I am simulating the dependence of $\Lambda$, as defined in google's Exponential suppression of bit or phase errors with cyclic error correction paper as a function of physical error rate.

The $\Lambda$ model states that $$ \epsilon_{L} = C/\Lambda^{(d+1)/2} $$ where $$ \Lambda \propto p_{th}/p, $$ and in practice, in this paper, $\Lambda$ is calculated as $$ \Lambda(d)=\hat{\epsilon}_L(d=3)/\hat{\epsilon}_L(d=5). $$ where $\hat{\epsilon}_L$ is taken from a fit over the decay curve as a function of number of rounds of the code. Specifically, I look at the error rate as the number of rounds increases, and fit it to: $$ P_{error}=0.5[1-(1-2\hat{\epsilon}_L)^{n_{rounds}}]. $$ To clarify, here is an example of a fit:

enter image description here

The above implies that $$ 1/\Lambda = \hat{\epsilon}_L(d=5)/\hat{\epsilon}_L(d=3) \propto p/p_{th} $$ and so $1/\Lambda$ should be linear in the physical error rate.

However, when I do the simulation, I get two interesting outcomes

  1. The threshold seems to be about a factor of 2 larger from the one obtained when I take a single $d$-dependent number of rounds.

  2. the simulation results, I get that $1/\Lambda$ behaves very nonlinearly when crossing the threshold. This means that if someone is performing and experiment above the threshold and wants to extrapolate it to below the threshold behavior, he will make an overly pessimistic estimate! Has this been observed/taken into account?

Below you can see the difference. On the left is the threshold and $1/\Lambda$ using fit on increasing number of rounds, and on the right is the threshold from rounds = 3 * distance - you can see that $1/\Lambda$ shoots past the threshold and only begins to distort when the error probability approaches 0.5.

enter image description here

So, what is the right way to simulate it? And does this nonlinear behavior of $1/\Lambda$ make sense? To me it seems like the fit approach for extracting $\hat{\epsilon}_L$ is the way to go, because the error rate is clearly rounds-dependent. So extracting the error rate gives you the rounds-independent parameters, much the same way you can extract resistivity from resistance by normalizing out the length of a conductor. However, this still leaves the question of validity of linear $1/\Lambda$ model open.


1 Answer 1


You are underappreciating how much of an approximation that equation is supposed to be.

the simulation results, I get that 1/Λ behaves very nonlinearly when crossing the threshold

Yes, everything is going to get weird as you approach the threshold. You're going to get an awful fit if you include data near there, and especially if you include data above threshold! You'll also find that the rule doesn't work very well at low code distances.

Getting a constant factor suppression in logical error rate per increase in code distance is just a rule of thumb. The suppression factor being proportional to the physical error rate is also a rule of thumb. These rules of thumb are only intended to be good enough to make rough plans around, not intended to be exacting predictions.

Has this been observed/taken into account?

Yes, if you talk with QEC researchers they will be aware that lambda distorts near threshold and at low code distances and at high distance if there are half distance errors and etc and etc. It's just a rule of thumb.

  • $\begingroup$ got it. Though, to be perfectly fair, I am basically working on the same prediction level as the paper. And, below the threshold it works quite nicely. Do you agree that the error rate from the fit is a more meaningful threshold indicator? $\endgroup$
    – Lior
    Feb 12, 2022 at 17:07
  • $\begingroup$ @Lior I think it's a good enough model for simple extrapolations and for planning. It's okay to be off by a factor of 10 because you just add 2 or 4 more code distance. Ultimately what matters is coding rate for a target logical error rate, but we won't be able to measure that directly for awhile. $\endgroup$ Feb 12, 2022 at 17:13

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.