I need to know where the $transpose\ map$transpose map, as a linear map that is positive and trace preserving on the trace-class but not completely positive, appears in mathematical physics. In mathematics I can track it back to the seminal papers of W.B. Arveson: Subalgebras of $C^*$-algebras, Acta Math. 123(1969), 141--224, and Subalgebras of $C^*$-algebras. II, Acta Math. 128(1972), 271--308.
I believe that this example is playing and important role in mathematical physics as well.
To be more specific, in the following I briefly describe what I am speaking about.
Let $\mathcal H$ be an arbitrary Hilbert space with dimension at least $2$, for which we fix an orthonormal basis $\{e_j\}_{j\in\mathcal{J}}$. We consider the $conjugation$conjugation operator $J\colon \mathcal H\rightarrow\mathcal H$ defined by $Jh=\overline h$, where for arbitrary $h=\sum_{j\in\mathcal{J}} h_j e_j$ we let $\overline h=\sum_{j\in\mathcal{J}}\overline{h}_j e_j$. Then $J$ is conjugate linear, conjugate selfadjoint, that is, it has the following property \begin{equation}\label{e:lap} \langle Jh,k\rangle=\langle Jk,h\rangle,\quad h,k\in\mathcal{H}, \end{equation} isometric, and $J^2=I$.
Further on, let $\Phi\colon \mathcal{B}(\mathcal{H})\rightarrow\mathcal{B}(\mathcal{H})$ be defined by $\Phi(S)=JS^*J$, for all $T\in\mathcal{B}(\mathcal{H})$. It is easy to see that $\Phi$ is isometric, that is, $\|\Phi(S)\|=\|S\|$ for all $S\in\mathcal{B}(\mathcal{H})$, and that $\Phi(I)=I$. On the other hand, if $S\in\mathcal{B}(\mathcal{H})^+$ then \begin{equation*} \langle \Phi(S)h,h\rangle=\langle JSJh,h\rangle=\langle Jh,SJh\rangle=\langle SJh,Jh\rangle\geq 0,\quad h\in\mathcal{H}, \end{equation*} hence $\Phi$ is positive. Let us also observe that, with respect to the matrix representation of operators in $\mathcal B(\mathcal H)$ associated to the orthonormal basis $\{e_j\}_{j\in\mathcal J}$, $\Phi$ is the $transpose\ map$transpose map: if $T$ has the matrix representation $[t_{i,j}]_{i,j\in\mathcal J}$ then $\Phi(T)$ has the matrix representation $[t_{j,i}]_{j,i\in\mathcal J}$.
It can be proven that $\Phi$ leaves $\mathcal{B}_1(\mathcal{H})$ invariant and the corresponding restriction map
$\mathcal B_1(\mathcal H)\rightarrow\mathcal B_1(\mathcal H)$ is $stochastic$stochastic in the sense that it is positive and trace preserving.
Finally, $\Phi$ is not completely positive, more precisely, it is not even $2$-positive. To see
this, we consider the $matrix\ units$matrix units $\{E_{i,j}\}_{i,j\in\mathcal J}$, that is, for any
$i,j\in\mathcal J$, $E_{i,j}$ denote
the rank $1$ operator on $\mathcal H$ with $E_{i,j}e_j=e_i$ and $E_{i,j}e_k=0$ for all
$k\neq j$ and observe that $\Phi(E_{i,j})=E_{j,i}$.
Since $\dim\mathcal H\geq 2$, there exist $i,j\in\mathcal J$ with $i\neq j$.
Then, consider the positive finite rank operator in $M_2(\mathcal B_1(\mathcal H))$ defined by
\begin{equation*}E=\left[\begin{matrix} E_{i,i} & E_{i,j} \\ E_{j,i} & E_{j,j}\end{matrix}\right]
\end{equation*}
and observe that
\begin{equation*} \Phi_2(E)=\left[\begin{matrix} \Phi(E_{i,i}) & \Phi(E_{i,j}) \\ \Phi(E_{j,i}) &
\Phi(E_{j,j})\end{matrix}\right]=\left[\begin{matrix} E_{i,i} & E_{j,i} \\ E_{i,j} & E_{j,j}\end{matrix}\right]
\end{equation*} which is not positive. Therefore,
$\Phi$ is a stochastic map but not a $quantum\ channel$quantum channel (completely positive and trace preserving).
