I was skimming through the [Google quantum supremacy paper](https://www.inverse.com/article/59507-full-quantum-supremacy-paper) but got stuck on this section:
> 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.
Questions:
1. If $\mathcal{F}_{XEB} = 1$ when $x_i$ are sampled from the correct distribution then $\left< P(x_i) \right>_i$ must be $2^{1-n}$ (for any quantum circuit). That seems to restrict the output probability distributions of all quantum circuits to rather high entropy distributions. This is not what I would have expected for any kind of computation. Is this correct?
2. If $P(x_i)$ is the probability of bitstring $x_i$ computed for the ideal quantum circuit, then how can $\left< P(x_i) \right>_i$ equal $1/2^n$ when the bitstrings $x_i$ are sampled from the uniform distribution? This is the result we would see if $P(x_i)$ had been $1/2^n$ for all $x_i$ (i.e. the probability mass function for the uniform distribution), but it is explicitly stated that $P(x_i)$ is the probability of bitstring $x_i$ computed for the ideal quantum circuit. Intuitively it seems to me more reasonable to assume that $\left< P(x_i) \right>_i$ with $x_i$ sampled from the uniform distribution would be much closer to zero, because if $P(x_i)$ concentrates much of its probability mass to a relatively low number of bitstrings (as I would assume computation to do) then $P(x_i) = 0$ for almost all bitstrings from the uniform distribution.
3. How can the value of $\mathcal{F}_{XEB}$ correspond to "the probability that no error has occurred while running the circuit"? This sounds so simple/simplistic to me that it's hard to believe.
**UPDATE** 2019-10-23: The article [Quantum supremacy using a programmable superconducting processor](https://www.nature.com/articles/s41586-019-1666-5) 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?