Measurement compatibility, dilation and identifiable tomography

Summing the product law [A1] over the other $n-1$ coordinates gives $\mathbb{P}(X_k=1)=p{\prod }_{j\mathrm{\ne }k}(p+(1-p))=p$, so $\mathbb{E}{X}_k=p$. Put $S={\sum }_{k=1}^n(X_k-p)$. By [F1], $\mathbb{E}{e}^{\lambda (X_k-p)}\le e^{{\lambda }^2\mathrm{/}8}$ for every real $\lambda$. Expanding the finite expectation over $\{0,1{\}}^n$ and using [A1] factors it into the product of these $n$ expectations. Hence $\mathbb{E}{e}^{\lambda S}\le e^{n{\lambda }^2\mathrm{/}8}$.

For $\lambda >0$, $S\ge nt$ implies $e^{\lambda S}\ge e^{\lambda nt}$. For any nonnegative $W$ and $c>0$, $W\ge c{\mathbf{1}}_{\{W\ge c\}}$, so $\mathbb{P}(W\ge c)\le \mathbb{E}W\mathrm{/}c$. Apply this to $W=e^{\lambda S}$ and then step 1.1 to get $\mathbb{P}(f-p\ge t)\le e^{-\lambda nt+n{\lambda }^2\mathrm{/}8}$. Setting $\lambda =4t$ gives $e^{-2n{t}^2}$.

Apply the same argument to $-S$, using step 1.1 at $-4t$; it gives $\mathbb{P}(p-f\ge t)\le e^{-2n{t}^2}$. The union bound follows from ${\mathbf{1}}_{A\cup B}\le {\mathbf{1}}_A+{\mathbf{1}}_B$, so adding the two tails proves the claim. Endpoints $p=0,1$ are included by [A1]: then every $X_k=p$ almost surely and the left side is zero for $t>0$.

By [F1], each $I_a$ is completely positive in the finite matrix sense. Apply [F2] to obtain a finite Kraus family $(K_{a,r}{)}_r$ for it, allowing the empty family for a zero map. Since $\mathcal{A}$ is finite, these finitely many choices require only finite choice. Put $q={\sum }_a{q}_a$, where $q_a$ is the number of Kraus operators for outcome $a$.

The combined map ${\sum }_a{I}_a$ has Kraus family $(K_{a,r}{)}_{a,r}$ and is trace-preserving by [A1] and [F1]. Thus [F2] gives ${\sum }_{a,r}{K}_{a,r}^{\ast }{K}_{a,r}=I_d$. Since $d\ge 1$, this identity implies $q\ge 1$.

Let $(e_{a,r}{)}_{a,r}$ be the standard orthonormal basis of ${\mathbb{C}}^q$ indexed by these labels. Define $Vx={\sum }_{a,r}{K}_{a,r}x\otimes e_{a,r}$ and $Q_a={\sum }_r\mathrm{\mid }{e}_{a,r}⟩⟨e_{a,r}\mathrm{\mid }$. Then $V^{\ast }V={\sum }_{a,r}{K}_{a,r}^{\ast }{K}_{a,r}=I_d$, so $V$ is an isometry. Disjoint label sets make the $Q_a$ pairwise orthogonal and their sum $I_q$.

Expanding the sandwich gives ${\sum }_{r,s}{K}_{a,r}X{K}_{a,s}^{\ast }\otimes \mathrm{\mid }{e}_{a,r}⟩⟨e_{a,s}\mathrm{\mid }$ with both labels restricted to outcome $a$. The partial trace sends each register matrix unit $\mathrm{\mid }{e}_{a,r}⟩⟨e_{a,s}\mathrm{\mid }$ to ${\delta }_{rs}$, leaving ${\sum }_r{K}_{a,r}X{K}_{a,r}^{\ast }=I_a(X)$ by step 1.1. This proves the claimed formula.

