Theorem: Conserving thermal nonidentifiability

The maps (eq:V1) and (eq:V0) are isometries on the prepared subspaces and conserve the stated total energy. Their accessible channels are

For every operator $X$,

Both models have emitted-energy probability $(n+1)d_n$ in sector $n$. Nevertheless,

Their recovery errors differ whenever $p>0$.

Proof. Orthogonality of the residual subspaces in (eq:V1) gives

Equation (eq:V0) is isometric because ${\sum }_n(n+1)d_n=1$. Distinct key inputs are orthogonal. Every term has radiation energy $n\omega$ and residual energy $M_{\star }-n\omega$, hence total energy $M_{\star }$.

Tracing the orthogonal residual labels and $K$ gives (eq:N1) and (eq:N0). The Pauli twirl satisfies

Each paired rail therefore contributes $d_n\mathrm{Tr}(X)J_{n,k}{I}_C{J}_{n,k}^{†}$ to the late marginal. Adding the unpaired vectors proves (eq:samemarginals). The sector probabilities and energy distributions follow.

For ${\mathcal{N}}_1$, read the orthogonal pair/failure flag and the early key. On a paired rail, invert $J_{n,k}$ and then $P_r$; on failure, prepare $I_C/2$. This decoder gives

The qubit depolarizing replacement is the Pauli twirl. Thus

by the triangle inequality after separating the identity Pauli term. Equality follows by applying the two channels to half of a normalized Bell state: its overlap with the replaced state is $1/4$. This proves the upper bound in (eq:toyerrors).

For any decoder, the success contribution to Bell overlap is at most $p$. The failure output is independent of the input, so its Bell overlap is exactly $(1-p)/4$. Testing the Bell projector therefore gives a lower bound $3(1-p)/4$ on the recovery error. For ${\mathcal{N}}_0$, every decoded channel is a replacement channel; the same Bell test gives $3/4$, attained by output $I_C/2$.