Checkpoint: Estimate Gate Fidelity

This checkpoint ties the module together. You have an RB survival curve measured on a single-qubit device, and your job is the thing benchmarking exists to do: turn that raw data into a gate fidelity.

The data

A randomized benchmarking experiment on one qubit ( $d = 2$ ) reports the average survival probability of $|0\rangle$ as a function of the Clifford sequence length $m$ . The decay follows the standard model

\bar F(m) = A\,p^{\,m} + B,

where $p$ is the per-Clifford depolarizing parameter and $B = 1/d = 0.5$ is the fully-depolarized floor for one qubit. You are handed two clean points on this curve:

\bar F(1) = 0.99, \qquad \bar F(5) = 0.95196.

From the decay to $p$

Subtract the floor $B$ from each point to expose the pure exponential $A p^m$ :

\frac{\bar F(m_2) - B}{\bar F(m_1) - B} = \frac{A\,p^{\,m_2}}{A\,p^{\,m_1}} = p^{\,m_2 - m_1}.

The amplitude $A$ — which carries the SPAM errors — cancels, exactly the robustness RB promises. Solve for $p$ :

p = \left(\frac{\bar F(m_2) - B}{\bar F(m_1) - B}\right)^{\!1/(m_2 - m_1)}.

From $p$ to the fidelity

The per-Clifford error rate for dimension $d$ is

r = \frac{(d-1)(1-p)}{d},

and the average gate fidelity per Clifford is its complement,

F = 1 - r = 1 - \frac{(d-1)(1-p)}{d}.

For one qubit, $d = 2$ , so $r = (1-p)/2$ and $F = (1+p)/2$ .

Try it

Given $\bar F(1) = 0.99$ , $\bar F(5) = 0.95196$ , $B = 0.5$ , and $d = 2$ :

Extract $p$ from the two survival points using the ratio formula.
Convert $p$ to the average gate fidelity $F = (1+p)/2$ .
Return $F$ .

The grader checks your returned fidelity against the value derived from the same formulas (it should come out essentially $0.99$ ).

Run your code to see the quantum state.

The data

From the decay to ppp

From ppp to the fidelity

Try it

From the decay to $p$

From $p$ to the fidelity