Measurements, POVMs, and Instruments

Fix $a$. Set ${\tau }_a=P_a\rho P_a$. Self-adjointness of $P_a$ and $\rho$ gives $⟨{\tau }_{a}x,y⟩=⟨\rho P_{a}x,P_{a}y⟩=⟨x,{\tau }_{a}y⟩$ for all $x,y$, so ${\tau }_a$ is self-adjoint. It is nonnegative, since $⟨{\tau }_{a}x,x⟩=⟨\rho P_{a}x,P_{a}x⟩\ge 0$ for every $x$ by [F6]–[F7]. By [F2], ${\tau }_a$ has nonnegative diagonal entries in an orthonormal basis, so [F5] gives $\mathrm{tr}({\tau }_a)\ge 0$. By [F3] and $P_a^2=P_a$, $p_{\rho }(a)=\mathrm{tr}(\rho P_a)=\mathrm{tr}(P_a\rho P_a)=\mathrm{tr}({\tau }_a)\ge 0$.

By [F4], the resolution of the identity in [F1], and [F6], ${\sum }_a{p}_{\rho }(a)=\mathrm{tr}(\rho {\sum }_a{P}_a)=\mathrm{tr}(\rho I_H)=\mathrm{tr}(\rho )=1$.

For each $a$, $1-p_{\rho }(a)={\sum }_{b\mathrm{\ne }a}{p}_{\rho }(b)\ge 0$ by step 1.1, with the empty sum interpreted as $0$ when $\mathrm{\Omega }=\{a\}$. Therefore every weight is at most one and, by step 1.1, belongs to $[0,1]$.

Choose finite bases of $H$ and $K$, which exist by [F7] and are linearly independent by the basis definition; apply [F9] to obtain orthonormal bases $(u_{\alpha })$ and $(v_{\beta })$ with the same spans. By [F1]–[F3], $(u_{\alpha }\otimes e_j{)}_{\alpha ,j}$ and $(v_{\beta }\otimes e_j{)}_{\beta ,j}$ are orthonormal bases of $H\otimes R$ and $K\otimes R$. For each $j,\alpha$, uniquely expand $X(u_{\alpha }\otimes e_j)={\sum }_{i<r,\beta }{c}_{\beta i;\alpha j}{u}_{\beta }\otimes e_i$. Define $X_{ij}{u}_{\alpha }={\sum }_{\beta }{c}_{\beta i;\alpha j}{u}_{\beta }$ and extend linearly from the basis of $H$. Linearity of $X$ gives the required block formula on every $x\in H$. Any other family satisfying it has the same coefficients on every product-basis vector, hence each block agrees on the basis $u_{\alpha }$ and is equal.

On each product-basis vector $v_{\beta }\otimes e_j$, prescribe the image ${\sum }_i\mathrm{\Phi }(X_{ij})v_{\beta }\otimes e_i$. Since these product vectors are a basis, there exists a unique linear operator $Y$ on $K\otimes R$ with those values. By linearity this gives the displayed rule for all $y\in K$. Define ${\mathrm{\Phi }}^{(r)}(X)=Y$. Uniqueness of the blocks from step 1.1 and uniqueness of an operator from its values on a basis show that this definition is unambiguous.

For $X,X^{\mathrm{\prime }}\in \mathcal{L}(H\otimes R)$, their block families are respectively $[X_{ij}]$ and $[X_{ij}^{\mathrm{\prime }}]$. The defining action on the product basis shows that the blocks of $sX+t{X}^{\mathrm{\prime }}$ are $s{X}_{ij}+t{X}_{ij}^{\mathrm{\prime }}$ for any scalars $s,t\in \mathbb{C}$. Applying the linearity of $\mathrm{\Phi }$ from [F6] gives blocks $s\mathrm{\Phi }(X_{ij})+t\mathrm{\Phi }(X_{ij}^{\mathrm{\prime }})$, which are exactly the blocks of $s{\mathrm{\Phi }}^{(r)}(X)+t{\mathrm{\Phi }}^{(r)}(X^{\mathrm{\prime }})$. Equality on a basis yields ${\mathrm{\Phi }}^{(r)}(sX+t{X}^{\mathrm{\prime }})=s{\mathrm{\Phi }}^{(r)}(X)+t{\mathrm{\Phi }}^{(r)}(X^{\mathrm{\prime }})$. When $r=1$, identify $R={\mathbb{C}}^1$ with the tensor unit $\mathbb{C}$ by $1\mapsto e_0$, and define $u_H:H\to H\otimes R$ and $u_K:K\to K\otimes R$ by $u_{H}x=x\otimes e_0$ and $u_{K}y=y\otimes e_0$. These are the corresponding canonical unit identifications of [F8]. By [F3], they send the orthonormal bases chosen in step 1.1 to the corresponding product bases, and [F1]–[F2] show they preserve inner products; hence they are unitary. The sole block equations are $X{u}_H=u_H{X}_{00}$ and ${\mathrm{\Phi }}^{(1)}(X)u_K=u_K\mathrm{\Phi }(X_{00})$, so $X_{00}=u_H^{-1}X{u}_H$ and $u_K^{-1}{\mathrm{\Phi }}^{(1)}(X)u_K=\mathrm{\Phi }(X_{00})=\mathrm{\Phi }(u_H^{-1}X{u}_H)$. This is the original map under these identifications.

