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The one-body reduction trace law of the finite-mode permutation sectors

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 $d \geq 1$ (Hilbert space), let $n \geq 1$ , let $H^{\otimes n}$ be the $n$ -fold completed tensor power with $H^{\otimes 0} : = C$ , $H^{\otimes 1} : = H$ and $H^{\otimes (k + 1)} : = H^{\otimes k} \hat{\otimes} H$ , let $(U_{π})_{π \in S_{n}}$ be the permutation unitaries and let $P_{σ}^{(n)}$ , $σ \in {+, -}$ , be the sector projections (Permutation sector projections on the n-fold tensor power), with sector space $H_{n}^{σ} : = ran P_{σ}^{(n)} \subseteq H^{\otimes n} .$ For $O \in B (H)$ and $i \in {1, \dots, n}$ let $O^{(i)} \in B (H^{\otimes n})$ be the operator of clause (a) below, and put $Σ (O) : = \sum_{i = 1}^{n} O^{(i)} \in B (H^{\otimes n}) .$ Identify $H^{\otimes n}$ with $H \hat{\otimes} H^{\otimes (n - 1)}$ by the surjective linear isometry $J_{n}$ of clause (b) below, and let ${Tr}_{H^{\otimes (n - 1)}}$ be the partial trace of Finite partial trace, Schmidt decomposition and finite no-signalling clause (a) over the second factor of that identification. Then:

(a) slot operators: for every $O \in B (H)$ and every $i \in {1, \dots, n}$ there is a unique $O^{(i)} \in B (H^{\otimes n})$ with $O^{(i)} (v_{1} \otimes \dots \otimes v_{n}) = v_{1} \otimes \dots \otimes v_{i - 1} \otimes O v_{i} \otimes v_{i + 1} \otimes \dots \otimes v_{n}$ for all $v_{1}, \dots, v_{n} \in H$ ; it satisfies $∥ O^{(i)} ∥ \leq ∥ O ∥$ and $(O^{*})^{(i)} = (O^{(i)})^{*}$ ; in particular $O = O^{*}$ gives $O^{(i)} = (O^{(i)})^{*}$ ;

(b) parenthesisation: the assignment $v_{1} \otimes \dots \otimes v_{n} \mapsto v_{1} \otimes (v_{2} \otimes \dots \otimes v_{n})$ extends to a surjective linear isometry $J_{n} : H^{\otimes n} \to H \hat{\otimes} H^{\otimes (n - 1)}$ ; for $n = 1$ the empty product is $1 \in C$ and $J_{1} u = u \otimes 1$ ;

(c) permutation invariance: $U_{π} Σ (O) U_{π}^{*} = Σ (O)$ for every $π \in S_{n}$ ; consequently $Σ (O)$ preserves both sector spaces and commutes with $P_{σ}^{(n)}$ : $Σ (O) P_{σ}^{(n)} = P_{σ}^{(n)} Σ (O) = P_{σ}^{(n)} Σ (O) P_{σ}^{(n)};$

(d) trace law: for every density operator $ρ$ on the sector space $H_{n}^{σ}$ (Density operators and scalar spectral probability clause (a)), let $j : H_{n}^{σ} ↪ H^{\otimes n}$ be the isometric inclusion and define its zero extension $\tilde{ρ} : = j ρ j^{*} \in B (H^{\otimes n})$ ; then for every $O \in B (H)$ one has $tr (\tilde{ρ} Σ (O)) = n tr (γ O)$ , where $γ : = {Tr}_{H^{\otimes (n - 1)}} (J_{n} \tilde{ρ} J_{n}^{- 1}) \in B (H)$ ;

(e) the reduced operator is a density operator: the operator $γ$ of clause (d) is a density operator on $H$ ; it is independent of the choice of slot in the sense that it equals the partial trace of $\tilde{ρ}$ over the other $n - 1$ slots in the parenthesisation in which the retained slot is the one the display names; for $n = 1$ , $γ = ρ$ . If $H_{n}^{σ} = {0}$ , there is no density operator of trace one on that sector, so clauses (d)-(e) make no assertion about a sector preparation;

(f) scope: clauses (a)-(e) are exact operator identities on the finite-dimensional labelled tensor powers; no physical system, particle species, preparation or measurement is asserted, and the parameter $n$ is a dimension count of the construction, not an observed particle number.

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 finite-dimensional complex Hilbert space $H$ of dimension $d \geq 1$ , the integer $n \geq 1$ , the completed tensor powers $H^{\otimes n}$ with permutation unitaries $U_{π}$ and sector projections $P_{σ}^{(n)}$ , the operator $O \in B (H)$ , the slot index $i$ , and the density operator $ρ$ on $H_{n}^{σ}$ with its zero extension $\tilde{ρ}$ .

