Is un-computing $U$ a good proxy for circuit fidelity?

Question

I'm trying to estimate the fidelity of some family of unitaries $U(\theta)$ implemented on a noisy quantum computer. To do so, I start from an uninitialized state $|0\rangle$ and run the circuit $U(\theta)^\dagger U(\theta)$, and then measure the probability of returning to the all-zeros bitstring as

$$ P_\theta (0) = |\langle 0| \mathcal{E}_{U(\theta)^\dagger} \mathcal{E}_{U(\theta)} |0 \rangle |^2 $$

where $\mathcal{E}_U$ is the (noisy) process that results from trying to run $U$. If $P(0) \neq 1$ then I have some idea that my operation ran with lower fidelity. But I would like to know more about how well this kind of reasoning generalizes:

• How well does $P_\theta(0)$ track the fidelity of this process, $F(\mathcal{E}_{U(\theta)} , U(\theta))$, if at all?
• Can I expect any kind of relationship between $P_\theta(0)$ and $P_\theta(\psi) \equiv |\langle \psi| \mathcal{E}_{U(\theta)^\dagger} \mathcal{E}_{U(\theta)} |\psi \rangle |^2$, i.e. the same experiment repeated for a different input state?
• Can I expect any kind of relationship between $P_\theta(0)$ and $P_\phi (0)\equiv |\langle 0| \mathcal{E}_{U(\phi)^\dagger} \mathcal{E}_{U(\phi)} |0 \rangle |^2$ i.e. the same experiment repeated on a similar unitary $U(\phi)$

I can think of specific counterexamples that show that some of these relationships fail to hold but ideally there would be a way to assess the likelihood of this being a good (or bad!) metric.

Answer 1

Here are my thoughts.

1. I would think the final probability gives the histogram that can help calculate the average fidelity of the circuit. The machine itself does not recognize that it is implementing an identity operator, therefore whatever you get after running the circuit can be used to track the fidelity. (I hope I understand your question correctly.) If you are using some circuit optimizer that can reduce the circuit into identity before running, in the end of course you would get nothing useful to get the fidelity since you will always get 100% fidelity.

2. Since $P_\theta(\psi) = |\langle \psi| \varepsilon_{U(\theta)^\dagger}\varepsilon_{U(\theta)}|\psi\rangle|^2 =|\langle 0| \varepsilon_{U(\phi)^\dagger}\varepsilon_{U(\theta)^\dagger}\varepsilon_{U(\theta)}\varepsilon_{U(\phi)}|0\rangle|^2$, I believe it has the similar behavior as having $|0\rangle$ as input.

3. Generally the longer circuit you run, the lower fidelity you get. So I would expect they have the similar behavior but without the same fidelity, since the actual implementation of the unitary would have different depth of the circuit.

I'd like to know what are the counterexamples that you mentioned.