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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

glS
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ironmanaudi
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    please mark cross-posts as such – glS Mar 20 '24 at 15:45
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    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). – Norbert Schuch Mar 21 '24 at 06:52
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    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 $(1-\lambda_j){\rm tr}(\Phi_j)=0$ – Frederik vom Ende Mar 21 '24 at 07:58
  • @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? – ironmanaudi Mar 24 '24 at 08:22
  • @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) – Frederik vom Ende Mar 24 '24 at 16:17
  • @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. – ironmanaudi Mar 25 '24 at 07:38

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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.

Norbert Schuch
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    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. – Frederik vom Ende Mar 21 '24 at 06:48
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    @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. – Norbert Schuch Mar 21 '24 at 06:51
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    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 – Markus Heinrich Mar 21 '24 at 08:50
  • @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. – ironmanaudi Mar 23 '24 at 14:26