When decomposing an arbitrary SU(2) matrix into elements from a universal gate set (i.e. with the Solovay-Kitaev Theorem) we try to get the lowest error possible, which leads to 'chains' of these elementary gates becoming very long.

E.g. some arbitrary SU(2) might be approximated with the decomposition THTSHTHTHT (this is just a random sequence from the {H,T,S} universal gate set that I have typed).

But when we implement this chain in a physical quantum computer each gate will have a non-zero error associated with it. This means a longer chain length both reduces the total error (through higher precision estimation) and increases it (through each gate in the chain contributing to the total error).

My question is: how can the error propagation of the chain length be calculated. For example if all gates have a fidelity of 99.99%, is the total error of a chain of length 10 just 0.9999^(10)? I have been told that the worst case scenario would be to add the errors of each gate linearly, but I am confused to how this works.

I am trying to optimise the chain length to reduce the total error (balancing the longer chain length leading to a more precise approximation, and the longer chain length increasing error through imperfect gates in the chain).

Thanks for any help and sorry if the wording of the question is poor.


1 Answer 1


Usually, you use such a finite gate set in a fault tolerant scenario, acting on logical qubits. If you're working with physical qubits, you can often make the single-qubit gate directly. But in the fault-tolerant scenario, you would usually do a round of error correction between each elementary gate. It is the structure of the fault-tolerant error correction that prevents those errors from accumulating. So I don't think the sort of calculation that you're talking about is particularly relevant.

  • $\begingroup$ Thanks for your reply. But what about in near term applications where we don't have access to any logical qubits, but still want to perform arbitrary gates on the physical qubits? $\endgroup$
    – shuey97
    Commented Feb 12, 2021 at 16:27
  • $\begingroup$ Then most systems let you apply an arbitrary single-qubit gate directly, and you don't need to mess about with decomposing into H and T. $\endgroup$
    – DaftWullie
    Commented Feb 13, 2021 at 12:54

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