From this paper, it has been nicely shown that the number of perfect $T$ gates required to simulate arbitrary single-qubit gates grows linearly with $\log(1/\epsilon)$, where $\epsilon$ is the error tolerance between the target gate and the simulated gate. Particularly, the linear dependence for any $R_Z(\theta)$ is $N\approx3.067\log_2(1/\epsilon)-4.322$.

What if the $T$ gates available is no longer perfect? Say the available $T'$ gates have fidelity $(1-\epsilon_g)$ from the perfect gate, how does the gate infidelity affect the performance of gate synthesis?

EDIT: Any related reference about how the T gate error is accumulated during the synthesis is appreciated.


1 Answer 1


If your T gates aren't perfect, then you will need to do state distillation to make them good enough that the error you accumulate from using $N$ of them is less than the error you incur from approximating the target rotation with Clifford+T. So your question really comes down to how expensive is state distillation.

If you don't distill, then you simply hit a lower bound on the precision you can achieve because as the rotation-approximation error goes down from increasing the sequence length the bad-T error will increase. At some point the bad-T error will exceed the rotation-approximation error, setting your no-distillation precision floor.

  • $\begingroup$ Do you have any references that cover how bad-T errors accumulate during the gate synthesis? $\endgroup$
    – Yunzhe
    Mar 5 at 7:26
  • $\begingroup$ @Yunzhe I don't. $\endgroup$ Mar 5 at 9:06

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