On the definition of positive linear superoperators on Hilbert spaces Consider a linear map between linear maps of a Hilbert space, $\mathcal{E}: \mathcal{L}(\mathcal{H})\to\mathcal{L}(\mathcal{H})$. The standard definition I have encountered is that $\mathcal{E}$ is positive if $$\mathcal{E}(A) \geq 0 \quad \forall A\geq 0.$$
An element $A$ in $\mathcal{L}(\mathcal{H})$ is in turn said to be positive if
$$\left<{\psi, A(\psi)}\right>\geq 0 \quad \forall \psi \in \mathcal{H}.$$
(Obs: I am thinking finite dimensions for now, because it is a very basic question, but feel free to discuss infinite dim too.)
Then, I wondered, since $\mathcal{L}(\mathcal{H})$ is also a Hilbert space, an analogous to the second definition also exists for linear operators in $\mathcal{L}(\mathcal{L}(\mathcal{H}))$ if we use the Hilbert-Schmidt inner product $\left<A,B\right>= \operatorname{Tr}(A^* B)$ in $\mathcal{L}(\mathcal{H})$, right?
With this, we would have that a map in $\mathcal{L}(\mathcal{L}(\mathcal{H}))$ is positive if
$$\operatorname{Tr}(B^*\mathcal{E}(B))\geq 0 \quad \forall B \in \mathcal{L}(\mathcal{H}).$$
My question is: is this equivalent to taking positive maps to positive maps? I do not know how to prove the statement or even if is true. I also haven’t found that on the internet, and I realize that the first definition and some others equivalent to it are always used, with no mention to this. Is the equivalence valid at least in the simplest case?
 A: In the fields of mathematics and quantum information science, the first concept is referred to as a positive map, although I believe that this terminology is misleading and we should instead refer to it as a positive preserving map (as I shall use in the following).
Indeed, the two concepts are completely distinct: a map can be positive preserving but not positive, positive preserving and positive, not positive preserving but positive, or neither positive preserving nor positive.
A simple example of a positive preserving but not a positive map has already been given by @Tobias Fünke.
To see the converse, first it is easy to see that a linear map ${\mathcal E}$ defined by ${\mathcal E}(X) := \mathrm{Tr}(A^\ast X) A$ with any $A \in {\mathcal L}({\mathcal H})$ is positive, as for all $B \in {\mathcal L}({\mathcal H})$, we have $\mathrm{Tr}(B^\ast {\mathcal E}(B)) = |\mathrm{Tr}(A^\ast B)|^2 \ge 0$. However, it is not necessary for ${\mathcal E}$ to be positive preserving unless $A$ is positive. For instance, if we let $A = |\phi_0 \rangle \langle \phi_0| - |\phi_1 \rangle \langle \phi_1|$ where $\phi_0, \phi_1 \in {\mathcal H}$ is an orthonormal system, then ${\mathcal E}(|\phi_0 \rangle \langle \phi_0 |) = \langle \phi_0, A \phi_0 \rangle A = A$, which is not positive as $A$ is not a positive operator (having an eigenvalue $-1$). Hence, this gives an example of positive but not positive preserving.
On the other hand, if we choose $A$ to be positive, the example gives a map that satisfies both definitions simultaneously. In fact, the example even gives a completely positive (preserving) map for any positive $A$. To see this, note that the Choi-Jamiolkowski isomorphism $\sum_{i,j}{\mathcal E}(|i \rangle \langle j|) \otimes |i \rangle \langle j|$ (with a fixed orthonormal basis ${ |i \rangle }$) is given by $A \otimes \overline{A}$ where $\overline{A}$ is the complex conjugate of $A$ (defined as $\overline{A} = \sum_{i,j} \langle i, A j\rangle |i\rangle \langle j|$). It is a well-known fact that complete positivity is equivalent to the positivity of the Choi-Jamiolkowski matrix. Moreover, if $A$ is positive, then $\overline{A}$ is also positive and $A \otimes \overline{A}$ is positive.
A: The relation you speculate about is wrong not only for specific counterexamples, but it is wrong generically.
Namely, positive matrices have a positive spectrum, while a general completely positive superoperator (which is a special case of a positive superoperator) has a complex spectrum (with some constraints). Thus, any random positive map, or completely positive map, will be a counterexample.
Conversely, a positive spectrum of a superoperator is not enough to correspond by a positive matrix: You will need in addition that it is hermitian. Then, indeed, positivity of the corresponding matrix follows.
A: I don't think that the converse statement holds true. Here is a counter example:
For $B, U\in\mathcal L(\mathcal H)$, let $\mathcal E(B):=U BU^*$. This defines a linear operator and it holds that if $B\geq 0$, then $\mathcal E(B) \geq 0$ on $\mathcal H$.
Now choose $B:=|\phi\rangle \langle \psi|$ for $\phi, \psi \in \mathcal H$ orthonormal (which means that $B\,\varphi:=\langle \psi,\varphi\rangle\, \phi$ for $\varphi \in \mathcal H$) and further pick $U:= -|\phi\rangle \langle \phi| + |\psi\rangle \langle \psi|$. With these choices, we find $$\mathcal E(B) \,\varphi  = -\langle \psi,\varphi\rangle  \phi $$ and hence
$$\mathrm{Tr}\,B^* \mathcal E(B) =   \mathrm{Tr}\, |\psi\rangle\langle\phi|\, \mathcal E(B)  = \langle \phi,\mathcal E(B)\,\psi\rangle = -1 \quad .$$
Thus, we have shown that there exists a linear map $\mathcal E$ on $\mathcal L(\mathcal H)$ which preserves positivity but does not fulfill the last inequality of the question.

As the other answer points out, the other direction does not hold either. However, as the OP suggested, we have that
$$ \mathrm{Tr}\, A\,\mathcal E(B) \geq 0 \quad \forall A,B\geq 0  \Longleftrightarrow \mathcal E(B)\geq 0 \quad \forall B\geq 0\quad .$$
To see this, note that if $\mathcal E$ preserves positivity, i.e. if $\mathcal E(B)\geq 0$ for $B\geq 0$, then for all $A\geq 0$ it holds that $\sqrt A\, \mathcal E(B)\, \sqrt A \geq 0$ and thus $ 0 \leq \mathrm{Tr}\,\sqrt A\, \mathcal E(B)\, \sqrt A = \mathrm{Tr}\, A\, \mathcal E(B)$ for all $B\geq 0$. The converse direction follows by noticing that the inequality in particular holds for all positive $A$ of the form $A_\psi:=|\psi\rangle\langle \psi|$, which yields $\langle \psi,\mathcal E(B)\,\psi\rangle \geq 0$ and hence $\mathcal E(B)\geq 0$, for all positive $B$.
A: Usually one endows $\mathcal{L}(\mathcal{H})$ with the operator-norm, i.e.
$$ \| T \|_{\operatorname{op}} := \sup_{\| x \| = 1} \| Tx \|.$$
Then $\| \cdot \|_{\operatorname{op}}$ does not fulfill the parallelogram identity, so it is not induced by an inner product. However, since in finite dimensions all norms are equivalent, its topology is the same as the topology of a Hilbert space. The most natural inner product is then given by $\langle A, B \rangle := \operatorname{tr}(A^*B)$.
However in infinite dimensions not all norms are equivalent, and $\mathcal{L}(\mathcal{H})$ will not be isomorphic to a Hilbert space. Also, you cannot define the trace for all bounded linear maps on $\mathcal{H}$, so the above definition of an inner product will also not carry over.
