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It is known that quantum channels, being CPTP maps, map density operators to density operators. And thus, they can be seen as superoperators. Similar to operators, where eigenstates and eigenvalues can be derived, one can also define the eigen-operators $\Phi_j$ (typically, being a mixed state) and eigenvalues of quantum channels: $$\mathbb{N}(\Phi_j)=\lambda_j \Phi_j.$$

See page 3 of this lecture note for the deduction. Given these similarities between operators (e.g., Hermitian operators) and quantum channels, the question is what can we say about the properties of their eigendecomposition? Specifically, is the basis of the superoperator $\{\Phi_j\}$ a complete basis? Namely, does $\sum_j\Phi_j=I$ hold? Besides, do the elements in the basis orthogonal to each other? Here, the orthogonality may not be directly followed from the state vector, but maybe something like Hilber-Schmidt orthogonal, i.e., $\mathrm{tr}(\Phi_i\Phi_j)=\delta_{ij}$.

PS: I would be really grateful if someone could point me to some literature regarding this topic.


Crossposted from Physics.SE

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    $\begingroup$ Note that your eigenvectors (or "-operators") are typically not mixed states (i.e. not positive (otherwise they could never be orthogonal!), and typically not even hermitian). $\endgroup$ Mar 21 at 6:52
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    $\begingroup$ What is more, the eigenvectors of a trace-preserving map to any eigenvalue $\lambda\neq 1$ necessarily have trace zero: taking the trace of the eigenvalue equation yields $(1-\lambda_j){\rm tr}(\Phi_j)=0$ $\endgroup$ Mar 21 at 7:58
  • $\begingroup$ @FrederikvomEnde Nice answer. Does this mean that if one expands a density operator in the eigenbasis of a hermitian CPTP map (diagonalizable and forms a complete basis), the expanded coefficient for the $\lambda=1$ component (assume it is of trace 1) must also be 1? $\endgroup$ Mar 24 at 8:22
  • $\begingroup$ @ironmanaudi Counterexample: the dephasing channel for $p=1$ which is Hermitian and has two-fold eigenvalue 1 (with eigenspace ${\rm span}(|0\rangle\langle 0|,|1\rangle\langle 1|\}$ and two-fold eigenvalue $0$ (with eigenspace ${\rm span}\{|0\rangle\langle 1|,|1\rangle\langle0|\}$), i.e. it is diagonal in the standard basis. However, the corresponding entries ${\rm tr}(|0\rangle\langle 0|\rho)=\langle0|\rho|0\rangle$ and $\langle 1|\rho|1\rangle$ of a density operator can, of course, be anything between 0 and 1 (as long as they add up to 1) $\endgroup$ Mar 24 at 16:17
  • $\begingroup$ @FrederikvomEnde I think now it is safe to say that the expansion must contain a convex combination (probability combination) of steady states (one with eigenvalue 1). When the steady state is non-degenerate, then, the coefficient of the steady state must be 1. $\endgroup$ Mar 25 at 7:38

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As you observe correctly, $\mathbb N$ is a linear map. Thus, the same holds as for any eigendecomposition of linear maps.

In particular, there need not be a complete basis of eigenvectors (there can be Jordan blocks), and eigenvectors are not orthogonal.

To get a complete eigenbasis, you need a diagonalizable map $\mathbb N$, and for orthogonality, it needs to be normal -- just as for any linear map.

If you want a good source on quantum channels, what I can recommend are Michael Wolf's lecture notes Quantum Channels and Operations: Guided Tour.

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    $\begingroup$ To complement this already spot-on answer let me link OP to this math.SE post of examples of channels with Jordan blocks, for quick reference. $\endgroup$ Mar 21 at 6:48
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    $\begingroup$ @FrederikvomEnde Thanks. It is important to point this out -- examples of quantum channels with Jordan blocks are plentiful, whereas one usually things of matrices with Jordan blocks as rather exotic. $\endgroup$ Mar 21 at 6:51
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    $\begingroup$ Great answer. Let me add some obvious examples for normal quantum channels: (i) Unitary channels, (ii) Pauli channels, or more generally convex mixtures of unitary channels in an Abelian group, (iii) mixed unitary channels using any measure that is invariant under inversion $\endgroup$ Mar 21 at 8:50
  • $\begingroup$ @NorbertSchuch Thanks for the wonderful answer! I am still curious given a CPTP map are there other spectral properties that we can utilize so that one can set them apart from other more general types of maps? While I know that the eigenvalues satisfy $|\lambda_i|\leq1$ and Frederik mentioned the traceless property of TP maps, it seems like we cannot set apart completely positive maps from, say positive maps, based solely on spectral properties. $\endgroup$ Mar 23 at 14:26
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At the risk of saying something trivial I'd like to add to Norbert's great answer and point out that normal channels are necessarily unital, i.e. to have an orthonormal basis of eigenvectors it is necessary that $\mathbb N(I)=I$.

To see this assume that we are given a normal ($\mathbb N\mathbb N^\dagger=\mathbb N^\dagger\mathbb N$) channel $\mathbb N$. Because $\mathbb N$ is completely positive, it preserves Hermiticity which is equivalent to $\mathbb N^\dagger=\mathbb N^*$ where the latter symbol denotes the dual channel defined via ${\rm tr}(B\mathbb N(A))={\rm tr}(\mathbb N^*(B)A)$ for all $A,B$. With this in mind we compute the purity of the state $\mathbb N(\tfrac{I}n)$: \begin{align*} {\rm tr}\big((\mathbb N(\tfrac{I}n))^2\big)&=\tfrac1{n^2}{\rm tr}\big( \mathbb N(I)\mathbb N(I) \big)\\ &=\tfrac1{n^2}{\rm tr}\big( (\mathbb N^*\circ\mathbb N)(I)I \big)\\ &=\tfrac1{n^2}{\rm tr}\big( (\mathbb N\circ\mathbb N^*)(I)I \big)\\ &=\tfrac1{n^2}{\rm tr}\big( \mathbb N(\mathbb N^*(I)) \big)\\ &=\tfrac1{n^2}{\rm tr}(\mathbb N^*(I)I )\\ &=\tfrac1{n^2}{\rm tr}(\mathbb N(I) )\\ &=\tfrac1{n^2}{\rm tr}(I)=\tfrac1n \end{align*} In the 4th and 7th step we used that $\mathbb N$ is trace preserving. Either way the purity is well known to take its minimal value $\frac1n$ if and only if the state itself is maximally mixed (this follows, e.g., from uniqueness in the Hilbert projection theorem as purity is just the Hilbert-Schmidt norm on the compact convex set of quantum states). Thus we showed $\mathbb N(\tfrac{I}n)=\tfrac{I}n$ so by linearity $\mathbb N$ is unital, as claimed.

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