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Much effort has gone into optimizing T gates (distillation, factories, and numerous circuit optimizations).

I am aware that non-Clifford gates are necessary to attain quantum speedup, and the T gate, which is basically $R_Z\left(\frac{\pi}{4}\right)$, is probably the simplest non-Clifford gate.

Why is it so difficult to implement a T gate? I have seen references to the error correction being more difficult (non-transversal?)

The Pauli gates ($\pi$ rotation around the Bloch axes) are considered simple. Is that due to simpler error correction as well?

In super-conducting qubits, after they are calibrated, they are manipulated with pulses and the parameters (duration, frequency, phase, amplitude) of the pulse control the rotation. Why is it harder to create a $R_Z\left(\frac{\pi}{4}\right)$ rotation than a $R_Z\left(\pi\right)$ rotation? (and might it be easier in different qubit implementations? e.g., silicon, photonics, ...)?

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    $\begingroup$ Does this answer your question? Why do we care about the number of $T$ gates in a quantum circuit? $\endgroup$
    – Tristan Nemoz
    Jun 15, 2023 at 8:38
  • $\begingroup$ Thanks @TristanNemoz, the link is helpful. If I understand correctly, we could have used a pi/4 X rotation instead of the T gate which rotates about the Z axis as the non-clifford gate. I understand that T gates are the bottleneck when it comes to fault tollerant threshold. I don't get the details of the tranversal error correction and their effect on T gates. $\endgroup$
    – inq
    Jun 16, 2023 at 0:10

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A few misconceptions here.

The $T$ gate is necessary to obtain universality, not speed-up.

Also, running a $T$ gate is not harder than other single-qubit gates. On contrary, it is among the most reliables.

The complexity is not given by the gate itself, but by the domain space that we can reach with combination of Clifford and non-Clifford gates.

Since we need to define and implement fault-tolerant computation, the distinction between Clifford and non-Clifford becomes convenient. As we can make Clifford gates resistant to noise by means of very efficient correction schemes. While fault-tolerant non-Clifford gates are achieved by means of distillation.

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    $\begingroup$ "The $T$ gate is necessary to obtain universality, not speed-up". Well, since Clifford circuits are efficiently simulated, I'd also argue that it is necessary to get Quantum speedup (at least with this set of universal gates). $\endgroup$
    – Tristan Nemoz
    Jun 15, 2023 at 8:38
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    $\begingroup$ @TristanNemoz Fair observation. But here's a counter-example: consider a brute-force simulator for Clifford+T circuits. Implement a better simulator (still inefficient). Speed-up is achieved and there is no relation with gate set assumptions. $\endgroup$ Jun 15, 2023 at 10:19
  • $\begingroup$ Ah I think I get your point. It seems like we're using different definitions for speedup. If I'm not mistaken, you talk about a speedup from a complexity point of view, while I consider it from a practical point of view. So at first, I didn't get why you could say that you've achieved speedup with an inefficient simulator $\endgroup$
    – Tristan Nemoz
    Jun 15, 2023 at 11:20

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