Measurement law of a coherent sampler

Since $V$ is unitary and $\mathrm{\mid }0⟩$ is a unit vector, $\mathrm{\mid }\psi ⟩=V\mathrm{\mid }0⟩$ has norm one. Thus [F5] supplies the density operator $\rho =\mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid }$ required by [F4].

For each sample label $x$, let $P_x=\mathrm{\mid }x⟩⟨x\mathrm{\mid }\otimes I_{\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{u}\mathrm{e}}\otimes I_{\mathrm{w}\mathrm{o}\mathrm{r}\mathrm{k}}$. The distinct orthonormal sample labels in [F2] make these mutually orthogonal projections. If the labels do not exhaust the sample basis, include the residual projection $P_{\perp }=I-{\sum }_{x\in X}{P}_x$; it satisfies $P_{\perp }\mathrm{\mid }\psi ⟩=0$ by [F1], so its measurement probability is zero. By [F3], the projection of $\mathrm{\mid }\psi ⟩$ is $P_x\mathrm{\mid }\psi ⟩=\sqrt{p_x}\mathrm{\mid }x⟩\mathrm{\mid }g(x)⟩\mathrm{\mid }{w}_x⟩$, so $\mathrm{\parallel }{P}_x\psi {\mathrm{\parallel }}^2=p_x$.

By cyclicity [F8], $\mathrm{tr}(\rho P_x)=\mathrm{tr}(P_x\mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid })$. The rank-one map here is $\mathrm{\mid }{P}_x\psi ⟩⟨\psi \mathrm{\mid }$; [F6]–[F7] make its trace $⟨P_x\psi ,\psi {⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}$. Since $P_x$ is an orthogonal projection, self-adjointness gives $⟨P_x\psi ,(I-P_x)\psi ⟩=⟨\psi ,P_x(I-P_x)\psi ⟩=0$. Thus $⟨P_x\psi ,\psi ⟩=\mathrm{\parallel }{P}_x\psi {\mathrm{\parallel }}^2=p_x$ by step 1.2. The measurement definition [F4] therefore assigns probability $p_x$.

Measuring value as well refines the event $x$ by a computational-basis value outcome. The sample component $\mathrm{\mid }x⟩$ is orthogonal to every other sample label, and its value register is the single basis vector $\mathrm{\mid }g(x)⟩$; summing over work outcomes leaves squared norm $p_x\mathrm{\parallel }{w}_x{\mathrm{\parallel }}^2=p_x$. Distinct work states are not used to separate sample labels.