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 $\mathcal{H}=Z$.

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    $\begingroup$ "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. $\endgroup$ – DaftWullie May 13 at 11:19

If two operators $A$ and $B$ commute then we can always write $e^{i(A+B)t} = e^{iAt}e^{iBt}$, so we don't need to worry about the Trotterization. Otherwise if $A$ and $B$ don't commute, then $e^{i(A+B)t} \ne e^{iAt}e^{iBt}$ and that's why we will need to apply the Trotterization procedure. Both (in)equalities can be proved with the Taylor series.

Now about how to obtain the ground state of the $H=Z$ Hamiltonian.

If we will apply $e^{iHt} = R_z(-2t)$ to an arbitrary state $|\psi\rangle = \alpha |0\rangle + \beta |1\rangle$ we will obtain only some relative phase (desregarding the global phase):

$$R_z(-2t) |\psi\rangle = \alpha |0\rangle + e^{-it}\beta |1\rangle$$

So, by just applying the $R_z(-2t)$ on some fixed state we will not succeed. One way for obtaining the ground state of the Hamiltonian is using the VQE algorithm. Here is the circuit that we will need:


With this circuit, one has a possibility to obtain all one-qubit states in the Bloch sphere, if the initial state is $|0\rangle$. For each given $\theta_1$ and $\theta_2$, the circuit will run $N$ times and we will measure the expectation value of the Hamiltonian $\langle H \rangle = \langle Z \rangle = \frac{N_0 - N_1}{N}$, where $N_0$ is the number of measured $|0\rangle$s and $N_1$ is the number of measured $|1\rangle$s. With some optimization method we will change $\theta$s in order to minimize $\langle Z \rangle$. After the optimization is over (we have found the state for which $\langle Z \rangle$ is minimal: in this case $\langle Z \rangle = -1$ is the minimal value), the circuit with the final $\theta$s can recreate the ground state of the Hamiltonian (the $|1 \rangle$ state, because $\langle 1| Z |1 \rangle = -1$). Note, that I haven't used the circuit for the Hamiltonian simulation $e^{iHt}$.

I hope and I am interested to see an answer that will use Adiabatic state preparation algorithm for the same job.

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    $\begingroup$ Thanks for the answer. The circuit seems like to be an implementation of any single qubit gate (I also infred this from ...we can obtain all one-qubit states on Bloch sphere..). So, I am a little bit confused why you do not used the exponential of matrix $Z$. What about $Rz$ gate? Additionally, what about Hamiltonian $\mathcal{H} = X$? Is the circuit the same but we measure in Hadamard basis? $\endgroup$ – Martin Vesely May 14 at 10:16
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    $\begingroup$ Yes, we can just use only one $u3$ gate. For $H = X$ we can use the same circuit: in fact, we can use it for any $H = a I +bX + cZ + d Y$, like it was implemented in my tutorial with a similar circuit (I used $R_x R_y$). And yes we can do the measurements in $X$ basis for $H = X$, only one Hadamard will be added to the circuit. github.com/DavitKhach/quantum-algorithms-tutorials/blob/master/… $\endgroup$ – Davit Khachatryan May 14 at 10:40
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    $\begingroup$ For VQE we should have the decomposition of the Hamiltonian into Pauli terms because we are interested to find the expectation values of each Pauli $\langle H \rangle = a\langle I \rangle+b\langle X \rangle+c\langle Z \rangle+d\langle Y \rangle$. So, it is like we are doing tomography, but not for all possible Pauli terms (for those which are in the $H$'s decomposition), because it will not be efficient algo if you will need to do full tomography. $\endgroup$ – Davit Khachatryan May 14 at 10:54
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    $\begingroup$ So, we can avoid using the circuit for $e^{iHt}$, but also I guess (my guess is motivated from QAOA approach) for more complicated Hamiltonians it is possible that the $e^{iHt}$ will be helpful :) $\endgroup$ – Davit Khachatryan May 14 at 10:58
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    $\begingroup$ Thanks for help, understand now. $\endgroup$ – Martin Vesely May 14 at 13:42

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