Density operators and scalar spectral probability

[Expansion of a density operator] Let $\rho$ be a density operator. By [F4] and [F1] it is compact, self-adjoint and positive, so [F3] applies: its positive eigenvalues ${\mu }_1>{\mu }_2>\cdots$ have finite multiplicities, accumulate only at $0$, and $\rho x={\sum }_k{\mu }_k{P}_{k}x$ with $P_k$ the orthogonal projection onto $E_{{\mu }_k}(\rho )$. Each $E_{{\mu }_k}(\rho )$ is finite-dimensional, so it has an orthonormal basis by Every finite-dimensional real or complex inner product space has an orthonormal basis; using Countable Choice to select the finite bases and listing their vectors in a finite or countable sequence $(u_n)$ with the corresponding eigenvalues $w_n>0$ indexed by a finite initial segment or the positive integers as appropriate and enumerated by [F3] (the multiset is finite or countably infinite) gives an orthonormal family with $\rho x={\sum }_n{w}_n⟨x,u_n⟩u_n$, that is, $\rho ={\sum }_n{w}_n{T}_{u_n,u_n}$ in operator norm, and [F4] gives $\mathrm{tr}\rho ={\sum }_n{w}_n=\mathrm{\parallel }\rho {\mathrm{\parallel }}_1=1$. Their rank-one trace norms sum to one by [F1], so the partial sums also converge in trace norm by completeness [F2]; that limit equals $\rho$ because both limits agree in operator norm.

[Weighted sums are density operators] Conversely let $(u_n)$ be orthonormal and $w_n\ge 0$ with ${\sum }_n{w}_n=1$. The series ${\sum }_n{w}_n{T}_{u_n,u_n}$ is nuclear with ${\sum }_n{w}_n\mathrm{\parallel }{u}_n\mathrm{\parallel }\mathrm{\parallel }{u}_n\mathrm{\parallel }={\sum }_n{w}_n<\mathrm{\infty }$, and its trace norms sum to one by [F1]. Completeness of trace class [F2] therefore makes its partial sums converge in trace norm to a trace-class operator $\rho$ with $\mathrm{\parallel }\rho {\mathrm{\parallel }}_1\le 1$; this also gives operator-norm convergence; it is self-adjoint, and $⟨\rho x,x⟩={\sum }_n{w}_n\mathrm{\mid }⟨x,u_n⟩{\mathrm{\mid }}^2\ge 0$ by continuity of the pairing and nonnegativity of each term, so $\rho$ is positive; [F1] and [F2] give $\mathrm{tr}\rho ={\sum }_n{w}_n⟨u_n,u_n⟩={\sum }_n{w}_n=1$ by linearity and absolute convergence. Hence $\rho$ is a density operator. Distinct orthonormal bases of a single positive eigenspace produce different families $(u_n)$ with the same sum, so expansions are not unique.

[The scalar law of a PVM] Let $E$ be a PVM on $(X,\mathrm{\Sigma })$ and let $\rho ={\sum }_n{w}_n{T}_{u_n,u_n}$ be an expansion from step 1.1 (any expansion gives the same value below by step 1.2 and [F2]). For $B\in \mathrm{\Sigma }$, $T_{u_n,u_n}E(B)=T_{u_n,E(B)u_n}$, so by linearity of the trace and trace-norm convergence of the series (traced against the bounded $E(B)$), $\mathrm{tr}(\rho E(B))={\sum }_n{w}_n\mathrm{tr}(T_{u_n,E(B)u_n})={\sum }_n{w}_n⟨u_n,E(B)u_n⟩={\sum }_n{w}_n\mathrm{\parallel }E(B)u_n{\mathrm{\parallel }}^2$, the series converging absolutely with value in $[0,1]$ because the finite subsums are bounded by $\sum w_n\mathrm{\parallel }{u}_n{\mathrm{\parallel }}^2=1$ by [F5]. Setting $B=X$ gives ${\mu }_{\rho }(X)={\sum }_n{w}_n=1$. If $B_k$ are pairwise disjoint with union $B$, then for each $n$ the identity $\mathrm{\parallel }E(B)u_n{\mathrm{\parallel }}^2={\sum }_k\mathrm{\parallel }E(B_k)u_n{\mathrm{\parallel }}^2$ holds by strong additivity of $E$ and continuity of the norm (Projection valued measure); summing the resulting absolutely convergent nonnegative double series by [F7] gives ${\mu }_{\rho }(B)={\sum }_k{\mu }_{\rho }(B_k)$. Thus ${\mu }_{\rho }$ is a probability measure; for $\rho =T_{u,u}$ with $\mathrm{\parallel }u\mathrm{\parallel }=1$ step 1.2 applies with $w_1=1$, giving ${\mu }_{\rho }(B)=\mathrm{\parallel }E(B)u{\mathrm{\parallel }}^2$.

