Measuring the Fidelity of a MS Gate

We want to measure the fidelity of a 2 qubit Molmer Sorensen gate.

• MS gate

We want to measure the fidelity of a 2 qubit Molmer Sorensen gate.

1. Calculate the probability of each state:

   • Let $p_{00}$ be the probability of measuring the state $

     00\rangle$.

   • Let $p_{11}$ be the probability of measuring the state $

     11\rangle$.

2. Calculate the standard error:
   • The standard error for a binomial distribution is given by $\sqrt{\frac{p(1-p)}{n}}$, where $p$ is the probability of the state and $n$ is the number of repetitions.

Example Calculation

Assume you measured the following counts for each state:

• $n_{00}$ = number of times $

  00\rangle$ was measured

• $n_{11}$ = number of times $

  11\rangle$ was measured

The probabilities are: $p_{00} = \frac{n_{00}}{625}$ $p_{11} = \frac{n_{11}}{625}$

The standard errors are: $\text{SE}{00} = \sqrt{\frac{p{00}(1 - p_{00})}{625}}$ $\text{SE}{11} = \sqrt{\frac{p{11}(1 - p_{11})}{625}}$

Example with Hypothetical Data

Let’s say you measured $n_{00} = 156$ and $n_{11} = 200$:

$p_{00} = \frac{156}{625} = 0.2496$ $p_{11} = \frac{200}{625} = 0.32$

The standard errors would be: $\text{SE}{00} = \sqrt{\frac{0.2496 \times (1 - 0.2496)}{625}} \approx 0.0175$ $\text{SE}{11} = \sqrt{\frac{0.32 \times (1 - 0.32)}{625}} \approx 0.0188$

Given that the populations sum to 1, the covariance between $p_{00}$ and $p_{11}$ can be calculated as:

$ \text{Cov}(p_{00}, p_{11}) = -\frac{p_{00} p_{11}}{n} $

where $n$ is the total number of measurements (625 in this case).

The error of the combined population $p_{00} + p_{11}$ is then given by:

$ \text{SE}{\text{combined}} = \sqrt{\text{SE}{00}^2 + \text{SE}{11}^2 + 2 \cdot \text{Cov}(p{00}, p_{11})} $

Example Calculation

Using the hypothetical data from the previous example:

• $n_{00} = 156$
• $n_{11} = 200$
• $p_{00} = \frac{156}{625} = 0.2496$
• $p_{11} = \frac{200}{625} = 0.32$
• $\text{SE}_{00} = \sqrt{\frac{0.2496 \times (1 - 0.2496)}{625}} \approx 0.0175$
• $\text{SE}_{11} = \sqrt{\frac{0.32 \times (1 - 0.32)}{625}} \approx 0.0188$
• $\text{Cov}(p_{00}, p_{11}) = -\frac{0.2496 \times 0.32}{625} \approx -0.0001278$

The error of the combined population $p_{00} + p_{11}$ is:

$ \text{SE}_{\text{combined}} = \sqrt{(0.0175)^2 + (0.0188)^2 + 2 \cdot (-0.0001278)} \approx \sqrt{0.00030625 + 0.00035344 - 0.0002556} \approx \sqrt{0.00040409} \approx 0.0201 $