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The quantum volume metric $V_Q$ is a proposed metric for quantifying and comparing the performance of quantum computers1. The quantum volume is defined as

$$V_Q = \max_{n<N} \left(\min\left[n, d(n)\right]^2\right),$$

where $n$ is the number of qubits used (out of the maximum $N$ available qubits on the device), $d(N)\simeq 1/nε_{\text{eff}}$ is the effective/achievable circuit depth, and $\epsilon_{\text{eff}}$ is the effective/average error rate of a random $SU(4)$ gate between any two qubits (including additional swap gates if the device is not fully connected)[2].

Now setting aside the question of whether $V_Q$ is a "good" metric (asked previously here), can an estimate be made for what $V_Q$ is needed in order to have a single logical qubit? More generally, what is the number of logical qubits a device with $V_Q$ can support?

Edit: By logical qubit I mean in the sense of fault-tolerant operation, i.e. if I have $n$ logical qubits I can apply any arbitrary sequence of gates to the $n$ qubits and end up with the ideal quantum state with bounded error (e.g. probability greater than 2/3).

NOTE: Cross-posted on physics.SE.

References
1. Bishop, Lev S., et al. "Quantum volume." Quantum Volume. Technical Report (2017) (PDF)
2. Moll, Nikolaj, et al. "Quantum optimization using variational algorithms on near-term quantum devices." Quantum Science and Technology 3.3 (2018): 030503
3. Cross, Andrew W., et al. "Validating quantum computers using randomized model circuits." arXiv preprint arXiv:1811.12926 (2018).

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    $\begingroup$ That depends on what requirements you have of the reliability of your logical qubit. There's no reason why you couldn't have one physical qubit to represent one logical qubit --- so long as you don't mind your 'logical' qubit having a lot of noise on it. Because you can almost never absolutely eliminate noise even in theory, the question then is just one of how little noise you want. (To say nothing of the impact of the connectivity of the logical qubits on the error correction procedures available to you.) $\endgroup$
    – Niel de Beaudrap
    Commented May 9, 2019 at 8:19
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    $\begingroup$ Consider an extreme case when you have one logical qubit and you don't really want to do much with it. You don't really care about noise because you're never performing any complicated operations on it. $\endgroup$
    – AHusain
    Commented May 9, 2019 at 17:18
  • $\begingroup$ I meant "logical qubit" in the sense of thresholds for fault-tolerant computing. I've edited the question to be more explicit on this point $\endgroup$
    – Punk_Physicist
    Commented May 9, 2019 at 21:45

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Quantum volume is a bad metric for this purpose.

For example, suppose you have a ten thousand by ten thousand grid of qubits with a gate error rate of 1 in one thousand. The quantum volume of this grid is basically 0, because if you pick two qubits at random they will on average be more than one thousand steps apart. So an error will almost certainly occur as they are routed together in order to interact, and so you can't even perform 1 randomly chosen interaction.

But quantum error correction doesn't require randomly chosen interactions. The surface code only performs local interactions, and as a result the 10k x 10k with 10^-3 error grid that I described would be easily capable of large scale fault tolerant quantum computations over thousands of logical qubits.

As you make the grid bigger, the quantum volume at the physical level will continue to get worse and worse as the quantum volume at the logical level continues to get better.

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    $\begingroup$ I think you misunderstand how $V_Q$ is calculated. As described in Ref 2 for a square connected grid your effective error scales as $\epsilon_{\text{eff}} \sim \sqrt{n}\epsilon$. Now finding the number of qubits $n<N=10,000^2$ that maximizes the quantum volume gives $V_Q \sim \epsilon^{-4/5} \approx 250$, not zero as you claim. Now if you make the grid bigger, $d(N)$ gets worse, but $V_Q$ stays the same. This is the point of the maximization over number of qubits, to make this metric not sensitive to the issue you point out. $\endgroup$
    – Punk_Physicist
    Commented May 9, 2019 at 22:57
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The Quantum Volume is a benchmark for near term, noisy quantum systems. Indeed, like other random unitary benchmarks, you need to be able to sample the ideal distribution. This distribution comes from classical simulations, so your limited to about the ~40 or so qubit limit. However, the Quantum Volume itself was designed to benchmark not only the quantum device itself, but also things like control electronics, measurements, circuit compilation methods, etc, while encoding all of this in a single numerical value. Currently, IBM has shown Quantum Volume numbers for at most four qubits, 2^4=16.

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  • $\begingroup$ I totally agree that the goal of this metric is to benchmark NISQ devices (and the approximate limits of verification via classical simulation), but it is not obvious to me that this means you can't use this metric as an estimate for deciding when you have a good enough system to use the resource for error correction. $\endgroup$
    – Punk_Physicist
    Commented May 15, 2019 at 21:06

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