Choose a finite basis of $H$, which exists by [F18] and is linearly independent by the basis definition, and apply [F17] to obtain an orthonormal basis of $H$. With the standard orthonormal basis of $R$ from [F4], [F6] gives an orthonormal product basis of $H\otimes R$. For elementary tensors, [F7] and the tensor-map formula [F15] give $Q_a(x\otimes e_c)={\delta }_{j(a),c}x\otimes e_{j(a)}$. Therefore $Q_a^2=Q_a$, $Q_a{Q}_b=0$ when $a\mathrm{\ne }b$, and ${\sum }_a{Q}_a=I$ on every product-basis vector; equality on that basis gives these operator identities. For elementary tensors $x\otimes e_c,y\otimes e_d$, orthonormality and the product inner product give $⟨Q_a(x\otimes e_c),y\otimes e_d⟩={\delta }_{j(a),c}{\delta }_{j(a),d}⟨x,y⟩=⟨x\otimes e_c,Q_a(y\otimes e_d)⟩$. Sesquilinearity and spanning by [F6] extend this identity to all vectors, so uniqueness of adjoints [F11] gives $Q_a^{\ast }=Q_a$. By [F2], $(Q_a)$ is a projective measurement.

Put $M_a=E_a^{1\mathrm{/}2}$ using [F9]. Each summand $x\mapsto M_{a}x\otimes e_{j(a)}$ is linear by [F15], so $V$ is linear. For $x,y\in H$, orthogonality of the register labels and the product inner product give $⟨Vx,Vy⟩={\sum }_a⟨M_{a}x,M_{a}y⟩=⟨x,\left({\sum }_a{M}_a^{\ast }{M}_a\right)y⟩=⟨x,y⟩$, because $M_a$ is self-adjoint by [F8]–[F9], so $M_a^{\ast }{M}_a=E_a$, and [F1] gives ${\sum }_a{E}_a=I_H$. The adjoint identities [F10]–[F11] also give $⟨Vx,Vy⟩=⟨x,V^{\ast }Vy⟩$. For fixed $y$, set $x=(V^{\ast }V-I_H)y$; then $\mathrm{\parallel }x{\mathrm{\parallel }}^2=0$, so positive definiteness [F13] implies $(V^{\ast }V-I_H)y=0$. Since $y$ was arbitrary, $V^{\ast }V=I_H$.

Fix $b\in \mathrm{\Omega }$. Step 1.1's action formula and the definition of $V$ give $Q_{b}Vx=M_{b}x\otimes e_{j(b)}$. For all $x,y\in H$, the adjoint identity [F10], conjugate symmetry [F13], and product inner product [F5] yield $⟨V^{\ast }{Q}_{b}Vx,y⟩=⟨Q_{b}Vx,Vy⟩=⟨M_{b}x,M_{b}y⟩=⟨x,E_{b}y⟩$, where the last equality uses self-adjointness from [F8]–[F9], which gives $M_b^{\ast }{M}_b=E_b$. If $T=V^{\ast }{Q}_{b}V-E_b$, then $⟨Tx,y⟩=0$ for every $x,y$. Taking $y=Tx$ gives $\mathrm{\parallel }Tx{\mathrm{\parallel }}^2=0$, so [F13] implies $Tx=0$ for every $x$; hence $V^{\ast }{Q}_{b}V=E_b$.

