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Permutation-restricted self-adjointness

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a complex Hilbert space, let $H$ be a self-adjoint operator on $H$ (Symmetric, self-adjoint and essentially self-adjoint operators), let $N \geq 1$ , and let $(U_{π})_{π \in S_{N}}$ be unitary operators on $H$ with $U_{π} U_{σ} = U_{π σ}$ and $U_{π}^{*} = U_{π^{- 1}}$ for all $π, σ \in S_{N}$ (The Hilbert-space adjoint of a bounded operator). Define $P_{+} : = \frac{1}{N!} \sum_{π \in S_{N}} U_{π}, P_{-} : = \frac{1}{N!} \sum_{π \in S_{N}} sgn (π) U_{π} .$ Assume that for every $π \in S_{N}$ , $U_{π} D (H) \subseteq D (H) and U_{π} H ψ = H U_{π} ψ (ψ \in D (H)) .$ Then:

(a) $P_{+}$ and $P_{-}$ are orthogonal projections on $H$ ;

(b) for every $z \in ρ (H)$ one has $U_{π} (H - z)^{- 1} = (H - z)^{- 1} U_{π}$ and hence $P_{\pm} (H - z)^{- 1} = (H - z)^{- 1} P_{\pm}$ ;

(d) if $ran P_{\pm} \neq {0}$ , then the restriction $H_{\pm} : = H ∣_{D (H) \cap ran P_{\pm}}$ is a self-adjoint operator on the Hilbert space $ran P_{\pm}$ (Range criterion for self-adjointness);

(e) application: under the additional Axiom of Choice (The Axiom of Choice), let $N \geq 1$ , $m > 0$ , $ℏ > 0$ , let $r \geq 1$ , let $V_{e x t}, W : R^{3} \to R$ be measurable and bounded, and assume $W (- z) = W (z)$ for every $z \in R^{3}$ . Put $V (X) : = \sum_{i = 1}^{N} V_{e x t} (x_{i}) + \sum_{i < j} W (x_{i} - x_{j})$ on $R^{3 N}$ , let $K$ be the free kinetic operator of Bounded fixed-N operator with all masses $m_{i} = m$ , let $Θ : L^{2} (R^{3})^{\otimes N} \to L^{2} (R^{3 N})$ be the iterated Euclidean product unitary (Euclidean L² product unitary: L²(R^a)⊗̂L²(R^b) ≅ L²(R^{a+b}), Hilbert tensor construction: inner product, completion and product bases), let $U_{π}$ be the permutation unitaries of Permutation sector projections on the n-fold tensor power and set $U_{π}^{p h y s} : = Θ U_{π} Θ^{- 1}, P_{\pm}^{p h y s} : = Θ P_{\pm}^{(N)} Θ^{- 1} .$ Then $H : = K + M_{V}$ satisfies the hypotheses above for the representation $U_{π}^{p h y s}$ , so $P_{\pm}^{p h y s}$ reduce $H$ and the restrictions of $H$ to the symmetric and antisymmetric sectors are self-adjoint on those sectors.

No spin-statistics statement, no identification of a physical species with a sector, and no claim about bound states or spectra is made.

Facts & Assumptions

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. Given: the complex Hilbert space $H$ , the self-adjoint operator $H$ , the unitary representation $(U_{π})$ of $S_{N}$ , its averages $P_{\pm}$ , numbers $z \in ρ (H)$ and vectors $ψ, φ \in H$ .

[F1]

By The finite symmetric group $S_{n}$ , one-line notation, and cycle notation, $S_{N}$ consists of the bijections of an $N$ -element set, so it has $N!$ elements by choosing successive images in $N, N - 1, \dots, 1$ ways. Its sign map is a homomorphism and is surjective for $N \geq 2$ (The sign is a homomorphism $S_{n} \to {+ 1, - 1}$ , surjective exactly when $n \geq 2$ ); choose $τ$ with $sgn (τ) = - 1$ . Left multiplication by $τ$ pairs the terms of $\sum_{π \in S_{N}} sgn (π)$ with their negatives, so this sum is zero. Group averaging now gives $P_{+}^{*} = P_{+}$ , $P_{+}^{2} = P_{+}$ and likewise $P_{-}^{*} = P_{-}$ , $P_{-}^{2} = P_{-}$ ; for $N = 1$ , both averages equal the identity. The Hilbert adjoint and bounded-operator conventions are The Hilbert-space adjoint of a bounded operator and A bounded linear operator between normed spaces.

[F2]

For a self-adjoint $H$ and $z \in ρ (H)$ the resolvent $R (z) : = (H - z)^{- 1}$ is a bounded bijection of $H$ onto $D (H)$ with inverse $H - z$ ; $D (H)$ is dense (Symmetric, self-adjoint and essentially self-adjoint operators, A bounded linear operator between normed spaces).

[F3]

Range criterion: a densely defined symmetric operator $S$ is self-adjoint if and only if it is closed and $ran (S - i) = ran (S + i) = H$ ; equivalently if it is closed, symmetric and both $S \pm i$ are surjective (Range criterion for self-adjointness).

