What are the singular values of a quantum channel?

Question

I have tried to find the explicit definition of them but was not able to. My guess is that they are eigenvalues of the superoperator $\Phi^{\ast}(\Phi)$, where $\Phi$ is the channel and $\Phi^{\ast}$ is its adjoint.

Then, if the channel is expressed in a matrix form, it is about the eigenvalues of matrix $F^{\dagger} \cdot F$. Numerically (by sampling over randomly generated channels), I have found that the largest singular eigenvalue $\sigma_1$ is always larger than $1$ (which is the largest eigenvalue, $\lambda_1 = 1$, of $F$). It is no problem in general, since for a generic matrix $\sigma_1 \ge \vert\lambda_1\vert$. However, I cannot understand how does it fit the following statement from the paper by R. Kukulski et al:

"Let us now discuss singular values of a random superoperator $\Phi$. The leading singular value is equal to unity, which is a consequence of preservation of trace, \begin{equation} \langle \chi | \Phi | \chi \rangle = \langle \chi | \chi \rangle = 1, \end{equation} where $| \chi \rangle = | \rho_{\rm inv} \rangle \rangle/ \Vert\rho_{\rm inv} \Vert_2$ represents the normalized vector of length $d^2$ corresponding to the invariant state of the map, $\rho_{\rm inv}=\Phi(\rho_{\rm inv})$."

So, the question is: What is meant in this excerpt?

Answer 1

As usual, the singular values of an operator $\phi$ are the square roots of the eigenvalues of the positive semi-definite operator $\phi^\dagger \phi$ (or $\phi^*\phi$ if you prefer the $*$ for the adjoint). Of course, this applies in particular to linear operators on the Hilbert space $L(\mathcal H)$, i.e. superoperators.

That being said, the cited statement is at best misleading. Perhaps, the "random" is used as an argument that $\phi$ should be diagonalizable by a density argument (although it is not quite clear why this should hold for channels, as we deal with real matrices). Note that the given argument does not imply the statement for an arbitrary TP map $\phi$. It holds for "eigenvalue" instead of "singular value".

In general, the largest singular value (i.e. the spectral norm $\| \cdot \|_\infty$) of a (trace-preserving) quantum channel can be larger or equal to one. More precisely, we have $1 \leq \| \phi \|_\infty \leq \sqrt{d}$ where $d=\dim\mathcal H$ is the Hilbert space dimension. To see this, note that we have $$ \| \phi \|_\infty \leq \sqrt{d} \| \phi \|_\diamond\,, $$ which is tight for a replacement channel $\phi: \, X \mapsto \mathrm{tr}(X)|\psi\rangle\langle\psi|$ where $\psi$ is an arbitrary pure state. On the other hand, every CPTP map has a fixed point (see e.g. Watrous Thm. 4.24), which gives the lower bound on $\| \phi \|_\infty$. In fact, the spectral radius of any such $\phi$ is 1 (Watrous Prop. 4.26).

Finally, note that $\| \phi \|_\infty=1$ if and only if $\phi$ is unital (Watrous Thm. 4.27).

Watrous' "The Theory of Quantum Information" is a good reference on such questions.