[F1]

Tensor powers. $H^{\otimes n}$ is a complex Hilbert space; the elementary tensors have dense linear span; $∥ v_{1} \otimes \dots \otimes v_{n} ∥ = ∥ v_{1} ∥ \dots ∥ v_{n} ∥$ and $⟨ v_{1} \otimes \dots \otimes v_{n}, w_{1} \otimes \dots \otimes w_{n} ⟩ = \prod_{k} ⟨ v_{k}, w_{k} ⟩$ ; for orthonormal families spanning $H$ the corresponding elementary tensors form an orthonormal family with dense span (Hilbert tensor construction: inner product, completion and product bases clauses (b), (c), (d)).

[F2]

Parenthesisation. For every parenthesisation the iterated tensor product represents multilinear maps, and different parenthesisations are connected by the unique isomorphism preserving pure tensors (Finite iterated tensor products represent multilinear maps independently of parenthesization).

[F3]

Bounded tensor maps. For complex Hilbert spaces $V, W$ and $S \in B (V)$ , $T \in B (W)$ there is a unique $S \otimes T \in B (V \hat{\otimes} W)$ with $(S \otimes T) (u \otimes v) = S u \otimes T v$ , $∥ S \otimes T ∥ \leq ∥ S ∥ ∥ T ∥$ , $(S \otimes T)^{*} = S^{*} \otimes T^{*}$ , and for endomorphisms $S \otimes I$ and $I \otimes T$ commute (Bounded tensor maps: extension, norm estimate, adjoint and commutation clauses (a), (b), (c)).

[F4]

Permutation unitaries and sectors. $U_{π}$ is unitary with $U_{π} (v_{1} \otimes \dots \otimes v_{n}) = v_{{\hat{π}}^{- 1} (1)} \otimes \dots \otimes v_{{\hat{π}}^{- 1} (n)}$ and $U_{π} U_{σ} = U_{π σ}$ ; $U_{π} P_{+}^{(n)} = P_{+}^{(n)} U_{π} = P_{+}^{(n)}$ and $U_{π} P_{-}^{(n)} = P_{-}^{(n)} U_{π} = sgn (π) P_{-}^{(n)}$ ; the range of $P_{+}^{(n)}$ is the subspace of vectors fixed by every $U_{π}$ , and the range of $P_{-}^{(n)}$ is the subspace of vectors transforming by $sgn$ (Permutation sector projections on the n-fold tensor power clauses (a), (b), (c)).

[F5]

Partial trace. For endomorphisms $R$ of $V \otimes W$ with $V, W$ finite-dimensional, there is a unique endomorphism ${Tr}_{W} R$ of $V$ with $tr (({Tr}_{W} R) A) = tr (R (A \otimes I_{W}))$ for every $A \in End (V)$ , independent of the orthonormal basis used in its construction; if $R \geq 0$ and $tr R = 1$ then ${Tr}_{W} R \geq 0$ and $tr {Tr}_{W} R = 1$ (Finite partial trace, Schmidt decomposition and finite no-signalling clauses (a), (b)).

[F6]

Density operators. A density operator on a nonzero complex Hilbert space is a self-adjoint positive trace-class operator of trace one (Density operators and scalar spectral probability clause (a), Trace class operator, Trace of a trace class operator).

[F7]

Trace laws. The trace of an endomorphism of a finite-dimensional space is basis independent, linear, and $tr (X Y) = tr (Y X)$ (The basis-independent trace of an endomorphism of a finite-dimensional vector space, Trace is a linear functional on $M_{n} (F)$ ); for a trace-class operator $T$ and bounded $S$ one has $tr (S T) = tr (T S)$ (Cyclicity of the trace).

[F8]

Agreement on dense sets. Two bounded linear operators from a normed space into a Banach space that agree on a dense subspace are equal (A bounded linear map from a dense normed subspace into a Banach space extends uniquely with the same norm).

[F9]

Unitary conjugation preserves the operator norm. If $U$ is unitary then $∥ U T U^{*} ∥ = ∥ T ∥$ (Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A bounded linear operator between normed spaces).

Proof

technique · direct; construct the slot operators by conjugating the first-slot tensor map with a permutation unitary, transport the partial trace through the parenthesisation isometry, and reduce every slot to the first by permutation invariance and cyclicity

Given: the data of the statement and [F1]-[F9].