[Moments for a self-adjoint operator, under AC] Assume the Axiom of Choice and let $\psi \in D(A)$ with $\mathrm{\parallel }\psi \mathrm{\parallel }=1$ and $E_A$ the PVM of $A$. By [F6], $\int {\lambda }^{2}d{E}_{A,\psi }<\mathrm{\infty }$ and, for the bounded measurable truncations ${\lambda }_n:=\lambda {\mathbf{1}}_{\{\mathrm{\mid }\lambda \mathrm{\mid }\le n\}}$, ${\mathrm{\Phi }}_{E_A}({\lambda }_n)\psi \to A\psi$ in norm. The bounded quadratic identity gives $\mathrm{\parallel }{\mathrm{\Phi }}_{E_A}({\lambda }_n)\psi {\mathrm{\parallel }}^2=\int {\lambda }^2{\mathbf{1}}_{\{\mathrm{\mid }\lambda \mathrm{\mid }\le n\}}d{E}_{A,\psi }$. These nonnegative measurable integrands increase to ${\lambda }^2$, so monotone convergence [F7] and norm continuity give $\int {\lambda }^{2}d{E}_{A,\psi }={\lim }_n\mathrm{\parallel }{\mathrm{\Phi }}_{E_A}({\lambda }_n)\psi {\mathrm{\parallel }}^2=\mathrm{\parallel }A\psi {\mathrm{\parallel }}^2$. Since $\psi \in D(A)$ the integral $\int \mathrm{\mid }\lambda \mathrm{\mid }d{E}_{A,\psi }$ is finite, because $\mathrm{\mid }\lambda \mathrm{\mid }\le (1+{\lambda }^2)\mathrm{/}2$ and $\int d{E}_{A,\psi }=1$; the same truncations ${\lambda }_n$ satisfy $\mathrm{\mid }{\lambda }_n\mathrm{\mid }\le \mathrm{\mid }\lambda \mathrm{\mid }$ and ${\lambda }_n\to \lambda$ pointwise, so Dominated convergence applied to the finite measures $E_{A,\psi }$ gives $\int \lambda d{E}_{A,\psi }={\lim }_n\int {\lambda }_{n}d{E}_{A,\psi }$. Pairing the identity $A\psi ={\lim }_n{\mathrm{\Phi }}_{E_A}({\lambda }_n)\psi$ (the defining truncation limit of Integral of a measurable function against a projection-valued measure) with $\psi$ and using $⟨{\mathrm{\Phi }}_{E_A}({\lambda }_n)\psi ,\psi ⟩=\int {\lambda }_{n}d{E}_{A,\psi }$ from [F6] yields $⟨A\psi ,\psi ⟩=\int \lambda d{E}_{A,\psi }(\lambda )$. Finally [F6] applied to $f=\mathrm{\mid }\lambda {\mathrm{\mid }}^{1\mathrm{/}2}$ gives $\int \mathrm{\mid }\lambda \mathrm{\mid }d{E}_{A,\psi }<\mathrm{\infty }$ exactly when $\psi \in D(\mathrm{\mid }A{\mathrm{\mid }}^{1\mathrm{/}2})$, and in that case the mean integral is finite.