Suppose all $P_i$ commute with all $Q_j$. Then $G_{ij}=P_i{Q}_j$ is self-adjoint, since $(P_i{Q}_j{)}^{\ast }=Q_j{P}_i=P_i{Q}_j$, and is idempotent, since $(P_i{Q}_j{)}^2=P_i^2{Q}_j^2=P_i{Q}_j$. For any $x$, $⟨G_{ij}x,x⟩=⟨G_{ij}x,G_{ij}x⟩\ge 0$, so each is positive; the sum identity makes each bounded above by $I_H$. Finite summation gives ${\sum }_{i,j}{G}_{ij}=({\sum }_i{P}_i)({\sum }_j{Q}_j)=I_H$, and the two marginal sums are $P_i$ and $Q_j$. So this family is a joint POVM.

Conversely, suppose a joint POVM $(G_{ij})$ exists. For each pair, $0\le G_{ij}\le P_i$ because $P_i-G_{ij}={\sum }_{k\mathrm{\ne }j}{G}_{ik}\ge 0$, and similarly $0\le G_{ij}\le Q_j$. If $x\in \ker P_i$, then $0\le ⟨G_{ij}x,x⟩\le ⟨P_{i}x,x⟩=0$. By [F2], $⟨G_{ij}x,x⟩=\mathrm{\parallel }{G}_{ij}^{1\mathrm{/}2}x{\mathrm{\parallel }}^2$, hence $G_{ij}x=0$ and $G_{ij}(I-P_i)=0$. Taking adjoints gives $(I-P_i)G_{ij}=0$, so $G_{ij}=P_i{G}_{ij}{P}_i$. The same argument yields $G_{ij}=Q_j{G}_{ij}{Q}_j$.

Using $P_i={\sum }_k{G}_{ik}$ and $Q_j={\sum }_{\mathrm{\ell }}{G}_{\mathrm{\ell }j}$, we obtain $P_i{Q}_j={\sum }_k{G}_{ik}{Q}_j$. The support relation $G_{ik}=Q_k{G}_{ik}{Q}_k$ from step 1.2 makes every term with $k\mathrm{\ne }j$ zero, and the $k=j$ term is $G_{ij}$. Therefore $P_i{Q}_j=G_{ij}$. Since $G_{ij}$ is self-adjoint, taking adjoints gives $Q_j{P}_i=G_{ij}=P_i{Q}_j$. This proves commutation and uniqueness.

For Hermitian $A,B$, $\overline{\mathrm{Tr}(AB)}=\mathrm{Tr}((AB{)}^{\ast })=\mathrm{Tr}(BA)=\mathrm{Tr}(AB)$ by [F2]. The trace is additive and real-homogeneous, so the pairing is real-valued, symmetric, and real-bilinear.

Diagonalize $A\in {\mathsf{H}}_d$ by [F3] with real eigenvalues ${\lambda }_1,\dots ,{\lambda }_d$. Then $\mathrm{Tr}(A^2)={\sum }_j{\lambda }_j^2\ge 0$, with equality exactly when all ${\lambda }_j=0$, hence exactly when $A=0$. This proves positive definiteness and the displayed square-root formula.

For $c\in {\mathbb{R}}^m$, bilinearity gives $c^{\mathsf{T}}Gc=\mathrm{Tr}((Sc{)}^2)=\mathrm{\parallel }Sc{\mathrm{\parallel }}_{\mathrm{H}\mathrm{S}}^2\ge 0$. For $X\in {\mathsf{H}}_d$, the vector $v$ with coordinates $v_a=\mathrm{Tr}(F_{a}X)$ satisfies $⟨Sc,X{⟩}_{\mathrm{H}\mathrm{S}}=c^{\mathsf{T}}v$ for every $c$, so it is $S^{\ast }X$. By [F2], $v_a=\mathrm{Tr}(X{F}_a)$; hence the matrix of $S^{\ast }S$ is precisely $G$. The equality $⟨c,Gc⟩=\mathrm{\parallel }Sc{\mathrm{\parallel }}_{\mathrm{H}\mathrm{S}}^2$ implies $\ker G=\ker S$: $Sc=0$ gives $Gc=0$, while $Gc=0$ gives $\mathrm{\parallel }Sc{\mathrm{\parallel }}_{\mathrm{H}\mathrm{S}}^2=0$ and hence $Sc=0$ by step 2.1. By [F4], $\mathrm{\mid }⟨A,B{⟩}_{\mathrm{H}\mathrm{S}}\mathrm{\mid }\le \mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{H}\mathrm{S}}\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{H}\mathrm{S}}$; expanding $\mathrm{\parallel }A+B{\mathrm{\parallel }}_{\mathrm{H}\mathrm{S}}^2$ and applying this bound gives $\mathrm{\parallel }A+B{\mathrm{\parallel }}_{\mathrm{H}\mathrm{S}}\le \mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{H}\mathrm{S}}+\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{H}\mathrm{S}}$. Thus the formula is a norm.

