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The one-body readout law: sector means of summed bounded one-particle quantities and their reduced forms

Physical scope

The conditional one-body readout law of the declared finite-mode many-body model: for a bounded self-adjoint one-particle quantity the summed sector operator is an assigned sharp observable, its mean in every declared sector preparation exists and is N times the pairing of the one-body reduced preparation with the quantity, and for declared Slater and orthonormal-mode symmetric preparations the reduced preparation is the equal-weight mixture of the occupied modes. The theorem is conditional on P1, P2 and P6, the adopted finite-mode idealization and the declared preparation; it asserts no measured mean, no individual-particle property, no separability of the sector state and no spatial (infinite-mode) reduction.

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.

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) as the choice hypotheses of the cited tensor, permutation and trace suppliers. Work in the finite-mode realization of The declared one-body reduced operator and the one-body readout of a sector preparation: one-particle space $K$ of finite dimension $d \geq 1$ with modes $(f_{1}, \dots, f_{d})$ , integer $N \geq 1$ with nonzero sector space $H_{N}^{σ} = ran P_{σ}^{(N)}$ , and the slot operators $O^{(i)}$ , sum $Σ (O)$ and one-body reduced preparation $γ_{ρ}$ of that definition. Let $O \in B (K)$ be self-adjoint with declared unit $u_{O}$ in the sense of Dimensions, SI unit conventions and quantity types of the framework. Then:

(a) the readout is an assigned observable: the sector operator $A_{O} = Σ (O) ∣_{H_{N}^{σ}}$ is bounded and self-adjoint, so the pair $(A_{O}, u_{O})$ is an assigned sharp observable of the sector in the sense of Assigned sharp observables and the outcome law of the framework clause (a);

(b) the mean law: for every declared sector preparation $ρ$ the mean of the outcome law of $(A_{O}, u_{O})$ in $ρ$ exists and equals $m_{ρ} (A_{O}) = N tr (γ_{ρ} O);$

(c) product-type preparations: let $u_{1}, \dots, u_{N} \in K$ be pairwise orthonormal. For $σ = -$ and $N \geq 2$ , let $Ψ : = S (u_{1}, \dots, u_{N})$ be the Slater vector of Slater direct and exchange calculation clause (a); for $σ = -$ and $N = 1$ , put $S (u_{1}) : = u_{1}$ and $Ψ : = u_{1}$ . For $σ = +$ , put $Ψ : = \sqrt{N!} P_{+}^{(N)} (u_{1} \otimes \dots \otimes u_{N})$ . In either case $Ψ$ is a unit vector of the sector. For $ρ : = T_{Ψ, Ψ}$ one has $γ_{ρ} = N^{- 1} \sum_{j = 1}^{N} T_{u_{j}, u_{j}}$ and $m_{ρ} (A_{O}) = \sum_{j = 1}^{N} ⟨ O u_{j}, u_{j} ⟩ .$

(d) scope: the theorem is conditional on the adopted sector of P6, on the declared finite-mode idealization and on the declared preparation; it asserts no measured mean, no property of an individual labelled particle, no separability of $ρ$ , no result for spatial infinite-mode models, and no sampling or detection model. The equality $m_{ρ} (A_{O}) = N tr (γ_{ρ} O)$ is exact within the model.

Facts & Assumptions

Given: the finite-mode sector data of the statement, the self-adjoint $O \in B (K)$ with unit $u_{O}$ , and a declared sector preparation $ρ$ .

[F1]

Sector vocabulary. $H_{N}^{σ} = ran P_{σ}^{(N)}$ is a closed subspace of $K^{\otimes N}$ ; every permutation unitary $U_{π}$ preserves it and acts there as the scalar ${sgn}_{σ} (π)$ ; a declared sector preparation is a density operator on $H_{N}^{σ}$ (Declared species sectors and their preparations: the framework's reading of P6 clauses (c)-(d), Permutation sector projections on the n-fold tensor power clause (c)).

