventions and quantity types of the framework. Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), the choice hypothesis of the cited tensor construction.

(b) Declared species and sector. A declared species sector of the model consists of: a nonzero separable complex Hilbert space $H$ , the one-particle space of the species, as assigned in P1; a mass parameter $m > 0$ in kg; a spin label $s \in {0, \frac{1}{2}, 1, \frac{3}{2}, \dots}$ and the internal factor $V_{s} = C^{2 s + 1}$ of P6 clause (ii), with the orbital factor $H_{s p}$ the factor carrying no internal degree of freedom when the one-particle space is given as a completed product $H_{s p} \hat{\otimes} V_{s}$ ; for the massive spin- $s$ model of The massive spin-s particle model of the framework: state space, operators, units and frame readings the one-particle space is $L^{2} (R^{3}; C^{2 s + 1})$ , whose spin components are the internal factor of P6 clause (ii) in the reading of that definition; a statistics choice $σ \in {+, -}$ , where $+$ declares the species bosonic and $-$ declares it fermionic; and, for a charged species, a charge parameter $q \in R$ in coulombs. The parameters and the statistics choice are empirical model assignments with their uncertainties, in the sense of P6 clause (iii); no value and no statistics choice is derived by this definition.

(c) The $n$ -particle sector. Fix $n \geq 1$ . The labelled $n$ -particle space of the species is the $n$ -fold completed tensor power $H^{\otimes n}$ (Hilbert tensor construction: inner product, completion and product bases), whose tensor factors are mathematical labels and not observable particle identities, equipped with the permutation unitaries $(U_{π})_{π \in S_{n}}$ and the sector projections $P_{σ}^{(n)}$ (Permutation sector projections on the n-fold tensor power), where $P_{+}^{(n)}$ is the symmetrizing projection and $P_{-}^{(n)}$ the alternating projection. The $n$ -particle sector space of the declared species is $H_{n}^{σ} : = ran P_{σ}^{(n)} \subseteq H^{\otimes n},$ a closed linear subspace carrying the inner product of $H^{\otimes n}$ ; on it every permutation acts by the scalar $U_{π} ∣_{H_{n}^{σ}} = {sgn}_{σ} (π) {i d}_{H_{n}^{σ}}, {sgn}_{+} (π) = 1, {sgn}_{-} (π) = sgn (π),$ by clause (c) of the cited permutation lemma. For $n = 1$ one has $H_{1}^{+} = H_{1}^{-} = H$ .

(d) Declared sector preparations. A declared sector preparation of $n$ particles of the species is a state of the framework on the sector space, that is a density operator $ρ \in D (H_{n}^{σ})$ in the sense of The state space: density operators, pure and mixed states, and purity and clause (a) of The state space of the framework: preparations, pure rays and density operators, read as the state assigned to the preparation by P1 clause (iii). A pure sector preparation is a pure state, i.e. the rank-one operator of a unit ray of $H_{n}^{σ}$ ; a statistical mixture of sector preparations is a convex combination. The sector is part of the declaration of the species: a model that declares one statistics choice does not switch it within the same declaration. When a sector preparation is paired with an operator or a spectral projection on the labelled space, use its canonical zero extension $\tilde{ρ} : = j ρ j^{*}$ , where $j : H_{n}^{σ} ↪ H^{\otimes n}$ is the isometric inclusion and $j^{*}$ is the orthogonal projection viewed as a map into the sector. This extension is a density operator on the labelled space: by clause (a) of Density operators and scalar spectral probability, $ρ = \sum_{k} w_{k} T_{u_{k}, u_{k}}$ in trace norm for an orthonormal family in the sector and nonnegative weights summing to one. The family $(j u_{k})$ is orthonormal in the labelled space, so the same clause gives the density $\sum_{k} w_{k} T_{j u_{k}, j u_{k}}$ there; its action is $j ρ j^{*}$ , since both agree on the sector and vanish on its orthogonal complement. If the sector is zero, there is no density preparation of trace one and this convention is vacuous.

(e) Sector quantities. Let $A$ be a linear operator on the labelled space $H^{\otimes n}$ with domain $D (A)$ . The model reads $A$ as a quantity of the declared sector only when $A$ preserves the sector, $P_{σ}^{(n)} D (A) \subseteq D (A), A (D (A) \cap H_{n}^{σ}) \subseteq H_{n}^{σ},$ in which case the sector restriction $A_{σ} : = A ∣_{H_{n}^{σ}}$ with domain $D (A) \cap H_{n}^{σ}$ is part of the declaration and is the quantity actually read on sector preparations; For any bounded $A$ , commutation with $P_{σ}^{(n)}$ implies sector preservation; for bounded self-adjoint $A$ the converse also holds. Indeed, if such an $A$ preserves the sector, $y$ in its orthogonal complement and $x$ in the sector satisfy $⟨ A y, x ⟩ = ⟨ y, A x ⟩ = 0$ , so the complement is invariant as well; splitting each vector into its two orthogonal components gives $A P_{σ}^{(n)} = P_{σ}^{(n)} A$ . For an arbitrary bounded operator, invariance alone need not preserve the complement and is not asserted equivalent to commutation. An operator that does not preserve the sector is not read on the sector; in particular no quantity of a single labelled factor is assumed to define a quantity of the sector unless it preserves it. If $A_{σ}$ is self-adjoint, its outcome law in a sector preparation is the law of Assigned sharp observables and the outcome law of the framework read on $H_{n}^{σ}$ .

