# Is the trace of a positive map always positive?

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.

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

• Is there an intuition what the trace of a CP map quantifies? Commented Jun 16 at 11:36
• @NorbertSchuch The only interpretation I'm aware of is the entanglement fidelity $F_e(\Phi,\rho):=\langle\psi_\rho|\Phi(|\psi_\rho\rangle\langle\psi_\rho|)|\psi_\rho\rangle$ with respect to some state $\rho$ (where $\psi_\rho$ is any purification of $\rho$)—as first defined by Schumacher—which evaluates to the trace of the map $X\mapsto\Phi(\rho X\rho)$, cf. also Remark 1 in my preprint. So the trace of a CP map would be proportional to its entanglement fidelity w.r.t. the maximally mixed state. Commented Jun 16 at 12:26

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 $$\lambda\to\infty$$.

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.

• nice! this begs the question though: is there some characterisation for when a (Hermitian preserving) quantum map is positive semidefinite (as a linear operator)? Hermitianity is equivalent to Hermitian-preserving... isn't there anything similar to be said in general for positivity (as a linear operator, that is, equivalently, positive semidefiniteness of the natural representation)?
– glS
Commented May 2 at 12:17
• Good question! At first I thought it's just the Hadamard channels, my (faulty) reasoning being that $K(\Phi)=UDU^\dagger\Leftrightarrow U^\dagger K(\Phi)U=D$ implies that $D$ has to be a channel, hence Hadamard. However, it can of course happen that neither $U$ nor $D$ correspond to channels, e.g., for the channel $$\begin{pmatrix}x&y\\z&w\end{pmatrix}\mapsto\begin{pmatrix}(x+w)/2&y/2\\z/2&(x+w)/2\end{pmatrix}\,.$$ So instinctively I'm not sure what conditions one would need here (beyond the obvious "$\Phi$ has to be unital") Commented May 2 at 19:00