A metric called the "quantum volume" has been proposed to somehow compare the utility of different quantum computing hardware. Roughly speaking, it measures their worth by the square of the maximum depth of quantum computations it permits but limits its value to the square of the qubits involved. This limit is justified by wanting to forestall "gaming" of the system by optimizing towards few qubits. One reference is https://arxiv.org/abs/1710.01022.

I am concerned that this measure, as good as it may be for noisy near-term quantum computing devices, hides the actual quality advances for more advanced quantum computers (those with high quantum gate fidelity). The question is: Is this concern justified?

The argument behind my concern is the assumption that potential killer applications for quantum computers, for example quantum chemical calculations, will require computations with a gate depth much larger than the (potentially modest) number of qubits required. In this case, the "quantum volume" would be limited to the square of the number of qubits, regardless of whether one quantum computer (with particularly high fidelity) permits an essentially unlimited depth or whether it only allows the bare minimum gate depth to achieve the limitation of the "quantum volume" to the square of the number of qubits. One aspect of my question is: Is this argument correct?

This question has an open bounty worth +100 reputation from pyramids ending in 2 days.

This question has not received enough attention.

There is one reasonably good answer already. Yet I feel there might be better (future) answers that explore in more detail how good or bad the "quantum volume" captures relevant performance at the physical qubit level. I'd like to reward such an answer if it materializes.

As a start, you might want to look at https://arxiv.org/abs/1605.03590, which lays out conservative (i.e., high) qubit and gate requirements for a meaningful quantum chemistry calculation under some pretty reasonable assumptions. The estimates there are on the order of $10^{15}$ total logical gates (not gate depth) over roughly 100 logical qubits, which means that the gate depth must be on the order of $10^{13}$ or higher (I'm looking at the nested counts).

So at the logical level, you're correct: you don't need $10^{13}$ qubits and $10^{13}$ gate depth to run nitrogenase. The two metrics, qubit count and gate depth, are not really equivalent: to run a real problem, I need on the order of a hundred qubits, but ten quadrillion gate depth.

That's not the whole picture, though. It's on the order of a hundred logical qubits, and logical gate depth of 10^13. Quantum error correction is essentially all about trading physical qubit count to get better logical gate depth. As you can see in table II in the paper, the logical-to-physical ratio ranges from 17,000/1 to 300/1 as the physical qubits get better (i.e., as the physical gate depth increases). Again from table II, a physical gate depth of $10^3$ leads to needing $10^9$ physical qubits, while a physical gate depth of $10^9$ only requires $10^6$ physical qubits.

The "quantum volume" measurement still doesn't seem quite right to me at this scale, though. I think a measurement more on the order of the product of the physical gate depth and the square of the physical qubit count is more accurate; for the three cases in table II, which represent equally powerful (in some sense) quantum computers, this value is roughly constant across the three columns. It also matches the rule of thumb that the number of qubits in a distance $d$ QEC code scales as $d^2$.

The one thing this leaves out is that the computer with $10^9$ physical gate depth will run your chemistry simulation much faster than the computer with $10^3$ physical gate depth because the computational and wall-clock overhead of QEC will be much lower. You could come up with a more complicated formula to take this into account, if you like.

  • There's been significant progress in the past few years on top of the paper you've linked. arxiv.org/abs/1805.03662 gives estimates that are tens of millions times better. – Craig Gidney Aug 10 at 5:16
  • Agreed; both in algorithms, and in T distillation, and in QEC. As I said, the estimates in the nitrogenase paper are high -- but the tables are nice and clear :-), and it's focused on one problem and so (perhaps) easier to follow. – Alan Geller Aug 11 at 6:30

Quantum volume is likely only useful as a metric for small noisy computers.

It’s impossible to invent any single-number metric that’s ideal for all tasks. Even with classical computers, metrics such as Dhrystone or Windows Performance Index are at best suggestive at predicting performance on real-world tasks. Conversely, giving more than one number can potentially be much more informative. Within the quantum volume framework, I suggest when characterizing a QPU to give quantum volume as the “executive summary” but also quote for a range of different qubit numbers $N$ the model circuit depths $d(N)$. Comparing the $d(N)$ to the needed depth and qubits will be predictive, at least to the extent that the killer apps resemble the model circuit sequences of parallel random $SU(4)$ on random pairs of qubits.

The quantum volume is about correctly implementing the model circuits, thus measuring it involves simulating those circuits to compare the output of the QPU against the ideal results. Simulation is practical only for relatively few qubits or low depth, so it is only possible to measure the quantum volume for small/noisy devices (without additional assumptions). Fortunately, when width/depth reaches the limit of simulation (very roughly around $N\simeq d\simeq 50$), this is when the noise must necessarily be low enough that we could begin to use such a device to implement logical qubits. Defining appropriate metrics for logical qubits is an open question. The emphasis switches from “Can this algorithm run at all?” to “How long will this algorithm take?” and metrics will surely be very different, involving the logical gate time.

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.