Theorem: Snapshot determination and adjoint containment

For CPTP maps $\mathcal{N}:\mathcal{B}(H)\to \mathcal{B}(Q)$ and $\mathcal{T}:\mathcal{B}(H)\to \mathcal{B}(K)$, the following are equivalent:

1. For all states $\rho ,\sigma$, $\mathcal{N}(\rho )=\mathcal{N}(\sigma )$ implies $\mathcal{T}(\rho )=\mathcal{T}(\sigma )$.

2. $\ker \mathcal{N}\subseteq \ker \mathcal{T}$ as complex linear subspaces.

3. $\mathrm{ran}{\mathcal{T}}^{†}\subseteq \mathrm{ran}{\mathcal{N}}^{†}$.

4. There is a linear map $L$ on $\mathrm{ran}\mathcal{N}$ such that $\mathcal{T}=L\mathcal{N}$.

No positivity or complete positivity of $L$ follows.

Proof. Let $X=X^{†}\in \ker \mathcal{N}$. Trace preservation gives $\mathrm{Tr}X=0$. For sufficiently small $t>0$, both $I_H/\dim H+tX$ and $I_H/\dim H-tX$ are states. Condition (i) implies $\mathcal{T}(X)=0$. Decomposing a general kernel element into Hermitian and anti-Hermitian parts proves (ii). The converse is immediate. Finite-dimensional orthogonal-complement duality proves (ii)$\iff$(iii). For (ii)$\Rightarrow$(iv), set $L(\mathcal{N}(X))=\mathcal{T}(X)$; the kernel inclusion makes this well defined. The remaining implication is immediate.