(f) Status and scope. Clauses (a)-(e) fix the vocabulary with which the framework reads the adopted P6 assumptions: they add no assumption to P1 and P6, and the permutation invariance of sector preparations and of sector readouts is a conditional consequence proved on this page (Permutation invariance of sector preparations and the indistinguishability of relabelled readouts), not a content of this definition. The definition asserts no spin-statistics theorem, no dynamics, no measurement instrument and no empirical fact; whether a laboratory ensemble is described by a declared sector preparation is a tested identification with uncertainty under the preparation and apparatus postulate P7, not a consequence of the declaration.

Remarks

Labels versus particles. The tensor factors of $H^{\otimes n}$ are bookkeeping slots, not particle identities: by clause (c) the model assigns the same sector state to every relabelling, and clause (i) of P6 records that the labels carry no physical content.
Why the sector restriction is stated for operators. The rule of clause (e) is the one recorded in the owner direction for permutation sectors: a bounded quantity restricts to a sector only when it preserves the sector. For $n = 1$ , both sector projections are the identity, so the assigned one-particle position and momentum operators preserve both sectors and their domains. For $n \geq 2$ , an operator on a single labelled factor need not preserve either sector; clause (e), including its domain condition, must be checked before reading it as a sector quantity.

DefinitionDefinition: AI-adaptedProof: Not applicableOpen item page →

The declared finite-mode model and its occupation readouts

Physical scope

The declared finite-mode occupation model of an identical-particle sector: a declared species sector, a finite declared orthonormal mode family, the occupation readouts and their joint outcome law, and the total-occupation identity. The finite mode family is a declared idealization of the model, its identification with any laboratory mode set is a tested physical identification with uncertainty, and the definition asserts no measured count, no dynamics and no empirical result.

Definition

(a) Adopted readings. Adopt a declared species sector in the vocabulary of Declared species sectors and their preparations: the framework's reading of P6, the observable and probability postulate Observable and probability postulate P2: self-adjoint observables, spectral probability and finite unsharp outcomes (P2) in the reading of Assigned sharp observables and the outcome law of the framework, and the quantities and units of Dimensions, SI unit conventions and quantity types of the framework. Assume the Axiom of Countable Choice and the Axiom of Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Axiom of Choice). The latter is inherited from the general spectral-PVM construction in P2 and clause (b) of the observable definition used in clause (c) below; the finite joint-eigenspace decomposition in clause (d) requires no additional choice.

(b) The declared finite-mode model. Fix $d \geq 1$ , a finite-dimensional complex Hilbert space $K$ of dimension $d$ (Hilbert space) with a declared orthonormal basis $(f_{1}, \dots, f_{d})$ , the declared modes of the model, and an integer $n \geq 1$ . The declared finite-mode model of the species with these data is the declared species sector of clause (a) whose one-particle space is $K$ , together with its $n$ -particle sector space $H_{n}^{σ} = ran P_{σ}^{(n)} \subseteq K^{\otimes n}$ and the declared sector preparations of Declared species sectors and their preparations: the framework's reading of P6 clauses (c)-(d). The declaration is an idealization: it asserts that the model confines the one-particle content to the $d$ declared modes. No claim is made that a species has finitely many modes; whether the declared modes describe a laboratory mode family is a tested identification with uncertainty under the preparation and apparatus postulate P7, not a consequence of this definition.

(c) The occupation readouts. For $j \in {1, \dots, d}$ let $N_{j} : K^{\otimes n} \to K^{\otimes n}, N_{j} x : = \sum_{i \in {1, \dots, d}^{n}} ∣ i ∣_{j} ⟨ x, f_{i_{1}} \otimes \dots \otimes f_{i_{n}} ⟩ f_{i_{1}} \otimes \dots \otimes f_{i_{n}}, ∣ i ∣_{j} : = # {k : i_{k} = j},$ be the occupation operator of Slater and symmetric occupation-number bases of the finite-dimensional permutation sectors clause (d) for the basis $(f_{1}, \dots, f_{d})$ . By that clause $N_{j}$ is bounded and self-adjoint, the $N_{j}$ commute pairwise, and $N_{j}$ preserves the sector $H_{n}^{σ}$ ; the sector occupation readout is the restriction $N_{j}^{σ} : = N_{j} ∣_{H_{n}^{σ}},$ a bounded self-adjoint endomorphism of $H_{n}^{σ}$ with spectrum contained in ${0, 1, \dots, n}$ . The model reads $(N_{j}^{σ}, particle count)$ as an assigned sharp observable of the declared sector in the sense of Assigned sharp observables and the outcome law of the framework clause (a): its outcome law in a declared sector preparation $ρ$ is $μ_{ρ}^{N_{j}^{σ}} (B) = tr (ρ E_{N_{j}^{σ}} (B)), B \subseteq R Borel,$ the reading of P2 clause (ii) through the observable definition.

