# Necessity for discarding errored magic state distillation

The magic state distillation process ends with measuring the state of a logical qubit that is encoded in many (for example 15) data qubits.

Why does the distillation needs to be discarded and repeated if the final measurement is not a logical state of the logical qubit?

I thought that if the logical qubit code has a distance of 3, than one should be able to handle and correct an error.

The reason you discard them is because this makes distillation much more efficient.

For example, consider 15-to-1 T state distillation using the Reed-Solomon code. This code has distance 3, so you can either correct 1 error or detect 2 errors.

If you choose to correct 1 error, then you fail when there are a pair errors. There are $${15 \choose 2} = 105$$ such pairs of errors, so for an input failure chance $$p$$ your output failure chance will be $$105p^2$$. Your discard rate will be $$0$$.

If you choose to detect 2 errors, then you fail when there are a triplet of errors that exactly move you between code points. There are 35 such triplets. So for an input failure chance $$p$$ your output failure chance will be $$35p^3$$. Your discard rate will be around $$15p$$.

Example

Suppose your input error rate is $$p=10^{-3}$$ and you want to get to a target error rate of $$10^{-12}$$. When correcting your error rates progress as follows:

$$p \rightarrow 105p^2 \rightarrow 105^3 p^4 \rightarrow 105^7 p^8 \rightarrow 105^{15} p^{16}$$

$$10^{-3} \rightarrow \approx 10^{-4} \rightarrow \approx 10^{-6} \rightarrow \approx 10^{-10} \rightarrow \approx 10^{-18}$$

When detecting your error rates progress as follows:

$$p \rightarrow 35p^3 \rightarrow 35^4 p^9$$

$$10^{-3} \rightarrow \approx 10^{-8} \rightarrow \approx 10^{-21}$$

When using correction, you go through four stages of distillation, so you get 1 good output for every $$15^4=50625$$ inputs.

When using detection, you go through two stages of distillation. By far the dominant source of loss is a first stage failing. That happens about 1.5% of the time and forces you to get 15 more T states. Ultimately you get 1 good output for approximately every $$230$$ inputs.

So there you go. The reason we use detection instead of correction is because its a solid >100 times more efficient when considering plausible starting error rates.

• Great answer! So if we want to summarize it, it’s because each distillation step is exponentially more expensive than the previous, whereas trying again costs linearly, so we much rather try a current step multiple times and come out more certain than to move on to the next step as soon as possible?
– Lior
Commented Nov 25, 2022 at 5:38