# The transpose map in mathematical physics

I need to know where the 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 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 $$\label{e:lap} \langle Jh,k\rangle=\langle Jk,h\rangle,\quad h,k\in\mathcal{H},$$ 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: 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 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 $\{E_{i,j}\}_{i,j\in\mathcal J}$, 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 (completely positive and trace preserving).

A bipartite state $\rho$ living in $H_A \otimes H_B$ is separable (not entangled) iff $(I\otimes\Lambda)\rho$ is positive for any positive map $\Lambda$ on $H_B$.
So to detect entanglement one needs a to find a single positive map $\Lambda$ such that $(I\otimes\Lambda)\rho$ is no longer a state, i.e. is not positive. Note that completely positive maps (which remain positive when tensored with the identity) are useless for this purpose, they are "blind" to entanglement.
The transpose map plays the role of such an "entanglement witness" for a two-dimensional Hilbert space $H_A$, $H_B$ (and actually also if $H_A$ is two-dimensional and $H_B$ is three-dimensional, but not for higher dimensions).