(d) The joint occupation readout. The operators $N_{1}^{σ}, \dots, N_{d}^{σ}$ are simultaneously diagonalised by the sector bases of the cited theorem: the Slater basis in the fermionic case and the family $(m_{λ})$ in the bosonic case, with joint eigenvalues $1_{J}$ and $λ$ respectively. Write $Λ_{n} : = {λ \in N^{d} : λ_{j} \leq n, \sum_{j} λ_{j} = n}$ and, for $λ \in Λ_{n}$ , let $P_{λ}$ be the orthogonal projection of $H_{n}^{σ}$ onto the joint eigenspace of $(N_{1}^{σ}, \dots, N_{d}^{σ})$ for the eigenvalue vector $λ$ ; the cited theorem exhibits a line $C m_{λ}$ for every bosonic $λ \in Λ_{n}$ . In the fermionic sector there is a line $C f_{J}^{\land}$ exactly when $λ = 1_{J}$ for an $n$ -element subset $J$ ; if any entry of $λ$ is at least $2$ , the joint eigenspace is ${0}$ and $P_{λ} = 0$ . Thus the $P_{λ}$ , including these zero projections, are mutually orthogonal, their finite sum is the identity of the sector, and $N_{j}^{σ} = \sum_{λ \in Λ_{n}} λ_{j} P_{λ}$ . The model reads the joint occupation readout through the finite family $(P_{λ})_{λ \in Λ_{n}}$ : the joint law in a declared sector preparation $ρ$ is $μ_{ρ}^{N} (λ) : = tr (ρ P_{λ}) (λ \in Λ_{n}),$ and the marginal law of $N_{j}^{σ}$ is recovered by $μ_{ρ}^{N_{j}^{σ}} (B) = \sum_{λ : λ_{j} \in B} μ_{ρ}^{N} (λ)$ , because the eigenvalue decomposition of $N_{j}^{σ}$ groups the $P_{λ}$ by the value $λ_{j}$ .

(e) Total occupation. By clause (d) of the cited theorem, $\sum_{j = 1}^{d} N_{j}^{σ} = n {i d}_{H_{n}^{σ}}$ ; hence the total occupation $\sum_{j} N_{j}^{σ}$ has the degenerate outcome law $δ_{n}$ in every declared sector preparation, and the joint law is supported on $Λ_{n}$ : the total number of particles read by the model is $n$ with certainty.

(f) Status and scope. Clauses (a)-(e) fix the vocabulary of the model and read the cited spectral mathematics through P2 under the choice hypotheses of clause (a); they add no physical assumption beyond the sector declaration and the finite-mode idealization of clause (b). The occupation readouts are sector quantities in the sense of clause (e) of Declared species sectors and their preparations: the framework's reading of P6; the exclusion and occupation-count consequences are proved on this page (Pauli exclusion in the declared finite-mode occupation readout, with the bosonic contrast and Finite-mode occupation configurations: fermionic exclusion counts, bosonic counts and truncated dimensions). No measured count, no detector response, no dynamics and no empirical claim is made.

Remarks

Mode versus particle. $N_{j}^{σ}$ counts how many of the $n$ particles the model reads in the declared mode $f_{j}$ ; it is a collective readout of the sector, not a quantity of a labelled factor, and it is insensitive to which labels carry the occupied modes.
Choice hypotheses. The tensor construction and the sector projections carry the Axiom of Countable Choice. The reading of the general spectral construction through P2 additionally carries the Axiom of Choice; both hypotheses are declared in clause (a). The finite spectral decomposition of clause (d) uses no further choice.

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Permutation invariance of sector preparations and the indistinguishability of relabelled readouts

Physical scope

The conditional indistinguishability result of the framework for a declared identical-particle sector: every declared sector preparation is invariant under the permutation unitaries of the labelled space, and the outcome law of any assigned sharp observable of the labelled model equals that of its relabelled conjugate. The result is conditional on the adopted sector of P6 and on P1-P2; it asserts no measured invariance, no dynamics and no spin-statistics theorem.

Statement

Adopt the state and composition postulate P1 (State and composition postulate P1: Hilbert spaces, preparations and composite systems), the observable and probability postulate P2 (Observable and probability postulate P2: self-adjoint observables, spectral probability and finite unsharp outcomes) and the identical-particle and spin postulate P6 (Identical-particle and spin assignments P6: permutation sectors and finite-spin representations) in the readings of Declared species sectors and their preparations: the framework's reading of P6 and Assigned sharp observables and the outcome law of the framework, and assume the Axiom of Countable Choice and the Axiom of Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Axiom of Choice), the choice hypotheses of the cited spectral mathematics. Let a species sector with parameters $(H, m, s, σ)$ , integer $n \geq 1$ , sector space $H_{n}^{σ} = ran P_{σ}^{(n)}$ and permutation unitaries $(U_{π})_{π \in S_{n}}$ be declared, and let $ρ$ be a declared sector preparation. In formulas involving labelled-space operators, $ρ$ denotes its canonical zero extension from clause (d) of the species-sector definition; the same symbol denotes its restriction on the sector. Then:

