Permutation invariance of sector preparations and the indistinguishability of relabelled readouts

[Invariance] Let $x,y\in {\mathfrak{H}}_n^{\sigma }$. By [F1] one has $U_{\pi }^{\ast }x={\mathrm{sgn}}_{\sigma }(\pi {)}^{-1}x={\mathrm{sgn}}_{\sigma }(\pi )x$ and likewise for $y$, so $⟨U_{\pi }\rho U_{\pi }^{\ast }x,y⟩=⟨\rho U_{\pi }^{\ast }x,U_{\pi }^{\ast }y⟩={\mathrm{sgn}}_{\sigma }(\pi {)}^2⟨\rho x,y⟩=⟨\rho x,y⟩$ by the unitarity of $U_{\pi }$ and the sesquilinearity of the inner product. Every $U_{\pi }$ 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, $⟨U_{\pi }y,x⟩=⟨y,U_{\pi }^{\ast }x⟩=0$. The extended $\rho$ vanishes on the complement and maps into the sector, so the same holds for $U_{\pi }\rho U_{\pi }^{\ast }$. The established equality on the sector therefore gives $U_{\pi }\rho U_{\pi }^{\ast }=\rho$ on the entire labelled space; multiplying this identity on the left by $U_{\pi }^{\ast }$ and on the right by $U_{\pi }$ and using $U_{\pi }^{\ast }{U}_{\pi }=I$ gives $U_{\pi }^{\ast }\rho U_{\pi }=\rho$ as well. This proves (a).

[Sector-preserving case] Put $P:=P_{\sigma }^{(n)}$, $M:=\mathrm{ran}P$ and $A_{\sigma }:=A{\mathrm{\mid }}_{D(A)\cap M}$. Under (c), [F3] gives $E_A(B)=U_{\pi }{E}_A(B)U_{\pi }^{\ast }$ for every $\pi$; hence $E_A(B)$ commutes with every $U_{\pi }$ and with their finite weighted sum $P$ from [F1]. Thus $E_{\sigma }(B):=E_A(B){\mathrm{\mid }}_M$ 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 $APx=PAx$ for $x\in D(A)$ by the finite weighted sum and the domain equality $U_{\pi }D(A)=D(A)$. 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 $y\in D(A_{\sigma }^{\ast })\subset M$ and $z=A_{\sigma }^{\ast }y\in M$, then for every $x\in D(A)$, $⟨Ax,y⟩=⟨APx,y⟩=⟨Px,z⟩=⟨x,z⟩$. Therefore $y\in D(A^{\ast })=D(A)$ and $Ay=z$, so $D(A_{\sigma }^{\ast })\subseteq D(A_{\sigma })$ and symmetry gives equality. For $x\in M$ the scalar second-moment domain and spectral integrals of $E_{\sigma }$ equal those of $E_A$, so [F2]'s uniqueness identifies $E_{\sigma }$ as the spectral PVM of $A_{\sigma }$. Finally the density expansion in clause (d) of [F1], $\rho ={\sum }_k{w}_k{T}_{u_k,u_k}$ 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 ${\mu }_{\rho }^{A_{\sigma }}(B)={\mu }_{\rho }^A(B)={\mu }_{\rho }^{A_{\pi }}(B)$, proving (c).

[Relabelled laws] Fix $\pi$ and an assigned sharp observable $(A,u_A)$. Since $U_{\pi }$ is a bijective linear isometry of $H^{\otimes n}$ and $A$ is self-adjoint, [F3] gives that $A_{\pi }=U_{\pi }A{U}_{\pi }^{\ast }$ is self-adjoint on the domain $U_{\pi }D(A)$ with spectral measure $E_{A_{\pi }}(B)=U_{\pi }{E}_A(B)U_{\pi }^{\ast }$. Hence for every preparation $\rho$ and Borel $B$, ${\mu }_{\rho }^{A_{\pi }}(B)=\mathrm{tr}(\rho U_{\pi }{E}_A(B)U_{\pi }^{\ast })$ by [F2]. Because $\rho$ is trace class and $U_{\pi }{E}_A(B)$ is bounded, [F4] applied twice gives $\mathrm{tr}(\rho U_{\pi }{E}_A(B)U_{\pi }^{\ast })=\mathrm{tr}(U_{\pi }^{\ast }\rho U_{\pi }{E}_A(B))$; by step 1.1 the middle factor is $\rho$, so ${\mu }_{\rho }^{A_{\pi }}(B)=\mathrm{tr}(\rho E_A(B))={\mu }_{\rho }^A(B)$, 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.

[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.