In this neat answer by Markus Heinrich, it is shown that twirling an arbitrary quantum channel $\Lambda$ over the unitary group $U(d)$ yields a depolarizing channel $\tilde{\Lambda}$ given by $$ \tilde{\Lambda}(M) = \Pi_{U(d)}(\Lambda)(M) = (1-p)\mathrm{Tr}(M) \frac{I}{d} + p M, $$ where $M$ is a linear operator on the Hilbert space (and $p$ is some function of $\Lambda$, $M$, and $d$).

1st question: Is this "unitary twirling operation" considered something physically realizable? For example, it often seems to be the case that the depolarizing channel is used in simulations and calculations, etc; is this because if we are given an arbitrary channel we can simply start any algorithm by unitary-twirling in some physical manner and therefore we may as well assume we started with the depolarizing channel to begin with? The comment by Norbert Schuch to this answer seems to suggest that the answer is yes, but maybe someone can provide more details?

2nd question: As Heinrich points out in his answer, one can replace $U(d)$ by any group $G$ and take the "$G$-twirl" of a channel instead of the unitary twirl. Is the $G$-twirl operation consider something physical?


1 Answer 1


Question 1: I guess it depends what your understanding of "physical" is. In my understanding, everything you can do in the lab is physical. Thus, twirling is perfectly physical. Note that you do not have to implement the twirled channel $\tilde\Lambda$ directly. As you have to collect statistics anyway, you can simply conjugate the quantum channel $\Lambda$ by randomly drawn unitaries in every run. This will be operationally indistinguishable from implementing $\tilde\Lambda$.

The main reasons why depolarizing channels are used in simulations and calculations is simplicity (or, to be mean, ignorance). It is a case where explicit calculations can be performed, that's why it is used. In some cases, depolarizing noise might be a good approximation to the physical noise, but this should only be expected in the sense of "effective noise models" for more complicated circuits. Gate noise is typically not depolarizing; it will be biased and, to a large degree, coherent (of course that depends on the platform). IMO, there's a considerable gap in the literature on realistic noise models.

What one can do is schemes like randomized compiling, which take a circuit and essentially randomize certain gates to effectively twirl the gate noise (under some assumptions of course). This turns coherent noise into stochastic (incoherent) noise which may improve the overall performance of the circuit, but could also be advantageous for quantum error correction. Nevertheless, the effective noise is still not depolarizing, but some Pauli channel which may still be very complicated to characterize.

Question 2: Well, $G$ should be represented as a subgroup of $U(d)$, so it is as physical as a unitary twirl (just perform the appropriate subset of gates!). In practice, you would anyway replace the unitary twirl by e.g. a Clifford twirl, or even random circuits (the twirl does not change, but its implementation!). This is simply because a) sampling from the Haar measure is computationally hard and so is b) (approximately) compiling Haar-random unitaries.

  • $\begingroup$ Ok, great answer! But what if a G-twirled (or unitary-twirled) channel applied to certain inputs $M$ yields $\textbf{0}$? Since twirling is a projection of one channel onto another this "zero-rank" scenario certainly could happen for some group $G$. Naively it seems like such a channel wouldn't be allowed physically because if $M$ is a density matrix then the output doesn't have "purity 1", i.e., $Tr(M^2) = 0 \neq 1$. In contrast, such a scenario seems a priori valid given your answer. $\endgroup$ Commented Feb 19 at 13:35
  • 1
    $\begingroup$ @EricKubischta This can't happen since twirling maps channels to channels. If the input to the twirled channel is a density matrix, so is the output. In particular, it can't be zero. $\endgroup$ Commented Feb 22 at 8:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.