Measurement compatibility, dilation and identifiable tomography — Examples

Since $P_a^{\ast }=P_a=P_a^2$, the Lüders Kraus operators have $M_a^{L\ast }{M}_a^L=P_a$. For the other family, $M_a^{R\ast }{M}_a^R=(\mathrm{\mid }a⟩⟨+\mathrm{\mid })(\mathrm{\mid }+⟩⟨a\mathrm{\mid })=⟨+\mathrm{\mid }+⟩P_a=P_a$, since the displayed normalized vector satisfies $⟨+\mathrm{\mid }+⟩=1$. Thus both families are normalized, with the same outcome effects, because $P_0+P_1=I$.

By [F3], the one-operator Kraus forms give completely positive maps, and their sum is trace-preserving because the combined adjoint products sum to identity by step 1.1. Hence both are finite instruments in the sense of the A-page theorem, and each admits its stated outcome-register dilation by [F2].

On $\rho =P_0$, the Lüders update for outcome $0$ is $I_0^L(P_0)=P_0$, whereas the other update is $I_0^R(P_0)=\mathrm{\mid }+⟩⟨+\mathrm{\mid }$. These matrices differ: $P_0=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$ and $\mathrm{\mid }+⟩⟨+\mathrm{\mid }=\frac{1}{2}\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)$. The outcome probabilities are nevertheless identical, $\mathrm{Tr}(P_0{P}_0)=1$ and $\mathrm{Tr}(P_0{P}_1)=0$. This refutes the statement.

Direct multiplication gives ${\sigma }_j^2=I$, $\mathrm{Tr}({\sigma }_j)=0$, ${\sigma }_x{\sigma }_y=i{\sigma }_z$, ${\sigma }_y{\sigma }_z=i{\sigma }_x$, and ${\sigma }_z{\sigma }_x=i{\sigma }_y$; reversing the order negates each mixed product. Thus $\mathrm{Tr}({\sigma }_j{\sigma }_k)=2{\delta }_{jk}$. Since a Hermitian $2\times 2$ matrix has two real diagonal entries and one complex off-diagonal entry, ${\dim }_{\mathbb{R}}{\mathsf{H}}_2=4$; the four nonzero orthogonal matrices $I,{\sigma }_x,{\sigma }_y,{\sigma }_z$ form a basis. Write $\rho =aI+{\sum }_j{b}_j{\sigma }_j$ in this basis. Since $\mathrm{Tr}(I)=2$ and $\mathrm{Tr}({\sigma }_j)=0$, the trace-one condition gives $a=1\mathrm{/}2$; pairing with ${\sigma }_k$ gives $\mathrm{Tr}(\rho {\sigma }_k)=2b_k$. Thus the representation is unique with $r_k:=\mathrm{Tr}(\rho {\sigma }_k)=2b_k\in \mathbb{R}$. Write $r=(r_x,r_y,r_z)\in {\mathbb{R}}^3$ and $\mathrm{\parallel }r{\mathrm{\parallel }}_2=(r_x^2+r_y^2+r_z^2{)}^{1\mathrm{/}2}$. Positivity [A2] restricts $\mathrm{\parallel }r{\mathrm{\parallel }}_2\le 1$, since $(r_x{\sigma }_x+r_y{\sigma }_y+r_z{\sigma }_z{)}^2=\mathrm{\parallel }r{\mathrm{\parallel }}_2^{2}I$ and the eigenvalues of $\rho$ are $(1\pm \mathrm{\parallel }r{\mathrm{\parallel }}_2)\mathrm{/}2$.

Each ${\sigma }_j$ is Hermitian with eigenvalues $1,-1$ by step 1.1, so $F_{j,\pm }$ has eigenvalues $1\mathrm{/}3,0$ and is positive. For each $j$, $F_{j,+}+F_{j,-}=I\mathrm{/}3$; summing over $x,y,z$ gives $I$. Also $F_{j,+}-F_{j,-}={\sigma }_j\mathrm{/}3$, so each ${\sigma }_j$ lies in the real span of the effects. Since $I$ is their sum, the basis in step 1.1 shows that the effects span ${\mathsf{H}}_2$.

In the coordinate order $(x+,x-,y+,y-,z+,z-)$, direct multiplication and step 1.1 give $G_{(j,s),(k,t)}=\frac{1}{36}\mathrm{Tr}((I+s{\sigma }_j)(I+t{\sigma }_k))=\frac{1}{18}(1+st{\delta }_{jk})$ for $s,t\in \{+1,-1\}$. Thus the diagonal entries are $1\mathrm{/}9$, opposite outcomes on one axis have entry $0$, and distinct axes have entry $1\mathrm{/}18$. Let $u=(1,1,1,1,1,1)$ and $v_j=e_{j+}-e_{j-}$. Then $Gu=u\mathrm{/}3$, $G{v}_j=v_j\mathrm{/}9$, and $G$ vanishes on vectors $(s_x,s_x,s_y,s_y,s_z,s_z)$ with $s_x+s_y+s_z=0$. These mutually orthogonal subspaces have dimensions $1,3,2$ and span ${\mathbb{R}}^6$.

By [F2] and the eigenspaces in step 3.1, $G^{+}=3P_u+9{\sum }_j{P}_{v_j}$, where $P_u=u{u}^{\mathsf{T}}\mathrm{/}6$ and $P_{v_j}=v_j{v}_j^{\mathsf{T}}\mathrm{/}2$. The $e_{j+}$ column of $G^{+}$ is $u\mathrm{/}2+9v_j\mathrm{/}2$. Applying the canonical-dual formula [F3], using ${\sum }_{k,\pm }{F}_{k,\pm }=I$ and $F_{j,+}-F_{j,-}={\sigma }_j\mathrm{/}3$, gives $D_{j,+}=\frac{1}{2}I+\frac{3}{2}{\sigma }_j$; similarly $D_{j,-}=\frac{1}{2}I-\frac{3}{2}{\sigma }_j$. The probabilities are $p_{j,\pm }=\mathrm{Tr}(\rho F_{j,\pm })=(1\pm r_j)\mathrm{/}6$, so ${\sum }_{j,\pm }{p}_{j,\pm }{D}_{j,\pm }=\frac{1}{2}I+\frac{1}{2}{\sum }_j{r}_j{\sigma }_j=\rho$, in agreement with [F1].

The frequency vector with $f_{x,+}=1$ and all other entries zero is nonnegative and sums to one. Its output is $D_{x,+}=\frac{1}{2}I+\frac{3}{2}{\sigma }_x$, whose eigenvalues are $2$ and $-1$. It is Hermitian but not positive, so the linear-inversion procedure does not preserve positivity for every frequency vector.