# Are continuous probability distributions over quantum channels possible?

I am not an expert in the subject and apologize in advance for a strange question and (possible) abuse of the terminology.

I have learned that any convex combination of quantum channels (CPTP maps, following another nomenclature) is a quantum channel, $$T = \sum_{n=1} ^{M} p_i T_i$$, $$\sum_{n=1} ^{M} p_i = 1$$. Here is the question:

Is it possible to introduce a continuous probability distribution over a parametrized set of quantum channels? Then define an integral over the pdf (somehow) and and nominate the resulting quantum channel for the 'mean' of the set?

Quantum channels describe stochastic events, so it should be completely legit to consider continuous sets of events. Integrating over a continuous set of quantum channels with some distribution give you a valid quantum channel, describing the "average" effect in some sense.

Here's an example. Consider an $$X$$-rotation of $$90^\circ$$ (i.e., $$\pi/2$$-pulse) on a single qubit system, $$\begin{equation} X_{\pi/2} = \exp\left(-\frac{i}2\cdot\frac{\pi}2X \right). \end{equation}$$ When implementing this pulse in experiment, due to imperfect control, the rotating angle might be something like $$\pi/2+\delta$$ instead of $$\pi/2$$, where $$\delta$$ is an independent random variable for every instance of realization of the pulse, satisfying certain distribution $$p(\delta)$$ such as a zero-mean Gaussian. In this case, the average effect of the experimental implementation of $$X_{\pi/2}$$ can be modeled as \begin{aligned} \mathcal E(\rho) &=\int d\delta~p(\delta)~X_{\pi/2+\delta}\rho X_{\pi/2+\delta}^\dagger\\ &=\int d\delta~p(\delta)~\exp\left(-\frac{i}2\left(\frac{\pi}2+\delta\right)X \right)\rho\exp\left(\frac{i}2\left(\frac{\pi}2+\delta\right)X \right), \end{aligned} which can be understood as integration over the continuous parameterized set of unitary quantum channels $$\mathcal E_\delta(\rho) \equiv X_{\pi/2+\delta}(\rho)X^\dagger_{\pi/2+\delta}$$ according to $$p(\delta)$$. Note that, $$\mathcal E$$ is no longer a unitary channel.

PS: By expanding $$\rho$$ in the eigenbasis of Pauli $$X$$ you can simplify the expression of $$\mathcal E$$.

• Thank you, Senrui. Eventhough mixed-unitary channels are a special kind of channels, you explanation answers my question. I am also thinking that the average channels (in general case) can be found, e.g., by using Choi-Jamiolkowski correspondence, from the corresponding parametrized sets. Spectral properties of 'averaged' channels are of a special interest to me Sep 8, 2022 at 11:13

Sure. A standard example of this is the use of "twirling operations": given a channel $$\mathcal E$$, one can define $$\mathcal E_T(\rho) = \int dU\, U^\dagger\mathcal E(U\rho U^\dagger)U,$$ with the integral over the Haar measure over unitary operators. This is a convex combination of channels, hence $$\mathcal E_T$$ is a channel (which can be shown to always be a depolarising channel). See e.g. this classical Nielsen 2002 paper using it (or rather, reviewing its previous use by the Horodeckis in 1999).

See also this related post: Why does the twirl of a quantum channel give a depolarizing channel?

• Thank you, giS. Do you know any example which is not a mixed-unitary channel? Sep 8, 2022 at 11:13
• @trurl why do you say that this is a mixed-unitary channel? Here $\mathcal E$ is arbitrary, so the channels $\rho\mapsto U^\dagger \mathcal E(U\rho U^\dagger)U$ are not unitary in general
– glS
Sep 8, 2022 at 11:13
• such channels - defined as a convex combination of unitaries - are called "mixed-unitary channels" (see, e.g.,, arxiv.org/pdf/1902.03164.pdf, just a first reference I found). However, you are right, your example is not a mixed-unitary channel. Sep 8, 2022 at 11:18
• @trurl what "such channels" are you referring to? I know what mixed-unitary channels are, but I'm confused as to what's their relevance here
– glS
Sep 8, 2022 at 11:24
• mea culpa. Sorry for the confusion, your example is what I needed. And of course, it is not a unitary-mixed channel, as you correctly pointed out. Sep 8, 2022 at 11:35