Is the trace of a positive map always positive?

Question

Obviously, positive semi-definite operators always admit a positive trace as ${\rm tr}(A)=\|A\|_1\geq 0$ whenever $A\geq 0$. This motivates the following "lifted" question:

Given any positive, linear map $\Phi:\mathbb C^{n\times n}\to\mathbb C^{n\times n}$ is it true that ${\rm tr}(\Phi)\geq 0$?

For this recall that the trace of a linear map $\Phi:\mathbb C^{n\times n}\to\mathbb C^{n\times n}$ is defined to be ${\rm tr}(\Phi)=\sum_j{\rm tr}(G_j^\dagger\Phi(G_j))$ where $\{G_j\}_j$ is any orthonormal basis of $\mathbb C^{n\times n}$ (equipped with the Hilbert-Schmidt inner product). As an example one could choose the standard/computational basis $\{|j\rangle\langle k|\}_{j,k}$ and obtain the explicit expression ${\rm tr}(\Phi)=\sum_{j,k}\langle j|\Phi(|j\rangle\langle k|)|k\rangle$. Equivalently, the trace of $\Phi$ is of course equal to the trace of any matrix representation of $\Phi$—such as the natural representation or the Pauli transfer matrix—and the trace is also equal to the sum of all eigenvalues of $\Phi$. To give an example the transposition map $T$—the prime example of a positive but not completely positive map—has trace zero which is in agreement with the above question.

For the special case where $\Phi$ is completely positive the above statement holds as a consequence of the Kraus representation $\Phi=\sum_lK_l(\cdot)K_l^\dagger$: \begin{align*} {\rm tr}(\Phi)&=\sum_{j,k}\langle j|\Phi(|j\rangle\langle k|)|k\rangle\\ &=\sum_{l,j,k}\langle j|K_l|j\rangle\langle k|K_l^\dagger|k\rangle\\ &=\sum_l|{\rm tr}(K_l)|^2\geq 0 \end{align*} As an aside the trace of a channel also represents the mean operation fidelity (cf. Chapter 10.5 in Bengtsson & Zyczkowski's book "Geometry of Quantum States" / alt link) and it can be recovered as an expectation value via ${\rm tr}(\Phi)=\langle \eta|\mathsf C(\Phi)|\eta\rangle$ where $|\eta\rangle:=\sum_j|j\rangle\otimes|j\rangle$ is the (unnormalized) maximally entangled state and $\mathsf C(\Phi)$ is the (unnormalized) Choi matrix of $\Phi$, cf. Lemma 2 in this paper for a slightly more general statement. However, this proof technique doesn't really help us as it does not generalize to arbitrary positive maps.

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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

Consider the qubit map $\Phi(\rho):=\sigma_Y\rho^T\sigma_Y$, that is, $$ \Phi\begin{pmatrix}\rho_{11}&\rho_{12}\\\rho_{21}&\rho_{22}\end{pmatrix}=\begin{pmatrix}\rho_{22}&-\rho_{12}\\-\rho_{21}&\rho_{11}\end{pmatrix}\,. $$ From the definition it is obvious that $\Phi$ is positive (even trace preserving), and its Pauli transfer matrix reads $$ \mathcal P(\Phi)=\begin{pmatrix} 1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1 \end{pmatrix}\quad\Rightarrow\quad {\rm tr}(\Phi)={\rm tr}(\mathcal P(\Phi))=-2<0\,. $$ Interestingly, for qubits this is as small as the trace of a positive trace-preserving map can be: every such map is trace-norm contractive meaning all eigenvalues have to lie in the closed unit disk. Thus the smallest possible trace of such a map comes from one eigenvalue $1$ (necessary: fixed point) and the remaining three eigenvalues being all $-1$ which is exactly how the spectrum of the above $\Phi$ looks like.

Moreover, if we lift trace preservation and only care about positivity this allows us to construct a positive map with arbitrarily small trace by setting $\Phi_\lambda:=\lambda\cdot\Phi$ for any $\lambda\geq 0$ so ${\rm tr}(\Phi_\lambda)=-2\lambda\to-\infty$ as $\lambda\to\infty$.

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Other rather famous examples of positive maps which are not completely positive are the following:

• $\Phi\in\mathcal L(\mathbb C^{3\times 3})$ defined via $$ \Phi(X):=2{\rm tr}(X){\bf1}_3-2{\rm diag}(X_{33},X_{11},X_{22})-X $$ first found by Choi (in "Positive Semidefinite Biquadratic Forms", Linear Algebra Appl. 12 (1975), 95-100). The significance of this map is that it is the earliest known example of an indecomposable map (i.e. a map which cannot be written as $\Phi=\Psi_1+\Psi_2\circ{}^T$ for any $\Psi_1,\Psi_2$ completely positive). To connect it to the question at hand note that $\Phi$ has simple eigenvalues $3,i\sqrt3,-i\sqrt3$, and a $6$-fold eigenvalue $-1$; hence $$ {\rm tr}(\Phi)=3+i\sqrt3-i\sqrt3+6\cdot(-1)=-3 $$
• The reduction map $\Phi\in\mathcal L(\mathbb C^{n\times n})$, $\Phi(X):={\rm tr}(X){\bf1}_n-X$ first given in this paper by Horodecki and Horodecki (arXiv). One readily verifies that $\Phi$ has simple eigenvalue $n-1$ and the remaining $n^2-1$ eigenvalues are $-1$. Thus $$ {\rm tr}(\Phi)=n-1+(n^2-1)\cdot(-1)=n-1-n^2+1=-n(n-1) $$
• The Breuer-Hall map (from this and this paper, cf. also end of Section 5 in this paper) $\Phi\in\mathcal L(\mathbb C^{4\times 4})$, $\Phi(X):={\rm tr}(X){\bf1}_4-X-(\sigma_y\otimes{\bf1}_2)X^T(\sigma_y\otimes{\bf1}_2)^\dagger$. It has simple eigenvalue $2$, 10-fold eigenvalue $0$ and 5-fold eigenvalue $-2$ meaning ${\rm tr}(\Phi)=2-10=-8$.

NB: The re-scaled reduction map $\frac1{n-1}\Phi$ as well as the re-scaled Breuer-Hall map $\frac12\Phi$ are positive and trace-preserving and they both have trace $-n$ which is what I, personally, believe to be the smallest possible trace a PTP map in $n$ dimensions can have.