(a) invariance of the preparation: $U_{π} ρ U_{π}^{*} = ρ$ for every $π \in S_{n}$ ;

(b) indistinguishability of relabelled readouts: for every $π$ and every assigned sharp observable $(A, u_{A})$ of the labelled model (a densely defined self-adjoint operator on $H^{\otimes n}$ with a declared unit), the operator $A_{π} : = U_{π} A U_{π}^{*}$ with domain $U_{π} D (A)$ is an assigned sharp observable of the labelled model with the same unit, and for every Borel set $B \subseteq R$ $μ_{ρ}^{A_{π}} (B) = μ_{ρ}^{A} (B);$ for bounded $A$ this reads $tr (ρ U_{π} A U_{π}^{*}) = tr (ρ A)$ ;

(c) sector-preserving quantities: if $A$ is an assigned sharp observable that preserves the sector and satisfies $U_{π} A U_{π}^{*} = A$ on its domain for every $π$ , then $A$ induces the sector quantity of clause (e) of the sector definition and its law on sector preparations is the same as that of $A_{π} = A$ .

The theorem is conditional on the adopted sector: it establishes that within the model no assigned readout of the labelled space can distinguish the labels, and it asserts nothing about which statistics a physical species realizes.

Facts & Assumptions

Given: A declared species sector $(H, m, s, σ)$ , $n \geq 1$ , the sector space and permutation unitaries, a declared sector preparation $ρ$ , and an assigned sharp observable $A$ .

[F1]

Sector vocabulary. $H_{n}^{σ} = ran P_{σ}^{(n)}$ is a closed subspace of $H^{\otimes n}$ , every $U_{π}$ is unitary and preserves $H_{n}^{σ}$ , acting there as the scalar ${sgn}_{σ} (π)$ with ${sgn}_{+} = 1$ and ${sgn}_{-} = sgn$ ; a declared sector preparation is a density operator on $H_{n}^{σ}$ , extended by zero to the labelled space when used in ambient conjugations or traces, as defined in clause (d) (Declared species sectors and their preparations: the framework's reading of P6 clauses (c)-(d), Permutation sector projections on the n-fold tensor power clauses (a),(c)).

[F2]

Observable reading. An assigned sharp observable $(A, u_{A})$ has a unique spectral projection-valued measure $E_{A}$ on the Borel sets of $R$ , with $D (A) = {x : \int λ^{2} d E_{A, x} < \infty}$ and $A x = \int λ d E_{A} (λ) x$ , and the outcome law in the preparation $ρ$ is $μ_{ρ}^{A} (B) = tr (ρ E_{A} (B))$ (Assigned sharp observables and the outcome law of the framework clauses (a),(b), Observable and probability postulate P2: self-adjoint observables, spectral probability and finite unsharp outcomes).

[F3]

Unitary transport of spectral measures. If $U$ is a bijective linear isometry of a Hilbert space and $T$ a self-adjoint operator, then $T^{'} : = U T U^{- 1}$ with domain $U D (T)$ is self-adjoint and its spectral projection-valued measure is $E_{T^{'}} (B) = U E_{T} (B) U^{- 1}$ (Transport of spectral projection-valued measures along isometric bijections clauses (a),(b), Hilbert space).

[F4]

Cyclicity of the trace. If $T$ is trace class and $S$ bounded, then $S T$ and $T S$ are trace class with $tr (S T) = tr (T S)$ (Cyclicity of the trace).

[F5]

States. A state on a Hilbert space is a self-adjoint positive trace-class operator of trace one; pure states are the rank-one operators of unit rays (The state space of the framework: preparations, pure rays and density operators, The state space: density operators, pure and mixed states, and purity).

Proof

technique · direct; use the scalar action of the permutation unitaries on the sector to prove the invariance of the preparation, transport it through the spectral measure of a relabelled observable by the unitary transport lemma, and compare the outcome laws with the cyclic property of the trace

Given: The data of the statement and [F1]-[F5].

1.1F1F5algebra

[Invariance] Let $x, y \in H_{n}^{σ}$ . By [F1] one has $U_{π}^{*} x = {sgn}_{σ} (π)^{- 1} x = {sgn}_{σ} (π) x$ and likewise for $y$ , so $⟨ U_{π} ρ U_{π}^{*} x, y ⟩ = ⟨ ρ U_{π}^{*} x, U_{π}^{*} y ⟩ = {sgn}_{σ} (π)^{2} ⟨ ρ x, y ⟩ = ⟨ ρ x, y ⟩$ by the unitarity of $U_{π}$ and the sesquilinearity of the inner product. Every $U_{π}$ preserves the sector and, being unitary with inverse also preserving it, preserves its orthogonal complement: if $y$ is in that complement and $x$ is in the sector, $⟨ U_{π} y, x ⟩ = ⟨ y, U_{π}^{*} x ⟩ = 0$ . The extended $ρ$ vanishes on the complement and maps into the sector, so the same holds for $U_{π} ρ U_{π}^{*}$ . The established equality on the sector therefore gives $U_{π} ρ U_{π}^{*} = ρ$ on the entire labelled space; multiplying this identity on the left by $U_{π}^{*}$ and on the right by $U_{π}$ and using $U_{π}^{*} U_{π} = I$ gives $U_{π}^{*} ρ U_{π} = ρ$ as well. This proves (a).

