What are examples of quantum maps with complex eigenvalues?

Question

Chapter 6 of Michael Wolf's notes (MichaelWolf/QChannelLecture.pdf) discuss the structure of the spectrum of quantum maps and channels. However, it seems like the only explicit example given in the section is Example 6.1, which discusses the determinant of $T(\rho)=(\rho^{T_c}+I \operatorname{tr}(\rho))/(d+1)$.

One thing that can be used in this context, and is discussed in chapter 2 of the notes, is the relation between adjoint of the map and Hermitian conjugate of its matrix representation. Given a map $\Phi$, and denoting with $\hat\Phi$ its representation as a linear operator, we have $\Phi^\dagger=\hat\Phi^\dagger$, if $\Phi^\dagger$ is the adjoint map, defined as $\langle X,\Phi(Y)\rangle=\langle\Phi^\dagger(X),Y\rangle$ for all $X,Y$, and $\hat\Phi^\dagger$ the Hermitian conjugate taken with respect to some matrix representation of $\hat\Phi$. It follows that $\Phi$ has real eigenvalues iff it's self-dual. But if $\Phi(\rho)=\sum_a A_a \rho A_a^\dagger$ then $\Phi^\dagger(\rho)=\sum_a A_a^\dagger\rho A_a$. It follows that for any channel, $\Phi$ has real eigenvalues iff it has a Kraus decomposition in terms of Hermitian operators.

So a simple class of maps with nonreal eigenvalues are $\Phi(\rho)=A\rho A^\dagger$ with $A$ non-Hermitian. And in such cases $|u_i\rangle\!\langle u_j|$ is an eigenvector with eigenvalue $\lambda_i\bar\lambda_j$, if $A|u_i\rangle=\lambda_i|u_i\rangle$.

With this we can understand the spectrum of the simple class of maps $\rho\mapsto A\rho A^\dagger$. What are some other interesting examples where we can compute eigenvalues/eigenvectors?

Answer 1

You can slightly modify that example for $d=2$.

Take $\Phi(E_{11}) = (E_{11}+I)/3$, $\Phi(E_{22}) = (E_{22}+I)/3$, but $\Phi(E_{12}) = iE_{12}/3$, $\Phi(E_{21}) = -iE_{21}/3$, where $E_{ij}$ are matrix units.

It's easy to check that $\Phi$ is a quantum channel (the corresponding Choi matrix is positive). But it has complex eigenvalues $\pm i/3$. And $1/3,1$ since $\Phi(E_{11}-E_{22}) = (E_{11}-E_{22})/3$, $\Phi(I)=I$.

In general, if $\Phi(X) = \sum_i A_iXA_i^\dagger$ then we can consider vectorization, which gives $$ \hat \Phi \cdot {\rm vec}(X) = (\sum_i \overline{A_i} \otimes A_i) \cdot {\rm vec}(X). $$

So, the eigenvalues coincide with eigenvalues of $\sum_i \overline{A_i} \otimes A_i$.

Answer 2

Strictly speaking this is not an answer to what you asked, but in your question you claim that "for any channel, $\Phi$ has real eigenvalues iff it has a Kraus decomposition in terms of Hermitian operators" and the following is too long for a comment. Your argument rightly shows that Hermitian Kraus operators $\Rightarrow$ real eigenvalues, but that's far from necessary as the following simple example shows: Consider the qubit channel $\Phi$ defined via $$ \Phi(\rho):=\begin{pmatrix}\rho_{11}+\frac12\rho_{22}&0\\0&\frac12\rho_{22} \end{pmatrix}\,. $$ Its Choi matrix is diagonal so one obtains (non-Hermitian) Kraus operators $\{|0\rangle\langle 0|,\frac1{\sqrt2}|0\rangle\langle 1|,\frac1{\sqrt 2}|1\rangle\langle 1|\}$. Moreover, its superoperator reads $$ \hat\Phi=\begin{pmatrix}1&0&0&\frac12\\0&0&0&0\\0&0&0&0\\0&0&0&\frac12\end{pmatrix} $$ so $\sigma(\Phi)=\{1,\frac12,0\}\subset\mathbb R$. However, $\Phi$ does not admit any Hermitian set of Kraus operators because that would imply that $\Phi$ is unital (Hermitian Kraus operators give rise to a self-dual channel, and unital is the dual property of trace-preserving) which is obviously not true.