Computational-basis instruments and conditional states on two registers

Each basis projection $P_a$ and $Q_b$ is diagonal with entries $0$ or $1$, hence self-adjoint and idempotent; distinct projections in each family are orthogonal and each family sums to the identity. By [F8], their lifted maps $P_a\otimes I_B$ and $I_A\otimes Q_b$ are self-adjoint idempotents, and each lifted family sums to $I_{A\otimes B}$.

The product basis $(e_a\otimes g_b{)}_{a,b}$ is orthonormal by [F4]. Therefore the coordinates of $\psi$ satisfy ${\sum }_{a,b}\mathrm{\mid }{c}_{a,b}{\mathrm{\mid }}^2=1$. By [F7], $\rho =\mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid }$ has matrix $v{v}^{†}$, so for every coordinate vector $u$ its quadratic form is $u^{†}v{v}^{†}u=\mathrm{\mid }{v}^{†}u{\mathrm{\mid }}^2\ge 0$ and its trace is ${\sum }_{a,b}\mathrm{\mid }{c}_{a,b}{\mathrm{\mid }}^2=1$. Hence $\rho$ is a density operator by [F6].

Each branch map is complex-linear. For any finite reference register $R$, write an operator $Q$ on $(A\otimes B)\otimes R$ in the ancilla blocks of [F2]. Its amplified branch sends each block $Q_{ij}$ to $C{Q}_{ij}C$, so [F8] identifies the full block matrix as $Q^{\mathrm{\prime }}=(C\otimes I_R)Q(C\otimes I_R)$, where $C$ is the lifted projector from step 1.1. If $Q$ is nonnegative, then $Q^{\mathrm{\prime }}$ is self-adjoint and $⟨Q^{\mathrm{\prime }}u,u⟩=⟨Q(C\otimes I_R)u,(C\otimes I_R)u⟩\ge 0$ for every $u$ by [F3], so every amplification preserves nonnegative operators. Thus each branch is completely positive by [F2].

Write $v$ for the coordinate column of $\psi$ and $C$ for either lifted projector. By [F7], the corresponding branch is $Cv{v}^{†}C=(Cv)(Cv{)}^{†}$. For outcome $a$, $(P_a\otimes I_B)\psi ={\sum }_b{c}_{a,b}{e}_a\otimes g_b$, so the diagonal entry at each coordinate $(a,b)$ is $\mathrm{\mid }{c}_{a,b}{\mathrm{\mid }}^2$ and every other diagonal entry is zero; off-diagonal entries within the selected $a$-subspace can be nonzero. Its trace is ${\sum }_b\mathrm{\mid }{c}_{a,b}{\mathrm{\mid }}^2$. For outcome $b$, $(I_A\otimes Q_b)\psi ={\sum }_a{c}_{a,b}{e}_a\otimes g_b$, so the diagonal entry at each coordinate $(a,b)$ is $\mathrm{\mid }{c}_{a,b}{\mathrm{\mid }}^2$ and every other diagonal entry is zero; off-diagonal entries within the selected $b$-subspace can be nonzero. Its trace is ${\sum }_a\mathrm{\mid }{c}_{a,b}{\mathrm{\mid }}^2$. These traces are the asserted probabilities by [F1].

For every operator $R$, cyclicity of trace and idempotence give ${\sum }_a\mathrm{tr}((P_a\otimes I_B)R(P_a\otimes I_B))=\mathrm{tr}\text{\hspace{0.17em}⁣}(R{\sum }_a(P_a\otimes I_B))=\mathrm{tr}(R)\mathrm{.}$ The same calculation gives ${\sum }_b\mathrm{tr}((I_A\otimes Q_b)R(I_A\otimes Q_b))=\mathrm{tr}(R)$. Together with step 2.1 these are the finite-instrument conditions of [F1].

If a branch vector $w=C\psi$ has positive probability $p={\sum }_i\mathrm{\mid }{w}_i{\mathrm{\mid }}^2$ from step 2.2, its conditional operator in [F1] is $\mathrm{\mid }(w\mathrm{/}\sqrt{p})⟩⟨(w\mathrm{/}\sqrt{p})\mathrm{\mid }$. Its coordinate quadratic form is nonnegative and its trace is ${\sum }_i\mathrm{\mid }{w}_i{\mathrm{\mid }}^2\mathrm{/}p=1$, so it is a density operator by [F6]–[F7]; a zero-probability branch has no conditional state by [F1].