Completing instruments and failure branches

Suppose $\{{\mathcal{M}}_f{\}}_f$ is a complete instrument on the accessible interface and ${\mathcal{R}}_f$ is a CPTP return map for outcome $f$. The complete decoder is

For a failure branch, one may choose ${\mathcal{R}}_f(Y)=\mathrm{Tr}(Y)I_2/2$. The sum is CPTP because the instrument is complete.

Loss or leakage may require an enlarged input interface. For example, use ${\hat{H}}_j={\mathbb{C}}^2\oplus {H_j}^{\mathrm{flag}}$. Then the partial trace and decoder in the following equations are understood on those declared enlarged spaces. An unmodeled leakage subspace cannot be silently treated as a qubit input. Every flag branch must still be completed.

This is the finite complete-instrument distinction discussed in CheungVIIg,CheungEm05B,CheungEn02E. It does not require asserting results from another physical model.