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8

A conventional Hamiltonian is Hermitian. Hence, if it contains a non-Hermitian term, it must either also contain its Hermitian conjuagte as another term, or have 0 weight. In this particular case, since $Z\otimes X\otimes Y$ is Hermitian itself, the coefficient would have to be 0. So, if you're talking about conventional Hamiltonians, you've probably made a ...

6

For most functions $f(x)$, there is nothing better than calculating all the values. After all, for most functions, there is no better way of defining the function than giving its truth table. Probably, you want to talk about the relatively small fraction of cases in which the function $f(x)$ has some reasonably compact description. In that case, you should ...

4

Yes, you can decompose any Hamiltonian. For VQE purposes, any finite-dimensional Hamiltonian can be presented as a sum of terms which consist of tensor products of Pauli matrices (https://arxiv.org/abs/1304.3061):  H = \sum_{\alpha, i} h^{\alpha}_{i} \sigma^{\alpha}_{i} + \sum_{\alpha, \beta, i, j} h^{\alpha \beta}_{ij} \sigma^{\alpha}_{i} \sigma^{\beta}...

3

It's likely that this algorithm hasn't yet been implemented in QISKit / Q# / pyquil etc. It's also important to note that you would not be discovering new Mersenne primes with a quantum computer - the paper referenced says: We propose a quantum circuit that creates a pure state corresponding to the quantum superposition of all prime numbers less than $2^n$...

1

No. The computational basis is not necessarily the basis that diagonalizes the Hamiltonian. It also looks like you are confusing the X basis with the basis of the Hamiltonian. Advice: You should write it as $H = \sum \lambda_i | \lambda_i \rangle \langle \lambda_i |$ not with $x$ so you don't confuse with the X operator. The right notation will help you ...

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