Theorem: Exact operational embedding of the finite response space

The following hold.

1. Equations (eq:probe-prep) and (eq:binary) define complete instruments on every input, with every untouched reference.

2. For $p_k>0$,

   $$\mathbb{E}[B_{\ast }y\mid k,f]=f(\delta (k)).$$

3. The map taking a three-row real vector $v$ to the linear map given by the right side of (eq:centered) is injective when $\delta$ is surjective.

4. The centered operational contact image has rank two and

   $${\Delta }_m=\frac{2}{3}{\Delta }_{n_{{\ell }_a}}+\frac{2}{3}{\Delta }_{n_{{\ell }_b}}.$$

   Its raw-coordinate fitting family is ${\alpha }_0+{\ker }_{\mathrm{right}}{N}_S$.

Proof. The first instrument is obtained from a complete clock instrument, a fresh allocation, and a classically controlled unitary. The second is obtained by complete destructive readout of $Q$. Alternatively, each displayed branch is completely positive and the branch probabilities sum to $\mathrm{Tr}\rho$. The same arguments apply with $I_E$ appended.

Equation (eq:score) gives (ii). For (iii), if $v_i\ne 0$, choose a code $k$ with $\delta (k)={\beta }_i$ and input $P_k\otimes {\rho }_X$. The $y=+1$ block of the centered output is nonzero. Thus the induced linear map has trivial kernel.

Formula (eq:centered) is linear in $v$. Applying this injective linear map to Theorem thm:frozen proves (iv), including equality of the coefficient kernels.