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"how honest is the Chinese claim of quantum supremacy?" It's equally (or at least as) honest in comparison to Google's claim. In a comment to this answer at Quantum Computing Stack Exchange, Craig Gidney (who works at Google and was a co-author on Google's Quantum Supremacy paper), confirmed that the classical computer would have been 2^(20*7/4) = ...


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Quantum computers provide the possibility of simulating systems that are so mathematically complex that classical computers cannot practically be used. The difficulty comes from the fact that the electrical properties of materials, and other chemical systems, are governed by the laws of many body quantum mechanics, which contain equations that are extremely ...


15

How are quantum computers being used in drug design? They are not. Because useful quantum computers do not exist yet. Ask again in a couple decades :)


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Quantum computing has been extensively discussed as long as the first quantum computers have become a reality. As already pointed out, we are still at very early stages of such development. What we need to keep in mind is that a quantum computer needs a quantum computing algorithm. The latter can have few to no correlation with classical algorithms, such as ...


11

Quantum computers potentially could provide a far more efficient way to solve the electronic structure problem: finding the ground state energy (or other properties) of electrons in a compound. The most prominent quantum algorithms for this purpose are the Phase Estimation Algorithm (PEA) and the Variational Quantum Eigensolver (VQE). While PEA offers the ...


9

Realizing this Hamiltonian in a natural material I cannot imagine a material in which all nearest-neighbor spin-spin interactions can be adjusted arbitrarily at the same time. Spin-spin couplings that are stronger when the spins are closer together, and weaker when the spins are farther apart, can be adjusted by moving the spins relative to each other; so ...


8

As you said above, nitrogen has 2 s orbitals and one p orbital. However, one would typically freeze the chemically inactive 1s orbital, leaving you 5 electrons in 4 orbitals per atom, or 10 electrons in 8 orbitals for N$_2$; this is denoted $(10e,8o)$. Going beyond the minimal STO-3G basis, in addition to the occupied molecular orbitals (which are poorly ...


7

The angles $\phi$ and $\theta$ never contain a non-zero imaginary part (they are always purely real numbers). The general solution to Eq. 1 is: $$ \tag{1} |\Phi(t)\rangle = e^{-\textrm{i}H t} | \Phi(0)\rangle. $$ This can be written as: $$ \tag{2} \begin{bmatrix} a(t) \\ b(t) \end{bmatrix} = U \begin{bmatrix} a(0) \\ b(0) \end{bmatrix} . $$ where the ...


5

The idea in the FNO method is to do a MP2 calculation for the one-particle density matrix and diagonalize it to get the natural orbitals (NOs) and natural orbital occupation numbers (NOONs). So, starting from (usually) a Hartree-Fock reference wave function, you figure out which orbitals are the most strongly correlated, and which ones yield only small ...


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