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The probability that a depolarizing channel doesn't affect the information is usually assumed to be $1-3p$, while, for convenience, it is affected with same probability $p$ by any Pauli operator $X,Y,Z$.

Stepping to the error correction paradigms, it seems to be common to work only with operators $X$ and $Z$ as they can describe also a $Y$ operator (which, however results into a suppressed probability of $p^2$).

On contrary if one embeds in the framework also a probability $p$ for the $Y$ operator, this opens to the possibility, to "enhance" the noiseless probability to $$(1-p)^3 + p^3 = 1-3p + 3p^2.$$

I'm not sure if this is in contrast with the depolarizing channel, which states that the noiseless probability should be only $1-3p$. However, I want to argue that such an extra addendum $3p^2$ is not in contrast with the channel model, as this states that no noise occurs with probability $1-3p$, which doesn't necessarily mean that, if all the noise operators occurs simultaneously, they do not affect the logic of the system, even though a noise occurred.

In conclusion, I wonder whether such a reasoning is correct or I'm missing some fundamental concept regarding channels.

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I'm not sure what you mean by the probability of not losing information. The entropy of the qubit's mixed state after the depolarizing channel is higher than it was before the depolarizing channel. There's not a chance that it's higher, it's just higher. Having a chance of seeing the wrong thing is what information loss looks like.

As for the Y vs X*Z thing, you should be picking error channels that represent the hardware you are trying to model. Does that hardware produce bit flips and independently phase flips, but not native Y flips? Then do that. Does it not? Then don't. (An example of a system sorta like that is the logical errors experienced by a surface code. So if you're concatenating a code on top of the surface code you would potentially use such a model.)

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  • $\begingroup$ I just read something about that, thank you. However I'm not working with surface code, but just with the standard circuit model, in order to then try to witness my analysis on an IBM processor. As first doubt, assuming a non-native Y simplifies the analysis, but I don't know if this is too trivial for IBM hardware. As second doubt, I don't get what issues arise by introducing an independent Y noise, in the sense that it looks fine to me that an occurrence of XYZ is something that doesn't need to be corrected. $\endgroup$
    – Daniele Cuomo
    Commented May 9, 2022 at 15:28

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