I understand that both have 53 qubit devices, yet it is Google that has demonstrated quantum supremacy (although IBM refutes this!). I'm not sure if this is true but it seems like IBM cannot replicate the quantum circuits and measurements that Google has managed on their device. Please correct me if this assumption is wrong.

Reading the Google paper, the following line was significant

Our largest random quantum circuits have 53 qubits, 1,113 single-qubit gates, 430 two-qubit gates, and a measurement on each qubit, for which we predict a total fidelity of 0.2%.

That fidelity seems pretty good given that you've gone through so many gates. The data on IBM is a bit sketchy but this table seems to suggest that even their IBM-Q System One has 2-qubit gates with an average error of 1.69%. So after 400 odd gates, the fidelity drops below 0.1%. And that's not including the single qubit gates or the measurement at the end.

Is this the ace card that allows Google to demonstrate what they did? Have they out-engineered IBM in producing high fidelity gates or does the advantage come from something else?


1 Answer 1


Both IBM and Google unveiled 53-qubit processors. At this time, only Google published performance metrics such as 1- and 2-qubit gate errors. Until IBM publishes similar metrics we simply cannot even tell whether Google's processor outperforms IBM's.

What we can tell is that the connectivity of the two processors is different - Google's Sycamore processor has nearest-neighbor connectivity on a rectangular grid, while IBM's Rochester processor has a reduced connectivity on what appears to be an edge-centered hexagonal lattice. It will be interesting to see how the different connectivities play out with regards to performance benchmarks, such as quantum volume.

With regards to universality, an issue that was raised in the comments, Section VII.F in the supplement of the Google paper shows that the gate-set used in the experiment is universal:

The proof consists of two parts. First, we show that the CZ gate can be obtained as a composition of two fSim gates and single-qubit rotations. Second, we outline how the well-known proof that the H and T gates are universal for SU(2) [56] can be adapted for X1/2 and W1/2 . The conclusion follows from the fact that the gate set consisting of the CZ gate and SU(2) is universal [57].

To the extent that universality means "the processor uses a universal gate set", the Google processor is universal. If however, you rather define universality to require a fault-tolerant quantum processor, then no currently existing device can be considered as universal.


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