1.1F1F2F8algebra

[The parenthesisation isometry] For $n = 1$ , $J_{1} u = u \otimes 1$ is an isometry by [F1], and every tensor in $H \hat{\otimes} C$ equals $u \otimes 1$ , so it is surjective. For $n \geq 2$ , the map $(v_{1}, \dots, v_{n}) \mapsto v_{1} \otimes (v_{2} \otimes \dots \otimes v_{n})$ is multilinear from $H^{n}$ into $H \hat{\otimes} H^{\otimes (n - 1)}$ , so by [F2] applied to the two parenthesisations of the $n$ -fold algebraic tensor product it defines a linear map on the algebraic tensor product, and this linear map preserves inner products on elementary tensors by [F1]: $⟨ v_{1} \otimes \dots \otimes v_{n}, w_{1} \otimes \dots \otimes w_{n} ⟩ = \prod_{k} ⟨ v_{k}, w_{k} ⟩ = ⟨ v_{1}, w_{1} ⟩ ⟨ v_{2} \otimes \dots \otimes v_{n}, w_{2} \otimes \dots \otimes w_{n} ⟩$ . Since the elementary tensors span a dense subspace and both sides are complete, [F8] and this isometry identity give a linear isometry $J_{n}$ on the completed tensor powers, and the image is closed and contains all elementary tensors $v_{1} \otimes (v_{2} \otimes \dots \otimes v_{n})$ , whose span is dense by [F1]; hence $J_{n}$ is surjective. This proves (b).

1.2F1F3algebra

[The first-slot operator] By [F3] the operator $O \otimes I_{H^{\otimes (n - 1)}} \in B (H \hat{\otimes} H^{\otimes (n - 1)})$ is bounded with $(O \otimes I) (u \otimes w) = O u \otimes w$ . Transporting it through the isometry $J_{n}$ of clause (b) gives a bounded operator $A_{1} : = J_{n}^{- 1} (O \otimes I) J_{n}$ on $H^{\otimes n}$ with $A_{1} (v_{1} \otimes \dots \otimes v_{n}) = J_{n}^{- 1} (O v_{1} \otimes (v_{2} \otimes \dots \otimes v_{n})) = O v_{1} \otimes v_{2} \otimes \dots \otimes v_{n}$ for all $v_{1}, \dots, v_{n} \in H$ .

1.3F4F6F7algebra

[The zero extension and its symmetry] With $j$ the inclusion of the sector, $\tilde{ρ} = j ρ j^{*}$ restricts to $ρ$ on the sector and vanishes on its orthogonal complement. It is self-adjoint and positive because $⟨ \tilde{ρ} x, x ⟩ = ⟨ ρ j^{*} x, j^{*} x ⟩ \geq 0$ ; adjoining an orthonormal basis of the complement to one of the sector gives $tr \tilde{ρ} = tr ρ = 1$ , so it is a density operator by [F6]. Since $j j^{*} = P_{σ}^{(n)}$ and $U_{π} j = {sgn}_{σ} (π) j$ by [F4], $\tilde{ρ} = P_{σ}^{(n)} \tilde{ρ} P_{σ}^{(n)}$ and $U_{π} \tilde{ρ} U_{π}^{*} = \tilde{ρ}$ for every $π$ . A zero sector admits no trace-one endomorphism, so no $ρ$ exists in that case.

2.1F1F3F4F8F9step 1.2algebra

[Slot operators and the permutation invariance of their sum] Fix $i$ , choose $π \in S_{n}$ with $\hat{π} (1) = i$ and set $O^{(i)} : = U_{π} A_{1} U_{π}^{*} \in B (H^{\otimes n})$ ; [F9] and step 1.2 give $∥ O^{(i)} ∥ = ∥ O \otimes I ∥ \leq ∥ O ∥$ , and [F3], [F4] give $(O^{(i)})^{*} = U_{π} J_{n}^{- 1} (O^{*} \otimes I) J_{n} U_{π}^{*} = (O^{*})^{(i)}$ , so $O = O^{*}$ gives $O^{(i)} = (O^{(i)})^{*}$ . On elementary tensors, [F4] gives $U_{π}^{*} (v_{1} \otimes \dots \otimes v_{n}) = v_{\hat{π} (1)} \otimes \dots \otimes v_{\hat{π} (n)}$ ; step 1.2 replaces the first factor by $O v_{\hat{π} (1)} = O v_{i}$ ; and $U_{π}$ then moves that factor to slot $\hat{π} (1) = i$ , leaving every other factor in its original slot, so $O^{(i)} (v_{1} \otimes \dots \otimes v_{n}) = v_{1} \otimes \dots \otimes O v_{i} \otimes \dots \otimes v_{n}$ . If $π^{'}$ is another permutation with $\hat{π^{'}} (1) = i$ , then $τ : = π^{- 1} π^{'}$ has $\hat{τ} (1) = 1$ and $U_{τ} A_{1} U_{τ}^{*}$ agrees with $A_{1}$ on elementary tensors by [F1], [F3], hence equals it by [F8]; thus $U_{π^{'}} A_{1} U_{π^{'}}^{*} = U_{π} U_{τ} A_{1} U_{τ}^{*} U_{π}^{*} = U_{π} A_{1} U_{π}^{*}$ and $O^{(i)}$ is well defined, with the displayed elementary-tensor action. For $τ \in S_{n}$ one then has $U_{τ} O^{(i)} U_{τ}^{*} = U_{τ π} A_{1} U_{τ π}^{*} = O^{(\hat{τ} (i))}$ , and summing over $i$ with $\hat{τ}$ permuting ${1, \dots, n}$ gives $U_{τ} Σ (O) U_{τ}^{*} = Σ (O)$ for $Σ (O) : = \sum_{i = 1}^{n} O^{(i)}$ ; equivalently $Σ (O)$ commutes with every $U_{τ}$ . Since $P_{σ}^{(n)} = \frac{1}{n!} \sum_{π \in S_{n}} {sgn}_{σ} (π) U_{π}$ is a finite linear combination of the $U_{π}$ by [F4], $Σ (O)$ commutes with $P_{σ}^{(n)}$ , so it preserves $H_{n}^{σ}$ and its orthogonal complement and $Σ (O) P_{σ}^{(n)} = P_{σ}^{(n)} Σ (O) = P_{σ}^{(n)} Σ (O) P_{σ}^{(n)}$ ; this proves (a) and (c).