For all $z,w\in H\otimes R$, the adjoint identities [F10] and self-adjointness of $\rho$ from [F3], [F8] give $⟨V\rho V^{\ast }z,w⟩=⟨\rho V^{\ast }z,V^{\ast }w⟩=⟨V^{\ast }z,\rho V^{\ast }w⟩=⟨z,V\rho V^{\ast }w⟩$, where the last equality follows from [F10] and conjugate symmetry [F13]. Thus $V\rho V^{\ast }$ is self-adjoint. Also, for every $z$, $⟨V\rho V^{\ast }z,z⟩=⟨\rho V^{\ast }z,V^{\ast }z⟩\ge 0$ by [F3], [F8] and [F10], so it is nonnegative by [F8]. Also $\dim (H\otimes R)=r\dim H>0$ by [F4] and [F6], since $H$ is nonzero by [F18] and $r\ge 1$. In bases, rectangular trace cyclicity [F12] and [F14] give $\mathrm{tr}(V\rho V^{\ast })=\mathrm{tr}(V^{\ast }V\rho )=\mathrm{tr}(\rho )=1$ by step 1.2 and [F3]. Therefore $V\rho V^{\ast }$ is a density operator on the nonzero space $H\otimes R$.

Step 1.1 proves $(Q_a)$ is a projective measurement and step 2.2 proves $V\rho V^{\ast }$ is a density operator. Thus [F16] identifies its $b$-outcome weight as $\mathrm{tr}(V\rho V^{\ast }{Q}_b)$. Rectangular trace cyclicity [F12] and [F14] rewrite this as $\mathrm{tr}(V^{\ast }{Q}_{b}V\rho )$; step 2.1 makes that $\mathrm{tr}(E_b\rho )=\mathrm{tr}(\rho E_b)$, using [F12] once more on $H$. The final expression is the original POVM weight by [F1], proving the claimed equality.

Fix $a\in \mathrm{\Omega }$. Let $M_a=E_a^{1\mathrm{/}2}$ be the nonnegative square root supplied by [F2]. Since $M_a$ and $\rho$ are self-adjoint by [F2] and [F7]–[F8], for every $x,y\in H$ one has $⟨M_a\rho M_{a}x,y⟩=⟨\rho M_{a}x,M_{a}y⟩=⟨M_{a}x,\rho M_{a}y⟩=⟨x,M_a\rho M_{a}y⟩$, so $M_a\rho M_a$ is self-adjoint. For every $x\in H$, $⟨M_a\rho M_{a}x,x⟩=⟨\rho M_{a}x,M_{a}x⟩\ge 0$ by [F7]–[F8], so $M_a\rho M_a$ is nonnegative. By [F3] it is diagonal with nonnegative entries in an orthonormal basis, hence its trace is nonnegative by [F6]. By [F4] and $M_a^2=E_a$, $p_{\rho }(a)=\mathrm{tr}(\rho E_a)=\mathrm{tr}(M_a\rho M_a)\ge 0$.

Linearity of trace [F5] and the completeness relation in [F1] give ${\sum }_{a\in \mathrm{\Omega }}{p}_{\rho }(a)=\mathrm{tr}(\rho {\sum }_a{E}_a)=\mathrm{tr}(\rho I_H)=\mathrm{tr}(\rho )=1$ by [F7] and [F6].

For fixed $a$, $1-p_{\rho }(a)={\sum }_{b\in \mathrm{\Omega },b\mathrm{\ne }a}{p}_{\rho }(b)\ge 0$ by step 1.1; if $\mathrm{\Omega }=\{a\}$ this is the empty sum $0$. Thus $p_{\rho }(a)\le 1$, and together with step 1.1 each weight lies in $[0,1]$.

Set ${\tau }_a=P_a\rho P_a$. Self-adjointness of $P_a$ and $\rho$ gives $⟨{\tau }_{a}x,y⟩=⟨\rho P_{a}x,P_{a}y⟩=⟨x,{\tau }_{a}y⟩$ for all $x,y$, so ${\tau }_a$ is self-adjoint. For every $x\in H$, $⟨{\tau }_{a}x,x⟩=⟨\rho P_{a}x,P_{a}x⟩\ge 0$ by [F5] and [F6], so ${\tau }_a$ is nonnegative. Since $p_a>0$ is real, scaling the same pairing identity by $1\mathrm{/}{p}_a$ makes ${\rho }_a={\tau }_a\mathrm{/}{p}_a$ self-adjoint; its quadratic form is nonnegative as well, so ${\rho }_a$ is nonnegative.