By [F2], $G$ is real symmetric positive semidefinite, $G=S^{\ast }S$, and $\ker G=\ker S$. The coordinate formula for the adjoint is $(S^{\ast }X{)}_a=⟨F_a,X{⟩}_{\mathrm{H}\mathrm{S}}=\mathrm{Tr}(X{F}_a)$ by [F1] and [F2].

Regard the real symmetric positive semidefinite matrix $G$ as a complex Hermitian matrix and apply [F3]. Its eigenvalues are real. If $z=x+iy$ is in the complex eigenspace for a real eigenvalue $\lambda$, then realness of $G$ gives $Gx=\lambda x$ and $Gy=\lambda y$; thus that eigenspace is the complexification of its real eigenspace. Real eigenspaces for distinct eigenvalues are orthogonal, since $Gu=\lambda u$, $Gv=\mu v$, and symmetry give $(\lambda -\mu )u^{\mathsf{T}}v=0$. The real eigenspaces form a direct sum of dimension $m$: their complexifications are the orthogonal complex eigenspaces, which span ${\mathbb{C}}^m$ by [F3]. Each eigenvalue is nonnegative because, for a nonzero real eigenvector $x$ with $Gx=\lambda x$, $\lambda \mathrm{\parallel }x{\mathrm{\parallel }}_2^2=x^{\mathsf{T}}Gx\ge 0$. In each real eigenspace, take a finite basis $v_1,\dots ,v_k$ and set $u_1=v_1\mathrm{/}\mathrm{\parallel }{v}_1{\mathrm{\parallel }}_2$ and $u_j=(v_j-{\sum }_{i<j}(v_j^{\mathsf{T}}{u}_i)u_i)\mathrm{/}\mathrm{\parallel }{v}_j-{\sum }_{i<j}(v_j^{\mathsf{T}}{u}_i)u_i{\mathrm{\parallel }}_2$ for $2\le j\le k$; basis independence makes every denominator nonzero, and these vectors form a real orthonormal basis. These bases give a real spectral resolution $G={\sum }_{\lambda \ge 0}\lambda P_{\lambda }$ with real orthogonal projections whose sum is $I$. Put $H_0={\sum }_{\lambda >0}{\lambda }^{-1}{P}_{\lambda }$. Orthogonality gives $G{H}_{0}G=G$, $H_{0}G{H}_0=H_0$, and $G{H}_0=H_{0}G={\sum }_{\lambda >0}{P}_{\lambda }$, which are self-adjoint. Thus $H_0$ satisfies the four Penrose equations [F4], and uniqueness in [F5] gives $H_0=G^{+}$. Consequently $Q=G^{+}G={\sum }_{\lambda >0}{P}_{\lambda }$ is an orthogonal projection and $\ker Q=\ker G$, so $\mathrm{Ran}(I-Q)=\ker G=\ker S$ by [F2].

