## How to build a circuit for simulation of a simple Hamiltonian?

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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

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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.

1Thanks 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? – Martin Vesely – 2020-05-14T10:16:44.930

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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. https://github.com/DavitKhach/quantum-algorithms-tutorials/blob/master/variational_quantum_eigensolver.ipynb

– Davit Khachatryan – 2020-05-14T10:40:39.300

1For 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. – Davit Khachatryan – 2020-05-14T10:54:17.253

1So, 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 :) – Davit Khachatryan – 2020-05-14T10:58:53.490

1Thanks for help, understand now. – Martin Vesely – 2020-05-14T13:42:34.753