tl;dr: It is known that a single entanglement witness derived from a positive map is strictly weaker than the positivity criterion under the adjoint of said map, that is, $${\rm tr}(\mathsf C(\Lambda_{A\to B})\rho_{AB})<0\quad\Rightarrow\quad ({\rm id}_A\otimes \Lambda_{A\to B}^\dagger)(\rho_{AB})\not\geq 0$$ My question: If $A=B$, is there a direct implication $${\rm tr}(\mathsf C(\Lambda_{A\to A})\rho_{AA})<0\quad\Rightarrow\quad ({\rm id}_A\otimes \Lambda_{A\to A})(\rho_{AA})\not\geq 0$$ without taking the adjoint?
To add more context, this question is about the relation between different characterizations of separability; more precisely it is about the following famous equivalence, as first proven in this 1996 paper (arXiv):
- A bipartite state $\rho_{AB}$ is separable
- ${\rm tr}(W_{AB}\rho_{AB})\geq 0$ for all entanglement witnesses $W_{AB}$, i.e., for all $W_{AB}$ Hermitian which satisfy $\langle\psi_A\otimes\phi_B|W_{AB}|\psi_A\otimes\phi_B\rangle\geq 0$ for all $\psi_A,\phi_B$ (also known as "block positive")
- $({\rm id}_A\otimes \Lambda_{B\to A})(\rho_{AB})\geq 0$ for all positive maps $\Lambda_{B\to A}$ (which I call "positive-map criterion" for simplicity)
Also recall that 2. has an equivalent formulation via positive maps which can be found in many textbooks and lecture notes: $W_{AB}$ is an entanglement witness if and only if $W_{AB}=\mathsf C(\Lambda_{A\to B})$ for some positive map $\Lambda_{A\to B}$, with $\mathsf C$ denotes the usual Choi–Jamiołkowski isomorphism. This way 2. and 3. can be formulated purely via positive maps, and this makes for an interesting connection between these criteria, cf. p.15 in this review article by Gühne and Tóth (arXiv): If $|\Phi\rangle:=\sum_j|jj\rangle$ denotes the unnormalized maximally entangled state, then for all $\rho_{AB}$ \begin{align*} {\rm tr}(\mathsf C(\Lambda_{A\to B})\rho_{AB})&={\rm tr}\big((({\rm id}\otimes\Lambda_{A\to B})(|\Phi\rangle\langle\Phi|)\rho_{AB}\big)\\ &={\rm tr}\big(|\Phi\rangle\langle\Phi|({\rm id}\otimes\Lambda_{A\to B}^\dagger)(\rho_{AB})\big)\\ &=\langle\Phi|({\rm id}\otimes\Lambda_{A\to B}^\dagger)(\rho_{AB})|\Phi\rangle \end{align*} with $(\cdot)^\dagger$ the usual adjoint channel. Hence:
If a witness $W_{AB}=\mathsf C(\Lambda_{A\to B})$ detects entanglement of a state $\rho_{AB}$, then the adjoint $\Lambda_{A\to B}^\dagger$ detects this entanglement as well (because ${\rm tr}(\mathsf C(\Lambda_{A\to B})\rho_{AB})<0$ implies $({\rm id}_A\otimes \Lambda_{A\to B}^\dagger)(\rho_{AB})\not\geq 0$ by the above calculation).
If one wants to get rid of the adjoint one can obtain the following equivalent statement: If $W_{AB}^T=\mathsf C(\Lambda_{A\to B})^T$ detects entanglement of a state $\rho_{AB}$, then $\Lambda_{A\to B}\otimes{\rm id}$ detects this entanglement as well (because $\mathsf C(\Lambda_{A\to B})^T={\rm SWAP}\;\mathsf C(\Lambda_{A\to B}^\dagger)\;{\rm SWAP}$ as is readily verified). In this precise sense the witness criterion for a single witness is weaker than the corresponding positive-map criterion—despite the previously stated fact that the collection of all witnesses is equivalent to the collection of all positive maps in terms of entanglement detection. Still, this connection led me to the following
Question. If $A=B$, is the adjoint, resp. the transpose here really necessary? More precisely, given some positive map $\Lambda$ and a bipartite state $\rho$, does ${\rm tr}(\mathsf C(\Lambda)\rho)<0$ also imply $({\rm id}\otimes \Lambda)(\rho)\not\geq 0$? Or can it happen that, without the transpose, the witness $\mathsf C(\Lambda)$ is in some cases stronger than the positive map $\Lambda$?
My motivation behind this question is two-fold: on the one hand, this was the implication I naively expected because the connection between witnesses and positive maps is usually stated through Choi–Jamiołkowski (and rightfully so), but without the transpose/adjoint. Admittedly, when paying attention to domain and co-domain here it seems unlikely that this is true because the question itself only makes sense if the two Hilbert spaces coincide. On the other hand, however, I found multiple sources which, to me, made it seem like this is true:
- In the book "Geometry of Quantum States" by Bengtsson & Zyczkowski (alt link), on p.471 the following is written: "One might argue that in general a witness $W = [\mathsf C(\Phi)]/N$ is theoretically less useful than the corresponding map $\Phi$, since the criterion ${\rm tr}(\rho W)< 0$ is not as powerful as [...] $(\Phi \otimes \mathbb{1}) \rho \geq 0$."
- In the famous "Quantum Entanglement" review by the Horodeckis (2009) (arXiv), after Eq. (50) they state stating that "any single witness $W_\Lambda$ condition is much weaker than the condition given by the map. This is because the first is of scalar type, while the second represents an operator inequality condition."—but it seems that the transpose/adjoint is not invoked at any point throughout the paper. What is more, the paper goes on to claim that "it is not difficult to see [...] that the condition based on one map $\Lambda$ is equivalent to a continuous set of conditions defined by all witnesses of the form $W_{\Lambda,A} \equiv (A \otimes I)W_{\Lambda}(A^{\dagger} \otimes I)$ [with ${\rm rank}(A)>1$]".
However, neither source explicitly proves or demonstrates the direct implication for the original map I'm curious about, hence my question.
(This is a Q&A style question meant as a contribution to the list of counterexamples in quantum information)
