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Clifford+Toffoli+M is a universal gate set. What's a cheap way to approximately (or exactly) get a T state using this gate set? I know that once I have a T state I can catalyze more very easily, but how do I get the first one?

Here's my initial attempt. I tried some random stuff to make anything that overlaps with a T state, then copied it three times to produce three T state, expanded them into phase gradient states, then used kickback from adding into those states to do the T gates of a 15-to-1 factory. This produced a state with an infidelity below $10^{-4.5}$. But obviously this is suboptimal in its size and success rate:

enter image description here

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    $\begingroup$ Is there a motivation for this? $\endgroup$ Commented Feb 18 at 6:17
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    $\begingroup$ btw is clifford + toffoli properly universal or only computationally such as H + Toffoli? $\endgroup$ Commented Feb 18 at 6:59
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    $\begingroup$ @NorbertSchuch The circuit in the question is showing it can make a T state below the distillation threshold, so it's obviously universal via Clifford+T. $\endgroup$ Commented Feb 18 at 9:33
  • $\begingroup$ But it requires measurement/postselection, right? This seems a weaker notion of universality. BTW, what is CS? $\endgroup$ Commented Feb 18 at 10:51
  • $\begingroup$ @NorbertSchuch CS is controlled S. It can be performed via TOF+S+TOF using an ancilla. If the usual definition of universal doesn't include measurement or retrying until success, then I don't think that's the right definition. In practice you always have computational basis measurement and can always retry until success. In any case Toffolis can emulate all that stuff, as long as you have a supply of zero'd qubits to dump entropy into. $\endgroup$ Commented Feb 18 at 13:29

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Here's a much more efficient distillation circuit. By using a CS gate (which can be performed via Toffoli+S+Toffoli when an ancilla qubit is available) and some hadamards, plus some postselection, a passable T state can be made. Given two of these states, one can be destroyed to perform a check of the other. Arranging these checks into a binary tree produces a T state with super-exponentially low infidelity versus the number of levels.

Here's the circuit:

enter image description here

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