1.2F1F2F3F5algebra

[Sector-preserving case] Put $P : = P_{σ}^{(n)}$ , $M : = ran P$ and $A_{σ} : = A ∣_{D (A) \cap M}$ . Under (c), [F3] gives $E_{A} (B) = U_{π} E_{A} (B) U_{π}^{*}$ for every $π$ ; hence $E_{A} (B)$ commutes with every $U_{π}$ and with their finite weighted sum $P$ from [F1]. Thus $E_{σ} (B) : = E_{A} (B) ∣_{M}$ is a regular PVM on $M$ : projections, products, normalization and strong countable additivity restrict from $E_{A}$ , and its scalar measures at vectors of $M$ are unchanged. Also $A P x = P A x$ for $x \in D (A)$ by the finite weighted sum and the domain equality $U_{π} D (A) = D (A)$ . The domain $D (A) \cap M$ is dense in $M$ , since projecting any ambient approximating sequence in $D (A)$ produces one in that domain. The restriction is self-adjoint: it is symmetric; if $y \in D (A_{σ}^{*}) \subset M$ and $z = A_{σ}^{*} y \in M$ , then for every $x \in D (A)$ , $⟨ A x, y ⟩ = ⟨ A P x, y ⟩ = ⟨ P x, z ⟩ = ⟨ x, z ⟩$ . Therefore $y \in D (A^{*}) = D (A)$ and $A y = z$ , so $D (A_{σ}^{*}) \subseteq D (A_{σ})$ and symmetry gives equality. For $x \in M$ the scalar second-moment domain and spectral integrals of $E_{σ}$ equal those of $E_{A}$ , so [F2]'s uniqueness identifies $E_{σ}$ as the spectral PVM of $A_{σ}$ . Finally the density expansion in clause (d) of [F1], $ρ = \sum_{k} w_{k} T_{u_{k}, u_{k}}$ in trace norm with $u_{k} \in M$ , gives both traces as $\sum_{k} w_{k} ⟨ E_{A} (B) u_{k}, u_{k} ⟩$ , since the zero extension uses the same vectors in the ambient space. Hence $μ_{ρ}^{A_{σ}} (B) = μ_{ρ}^{A} (B) = μ_{ρ}^{A_{π}} (B)$ , proving (c).

2.1F2F3F4step 1.1algebra

[Relabelled laws] Fix $π$ and an assigned sharp observable $(A, u_{A})$ . Since $U_{π}$ is a bijective linear isometry of $H^{\otimes n}$ and $A$ is self-adjoint, [F3] gives that $A_{π} = U_{π} A U_{π}^{*}$ is self-adjoint on the domain $U_{π} D (A)$ with spectral measure $E_{A_{π}} (B) = U_{π} E_{A} (B) U_{π}^{*}$ . Hence for every preparation $ρ$ and Borel $B$ , $μ_{ρ}^{A_{π}} (B) = tr (ρ U_{π} E_{A} (B) U_{π}^{*})$ by [F2]. Because $ρ$ is trace class and $U_{π} E_{A} (B)$ is bounded, [F4] applied twice gives $tr (ρ U_{π} E_{A} (B) U_{π}^{*}) = tr (U_{π}^{*} ρ U_{π} E_{A} (B))$ ; by step 1.1 the middle factor is $ρ$ , so $μ_{ρ}^{A_{π}} (B) = tr (ρ E_{A} (B)) = μ_{ρ}^{A} (B)$ , which is (b) in the general form; for bounded $A$ the same computation with $A$ in place of $E_{A} (B)$ gives the displayed trace identity.

3.1step 1.1step 1.2step 2.1∎

[Assembly and scope] Step 1.1 proves (a), step 2.1 proves (b) and step 1.2 proves (c). The proof uses only the adopted sector of P6, the readings of P1 and P2, and the cited spectral and trace mathematics under the declared choice hypotheses; no measured invariance, no dynamics and no species assignment is asserted.

Remarks

What the theorem does and does not say. The invariance is a property of every state the model declares on a sector, and it is what makes the tensor labels unobservable inside the model; it does not say that the labels are meaningless in intermediate manipulations, and it does not by itself force a permutation-invariant preparation into a sector, which is a separate question not treated here.
Relation to the occupation readout. Every occupation readout of the declared finite-mode model is a sector quantity built from the operators $N_{j}$ , which commute with all $U_{π}$ because they are diagonal in the product basis and count index occurrences (Slater and symmetric occupation-number bases of the finite-dimensional permutation sectors clause (d)); the exclusion consequences for those readouts are the subject of the next theorem of this page.

Physics theoremStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

Pauli exclusion in the declared finite-mode occupation readout, with the bosonic contrast

Physical scope

The conditional exclusion statement of the declared finite-mode occupation model: in the adopted antisymmetric sector of P6 the joint occupation law is supported on configurations of occupations at most one, so the probability of reading two particles in one declared mode is zero in every declared sector preparation, while in the symmetric sector every occupation configuration of total n occurs. The statement is conditional on the declared sector and the finite-mode idealization; it asserts no measured occupancy, no apparatus claim and no derivation of a species' statistics.

