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?

  • $\begingroup$ Could you give us a few more details (perhaps an example calculation) about what you've done to express a channel in matrix form and find its eigenvalues? For me, reading what you say you've done currently, there seems to be a disconnect with what you should be doing, and I'm not sure if that's due to a misunderstanding on your part about the maths, or mine interpreting what you've written (what is $F$?). $\endgroup$
    – DaftWullie
    Commented Nov 6, 2023 at 8:01
  • $\begingroup$ @DaftWullie One can get a matrix corresponding to channel $\Phi$ by using a Hilbert-Schmidt basis. F. e., once can use the basis of delta matrices and this would correspond to a vectorization of operator (the channel acts on) and writing $\Phi$ as a matrix acting on the obtained supervector. This matrix is $F$. Then adjoint $\Phi^{\ast}$ is represented by the Hermitian-conjugate of the matrix, $F^{\dagger}$, and a concatenation of two channels corresponds to a multiplication of the corresponding matrices. $\endgroup$
    – trurl
    Commented Nov 6, 2023 at 21:48

1 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.

  • $\begingroup$ @ Markus Heinrich Thank you for clarification. "Note that the given argument does not imply the statement for an arbitrary TP map $\phi$. It holds for "eigenvalue" instead of "singular value"" instead of "singular value"" - this was indeed confusing. So it a relief to learn that it is not only me who is confused. " In fact, the spectral radius of any such $\phi$ is 1 " Perfect. So, perhaps, (and here I am guessing) what is meant is that, in the asymptotic limit, a random CPTP map (i.e., a channel) is unital with probability one [its fixed point is identity (normalized)]. . $\endgroup$
    – trurl
    Commented Nov 6, 2023 at 21:59
  • $\begingroup$ @trurl I had a look at the paper -- As indicated in my answer I think the statement is meant a.e. However, unital channels are not dense (more constraints!), thus I'm not sure what assumptions are (implicitly?) made here. $\endgroup$ Commented Nov 7, 2023 at 11:32
  • $\begingroup$ @ Markus Heinrich "unital channels are not dense (more constraints!)" - it depends on the definition of "density". If you sample a random CPTP map (f.e., by sampling a random Choi state or isometry), with probability tending to one in the asymptotic limit, the invariant state of the map will be the (normalized) identity. $\endgroup$
    – trurl
    Commented Nov 7, 2023 at 15:45
  • $\begingroup$ @trurl I wouldn't say that that it depends on the definition of density. Anyway, I see your point. $\endgroup$ Commented Nov 8, 2023 at 9:11

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