I am interested in the spectrum of unital quantum channels $\Phi$ (which act on finite dimensional spaces). The literature is extremely vast on such objects so I hope some experts can point me along the right direction or references.

I know that the spectrum (i.e. eigenvalues $\lambda$ appearing in the eigenvalue-eigenvector equation $\Phi(X) = \lambda X$ ) lies within the unit disk $|\lambda|\leq 1$. Because of unitality $\mathbb{I}$ is always an eigenvector with eigenvalue $1$ because $\Phi(\mathbb{I}) = \mathbb{I}$. There may be other eigenvectors that lie on the unit disk $|\lambda| = 1$ (called `peripheral eigenvectors/eigenvalues').

My question is what is known about the size of the gap of unital quantum channels, defined as $1 - |\lambda|$ such that $\lambda$ is the eigenvalue with largest magnitude that is not $1$? The physical reason why I am interested in because this sets the rate of convergence to the peripheral eigenvectors.

Perhaps what I am asking is too general, and it depends on the explicit quantum channel in question? In which case my question would then be, is there a way to estimate the magnitude of the gap? Like a variational principle of sorts?

  • $\begingroup$ Technically, spectral gap is ill-defined since the maximum of the set of eigenvalues other than one may not exist. Consider the identity channel. $\endgroup$ Nov 18, 2023 at 14:21
  • $\begingroup$ It should be possible to estimate the spectral gap of a given channel numerically using eigenvalue algorithms, such as Arnoldi iteration. $\endgroup$ Nov 18, 2023 at 14:27
  • $\begingroup$ Thanks! Sorry I realized my question was phrased in a slightly confusing manner. Obviously the spectral gap depends on the channel in question and there is no universal value for all unital channels that I might be giving the wrong impression of (I also take your point that the set of non-one eigenvalues may not exist, but in which case I would simply just define the gap as 1). I intended to ask is there an analytic way to bound the spectral gap given a channel. I suppose one can try to employ Arnoldi iteration "by hand" instead of numerically but I don't know how much mileage that will give $\endgroup$
    – nervxxx
    Nov 18, 2023 at 15:26

1 Answer 1


TL;DR: Spectral gap depends on the specific channel. Moreover, for any $g\in[0,1]$ there is a channel with spectral gap $g$.

Non-peripheral eigenoperators are traceless

We can make simple observations about properties of eigenoperators of $\Phi$ by hitting both sides of the eigenvalue equation $\Phi(A)=\lambda A$ with the trace to get $$ \mathrm{tr}(A)=\lambda\,\mathrm{tr}(A)\tag1 $$ where we used the fact that $\Phi$ is a channel and hence trace-preserving. Clearly, $(1)$ can only be true if $\lambda=1$ or $\mathrm{tr}(A)=0$, so all non-peripheral eigenoperators of $\Phi$ are traceless.

Prescribed spectral gap

This suggests a simple construction of a unital channel with a prescribed spectral gap. Consider depolarizing channel $$ \mathcal{D}_p(A)=pA+(1-p)\frac{\mathbb{I}}{d}\mathrm{tr}(A)\tag2 $$ where $d$ is the dimension of the Hilbert space and $p\in\big[-\frac{1}{d^2-1}, 1\big]$. Clearly, every non-zero traceless operator $A$ is an eigenoperator of $\mathcal{D}_p$ with eigenvalue $p$.

Thus, the spectral gap of a unital channel indeed depends on the specific channel. Moreover, for any $g\in[0,1]$ there is a unital channel with spectral gap $g$.


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