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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?

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  • $\begingroup$ How do you define the eigenvalue of a quantum channel? Thinking about it is a linear operator via the vec mapping? $\endgroup$
    – Rammus
    Feb 27, 2023 at 16:46
  • $\begingroup$ @Rammus yes, I mean thinking of it as a linear operator (via vec or in any other basis, I don't think that affects the spectrum). In other words, eigenvalues of a map $\Phi$ are $\lambda$ such that $\Phi(\rho)=\lambda\rho$ for some operator $\rho$. For example, if $\Phi(\rho)=A\rho A^\dagger$ and $Au=\alpha u$, then $uu^\dagger$ is an eigenvector for $\Phi$ with eigenvalue $\lambda=|\alpha|^2$. $\endgroup$
    – glS
    Feb 27, 2023 at 17:04
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    $\begingroup$ Ahh ok, yes I see. Well then take any unitary mapping $\rho \mapsto U \rho U^\dagger$. If $\{|v_i\rangle\}_i$ are the eigenvectors of $U$ with corresponding eigenvalues $\{\lambda_i\}_i$. Then $|v_i\rangle \langle v_j|$ is an eigenmatrix of the unitary channel with eigenvalue $\lambda_i \overline{\lambda_j}$. Note that like unitary matrices, unitary channels will have its eigenvalues on the unit circle. $\endgroup$
    – Rammus
    Feb 27, 2023 at 18:08
  • $\begingroup$ Side note: the link to Michael Wolf's lecture notes has expired, but here's a currently working link $\endgroup$ Mar 17 at 21:53

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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 $\sum_i \overline{A_i} \otimes A_i$.

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  • $\begingroup$ Do you know if anyone has studied the distribution of eigenvalues for random quantum channels? Some numerical sampling makes it look like the eigenvalues are uniform on some disc of radius related to the dimension and Kraus rank of the map (with the radius decaying to zero as the Kraus rank increases?). Could be a fun little exploration. $\endgroup$
    – Rammus
    Feb 27, 2023 at 20:32
  • $\begingroup$ The notes referenced in the question has some nice results. E.g. "the eigenvalues of every quantum channel are restricted to lie on the unit disc, there is one eigenvalue one and the others are real or come in complex conjugate pairs." And there are no other restrictions, except on their joint set. Can't say anything about their distribution. $\endgroup$
    – Danylo Y
    Feb 27, 2023 at 20:52
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    $\begingroup$ Found a reference that studied their distribution under random sampling, see section 5. $\endgroup$
    – Rammus
    Feb 27, 2023 at 21:21
  • $\begingroup$ Nice! So, the distribution of eigenvalues except 1 is indeed close to uniform on a disc that shrinks to 0 as the Kraus rank increases. $\endgroup$
    – Danylo Y
    Feb 28, 2023 at 7:56
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    $\begingroup$ It would appear so, although theres no proof yet as far as i can tell from the paper. $\endgroup$
    – Rammus
    Feb 28, 2023 at 8:03
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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.

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