# Understanding the heavy output problem [closed]

In this paper (or more pedagogically here) it is given a way to compute the Quantum Volume of a quantum computer (qc), specifically aimed at the IBM qcs.

In all this process I found some dark spots that I would like to understand. First of all we need the heavy outputs $$H_U$$ which can be obtained from a "classical" computation. Obviously in the implementation the transpiler will do some optimizations so from an original $$U$$ ($$m$$ qubits and depth $$d$$) we will get the transpiled $$U'$$. The probability of sampling a heavy output with this $$U'$$ is strangely

$$h_U = \sum_{x\in H_U} q_U(x).$$

1. Ok I buy that this quantity works, but why to sum it over all possibles $$x\in H_U$$ is still unclear to me.

2. The probability of observing a heavy output is then $$h_d = \int h_UdU$$. In the paper mentioned the Algorithm 1 computes $$h_d$$, right?

3. As far as I understand, this algorithm simply constructs the frequency plot between a realization of $$U$$ vs the number of heavy outputs found in a certain number of trial initial state $$|00\dots 0\rangle$$ ($$m$$-spins), right?

4. Then if 3 is correct, $$h_d$$ can be computed from the (normalized) histogram just constructed. However the authors give the formula

$$\frac{n_h-2\sqrt{n_h(n_s-n_h/n_c)}}{n_c n_s}>2/3,$$

to validate if $$h_d$$ has been achieved. How is this formula found? Is it some numerical (e.g. Simpson) rule?

1. On the other, one can measure how well the $$U'$$ resembles $$U$$ by computing the fidelity $$F_{\mathrm{avg}}(U,U')$$. Where does this quantity enters for computing $$V_Q$$? $$\log_2 V_Q =\mathrm{argmax}_m \; \mathrm{min}(m,d(m))$$ doesn't tell much...
• these are good questions but, as it stands, the post is way too broad. The stackexchange format is best-suited for laser-focused questions. Feel free to edit your post to focus it on a specific question. You can open different posts to ask different questions – glS Apr 21 at 10:41
• Please ask just one question. – user1271772 Apr 22 at 11:37