The synthesis map of the canonical dual is $D=S{G}^{+}$ by [F1]. For any $X\in {\mathsf{H}}_d$, the spanning hypothesis [A1] supplies $c\in {\mathbb{R}}^m$ with $X=Sc$. Then $D(S^{\ast }X)=S{G}^{+}{S}^{\ast }Sc=S{G}^{+}Gc=SQc=Sc=X,$ because $c-Qc\in \mathrm{Ran}(I-Q)=\ker S$ by step 2.1. The coordinates of $S^{\ast }X$ are $\mathrm{Tr}(X{F}_a)$ by step 1.1, which proves the stated sum.

Apply [F1] with $H=\rho$. Then ${\sum }_a{p}_a{D}_a=\rho$, so [F3] gives $X(p)=\rho$. The probabilities are in $[0,1]$ and sum to one by [F2].

For each sample $k$, ${\sum }_{a=1}^m{\mathbf{1}}_{\{Y_k=a\}}=1$, so $f_a\ge 0$ and ${\sum }_a{f}_a=1$, as required by [F3]. Fix an outcome $a$ and set $Z_k^a={\mathbf{1}}_{\{Y_k=a\}}$. For every $z\in \{0,1{\}}^n$, summing [A1] over the finite outcome tuples with ${\mathbf{1}}_{\{Y_k=a\}}=z_k$ gives $\mathbb{P}(Z_1^a=z_1,\dots ,Z_n^a=z_n)={\prod }_{k=1}^n{\sum }_{b:{\mathbf{1}}_{\{b=a\}}=z_k}{p}_b={\prod }_{k=1}^n{b}_{p_a}(z_k),$ since [F2] gives ${\sum }_b{p}_b=1$. Thus these indicators have the Bernoulli product law, and [F4] yields $\mathbb{P}(\mathrm{\mid }{f}_a-p_a\mathrm{\mid }\ge t)\le 2e^{-2n{t}^2}$. No independence between indicators for different outcomes is used.

By linearity, $X(f)-\rho =D(f-p)$. Set $C=({\sum }_a\mathrm{\parallel }{D}_a{\mathrm{\parallel }}_{\mathrm{H}\mathrm{S}}^2{)}^{1\mathrm{/}2}$. By [F7], this is the norm from a real inner product, so its triangle inequality follows from [F5] by expanding the square of a sum. Thus $\mathrm{\parallel }Dc{\mathrm{\parallel }}_{\mathrm{H}\mathrm{S}}\le {\sum }_a\mathrm{\mid }{c}_a\mathrm{\mid }\mathrm{\parallel }{D}_a{\mathrm{\parallel }}_{\mathrm{H}\mathrm{S}}\le C\mathrm{\parallel }c{\mathrm{\parallel }}_2$. Since $m\ge 1$, the unit sphere in ${\mathbb{R}}^m$ contains $e_1$, so the set $\{\mathrm{\parallel }Dc{\mathrm{\parallel }}_{\mathrm{H}\mathrm{S}}:\mathrm{\parallel }c{\mathrm{\parallel }}_2=1\}$ is nonempty; the bound makes it bounded above. Its supremum exists by [F6] and is the finite induced operator norm in the Statement. Scaling $c$ (and treating $c=0$ directly) gives $\mathrm{\parallel }Dc{\mathrm{\parallel }}_{\mathrm{H}\mathrm{S}}\le \mathrm{\parallel }D{\mathrm{\parallel }}_{2\to \mathrm{H}\mathrm{S}}\mathrm{\parallel }c{\mathrm{\parallel }}_2$, which proves the deterministic estimate.

A finite union bound gives $\mathbb{P}(\mathrm{\exists }a:\mathrm{\mid }{f}_a-p_a\mathrm{\mid }\ge t)\le 2m{e}^{-2n{t}^2}$. Set $t=\sqrt{\log (2m\mathrm{/}\delta )\mathrm{/}(2n)}$, which is positive since $2m\mathrm{/}\delta >1$; then the right side is $\delta$. On the complementary event, $\mathrm{\parallel }f-p{\mathrm{\parallel }}_2\le \sqrt{m}t$, and step 2.1 gives the asserted state-error bound.