Decomposition of 2-qubit Hamiltonian into standard gate set for QAOA

Question

I try to decompose ansatz into gate set in order to create a circuit in qiskit for QAOA algorithm.
I don't understand how represent  parametrized 2 qubit ansatz as circuit.
$ H{_B} = sum_{j=1}^{n} {sigma_j^x} $
$ H_{A} = frac{1}{2}sigma_z^1 + frac{1}{2}sigma_z^1otimessigma_z^2 $
Ansatz for p=1
$ left| gamma_1,beta_1 rightrangle = e^{-ibeta_1H_B} e^{-igamma_1H_A} left| ++ rightrangle$
It is clear how a circuit for $ e^{-ibeta_1H_B} $ looks like, but I stuck in decomposing $ e^{-igamma_1H_A} $(more precisely it second term) into  parametrised circuit acting on both qubits and depends on $gamma_1$
Any help would be appreciated as well as any insight on multiple qubit decomposition.

KAJ226 · Accepted Answer

Since $sigma_z^1 = I otimes Z$ and $sigma_z^1 otimes sigma_z^2 = Z otimes Z$ are commute with one another, that is
$$ [sigma_z^1 , sigma_z^1 otimes sigma_z^2 ] = sigma_z^1 cdot sigma_z^1 otimes sigma_z^2 - sigma_z^1 otimes sigma_z^2 cdot sigma_z^1 = boldsymbol{0} $$
we have that
$$ e^{igamma_1 H_a} = e^{i gamma frac{1}{2}(sigma_z^1  + sigma_z^1 otimes sigma_z^2 ) } = e^{i gamma frac{1}{2}sigma_z^1  } e^{i gamma frac{1}{2}sigma_z^1 otimes sigma_z^2  }  $$
and now note that $e^{i gamma frac{1}{2}sigma_z^1  }$ has circuit construction as: (look here and here page 7 and 8)

and similarly, $ e^{i gamma frac{1}{2}sigma_z^1 otimes sigma_z^2  } $ have the circuit construction as:

and put them together, we have the circuit construction for $e^{i gamma frac{1}{2}(sigma_z^1  + sigma_z^1 otimes sigma_z^2 ) }$ as:

Decomposition of 2-qubit Hamiltonian into standard gate set for QAOA

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