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)

  • $\begingroup$ Is there an intuition what the trace of a CP map quantifies? $\endgroup$ Commented Jun 16 at 11:36
  • $\begingroup$ @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. $\endgroup$ Commented Jun 16 at 12:26

1 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 $\lambda\to\infty$.

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.

  • $\begingroup$ 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)? $\endgroup$
    – glS
    Commented May 2 at 12:17
  • $\begingroup$ 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") $\endgroup$ Commented May 2 at 19:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.