Goal of Quantum Advantage Experiment: make a real-world quantum computer to do some task that is believed practically impossible for any classical computer
any task means doesn't have to be useful task
Google's Team: Oct. 2019, simulated 53-qubit (^ \times 9 = 54 but one qubit broke), 1000-gate quantum circuit.
Why is hard to build quantum computer: shielding from outside noise (since they become "trash" if outside noise get entangled with qubits)
It is not hard, though, to prevent accidental "measurement" because... there is no such thing called "measurement", all it really happen is entangle.
Hardness: - implementing 0-qubit "do nothing" gate is hard - implementing 1-qubit gate is not much harder - implementing 2-qubit gates is very hard
A typical setup is to have grids of qubit, and do pair-wise operation like reduction.
Today's computer has gate failure around p \llless 2^{-64} which is not a problem at all.
John von Neumann: universally \exists p_c > 0 such that the failure probability of a single classical gate p < \p_c \implies \text{can make any m-gate circuit into a fault-tolerant version with } O(m \log m) \text{ gates}
Lower bound is also \Omega(m \log m). The exact value of p_c depends on gate set, noise model, fault-tolerance scheme.
Quantum Fault-Tolerance Theorem: universally \exists p_q > 0 such that the failure probability of a single quantum gate p < \p_q \implies \text{can make any m-gate circuit into a fault-tolerant version with } O(m \text{poly}(\log m)) \text{ gates} where p_q \simeq 10^{-6}
However, 10^{-6} is very very challenging to achieve. With the fanciest, not a all realistic algorithm (ie. think about having 100+ extra physical qubits packed near one logical qubit for that gate to have such probability), we can only reach 10^{-3}, 10^{-2}
Steps:
However, Google expected P_Q is no where close to P_G, but P_C will perform even worse.
Arbitrary Unitrary: a arbitrary unitrary operation would likely to generate a heavy tail distribution on the output f(x) = C \cdot e^{-t}/N where x \in \{0, 1\}^n is string and N = 2^n is the number of possible strings.
Median x: P_G(x) \simeq \frac{\ln 2}{N}
Expected x: E_{x \sim P_G}[P_G(x)] = \frac{2}{N}
Declear Advantage
Some Criticism:
IBM claimed they can compute all 2^{53} values of P_G(x) in a few days, but they never did it
Classical computer may be able to do better than \frac{1}{N}
Several follow-up paper gave faster algorithm for computing P_G(x) for similar but slightly simpler circuit that models G.
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