Is the adjoint of a strictly positive channel again strictly positive?

Question

Building on the concept of positive definite operators${}^1$—denoted $A>0$—a linear map $\Phi:\mathbb C^{n\times n}\to\mathbb C^{k\times k}$ is called strictly positive if $\Phi(A)>0$ for all $A>0$. It is easy to see that this is a special case of what is commonly dealt with in quantum information as such maps are always positive in the usual sense (i.e. positive semi-definite operators are sent to positive semi-definite ones). What's more if a map in question is already positive, strict positivity is equivalent to the existence of some $A>0$ such that $T(A)>0$; this follows from a certain kernel property of positive maps as has basically been observed already by Bhatia on page 36 of his book "Positive Definite Matrices" (2007). For context, channels with this property frequently pop up in quantum thermodynamics as there, preservation of the Gibbs state ($\Phi(\frac{e^{-H/T}}{{\rm tr}(e^{-H/T})})=\frac{e^{-H/T}}{{\rm tr}(e^{-H/T})}>0$) is required.

Now it is well known that a linear map $\Phi$ is positive if and only if its adjoint map $\Phi^\dagger$ (defined via ${\rm tr}(B\Phi(A))={\rm tr}(\Phi^\dagger(B)A)$ for all $A,B$) is positive; this follows from the standard characterization that $X\geq 0$ if and only if${}^2$ ${\rm tr}(XY)\geq 0$ for all $Y\geq 0$. Similarly, complete positivity is equivalent to complete positivity of the adjoint channel. This begs the question:

Is a quantum channel (or general linear map) $\Phi$ strictly positive if and only if its adjoint channel $\Phi^\dagger$ is strictly positive?

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_{1: To write it out, $A\in\mathbb C^{n\times n}$ is called positive definite if $\langle x|A|x\rangle>0$ for all $x\in\mathbb C^n$ such that $x\neq 0$. Equivalently, $A$ is Hermitian and all eigenvalues are $>0$.}

_{2: A quick proof which should make this fact obvious: $\Rightarrow$ follows from ${\rm tr}(XY)={\rm tr}(\sqrt X\sqrt X\sqrt Y\sqrt Y)={\rm tr}((\sqrt Y\sqrt X)^\dagger(\sqrt Y\sqrt X))\geq 0$, while for $\Leftarrow$ one simply chooses $Y=|x\rangle\langle x|\geq 0$ for arbitrary $x$.}

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(This is a Q&A style question meant as a contribution to the list of counterexamples in quantum information)

Answer 1

The answer to this is two-fold: the adjoint of a strictly positive channel (in the Schrödinger picture, i.e. CPTP) is indeed always strictly positive, simply because the adjoint of any channel is unital (${\bf1}\mapsto{\bf1}$), hence strictly positive as observed above. However, for general linear maps this is false. For a counterexample consider $\Phi(X):=\langle 0|X|0\rangle{\bf1}$ which is completely positive and unital (hence a Heisenberg channel). In particular as previously stated, unitality implies that $\Phi$ is strictly positive. Now for its adjoint we use the definition $$ {\rm tr}(\color{blue}{\Phi^\dagger(B)}A)={\rm tr}(B\Phi(A))=\langle 0|A|0\rangle{\rm tr}(B)={\rm tr}\big( \color{blue}{{\rm tr}(B)|0\rangle\langle 0|}A \big) $$ to find that $\Phi^\dagger(\cdot)={\rm tr}(\cdot)|0\rangle\langle 0|$. Thus $\Phi^\dagger$ is a reset channel into a pure state meaning it is not strictly positive: $\langle 1|\Phi^\dagger({\bf1})|1\rangle=0$.

The deeper reason for why strict positivity cannot be universally preserved by the adjoint is as follows: First, lack of strict positivity is equivalent to the image of the map being a strict subalgebra of all matrices, cf. Theorem 3.5 in this paper of mine (arXiv); in our reset example this subalgebra is simply $\mathbb C\cdot|0\rangle\langle 0|$. Moreover, in Remark 3.2 of said paper it is shown that taking the adjoint translates this "defect" of the image (=subalgebra) of all $\Phi^\dagger(B)$ to the domain of $\Phi$ in the sense that the map is fully determined by its action on a subalgebra, and this is something that strict positivity cannot detect anymore. To go back to our example $\Phi$ is fully characterized by its action on the subalgebra $\mathbb C\cdot|0\rangle\langle 0|\subsetneq\mathbb C^{n\times n}$ which translates to $\Phi^\dagger(B)\in\mathbb C|0\rangle\langle 0|$ for all $B$.