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

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.

• This methodology will classify the identity operation as being a high fidelity controlled-absolutely-anything-at-all. Jan 23, 2021 at 0:38
• right, but if I get some low value of $P(0)$ shouldn't that also tell me that $U$ ran with low fidelity? Jan 23, 2021 at 0:51

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.