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Probably i found an additional solution: $$e^{-itH}=e^{-it(|x\rangle\langle \psi|+|\psi\rangle\langle x|)}=e^{-it(\alpha I+\sqrt{1-\alpha^2}X+\alpha Z)}=e^{-it\alpha}e^{-it(\sqrt{1-\alpha^2}X+\alpha Z)}$$ where we put the calculation basis through $|x\rangle$ and $|\psi\rangle$ the last exponent express the rotation on the Bloch sphere around vector $(\... 2 I assume this is not what was intended, but here's one method: You already know that you can approximately make$H'=2i\alpha(|x\rangle\langle\psi|-|\psi\rangle\langle x|)$by using products of terms such as$e^{i\delta|x\rangle\langle x|}$and$e^{i\delta|\psi\rangle\langle \psi|}$. So, we could simply convert$H'$into$H$:$\$ e^{i\pi/2|\psi\rangle\langle \...