[F2]

Slot operators and reduction. For $O \in B (K)$ and $i \in {1, \dots, N}$ the slot operator $O^{(i)}$ satisfies $O^{(i)} (v_{1} \otimes \dots \otimes v_{N}) = v_{1} \otimes \dots \otimes O v_{i} \otimes \dots \otimes v_{N}$ , is bounded with $∥ O^{(i)} ∥ \leq ∥ O ∥$ , and $(O^{(i)})^{*} = (O^{*})^{(i)}$ ; $Σ (O) = \sum_{i} O^{(i)}$ commutes with every $U_{π}$ and with $P_{σ}^{(N)}$ , preserves the sectors, and for every sector preparation $ρ$ with zero extension $\tilde{ρ}$ one has $tr (\tilde{ρ} Σ (O)) = N tr (γ_{ρ} O)$ , where $γ_{ρ}$ is a density operator on $K$ (The one-body reduction trace law of the finite-mode permutation sectors clauses (a), (c), (d), (e)).

[F3]

Observable reading and moments. An assigned sharp observable $(A, u_{A})$ has outcome law $μ_{ρ}^{A} (B) = tr (ρ E_{A} (B))$ ; for a bounded self-adjoint $A$ and every preparation $ρ$ the mean exists and equals $tr (ρ A)$ , and for a pure preparation of a unit vector $ψ$ it equals $⟨ A ψ, ψ ⟩$ (Assigned sharp observables and the outcome law of the framework clauses (a), (b), Mean, second moment and variance of an assigned observable in a preparation clauses (a), (d), Trace is a linear functional on $M_{n} (F)$ , Cyclicity of the trace).

[F4]

Slater and symmetric product vectors. For $N \geq 2$ and pairwise orthonormal $u_{1}, \dots, u_{N}$ the Slater vector $S (u_{1}, \dots, u_{N}) = N!^{- 1 / 2} \sum_{σ \in S_{N}} sgn (σ) u_{\hat{σ} (1)} \otimes \dots \otimes u_{\hat{σ} (N)}$ has norm one and equals $\sqrt{N!} P_{-}^{(N)} (u_{1} \otimes \dots \otimes u_{N})$ (Slater direct and exchange calculation clauses (a), (b)). For $N = 1$ , $S (u_{1}) : = u_{1}$ and $P_{\pm}^{(1)} = I$ give the same norm and projection identities directly. Setting $x : = u_{1} \otimes \dots \otimes u_{N}$ , orthonormality gives $⟨ P_{+}^{(N)} x, x ⟩ = \frac{1}{N!} \sum_{σ \in S_{N}} ⟨ U_{σ} x, x ⟩ = \frac{1}{N!} ⟨ x, x ⟩ = \frac{1}{N!}$ , so $Ψ : = \sqrt{N!} P_{+}^{(N)} x$ is a unit vector of the symmetric sector with $P_{+}^{(N)} Ψ = Ψ$ (Permutation sector projections on the n-fold tensor power clauses (b), (c), Hilbert tensor construction: inner product, completion and product bases).

[F5]

Rank-one pairing. $T_{u, v} x = ⟨ x, v ⟩ u$ is bounded with $tr (T_{u, v}) = ⟨ u, v ⟩$ and $T_{u, v}^{*} = T_{v, u}$ ; the map $(u, v) \mapsto T_{u, v}$ is linear in $u$ and conjugate-linear in $v$ (Rank-one operator calculus clauses (a), (b), (c)).

Proof

technique · direct; read the sector sum as a bounded observable, apply the supplied reduction identity, and evaluate the product-type cases by orthonormality of the mode family

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

1.1F2F3def. boundedalgebra

[The readout is a bounded self-adjoint sector observable] By [F2] each slot operator is bounded with $∥ O^{(i)} ∥ \leq ∥ O ∥$ , so $Σ (O)$ is bounded with $∥ Σ (O) ∥ \leq N ∥ O ∥$ ; since $O = O^{*}$ , [F2] gives $O^{(i) *} = O^{(i)}$ for every $i$ and $Σ (O)^{*} = Σ (O)$ . By [F2] $Σ (O)$ commutes with $P_{σ}^{(N)}$ , hence preserves $H_{N}^{σ}$ and its orthogonal complement, and the restriction $A_{O} = Σ (O) ∣_{H_{N}^{σ}}$ is a bounded self-adjoint operator on the sector space; with the unit $u_{O}$ the pair $(A_{O}, u_{O})$ is therefore an assigned sharp observable in the sense of [F3].

