# How do magic states circumvent the Eastin-Knill theorem?

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?
– AG47
Apr 12 at 8:48
• 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? Apr 12 at 9:31
• 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.
– AG47
Apr 12 at 10:49

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.