According to Wikipedia, the Honeywell computer has quantum volume ("QV") equal to 1024 and has 10 qubits. QV should not be more than $n^2$ squared, i.e. 100.

Is this any math error? Are the publicly given QV useless?

So it appears they are using the new (depth) definition, which to me seems to allow simulation and other rooms to inflate performance.


1 Answer 1


It is not an error. Under the subsection IBM's modified definition, they define the quantum volume as:

$\log_2 V_Q = \arg\max_{n \le N} \{\min [n, d(n)]\}$

That is, if you take the minimum of the number of qubits and the circuit depth over those qubits (i.e. $10$, because the circuit may well be deeper than the number of qubits, so the number of qubits is the limiting factor on Honeywell's device), that is equal to the $\log_2$ of the quantum volume $V_Q$, so $\log_2 1024 = 10$ or, alternatively, $2^{10} = 1024$.

As far as the utility of publicly given quantum volumes, in my mind they're not useless, rather they just require interpretation and a clear statement on which exact formula is being used. Other approaches to quantum volume exist; I really like this one paper that discusses volumetric benchmarks as a generic framework for understanding the capabilities of quantum hardware to implement specific circuit classes. One of the crucial ideas is that you can have circuits with many different shapes (among other properties) and, in practice, application circuits are pretty much never perfectly square or random.

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  • $\begingroup$ So basically we aren't given the fundamental information to do the calculations. $\endgroup$
    – D J Sims
    Commented Jul 24, 2021 at 0:21
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    $\begingroup$ I think you may want to re-read that comment and also realize that my confusion is in reference to your original comment "That sounds highly open to fraud, if you have to simulate what the depth is your simulation can be made in any way." It doesn't seem to me you're engaging in good faith discussion. Best of luck. $\endgroup$
    – Greenstick
    Commented Jul 24, 2021 at 1:15
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    $\begingroup$ You do need to simulate the circuits in order to know the heavy outputs. So there is a simulation piece to the metric. However, by the time you reach the limit of simulators you are probably at error rates where one can start to move to error correction. QV is a near term metric and, as formulated today, has a finite lifetime to it. $\endgroup$ Commented Jul 24, 2021 at 11:52
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    $\begingroup$ @PaulNation Yes, but that’s to establish the heavy output probabilities classically — the benchmark is to establish whether a quantum device can recapitulate them? Hope I’m not missing something. $\endgroup$
    – Greenstick
    Commented Jul 24, 2021 at 20:08
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    $\begingroup$ I cant measure the heavy outputs on a quantum machine if I don’t know which bitstrings are heavy. That is what the simulation is for. $\endgroup$ Commented Jul 25, 2021 at 0:57

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