Proposition: Reference-compatible cutoff

For any state ${\sigma }_{AR}$, with arbitrary reference $R$, put $p=\mathrm{tr}(Q_N{\sigma }_A)$. Then

If $\mathrm{tr}(N_{\mathrm{tot}}{\sigma }_A)\le E$, then $p\le E/(N+1)$. Without an input constraint, the cutoff's half-diamond distance from the identity embedding is one for every finite $N$.

Proof. Purify ${\sigma }_{AR}$ to $|\psi ⟩$ and set $|\varphi ⟩=(P_N\otimes I)|\psi ⟩$. The trace norm of $|\psi ⟩⟨\psi |-|\varphi ⟩⟨\varphi |$ is $\sqrt{4p-3p^2}\le 2\sqrt{p}$. Partial trace contracts it. The additional flag block has trace norm $p$. Markov's inequality gives the constraint estimate. A high-block input is sent to an orthogonal flag and attains distance one.