Statement

Adopt the observable and probability postulate P2 (Observable and probability postulate P2: self-adjoint observables, spectral probability and finite unsharp outcomes) and the identical-particle and spin postulate P6 (Identical-particle and spin assignments P6: permutation sectors and finite-spin representations) in the readings of Declared species sectors and their preparations: the framework's reading of P6 and The declared finite-mode model and its occupation readouts, and assume the Axiom of Countable Choice and the Axiom of Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Axiom of Choice). Declare a finite-mode model as in The declared finite-mode model and its occupation readouts with $d$ declared modes $f_{1}, \dots, f_{d}$ , $n \geq 1$ particles and statistics choice $σ$ , with sector space $H_{n}^{σ}$ , occupation readouts $N_{1}^{σ}, \dots, N_{d}^{σ}$ , joint projections $(P_{λ})_{λ \in Λ_{n}}$ and joint law $μ_{ρ}^{N} (λ) = tr (ρ P_{λ})$ . Then for every declared sector preparation $ρ$ :

(a) fermionic exclusion: if $σ = -$ , then $μ_{ρ}^{N_{j}^{-}} ([2, \infty)) = 0 (j = 1, \dots, d), tr (ρ N_{j}^{-} (N_{j}^{-} - I)) = 0,$ and the joint law is supported on the occupation configurations with entries in ${0, 1}$ : $μ_{ρ}^{N} ({λ \in Λ_{n} : λ_{j} \geq 2 for some j}) = 0;$

(b) no restriction in the symmetric sector: if $σ = +$ , then for every $λ \in W (n, d)$ the projection $P_{λ}$ is nonzero, and for the pure declared sector preparation of the ray of $m_{λ}$ the joint law gives weight one to $λ$ ; in particular for $n \geq 2$ and any $j$ the configuration with $λ_{j} = n$ occurs;

(c) total occupation: in either sector $\sum_{j = 1}^{d} N_{j}^{σ} = n {i d}_{H_{n}^{σ}}$ , so the total occupation has the degenerate law $δ_{n}$ and every outcome of the joint readout satisfies $\sum_{j} λ_{j} = n$ .

The exclusion in (a) is a conditional consequence of the adopted antisymmetric sector and of the declared finite-mode readout; the theorem asserts no measured occupancy, no detector behaviour and no derivation of the statistics of any species.

Facts & Assumptions

Given: A declared finite-mode model with $d$ modes, $n \geq 1$ particles, statistics choice $σ$ , sector space, occupation readouts, joint projections and a declared sector preparation $ρ$ .

[F1]

Sector readouts. $K$ is $d$ -dimensional with orthonormal basis $(f_{1}, \dots, f_{d})$ ; the operators $N_{j}$ of the occupation theorem are bounded self-adjoint, commute pairwise, satisfy $\sum_{j} N_{j} = n I$ and preserve both sector spaces, with $N_{j} f_{J}^{\land} = 1_{J} (j) f_{J}^{\land}$ and $N_{j} m_{λ} = λ_{j} m_{λ}$ , and the joint eigenspaces inside $ran P_{-}^{(n)}$ are the lines $C f_{J}^{\land}$ (indexed by $n$ -element subsets $J$ , eigenvalue vector $1_{J}$ ) and inside $ran P_{+}^{(n)}$ the lines $C m_{λ}$ (indexed by $λ \in W (n, d)$ , eigenvalue vector $λ$ ) (Slater and symmetric occupation-number bases of the finite-dimensional permutation sectors clauses (b)-(d), The declared finite-mode model and its occupation readouts clauses (c)-(d)).

[F2]

Observable reading. An assigned sharp observable $A$ has a unique spectral projection-valued measure $E_{A}$ and the outcome law $μ_{ρ}^{A} (B) = tr (ρ E_{A} (B))$ ; if a projection-valued measure $F$ satisfies $\int λ d F (λ) = A$ , then $F = E_{A}$ (Assigned sharp observables and the outcome law of the framework clauses (a),(b), Observable and probability postulate P2: self-adjoint observables, spectral probability and finite unsharp outcomes).

[F3]

Joint projectors. In the declared finite-mode model, the finite family $(P_{λ})_{λ \in Λ_{n}}$ consists of mutually orthogonal projections, sums to the identity of the sector, and satisfies $N_{j}^{σ} = \sum_{λ \in Λ_{n}} λ_{j} P_{λ}$ (The declared finite-mode model and its occupation readouts clause (d)).

[F4]

Laws and traces. For a preparation $ρ$ and projections $P_{λ}$ with $\sum_{λ} P_{λ} = I$ the numbers $tr (ρ P_{λ})$ are nonnegative and sum to one; a pure preparation of the unit ray of a vector $x \neq 0$ gives weight $∥ Q x ∥^{2} / ∥ x ∥^{2}$ to an orthogonal projection $Q$ (Density operators and scalar spectral probability clauses (a),(b), The state space of the framework: preparations, pure rays and density operators).

