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)


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 $\lambda\to\infty$.

  • $\begingroup$ 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

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