3.1F4F5F6F7step 1.1step 1.2step 1.3step 2.1algebra

[Reduction to one slot, the trace law and the reduced operator] By steps 1.3 and 2.1 and [F7], for $\hat{π} (1) = i$ one has $tr (\tilde{ρ} O^{(i)}) = tr (U_{π}^{*} \tilde{ρ} U_{π} A_{1}) = tr (\tilde{ρ} A_{1})$ . Set $R_{n} : = J_{n} \tilde{ρ} J_{n}^{- 1}$ , an endomorphism of $H \hat{\otimes} H^{\otimes (n - 1)}$ . Its partial trace $γ : = {Tr}_{H^{\otimes (n - 1)}} R_{n}$ is an endomorphism of $H$ by [F5]. Step 1.2 and [F5], [F7] give $tr (\tilde{ρ} Σ (O)) = n tr (\tilde{ρ} A_{1}) = n tr (R_{n} (O \otimes I)) = n tr (γ O)$ , proving (d). Unitary transport by $J_{n}$ preserves positivity and trace, so $R_{n}$ is positive of trace one; [F5] then makes $γ$ positive of trace one and hence a density operator on $H$ by [F6]. For the retained slot $i$ , use $J_{n} U_{π}^{*}$ as the parenthesisation map; its transported state is $J_{n} U_{π}^{*} \tilde{ρ} U_{π} J_{n}^{- 1} = R_{n}$ by step 1.3, so its partial trace is the same $γ$ . More generally the pairing $tr (γ O) = n^{- 1} tr (\tilde{ρ} Σ (O))$ for every $O$ uniquely determines $γ$ : testing against $O = T_{u, v}$ gives $tr (γ T_{u, v}) = ⟨ γ u, v ⟩$ by Rank-one operator calculus clause (e), which determines every matrix entry. For $n = 1$ , $J_{1} u = u \otimes 1$ , $P_{σ}^{(1)} = I$ and $Σ (O) = O$ , so this pairing gives $γ = ρ$ . This proves (e), including the one-particle boundary.

4.1step 1.1step 2.1step 3.1∎

[Conclusion and scope] Step 1.1 proves (b), step 2.1 proves (a) and (c), and step 3.1 proves (d) and (e); every argument used only the finite-dimensional tensor construction, the permutation sectors, the partial trace and the trace axioms, and no physical object, preparation or measurement entered, which is clause (f).

Remarks

Reading of the identities. The law $tr (\tilde{ρ} Σ (O)) = n tr (γ O)$ says that the mean of the sum of one-body slot operators in the sector preparation is $n$ times the trace pairing of a single one-particle operator with a reduced operator $γ$ on the one-particle space. Only the identity is asserted here; its physical reading requires the framework postulates and a declared preparation, and is stated separately.
Why the factor $n$ appears. The sum $Σ (O)$ has $n$ slot terms with equal pairings by permutation invariance of the sector state, and $γ$ is normalised once, $tr γ = 1$ ; the factor $n$ is the multiplicity of the sum, not a further normalisation.

The one-body reduction trace law of the finite-mode permutation sectors

Statement

Facts & Assumptions

Proof

Remarks

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