As far as I understood from a series of papers, minimizing the T-count in Clifford+T circuits is essential for fault-tolerant quantum computing:

While techniques such as magic state distillation and injection allow for fault-tolerant implementation of T gates, they typically require an order of magnitude more resources than Clifford gates

For example see here and here.

But do I understand correctly that while minimizing T-count (incl. in Clifford+T circuits) does not give a significant gain for the current IBM open quantum systems (real hardware, not simulations!), in particular, due to the fact that at the moment T gate (like T†, U1/P and RZ gates)

can be implemented virtually in hardware via framechanges (i.e. at zero error and duration)

See the documentation

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    $\begingroup$ I don't know the specifics of IBM hardware (hence this not being an answer), but yes, my general understanding is the T-gate optimisation is for fault-tolerant scenarios. At the level of physical qubits, you can directly implement any unitary, so why would you go for a long sequence of Hadamard + T to approximately synthesise it? I don't know a reason. $\endgroup$
    – DaftWullie
    Commented Nov 27, 2020 at 7:54
  • $\begingroup$ @DaftWullie: On IBM Q you can use U3 which allows to implement any one qubit operation, so you really do not have to do decomposition to H, S and T gates. However, only $U1(\theta)$ gate (in fact $Rz(\theta)$ gate up to global phase) and $Rx(\pi/2)$ are physically implemented one qubit gates. So, some decomposition to basic gate set is necessary and I would expect that $Rz$ for arbitrary angle is non-Clifford gate and hence number of such gates matters. $\endgroup$ Commented Nov 27, 2020 at 8:27
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    $\begingroup$ T gates, or any phase gate, is a virtual gate that comes for free today. This is different than the fault tolerant regime. $\endgroup$ Commented Nov 27, 2020 at 16:22
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    $\begingroup$ A small addition: IBM Q systems not the only quantum computers in which the Z-rotation gate is implemented virtually on the hardware level, there are others, for example, QSCOUT: "Single-qubit Z gates executed virtually by adjusting the reference clocks of individual qubits". Accordingly, there apparently also does not require the minimization of the T-count. $\endgroup$ Commented Mar 28, 2021 at 9:20

2 Answers 2


No, it doesn't make sense to minimize T count on NISQ hardware. As you note, T gates don't even have a cost because they are done virtually by updating the rotation frame of later operations. No reason to minimize things that are free! If anything, you should be maximizing their use.

  • $\begingroup$ Thank you very much for your reply! I suspected that it was so, but I was afraid that I did not understand something. With your authority you dispelled these fears! $\endgroup$
    – Psanfi
    Commented Apr 13, 2022 at 16:46

Every gate has a cost because each gate of NISQ computers are noisy so each one introduces a little bit of error. This means that if you use fewer gates to do the same task, you'll end up with less error. Also, coherence time is limited so one can only perform so many gates...meaning that if you reduce your gate-count, you are able to do more meaningful computation than if you didn't minimize the gate count.


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