# Unitary interaction term of two-qubit graph state

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.

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