[F5]

Configuration counts. No fermionic configuration has an entry $\geq 2$ ; for $n > d$ the antisymmetric sector is ${0}$ ; every $λ \in W (n, d)$ is a bosonic configuration, including the configuration with $λ_{j} = n$ for any $j$ (Finite-mode occupation configurations: fermionic exclusion counts, bosonic counts and truncated dimensions clauses (a),(c)).

Proof

technique · direct; diagonalise the sector restrictions of the occupation operators in the sector bases, identify their finite spectral measures, and read the support of the joint law off the eigenvalue vectors

Given: The data of the statement and [F1]-[F5].

1.1F1algebra

[Sector restrictions are self-adjoint] By [F1] the operator $N_{j}$ preserves $H_{n}^{σ}$ , and both the Slater family and the family $(m_{λ})$ consist of eigenvectors of $N_{j}$ with real eigenvalues and form a basis of the respective sector; a linear endomorphism of a finite-dimensional Hilbert space that is diagonal in an orthogonal basis with real eigenvalues is self-adjoint. Hence each $N_{j}^{σ} = N_{j} ∣_{H_{n}^{σ}}$ is a bounded self-adjoint endomorphism of $H_{n}^{σ}$ , which is the readout declared in clauses (c)-(d) of the model.

2.1F1F2F3step 1.1algebraconstruct

[Finite spectral measures and laws] Fix $σ \in {-, +}$ and $j \in {1, \dots, d}$ . By [F3], the finite family of projections $F_{j} (B) : = \sum_{\begin{array}{c} λ \in Λ_{n} \\ λ_{j} \in B \end{array}} P_{λ} (B \subseteq R Borel)$ is a projection-valued measure: $F_{j} (\emptyset) = 0$ and $F_{j} (R) = I$ because the $P_{λ}$ sum to $I$ ; each $F_{j} (B)$ is an orthogonal projection; intersections satisfy $F_{j} (B \cap C) = F_{j} (B) F_{j} (C)$ because the $P_{λ}$ are mutually orthogonal; and countable additivity is strong since only finitely many $λ$ occur and the Borel sets in a disjoint sequence contain each eigenvalue $λ_{j}$ at most once. Its scalar measures $μ_{x} (B) : = ⟨ F_{j} (B) x, x ⟩$ are finite and supported on the finite set $S_{j} : = {λ_{j} : λ \in Λ_{n}}$ , hence regular: for every Borel $B$ , the compact set $B \cap S_{j}$ and an open set $U$ with $U \cap S_{j} = B \cap S_{j}$ give $μ_{x} (B \cap S_{j}) = μ_{x} (B) = μ_{x} (U)$ . Its spectral integral is the finite sum $\int_{R} r d F_{j} (r) = \sum_{λ \in Λ_{n}} λ_{j} P_{λ} = N_{j}^{σ} .$ Thus $F_{j}$ is the spectral PVM of $N_{j}^{σ}$ by uniqueness of the spectral PVM in [F2]. The outcome law in $ρ$ is therefore $μ_{ρ}^{N_{j}^{σ}} (B) = tr (ρ F_{j} (B))$ . In the fermionic sector [F1] restricts the joint eigenvalue vectors to $1_{J} \in {0, 1}^{d}$ , so this spectral measure is supported on ${0, 1}$ ; in the symmetric sector it groups the lines $C m_{λ}$ by the value $λ_{j}$ . This identifies the readout laws of clauses (c)-(d) of the model.

3.1F1step 1.1step 2.1algebra

[Exclusion] Assume $σ = -$ . By step 2.1 the spectral measure of $N_{j}^{-}$ is supported on ${0, 1}$ , so $E_{N_{j}^{-}} ([2, \infty)) = 0$ and $μ_{ρ}^{N_{j}^{-}} ([2, \infty)) = tr (0) = 0$ ; moreover $N_{j}^{-} (N_{j}^{-} - I) = 0$ on the sector because the operator vanishes on the Slater basis by [F1], so its expectation $tr (ρ N_{j}^{-} (N_{j}^{-} - I))$ is zero. For the joint law, [F1] shows that the joint eigenspaces with an entry $\geq 2$ do not occur among the lines $C f_{J}^{\land}$ , so the sum of the corresponding projections $P_{λ}$ is zero and $μ_{ρ}^{N} ({λ : λ_{j} \geq 2 for some j}) = 0$ . This proves (a).

3.2F1F4F5step 2.1algebra

[Bosonic contrast] Assume $σ = +$ . For every $λ \in W (n, d)$ the vector $m_{λ}$ is nonzero and spans the joint eigenspace for $λ$ by [F1], so $P_{λ} \neq 0$ ; for the pure preparation of the ray of $m_{λ}$ the weight is $μ_{ρ}^{N} (λ) = tr (ρ P_{λ}) = ∥ P_{λ} x ∥^{2} / ∥ x ∥^{2} = 1$ for any unit representative $x$ of that ray by [F4]. By [F5] the configurations $λ$ with $λ_{j} = n$ lie in $W (n, d)$ , so they occur; this proves (b).

