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For a given circuit, we collect the measured bit-strings $\{x_i\}$ and compute the linear XEB fidelity [24-26, 29], which is the mean of the simulated probabilities of the bit strings we measured:

 

$$\mathcal{F}_{\text{XEB}} = 2^n\langle P(x_i)\rangle - 1 \tag {1}$$

 

where $n$ is the number of qubits, $P(x_i)$ is the probability of bitstring $x_i$ computed for the ideal quantum circuit, and the average is over observed bitstrings. Intuitively, $\mathcal{F}_\text{XEB}$ is correlated with how often we sample high probability bitstrings. When there are no errors in the quantum circuit, sampling the probability distribution will produce $\mathcal{F}_\text{XEB} = 1$. On the other hand, sampling from the uniform distribution will give $\langle P(x_i)\rangle_i = 1/2^n$ and produce $\mathcal{F}_\text{XEB} = 0$. Values of $\mathcal{F}_{XEB}$ between $0$ and $1$ correspond to the probability that no error has occurred while running the circuit.

For a given circuit, we collect the measured bit-strings $\{x_i\}$ and compute the linear XEB fidelity [24-26, 29], which is the mean of the simulated probabilities of the bit strings we measured:

$$\mathcal{F}_{\text{XEB}} = 2^n\langle P(x_i)\rangle - 1 \tag {1}$$

where $n$ is the number of qubits, $P(x_i)$ is the probability of bitstring $x_i$ computed for the ideal quantum circuit, and the average is over observed bitstrings. Intuitively, $\mathcal{F}_\text{XEB}$ is correlated with how often we sample high probability bitstrings. When there are no errors in the quantum circuit, sampling the probability distribution will produce $\mathcal{F}_\text{XEB} = 1$. On the other hand, sampling from the uniform distribution will give $\langle P(x_i)\rangle_i = 1/2^n$ and produce $\mathcal{F}_\text{XEB} = 0$. Values of $\mathcal{F}_{XEB}$ between $0$ and $1$ correspond to the probability that no error has occurred while running the circuit.

where $n$ is the number of qubits, $P(x_i)$ is the probability of bitstring $x_i$ computed for the ideal quantum circuit, and the average is over observed bitstrings. Intuitively, $\mathcal{F}_\text{XEB}$ is correlated with how often we sample high probability bitstrings. When there are no errors in the quantum circuit, sampling the probability distribution will produce $\mathcal{F}_\text{XEB} = 1$. On the other hand, sampling from the uniform distribution will give $\langle P(x_i)\rangle_i = 1/2^n$ and produce $\mathcal{F}_\text{XEB}$. Values of $\mathcal{F}_{XEB}$ between $0$ and $1$ correspond to the probability that no error has occurred while running the circuit.

where $n$ is the number of qubits, $P(x_i)$ is the probability of bitstring $x_i$ computed for the ideal quantum circuit, and the average is over observed bitstrings. Intuitively, $\mathcal{F}_\text{XEB}$ is correlated with how often we sample high probability bitstrings. When there are no errors in the quantum circuit, sampling the probability distribution will produce $\mathcal{F}_\text{XEB} = 1$. On the other hand, sampling from the uniform distribution will give $\langle P(x_i)\rangle_i = 1/2^n$ and produce $\mathcal{F}_\text{XEB} = 0$. Values of $\mathcal{F}_{XEB}$ between $0$ and $1$ correspond to the probability that no error has occurred while running the circuit.

UPDATE 2019-10-23: The article Quantum supremacy using a programmable superconducting processor has now been published in Nature, and the Supplementary Information is available there.

Having read section IV (XEB Theory) of the Supplemental Information I'd like to adjust my questions as follows:

1. Subsection C (Two limiting cases) derives this fact from the properties of probability distributions of the Porter-Thomas shape. The derivation looks correct to me. The answer here seems to be that my intuition (that the output distribution would be relatively low entropy) was simply wrong.

2. Subsection C also contains this passage: "[Suppose] bitstrings $q_i$ are sampled from the uniform distribution. In this case $P(q_i) = 1/D$ [where $D = 2^n$] for every i and $F_{XEB} = 0$." I see this as very problematic since the main text of the article makes the claim that "$P(x_i)$ is the probability of bitstring $x_i$ computed for the ideal quantum circuit". How can these two statements be reconciled?

3. Subsection V (Quantifying errors) contains a lengthy discussion of this. I can't say I understand it in full, but I'll give it the benefit of a doubt for now.

So, in summary, my question is now: The main article states that "$P(x_i)$ is the probability of bitstring $x_i$ computed for the ideal quantum circuit". However, Supplementary Information section IV.C seems to argue that if the "qubits are in the maximally mixed state" (i.e. the quantum computer doesn't work) then "the estimator [$F_{XEB}$] yields zero fidelity" since $P(x_i) = 1/2^n$ for every $i$ in this case. But then, in this case, $P(x_i)$ is clearly the probability of sampling bitstring $x_i$ from the non-ideal empirical distribution. Isn't this a contradiction?

As I see it either $F_{XEB}$ is computed such that $P(x_i)$ is the probability of $x_i$ being sampled from the ideal quantum circuit, or it is computed such that $P(x_i)$ is the probability of $x_i$ being sampled from the empirical non-ideal distribution. It can't be both. Which is it?