By [F3] and $P_a^2=P_a$, $\mathrm{tr}({\tau }_a)=\mathrm{tr}(P_a\rho P_a)=\mathrm{tr}(\rho P_a^2)=p_a$ by [F2]. Hence [F4] gives $\mathrm{tr}({\rho }_a)=p_a\mathrm{/}{p}_a=1$, so ${\rho }_a$ is a density operator by [F5].

For every $x\in H$, ${\rho }_{a}x=P_a(\rho P_{a}x)\mathrm{/}{p}_a$ is of the form $P_{a}y$ with $y=\rho P_{a}x\mathrm{/}{p}_a$. By [F7], every such vector lies in $\mathrm{im}(P_a)$, so $\mathrm{im}({\rho }_a)\subseteq \mathrm{im}(P_a)$.

Fix $\rho \in \mathcal{D}(H)$. By [F2], the weights $p_{\rho }(a)$ are nonnegative and sum to one, so [F3] gives a finite probability space on $\mathrm{\Omega }$. Its probability for an event $S$ is ${\sum }_{a\in S}{p}_{\rho }(a)$, which equals $K_E(\rho ,S)$ by [F1]. Thus the kernel value at each input is a probability measure on the finite outcome set.

Fix $S\subseteq \mathrm{\Omega }$ and use the ordered basis defining the matrix coordinates. In those coordinates, each summand is $\mathrm{tr}(\rho E_a)={\sum }_{i,j}{\rho }_{ij}(E_a{)}_{ji}$ by [F4] and the row-by-column product formula [F8]; it is a finite linear combination of coordinate projections $\rho \mapsto {\rho }_{ij}$. The projections are continuous in the matrix-coordinate topology, so each summand and the finite sum $g_S(\rho )=K_E(\rho ,S)$ are continuous.

For every open $U\subseteq \mathbb{R}$, continuity in step 1.2 makes $g_S^{-1}(U)$ open and hence Borel. The collection of subsets $B\subseteq \mathbb{R}$ for which $g_S^{-1}(B)$ is Borel is a $\sigma$-algebra, because inverse images preserve complements and countable unions; it contains every open set by the same continuity. It therefore contains the Borel $\sigma$-algebra, proving that $g_S$ is Borel measurable.

Let $\lambda \in [0,1]$ and $\rho ,\sigma \in \mathcal{D}(H)$. Put $T=\lambda \rho +(1-\lambda )\sigma$. Since $\lambda$ and $1-\lambda$ are real and $\rho ,\sigma$ are self-adjoint by [F6]–[F7], for every $x,y\in H$ one has $⟨Tx,y⟩=\lambda ⟨\rho x,y⟩+(1-\lambda )⟨\sigma x,y⟩=⟨x,Ty⟩$. Thus $T$ is self-adjoint; also $⟨Tx,x⟩=\lambda ⟨\rho x,x⟩+(1-\lambda )⟨\sigma x,x⟩\ge 0$, so $T$ is nonnegative by [F7]. It has trace one by [F6] and [F5], so it lies in $\mathcal{D}(H)$. Applying [F5] to the finite sum in [F1] gives $K_E(\lambda \rho +(1-\lambda )\sigma ,S)=\lambda K_E(\rho ,S)+(1-\lambda )K_E(\sigma ,S)$. This proves affinity.

For each $a$, complete positivity [F1]–[F2] at $r=1$ sends the nonnegative input $\rho$ from [F5] to a nonnegative ${\tau }_a:={\mathcal{I}}_a(\rho )$ in the sense of [F6]. By [F3], ${\tau }_a$ has nonnegative diagonal entries in an orthonormal basis, and [F4] therefore gives $p_a=\mathrm{tr}({\tau }_a)\ge 0$.

Apply the trace-preservation identity in [F1] to $X=\rho$. It gives ${\sum }_{a\in \mathrm{\Omega }}{p}_a=\mathrm{tr}(\rho )=1$ by [F5]. Thus the nonnegative branch weights sum to one.

For any $a$, $1-p_a={\sum }_{b\mathrm{\ne }a}{p}_b\ge 0$ by step 1.1, including the empty sum when $\mathrm{\Omega }=\{a\}$. Hence $p_a\in [0,1]$ for every outcome, and the weights form a finite probability distribution.