I am currently learning the basics of probabilistic error cancellation.

The idea, in summary, is the following:

We want to implement a quantum circuit having a noiseless (unitary) implementation $\mathcal{U}$, but, because of noise, we can only implement $\mathcal{E}=\mathcal{N} \circ \mathcal{U}$, where I assume $\mathcal{N}$ is a CPTP (Completely Positive Trace Preserving) operation that introduces noise in the computation (the noiseless case corresponds to $\mathcal{N}=\mathbb{I}$).

The idea behind probabilistic error cancellation is to try to find an inverse to $\mathcal{N}$ and to apply it. Ideally we then wish to implement the circuit: ( * )

$$ \mathcal{N}^{-1} \circ \mathcal{N} \circ \mathcal{U}=\mathcal{U} $$

Unfortunately, this inverse (if it exists!) will (i) not always correspond to a physical map (i.e. we loose the positivity condition for instance), (ii) will not always be implementable by the hardware.

Hence, the "trick" is to perform the following decomposition:

$$ \mathcal{N}^{-1}=\sum_{n} \alpha_n \mathcal{B}_n $$

where the $\{\alpha_n\}$ are some coefficients and $\{\mathcal{B}_n\}$ is a family of physical maps (i.e. CPTP) that I can exactly implement on the hardware. Hence, the $\{\mathcal{B}_n\}$ is typically a family of noisy maps (because my hardware is noisy).

My questions:

In order to be applicable, we need:

  • $\mathcal{N}^{-1}$ exists mathematically (not all maps have an inverse).
  • We can express $\mathcal{N}^{-1}$ as a linear combination of noisy operations that the hardware can actually do.

If I understood correctly the general idea, how are we sure that these conditions will be satisfied? Am I exactly expressing the limit in the applicability of the method?

( * ) I took an example where the noise map of the entire circuit can easily be found which is too complicated in general. However the idea can be generalized by introduced the noise map of each gate and writing a big sum (I skip these details here).

  • $\begingroup$ Do you mean we don't know the analytical form of $\mathcal{N}$? $\endgroup$
    – narip
    Commented Mar 11, 2023 at 12:12
  • $\begingroup$ @narip In the context of my question, we can assume we know perfectly the (analytical) description of $\mathcal{N}$. Thanks! $\endgroup$ Commented Mar 11, 2023 at 12:13
  • $\begingroup$ Then for the first one, since $\mathcal{N}$ is linear, we can easily check if it's invertible by writing it as a matrix. $\endgroup$
    – narip
    Commented Mar 11, 2023 at 12:15
  • 1
    $\begingroup$ @narip yes I agree. But my question is more to know if these two conditions are in practice two actual limitations of the method. In the papers I read these limitations were never explicitly written which I found weird (it seemed like we can always do P.E.C). Maybe there are good reasons to believe that these conditions are always satisfied. If so I would like to know these reasons. $\endgroup$ Commented Mar 11, 2023 at 12:18

1 Answer 1


What you are asking is "Do the implementable imperfect gates form a basis?". Unfortunately, you need a noise model assumption for this. Once you have a noise model assumption (which might be wrong!), then you can write the matrices for "the implementable imperfect gates" and then check if they form a basis. People don't do this check as often they confine their attention to Pauli noise.

This question is theoretically more complicated than it sounds. The basis check needs to be done in the super operator space. The set of super-operators corresponding to the set of noisy implementable basis gates will always fall short of spanning the full space.

But it practically works well. Because one of the noisy gates in your linear combination should be an approximation to the ideal. Hence, the available noisy basis gate kind of "fills up the cup". Said differently, the linear combination works well because there exists an approximation to the operation you want already and you simply want to use the Monte Carlo to improve that approximation. If the noisy basis gates had to do all the heavy lifting, then questions of span become critical.


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