3.3F1step 2.1algebra

[Total occupation] By [F1] one has $\sum_{j} N_{j} = n I$ on the whole tensor power, hence on the sector, and therefore $\sum_{j} N_{j}^{σ} = n {i d}_{H_{n}^{σ}}$ ; since the total occupation is the constant operator $n I$ , its outcome law assigns probability one to ${n}$ in every preparation, and applying the joint law of [F1] gives $\sum_{j} λ_{j} = n$ for every configuration of the support; this is (c).

4.1step 3.1step 3.2step 3.3∎

[Assembly and scope] Step 3.1 proves (a), step 3.2 proves (b) and step 3.3 proves (c). The proof is conditional on the declared sector of P6, the declared finite-mode readout and the cited spectral mathematics; no measured occupancy, no apparatus behaviour and no derivation of a species' statistics is asserted.

Remarks

The precise content of "exclusion". The theorem states the vanishing of the probability of reading an occupation $\geq 2$ in each declared mode and the vanishing of the expectation of the double-occupation operator $N_{j} (N_{j} - 1)$ ; it is a statement about the declared readout in every declared sector preparation, not a statement about particles avoiding each other in space.
Finite modes. The statement uses the finite-dimensional mode model of the definition; the counting facts for the index sets are those of the finite-mode count theorem. No infinite mode family and no continuum occupancy is treated.

RemarkRemark: AI-adaptedProof: Not applicableOpen item page →

Scope and empirical status of the identical-particles page

Physical scope

A scope and status remark for the identical-particles page: it separates the adopted sector assignments and finite-mode idealizations from the conditional deductions, records what the pair does not claim about nature, and states the empirical status of the model statements. It asserts no measurement, no derived statistics and no empirical result.

Scope of the page

Adopted assumptions. The page adopts the sector and species assignments of P6 in the reading of Declared species sectors and their preparations: the framework's reading of P6: which species is bosonic or fermionic, its mass, spin label and statistics, and, for a charged species, its charge, are declared model parameters; the symmetric or antisymmetric sector of the $n$ -particle space is part of that declaration. The finite-mode occupation model of The declared finite-mode model and its occupation readouts adds the declared finite mode family as an idealization. These are assumptions of the model, not consequences of the Hilbert-space formalism and not measurements.

Conditional results. The mathematical items of the page construct the two permutation sectors and their occupation bases (Slater and symmetric occupation-number bases of the finite-dimensional permutation sectors), count the occupation configurations (Finite-mode occupation configurations: fermionic exclusion counts, bosonic counts and truncated dimensions) and decompose a two-factor sector into opposite-symmetry summands (Regrouping two factors: the symmetry decomposition of a two-particle sector over an orbital and an internal factor). The physical results are conditional on the adopted sector: Permutation invariance of sector preparations and the indistinguishability of relabelled readouts proves that sector preparations and relabelled readouts are statistically indistinguishable, and Pauli exclusion in the declared finite-mode occupation readout, with the bosonic contrast proves the exclusion of double occupations in the antisymmetric sector together with the bosonic contrast. All of these are deductions from the stated postulates, idealizations and mathematical hypotheses.

What is not claimed. The page does not derive the spin-statistics connection: the relativistic theorem is outside this framework and the link between the spin label and the sector is a declared model assignment (clause (iv) of P6). It does not claim that the declared modes are an exact description of any apparatus or species, does not treat infinite mode families or the continuum of modes, and introduces no dynamics of the sector, no interaction and no instrument; the interaction example of the companion page is a conditional computation in a declared model. Nothing on the page attributes particle-like trajectories, spatial avoidance or a force to the exclusion statement.

Empirical status. The page records no measurement and no data. Whether some laboratory ensemble realizes a declared sector preparation, a declared mode family or a declared apparatus is a tested identification with uncertainty under the preparation and apparatus postulate Preparation and apparatus postulate P7: declared preparations, isolation and sampling assumptions; comparisons of model predictions with counts would test the combined model and apparatus assumptions, and agreement would be evidence, not a deductive proof of the model. The design's separate dedicated statistics treatment is a named destination of the later many-body page and is not silently claimed here.

5 · Examples, experiments and counterexamples

None yet.

Qm Identical Particles And Exclusion

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

The two-particle exchange operator and the symmetric and antisymmetric combinations

Statement

Facts & Assumptions

Proof

Remarks

Slater and symmetric occupation-number bases of the finite-dimensional permutation sectors

Statement

Facts & Assumptions

Proof

Remarks

Finite-mode occupation configurations: fermionic exclusion counts, bosonic counts and truncated dimensions

Statement

Facts & Assumptions

Proof

Remarks

Regrouping two factors: the symmetry decomposition of a two-particle sector over an orbital and an internal factor

Statement

Facts & Assumptions

Proof

Remarks

Declared species sectors and their preparations: the framework's reading of P6

Definition

Remarks

The declared finite-mode model and its occupation readouts

Definition

Remarks

Permutation invariance of sector preparations and the indistinguishability of relabelled readouts

Statement

Facts & Assumptions

Proof

Remarks

Pauli exclusion in the declared finite-mode occupation readout, with the bosonic contrast

Statement

Facts & Assumptions

Proof

Remarks

Scope and empirical status of the identical-particles page

Scope of the page

5 · Examples, experiments and counterexamples

Sources