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Consider the controlled phase gate $$U_{ab}(\varphi_{ab}) := e^{-i \varphi_{ab}H_{ab}}~~~~\text{where}~~~~H_{ab} := |1 \rangle^{a} \langle 1 | \otimes |1 \rangle^{b} \langle 1 |$$ is the two-qubit interaction. We can show that $$H_{ab} = \frac{1}{4}( I_{ab}-\sigma_{z}^{a}-\sigma_{z}^{b}+H_{ab}^{I})$$ where $H^{I}_{ab} = e^{-i \varphi_{ab} \sigma^a_z \sigma^b_z}$ and therefore $$U_{ab}(\varphi_{ab}) = e^{-\frac{i\varphi_{ab}}{4}}e^{\frac{i\varphi_{ab}}{4} \sigma^a_z}e^{\frac{i\varphi_{ab}}{4}\sigma^b_z}e^{-i \varphi_{ab} \sigma^a_z \sigma^b_z}.$$ Is it clear how it follows then that for $\varphi_{ab}= \pi$ we have $$U_{ab}(\pi) = |0\rangle^{a} \langle 0 | \otimes I^b + |1\rangle^{a} \langle 1 | \otimes \sigma^b_z,$$ where $I$ is the identity, as stated in Eq. (22) on page 13 of the paper "Entanglement in Graph States and its Applications (2006)" by M.Hein. Thanks for any assistance.

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Absolutely! Think about the eigenstates of $H_{ab}$: $|11\rangle$ has eigenvalue 1, while $|00\rangle, |01\rangle$ and $|10\rangle$ have eigenvalue 0. This tells us that the time evolution must be $$ U_{ab}(\varphi)=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & e^{-i\varphi} \end{bmatrix}. $$ When $\varphi=\pi$, you can easily check that this is what you want.

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  • $\begingroup$ Thanks for your quick response. I think I understand your reasoning, but should the last matrix entry not be $e^{-i \varphi}$? $\endgroup$
    – John Doe
    Apr 23, 2021 at 12:30
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    $\begingroup$ Yes, it should. $\endgroup$
    – DaftWullie
    Apr 23, 2021 at 12:52

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