I'm trying to understand magic states and how they circumvent the Eastin-Knill theorem. I understand that these magic states are used to implement non-Clifford gates but how are these magic states generated in the first place? From Bravyi and Kitaev's paper, my understanding is that one needs to be able to prepare certain error-prone states that cannot be obtain from just Clifford gates. In papers about magic state gadgets, it's always implied that you can create magic state like $|T\rangle$ for example, but how are these states generated in the first place?

  1. Do you still need non-Clifford $T$-gates, even if faulty, to create them? Or is there some way to use only Clifford gates to create these magic states?
  2. If one can generate these states without any non-Clifford gate, how is this possible without violating the Eastin-Knill theorem?
  3. If this is possible, doesn't that mean that we can efficiently simulate this on classical computers since we only need Clifford gates? Or is the overhead so large that magic states are not efficiently implementable on classical computers?
  • $\begingroup$ Does this answer your question? $\endgroup$
    – AG47
    Commented Apr 12 at 8:48
  • $\begingroup$ The bootstrapping still requires non-transversal gates so that means you still need them and can't avoid that. Or is there a way to do this differently? $\endgroup$
    – Karim
    Commented Apr 12 at 9:31
  • 1
    $\begingroup$ Note that I said might require. The question was not about transversal gates in particular. For the particular case of $T$-states, you probably cannot create them by measurement because $T$ is not Hermitian, but as the answer tells you, for some other magic states multi-qubit measurement can do the trick. $\endgroup$
    – AG47
    Commented Apr 12 at 10:49

1 Answer 1


The important distinction here is not between Clifford and non-Clifford. It's between transversal and non-transversal. The Eastin-Knill theorem simply says that you cannot create a universal set of gates using transversal operations. So, if you choose an error correcting code where you can implement all Clifford gates in a transversal manner, Eastin-Knill tells you that the way that you implement your T gate must be non-transversal. Note that transversal is sufficient for fault-tolerant gate constructions, but it's not necessary, so this restriction doesn't prevent fault tolerance, which is the ultimate goal of these constructions.

So, to make magic states (in a faulty way), you must use non-transversal gates (if you use gates), which means there must be a non-Clifford element. Very crudely, imagine you've some how produced lots of (physical) single-qubit magic states, but they're very faulty. You want to distil them into something better quality. You achieve this by some sort of comparative measurement. This comparison means interacting between the physical qubits in a way that violates the transversal structure.

However, there are other ways around your first question ("do you still need non-Clifford T gates"). But let me demonstrate this with a different gate set. There are error correcting codes for which the transversal set does not include Hadamard. Do you still need non-transversal Hadamard gates? No. The magic state that you'd have to prepare is the $|+\rangle$ state, but you do not need a Hadamard to produce it. You just need an $X$ measurement (which you generally have any way). You could imagine an equivalent solution for T gates.

  • $\begingroup$ Thanks for the clarification. In my question I implied that Clifford were transversal and non-Clifford non-transversal. To summarize, if you use gates, then you would still need non-transversal gates to magic (faulty) magic states. This is quite clear. But, you can use different approaches to get these magic states by some measurement. So that measurement somehow takes care of the non-transversality. Is that correct? $\endgroup$
    – Karim
    Commented Apr 12 at 9:23
  • 1
    $\begingroup$ Yes, that's right (it's a multi-qubit measurement, so violates transversality) $\endgroup$
    – DaftWullie
    Commented Apr 12 at 9:33

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.