I would like to know if there are any special properties of channels that permit a Kraus representation that includes an identity? That is, if I am given a Kraus representation of a CPTP map $\Phi$ for which one Kraus operator is $\sqrt{p} \,\mathbb{I}$ with $p > 0$ and no other Kraus operator is proportional to the identity, then I can express $\Phi$ as

$$ \Phi(\rho) = p \rho + \sum_{a} A_a \rho A_a^* \tag{1} $$ with $\sum_a A_a^* A_a = (1-p) \,\mathbb{I}$. Then can I make any interesting comments about $\Phi$?

This kind of noise seems interesting because $\Phi (\rho)$ written as a mixture with a term proportional to $\rho$ in its output as in Equation $(1)$ and so we can interpret the effect of the channel as "with probability $p$, nothing happened, otherwise with probability $(1-p)$ something nontrivial happened".

Certainly not all channels have this form, but maybe channels that do have this form share some other properties in common?


1 Answer 1


We can view $\Phi$ as a convex combination $\Phi = pI + (1-p)\Psi$ of the identity channel $I(\rho)=\rho$ and the channel

$$ \Psi(\rho)=\sum_a B_a\rho B_a^*\quad\text{where}\quad B_a=\frac{A_a}{\sqrt{1-p}}. $$

This indicates that some properties of $\Phi$ will depend on corresponding properties of $\Psi$. For example, it is not hard to show that $\Phi$ is unital$^1$ if and only if $\Psi$ is unital. More generally, $\Phi$ has the same fixed-points$^2$ as $\Psi$.

Other properties need to be checked for $\Phi$ independently. For example, $\Phi$ may preserve entanglement even if $\Psi$ is an entanglement-breaking channel. Also, it is easy to see that there is a floor for the average fidelity$^3$ of $\Phi$ regardless of $\Psi$, namely$^4$ $\overline{F}(\Phi)\ge p$.

Note that any channel $\Xi$ that cannot be written in the form $(1)$ can be approximated arbitrarily well by

$$ \Xi' = \epsilon I + (1-\epsilon)\Xi $$

which is of the form $(1)$, where $\epsilon$ is a sufficiently small positive number. In other words, the requirement $(1)$ defines a set which is not closed. Therefore, no property corresponding to a closed set of channels holds exclusively for channels of the form $(1)$.

$^1$ Channel is unital if it maps identity to identity. For example, depolarizing channel is unital and amplitude damping channel generally is not.
$^2$ All quantum channels have at least one fixed point by the Schauder fixed-point theorem.
$^3$ Defined as $\overline{F}(\Phi)=\int \langle\psi|\Phi\left(|\psi\rangle\langle\psi|\right)|\psi\rangle d\psi$.
$^4$ This naive bound can actually be improved using $\overline{F}(\Phi)=\frac{d F_e(\Phi)+1}{d+1}$ where $F_e$ is the entanglement fidelity and $d$ the dimension of the underlying Hilbert space.

  • $\begingroup$ Thank you for the answer. I'm trying to better understand the argument about the set of such channels not being closed. What do you mean by a set of channels being closed? Can one not use this same argument on, say, mixed unitary channels by saying any mixed unitary channel $\mathcal{\Phi}$ can be approximated well by some non mixed-unitary channel $(1 - \epsilon) \mathcal{\Phi} + \epsilon \mathcal{\Psi}$ where $\Psi$ is not mixed-unitary? $\endgroup$
    – forky40
    Commented Nov 5, 2021 at 15:16
  • $\begingroup$ A set $\mathcal{A}$ of channels is closed if the limit $\Phi$ of every convergent sequence of channels $\Phi_k\in\mathcal{A}$ is also in $\mathcal{A}$. Certainly not all properties correspond to closed sets of channels (some aren't even binary, e.g. channel capacity), but some do and then this way of thinking might be helpful. To be clear, I'm not making an argument for any specific claim, just offering an idea that might help since the problem is open-ended. The idea is inspired by the fact that physics is really a science of approximations, so considering limits is highly advisable :-) $\endgroup$ Commented Nov 5, 2021 at 17:23

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