IBM Qiskit QAOA gate implementation question

Question

In section $5.2$ of the QAOA chapter in Qiskit textbook, section $5.2$, state preparation uses the gate $U_{k,l}(\gamma) = e^{\frac{i \gamma}{2} (1-Z_k Z_l)}$. Later, in section $5.3$, this gate is given in terms of basic gates as $U_{k,l}(\gamma)=C_{u1}(−2 \gamma)_{k,l}u_1(\gamma)_k u_1(\gamma)_l$. Why is this the correct implementation?

Answer 1

Qiskit gates $u_1$ and $C_{u1}$ are

$$ u_1(\theta) = \begin{pmatrix} 1 & 0 \\ 0 & e^{i\theta} \end{pmatrix} \,\,\, C_{u1}(\theta) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & e^{i\theta} \end{pmatrix} $$

see here and here. Exponential of a normal matrix $A$ is the matrix $e^A$ with the same eigenbasis as $A$ and eigenvalues that are exponentials of the eigenvalues of $A$. Therefore, $u_1(\theta) = e^{\frac{i\theta}{2}(I-Z)}$. Also, note that

$$ I - Z_1 - Z_2 + Z_1Z_2 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 4 \end{pmatrix}, $$

so $C_{u1}(\theta) = e^{\frac{i\theta}{4}(I - Z_1 - Z_2 + Z_1Z_2)}$. Finally, diagonal matrices commute, so

$$ U_{k,l}(\gamma) = e^{\frac{i \gamma}{2}(I-Z_k Z_l)} = e^{-\frac{i\gamma}{2}(I-Z_k-Z_l+Z_k Z_l)}e^{\frac{i\gamma}{2}(I-Z_k)}e^{\frac{i\gamma}{2}(I-Z_l)} = C_{u1}(-2\gamma)_{k,l} u_1(\gamma)_k u_1(\gamma)_l. $$