1.2F1F2F4algebra

[Reduced preparation of product-type states] Let $u_{1}, \dots, u_{N}$ be pairwise orthonormal, put $x : = u_{1} \otimes \dots \otimes u_{N}$ and let $Ψ$ be the unit vector of clause (c) for $σ = -$ or $σ = +$ ; both satisfy $P_{σ}^{(N)} Ψ = Ψ$ by [F4], and both equal $\sqrt{N!} P_{σ}^{(N)} x$ . Since $Σ (O)$ commutes with $P_{σ}^{(N)}$ and $P_{σ}^{(N)}$ is self-adjoint, $⟨ Σ (O) Ψ, Ψ ⟩ = N! ⟨ Σ (O) P_{σ}^{(N)} x, P_{σ}^{(N)} x ⟩ = N! ⟨ P_{σ}^{(N)} Σ (O) P_{σ}^{(N)} x, x ⟩ = N! \cdot \frac{1}{N!} \sum_{τ \in S_{N}} {sgn}_{σ} (τ) ⟨ Σ (O) U_{τ} x, x ⟩$ , with ${sgn}_{+} = 1$ and ${sgn}_{-} = sgn$ . For fixed $τ$ , $⟨ Σ (O) U_{τ} x, x ⟩ = \sum_{i = 1}^{N} ⟨ O u_{{\hat{τ}}^{- 1} (i)}, u_{i} ⟩ \prod_{k \neq i} ⟨ u_{{\hat{τ}}^{- 1} (k)}, u_{k} ⟩$ by [F1] and the elementary-tensor pairing, and orthonormality makes every factor $δ_{{\hat{τ}}^{- 1} (k), k}$ vanish unless $τ = i d$ ; hence only the identity permutation contributes and $⟨ Σ (O) Ψ, Ψ ⟩ = \sum_{i = 1}^{N} ⟨ O u_{i}, u_{i} ⟩$ .

2.1F2F3step 1.1algebra

[The mean law] Let $ρ$ be a declared sector preparation with zero extension $\tilde{ρ}$ and let $A_{O}$ be the readout of step 1.1. Since $A_{O}$ is bounded, [F3] gives $m_{ρ} (A_{O}) = tr (ρ A_{O})$ , and by steps 1.1 and [F2], $tr (ρ A_{O}) = tr (\tilde{ρ} Σ (O)) = N tr (γ_{ρ} O)$ . This proves (a) and (b).

3.1F5step 1.2step 2.1algebra

[Product-type means and the reduced preparation] In the situation of step 1.2, the mean is $m_{ρ} (A_{O}) = ⟨ Σ (O) Ψ, Ψ ⟩ = \sum_{j = 1}^{N} ⟨ O u_{j}, u_{j} ⟩$ , while the mean law of step 2.1 gives $m_{ρ} (A_{O}) = N tr (γ_{ρ} O)$ . The identity extends from self-adjoint $O$ to every $O$ by writing $O = (O + O^{*}) / 2 + i (O - O^{*}) / (2 i)$ and using linearity of both pairings. Testing it against $O = T_{u, v}$ and using [F5], $tr (γ_{ρ} T_{u, v}) = ⟨ γ_{ρ} u, v ⟩$ and $N^{- 1} \sum_{j} ⟨ T_{u, v} u_{j}, u_{j} ⟩ = N^{- 1} \sum_{j} ⟨ u_{j}, v ⟩ ⟨ u, u_{j} ⟩$ ; the identity for all $u, v$ therefore gives $γ_{ρ} = N^{- 1} \sum_{j = 1}^{N} T_{u_{j}, u_{j}}$ , since two endomorphisms of a finite-dimensional space with equal pairings against every rank-one operator are equal. This proves (c).

4.1step 1.1step 2.1step 3.1∎

[Conclusion and scope] Steps 1.1-2.1 prove (a) and (b) and step 3.1 proves (c). Every quantity is a spectral datum of the declared finite-mode sector model and the argument used only P1, P2, P6 and the finite-dimensional tensor and trace calculus; no measured mean, individual-particle property, separability statement or spatial reduction entered, which is (d).

Remarks

The factor $N$ and the normalization of $γ_{ρ}$ . The reduced preparation is normalized as a probability state on the one-particle space, while $A_{O}$ sums $N$ slot terms; the product-type case shows the two conventions consistently: a Slater determinant of $N$ orthonormal modes has reduced preparation the equal-weight mixture of those modes.
No separability claim. The mean law holds for arbitrary sector preparations, entangled or not; nothing in the argument factors $ρ$ , and the product-type clause is a special case.

The one-body readout law: sector means of summed bounded one-particle quantities and their reduced forms

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources