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There are plenty of different variants, particularly with regards to the conditions on the Hamiltonian. It's a bit of a game, for example, to try and find the simplest possible class of Hamiltonians for which simulation is still BQP-complete. The statement will roughly be along the lines of: let $|\psi\rangle$ be a (normalised) product state, $H$ be a ...

11

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 ...

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One way order to perform Z rotations by arbitrary angles is to approximate them with a sequence of Hadamard and T gates. If you need the approximation to have maximum error $\epsilon$, there are known constructions that do this using roughly $3 \lg \frac{1}{\epsilon}$ T gates. See "Optimal ancilla-free Clifford+T approximation of z-rotations" by Ross et al. ...

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An approach for Hamiltonian simulation: Any Hermitian (Hamiltonian) matrix $H$ can be decomposed by the sum of Pauli products with real coefficients (see this thread). An example of 3 qubit Hamiltonian: $$H = 11 \sigma_z \otimes \sigma_z + 7 \sigma_z \otimes \sigma_x - 5\sigma_z \otimes \sigma_x \otimes \sigma_y$$ The final circuit for $e^{iHt}$ can be ...

8

Your question remains very unclear as to what it actually is that you want to calculate. There is no direct correspondence between a system Hamiltonian and the quantum state of the system. No matter what the Hamiltonian, any quantum state is a valid state of the system. Where a Hamiltonian comes in useful is, if you know the state at some time (say, $t=0$),...

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In each of the examples you mentioned, the task breaks very roughly down into two steps: finding a Hamiltonian that describes the problem in terms of qubits, and finding the ground state energy of that Hamiltonian. From that perspective, the Jordan–Wigner transform is a way to find a qubit Hamiltonian corresponding to a given fermionic Hamiltonian. Once you ...

7

The insight that suggests that sparse matrices are useful goes along the lines of: for any $H$, we can decompose it in terms of a set of $H_i$ whose individual components all commute (making diagonalisation straightforward), $$H=\sum_{i=1}^mH_i.$$ If the matrix is sparse, then you shouldn't need too many distinct $H_i$. Then you can simulate the ...

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Reformulating your question: How to perform Hamiltonian Simulation for a generic square matrix $A$? Quick answer: it is not possible. The goal of Hamiltonian Simulation (HS) is to find a quantum circuit (i.e. a succession of gates) that acts like $U(t) = e^{-iAt}$ on a quantum state. Here $U(t)$ needs to be unitary (because of the properties of quantum ...

6

Generally speaking, a realization of a quantum gate involves coherent manipulation of a two-level system (but this is nothing new to you, maybe). For example, you can use two long-lived electronic states in a trapped atom (neutral or ionized in vacuo) and use an applied electric field to implement single-qubit operations (see trapped ions or optical lattices,...

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Yes, in this special case the circuit will simplify as you suggest. The advantage of the circuit that was given is that it generalises more easily, and works for any $H$ which has $\pm 1$ eigenvalues. Here's a general form of the circuit for your reference: This essentially comes down to an issue of how you can reversibly compute a one-bit function \$f:\{0,1\...

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Given a simple quantum system, how do I derive its Hamiltonian? For quantum systems of continuous variables, the most common way to construct the Hamiltonian is to add the kinetic energy and potential energy, as described in this resource. The kinetic energy part is explained here, and various potential energy models are given here: which unfortunately I ...

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