What can we say about the eigendecomposition of quantum channels?

Question

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.

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Crossposted from Physics.SE

Answer 1

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.

Answer 2

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.