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I recently started learning about universal quantum computation using magic states, and I'm currently reading one of the early papers on the subject by Bravyi and Kitaev .

In the paper, they showed how copies of a single-qubit mixed state $\rho$ with relatively high fidelity between $\rho$ and a magic state could be "distilled" into a new mixed state with enhanced fidelity.

In particular, the paper had a nice geometric view of single-qubit mixed states. For example, those that were convex linear combinations of pure eigenstates of Pauli matrices lived within an octahedron inside the Bloch sphere. Such states (along with Clifford operations and Pauli measurements) could not be used for universal quantum computation. On the other hand, some of the mixed states outside the octahedron could be distilled to enhance their fidelity with the magic states.

However, it wasn't clear from the paper whether it might be the case that all mixed states outside of the octahedron could ultimately be distilled into magic states with some appropriate algorithm (the authors indeed flagged this as an interesting question). Is the answer now known?

To be explicit, can every $\rho = \frac{1}{2}(I + \rho_x \sigma^x + \rho_y \sigma^y + \rho_z \sigma^z)$ with $|\rho_x|+|\rho_y|+|\rho_z| > 1$ be distilled to a magic state?


I hope the answer is yes, because I would like it to be true that as soon as we can prepare a mixed state that is not a convex linear combination of eigenstates of Paulis, we have a resource that allows us to escape Gottesman-Knill.

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  • $\begingroup$ Of course escaping Gottesman-Knill is not necessarily the same as becoming universal! There could be more structure. $\endgroup$
    – DaftWullie
    Nov 17, 2023 at 7:47

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I've not kept sufficiently up to date with the most recent literature, however, here are some partial results:

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  • $\begingroup$ +1, this gives me a lot of food for thought. I have a couple comments. $\endgroup$
    – user196574
    Nov 17, 2023 at 18:47
  • $\begingroup$ I interpret the Veitch paper not as saying that there are magic states that can't enable universality, but rather that there are states outside the stabilizer polytope (the generalization of the qubit octahedron) that cannot be distilled to magic states. $\endgroup$
    – user196574
    Nov 17, 2023 at 19:31
  • $\begingroup$ In particular, these special non-stabilizer but non-universal mixed states outside the stabilizer polytope are those with a positive discrete Wigner function, and Clifford operators cannot introduce negativity into such Wigner functions, and so Clifford operations won't be able to distill such states into the desired magic states. See in particular page 4, Negativity is necessary for magic state distillation. $\endgroup$
    – user196574
    Nov 17, 2023 at 19:31
  • $\begingroup$ I emend my comment on 0411036 -- the protocol is indeed tight in directions that cross an octahedron edge (in those directions all states outside the octahedron can be converted into H-type magic states). $\endgroup$
    – user196574
    Nov 17, 2023 at 21:15
  • $\begingroup$ This depends a little bit on how you define "magic state". Some sources define "magic state" to be anything outside the octahedron. Others might define "magic state" to be those states that enable universality. $\endgroup$
    – DaftWullie
    Nov 20, 2023 at 7:44

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