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Consider very simple Hamiltonian $\mathcal{H} = Z = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}$. It has eigenvalues 1 and -1 with coresponding eigenstates $|0\rangle$ and $|1\rangle$, respectively. Hence, a ground state is $|1\rangle$.

For Hamiltonian simulation we need to construct a gate $U(t) = \mathrm{e}^{-i\mathcal{H} t}$. For our $\mathcal{H}$, this gate is $U(t) = Rz(-2t)$ gate.

To simulate the Hamiltonian we apply gate $U(\Delta t)$ several times to get from state $|\psi_0\rangle$ to state $|\psi_t\rangle$ where number of steps (or application of the Hamiltonian) is $t/\Delta t$. This is called Trotter method.

Since our gate $U$ is $z$ rotation which is additive, i.e. $Rz(\alpha)Rz(\beta) = Rz(\alpha+\beta)$, we do not have to bother about steps $\Delta t$ and simply apply $Rz(-2t)$.

I tried to apply $Rz$ gate on some states generated by Hadamard gate and $Ry$ gate with different angle $\theta$ (to have states in different superpositions) and then measure the outcome. I would expect that measured state should be ground state of Hamiltonian. But this was not the case. Probably I am missing something.

**So my question is how to build a circuit for finding the ground state of the Hamiltonian?** I would appreciate if you could provide a circuit for finding ground state of $\mathcal{H}=Z$.

1"I would expect that measured state should be ground state of Hamiltonian." Why would you expect this? If you evolve under a unitary, the weight of initial and final states being in a particular eigenvector are equal. – DaftWullie – 2020-05-13T11:19:01.943