3
$\begingroup$

I have been reading about the Eastin-Knill theorem, which states that no quantum error correction code can can transversely implement a universal gate set. In this context, for example, the surface code admits Clifford gates in a transversally, which do not form a universal gate set. A universal gate set can be Clifford+T, which is why usually the fault-tolerant T gate over a surface code can be implemented by means of magic state injection. This process requires to distill magic states with sufficiently good quality for being injected in the surface code logical qubit. This also presents some other requirements such as real-time decoding.

However, I have been reading about other possible ways of circumventing the problems that the Eastin-Knill theorem poses for executing arbitrary algorithms over QECs. This is were I crossed the term Shor's fault-tolerant Toffoli gate. However, I am a bit lost about how does this work. As far as I know the H+Toffoli set is universal and, thus, a code should not be able to admit both. Thus I do not see how this gate can circumvent Steane-Knill.

$\endgroup$
1
  • $\begingroup$ I am sorry about my previous comment, in which I said mistakenly that $H$ and $\text{Toffoli}$ do not substitute a universal gate set. This is wrong, and I apologise for any confusion. Eastin-Knill says that any quantum code with error correcting capabilities can not allow a universal set of gate to be transversally implemented. What does the paper you referenced say about the implementation? Is it transversal for both gates? $\endgroup$
    – JoJo P
    Feb 21 at 20:00

1 Answer 1

1
$\begingroup$

Shor's toffoli gate might be fault-tolerant, but that doesn't mean it's transversal. (It's not.) You should really think about this gate in the context of magic state distillation: you have to prepare a resource state and consume that resource, combined with the set of gates you have available, to make a new gate. It's just that the resource state is now a 3-qubit state rather than a 1-qubit state as it would be in the case of the T gate, so the performance is generally going to be much worse.

$\endgroup$
1
  • $\begingroup$ Ok, I guess I understand what you mean. So other than doing magic state factories for then injecting them, you would create "Toffoli" factories and then inject them. Yeah I agree it seems that it's a more complicated approach that should generally perform worse. Thanks! $\endgroup$ Feb 22 at 9:47

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.