# Is there a straightforward way to calculate the quantum volume for simple systems?

There are a few people who ask how to calculate IBM's "quantum volume", but it's not clear what the actual mathematical definition of this is. In the answers here, people just point to running some program, which is not helpful for understanding or comparing with other technologies.

Is there a straightforward way of calculating this for a simple system?

For example, if I have n-1qubit gates with efficiency of 1, m-2qubit gates with efficiency B, and complete connectivity between qubits - what would my quantum volume be?

I believe the calculation steps are quite complex to give a simple formula or even to follow it by hand.

Maybe this medium article and the cited paper gives you a better general understanding.
As you can see in the article and paper, QV takes into account various factors other than error rates e.g. calibration and circuit optimization.

IBM successfully ran this exact test to achieve a Quantum Volume of 64 on its 27 qubit “Montreal” system — and the test didn’t even require building a new device. Instead, the team incorporated improvements to the Qiskit compiler, refined the calibration of the two-qubit gates, and issued upgrades to the noise handling and readout based on tweaks to the microwave pulses and gates before they’re applied in the circuit.

Other paper

This metric takes into account all relevant hardware parameters. This includes the performance parameters (coherence, calibration errors, crosstalk, spectator errors, gate fidelity, measurement fidelity, initialization fidelity) as well as the design parameters such as connectivity and gate set. It also includes the software behind the circuit optimization.

The simplest explanation from medium article would be:
2 to the power of the depth of the largest square circuit (width=depth), which you were able to run on your device successfully with greater than two-thirds probability and confidence interval greater than 97.725%, gives you QV.

• okay so loosely speaking, a quantum volume of 64 suggests a 8 qubits and 8 gates at 97% fidelity (and 2/3 efficiency?) Oct 1, 2020 at 22:06
• Not completely! QV 64 means the largest circuit you've been able to run on your device had 6 qubits and depth 6. Depth 6 means the circuit had 6 computational septs (not gates). The probability and confidence here, refer to the correctness of the output bit-string of the circuit and not the gates. Oct 2, 2020 at 8:03
• What is the difference between a computational step and a gate? Oct 2, 2020 at 23:08
• By 6 computational steps I meant depth=6. quantumcomputing.stackexchange.com/questions/5769/…. Oct 3, 2020 at 8:11

To put in simple words, the quantum volume tells a maximal dimension of state space (Hilbert space) you can simulate on a quantum computer sucessfully. For example, quantum volume 64 means that you can simulate systems with 64 states, i.e. represened by 6 qubits ($$6 = \log_2 64$$). Moreover, maximal depth of your simulation (circuit) has to be less or equal 6.

Note that this is relatively rough description what the quantum volume is, however, I think it is sufficient for basic understanding what that means.

• Nice perspective! Oct 2, 2020 at 13:21