[F4]

The Euclidean product unitary $Θ : H_{1}^{\otimes N} \to L^{2} (R^{3 N})$ satisfies $Θ (f_{1} \otimes \dots \otimes f_{N}) (x) = f_{1} (x_{1}) \dots f_{N} (x_{N})$ ; the permutation unitaries act by $U_{π} (f_{1} \otimes \dots \otimes f_{N}) = f_{{\hat{π}}^{- 1} (1)} \otimes \dots \otimes f_{{\hat{π}}^{- 1} (N)}$ . The equal-mass free kinetic operator is the maximal Fourier multiplier $K = F_{2}^{*} M_{κ} F_{2}$ , $κ (ξ) = \frac{ℏ^{2}}{2 m} \sum_{i} ∣ ξ_{i} ∣^{2}$ , on $D (K) = W^{2, 2} (R^{3 N}; C)$ . With the supplier's $e^{- 2 π i X \cdot ξ}$ convention, it equals $- \frac{ℏ^{2}}{8 π^{2} m} \sum_{i} Δ_{i}$ there. The multiplier $V$ as in (e) is invariant under permutations of the coordinates because $W$ is even (Euclidean L² product unitary: L²(R^a)⊗̂L²(R^b) ≅ L²(R^{a+b}), Hilbert tensor construction: inner product, completion and product bases, Permutation sector projections on the n-fold tensor power, Bounded fixed-N operator).

Proof

1.1F1algebra

[The averages are projections] Since $π \mapsto U_{π}$ is a unitary representation and $U_{π}^{*} = U_{π^{- 1}}$ , $P_{+}^{*} = \frac{1}{N!} \sum_{π} U_{π^{- 1}} = P_{+}$ and $P_{+}^{2} = \frac{1}{(N!)^{2}} \sum_{π, σ} U_{π σ} = \frac{1}{N!} \sum_{τ} U_{τ} = P_{+}$ ; the same computation with the characters gives $P_{-}^{*} = P_{-}$ , $P_{-}^{2} = P_{-}$ . Hence both are orthogonal projections, which is (a).

1.2F2algebra

[Resolvent commutation] Fix $z \in ρ (H)$ and $φ \in H$ ; put $ψ : = R (z) φ \in D (H)$ . By hypothesis $U_{π} ψ \in D (H)$ and $(H - z) U_{π} ψ = U_{π} (H - z) ψ = U_{π} φ$ ; hence $U_{π} R (z) φ = R (z) U_{π} φ$ for every $φ$ , so $U_{π} R (z) = R (z) U_{π}$ ; averaging gives $P_{\pm} R (z) = R (z) P_{\pm}$ , which is (b).

1.3givenF1algebra

[Reduction] Put $χ_{+} (π) = 1$ and $χ_{-} (π) = sgn (π)$ . For $ψ \in D (H)$ , $P_{\pm} ψ = \frac{1}{N!} \sum_{π} χ_{\pm} (π) U_{π} ψ \in D (H)$ as a finite linear combination, and the commutation hypothesis gives $H P_{\pm} ψ = \frac{1}{N!} \sum_{π} χ_{\pm} (π) H U_{π} ψ = \frac{1}{N!} \sum_{π} χ_{\pm} (π) U_{π} H ψ = P_{\pm} H ψ$ , proving (c).

2.1F3step 1.3algebra

[Self-adjointness of the restrictions] Assume $ran P \neq {0}$ for $P = P_{\pm}$ and put $K : = ran P$ and $S : = H ∣_{D (H) \cap K}$ . Then $D (S) = P D (H)$ is dense in $K$ and $S$ is symmetric: for $φ, ψ \in D (S)$ , $⟨ S φ, ψ ⟩ = ⟨ H φ, ψ ⟩ = ⟨ φ, H ψ ⟩ = ⟨ φ, S ψ ⟩$ because $H$ is self-adjoint and $ψ = P ψ$ . $S$ is closed: if $ψ_{k} \in D (S)$ , $ψ_{k} \to ψ$ and $S ψ_{k} \to η$ with $ψ, η \in K$ , then $H$ closed gives $ψ \in D (H)$ and $H ψ = η$ , and $ψ \in K$ because $K$ is closed; hence $ψ \in D (S)$ , $S ψ = η$ . Finally $ran (S - i) = P ran (H - i)$ : indeed $(S - i) P ψ = P (H - i) ψ$ by step 1.3, and conversely every $P χ$ with $χ \in H$ equals $P (H - i) (H - i)^{- 1} χ = (S - i) P (H - i)^{- 1} χ$ . Since $H - i$ is onto $H$ ([F3] applied to the self-adjoint $H$ ), $ran (S - i) = K$ , and the same argument gives $ran (S + i) = K$ . By [F3], $S$ is self-adjoint on $K$ ; this proves (d).

3.1F4step 1.2step 1.3step 2.1algebra

[The permutation-invariant instance] Let $H = K + M_{V}$ as in (e). The unitaries $U_{π}^{p h y s} = Θ U_{π} Θ^{- 1}$ form a unitary representation of $S_{N}$ on $L^{2} (R^{3 N})$ ([F4]). The weight $κ (ξ) = \frac{ℏ^{2}}{2 m} \sum_{i} ∣ ξ_{i} ∣^{2}$ is invariant under permutations of the blocks $ξ_{i}$ , so $Θ U_{π} Θ^{- 1}$ maps $D (K)$ onto itself and intertwines $K$ with itself; and $V (π X) = V (X)$ gives $U_{π}^{p h y s} M_{V} U_{π}^{p h y s, - 1} = M_{V}$ on all of $L^{2}$ . Hence $U_{π}^{p h y s} D (H) = D (H)$ and $U_{π}^{p h y s} H = H U_{π}^{p h y s}$ on $D (H)$ , and step 2.1 applies to $P_{\pm}^{p h y s}$ , proving (e).

4.1step 1.1step 1.2step 1.3step 2.1step 3.1algebra∎

[Conclusion and scope] Steps 1.1-3.1 prove (a)-(e). The scope paragraph of the statement records that no spin-statistics assignment and no spectral conclusion is asserted; the symmetric and antisymmetric sectors are mathematical invariant subspaces of the specified operator.

Permutation-restricted self-adjointness

Statement

Facts & Assumptions

Proof

Depends on

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