In my reference, Page 259, Quantum Computation and Quantum Information by Nielsen and Chuang it is given that
Given the unitary operator $U(\Delta t)\equiv \exp(-i|\psi\rangle\langle\psi|\Delta t)\exp(-i|x\rangle\langle x|\Delta t)$ where $|x\rangle\langle x|=\dfrac{I+Z}{2}=\dfrac{I+\hat{z}.\vec{\sigma}}{2}$ where $\hat{z}=(0,0,1)$, and $|\psi\rangle\langle\psi|=\dfrac{I+\vec{\psi.\vec{\sigma}}}{2}$ where $\vec{\psi}=(2\alpha\beta,0,\alpha^2-\beta^2)$
Here $X,Y,Z$ are Pauli matrices with $\vec{\sigma}=(X,Y,Z)$ and $\alpha,\beta$ are real numbers, and we can also prove that $\exp(-i|\psi\rangle\langle\psi|\Delta t)=e^{-i\Delta t}|\psi\rangle\langle\psi|$ and $\exp(-i|x\rangle\langle x|\Delta t)=e^{-i\Delta t}|x\rangle\langle x|$, please refer to Hamiltonian Simulation Circuit for Grover's Search.
How do we obtain the expression $$U(\Delta t)=\Big(\cos^2(\Delta t/2)-\sin^2(\Delta t/2)\vec{\psi}.\hat{z}\Big)I\\-2i\sin(\Delta t/2)\bigg[\cos(\Delta t/2)\dfrac{\vec{\psi}+\hat{z}}{2}+\sin(\Delta t/2)\dfrac{\vec{\psi}\times\hat{z}}{2}\bigg].\vec{\sigma}$$ up to an unimportant global phase factor ?
Note: This is the Final Expression for the Unitary Operator for Grover's Search Hamiltonian Simulation
My Attempt
$$ U(\Delta t)=\exp(-i|\psi\rangle\langle\psi|\Delta t).\exp(-i|x\rangle\langle x|\Delta t)=e^{-i\Delta t}|\psi\rangle\langle\psi|.e^{-i\Delta t}|x\rangle\langle x|\\ =e^{-i\Delta t}.e^{-i\Delta t}(\dfrac{I+\vec{\psi}.\vec{\sigma}}{2})(\dfrac{I+\hat{z}.\vec{\sigma}}{2})=\frac{e^{-i\Delta t}}{2}\big(\cos(\Delta t)-i\sin(\Delta t)\big).\frac{1}{2}.\Big(I+(\vec{\psi}+\hat{z}).\vec{\sigma}+(\vec{\psi}.\vec{\sigma})(\hat{z}.\vec{\sigma})\Big)\\ $$ Making use of the identity $(\vec{a}.\vec{\sigma})(\vec{b}.\vec{\sigma})=(\vec{a}.\vec{b})I+i(\vec{a}\times\vec{b}).\vec{\sigma}$ $$ =\frac{e^{-i\Delta t}}{2}\big(\cos(\Delta t)-i\sin(\Delta t)\big).\frac{1}{2}.\Big(I+(\vec{\psi}+\hat{z}).\vec{\sigma}+(\vec{\psi}.\hat{z})I+i(\vec{\psi}\times\hat{z}).\vec{\sigma}\Big)\\ =\frac{e^{-i\Delta t}}{4}.\bigg[\color{red}{\cos(\Delta t)I-i\sin(\Delta t)(\vec{\psi}+\hat{z}).\vec{\sigma}+\cos(\Delta t)(\vec{\psi}.\hat{z})I+\cos(\Delta t)i(\vec{\psi}\times\hat{z}).\vec{\sigma}}-i\sin(\Delta t)I\\+\cos(\Delta t)(\vec{\psi}+\hat{z}).\vec{\sigma}+-i\sin(\Delta t)(\vec{\psi}.\hat{z})I-i\sin(\Delta t).i(\vec{\psi}\times\hat{z}).\vec{\sigma}\bigg]\\ =\frac{e^{-i\Delta t}}{4}.\bigg[\color{red}{(2\cos^2(\Delta t/2)-1)I-i2\sin(\Delta t/2)\cos(\Delta t/2)(\vec{\psi}+\hat{z}).\vec{\sigma}\\+(1-2\sin^2(\Delta t/2))(\vec{\psi}.\hat{z})I+i(1-2\sin^2(\Delta t/2))(\vec{\psi}\times\hat{z}).\vec{\sigma}}-i\sin(\Delta t)I+\cos(\Delta t)(\vec{\psi}+\hat{z}).\vec{\sigma}\\-i\sin(\Delta t)(\vec{\psi}.\hat{z})I+\sin(\Delta t)(\vec{\psi}\times\hat{z}).\vec{\sigma}\bigg]\\ $$ $$ =\frac{e^{-i\Delta t}}{4}.\bigg[\color{red}{2\cos^2(\Delta t/2)I}-I\color{red}{-2i\sin(\Delta t/2)\cos(\Delta t/2)(\vec{\psi}+\hat{z}).\vec{\sigma}}\\+\vec{\psi}.\hat{z}I\color{red}{-2\sin^2(\Delta t/2)\vec{\psi}.\hat{z}}I\color{red}{-2i\sin^2(\Delta t/2)(\vec{\psi}\times\hat{z}).\vec{\sigma}}+(\vec{\psi}\times\hat{z}).\vec{\sigma}-i\sin(\Delta t)I+\cos(\Delta t)(\vec{\psi}+\hat{z}).\vec{\sigma}\\-i\sin(\Delta t)(\vec{\psi}.\hat{z})I+\sin(\Delta t)(\vec{\psi}\times\hat{z}).\vec{\sigma}\bigg]\\ =\frac{e^{-i\Delta t}}{4}.\bigg[\color{red}{2\cos^2(\Delta t/2)I}\color{red}{-2i\sin(\Delta t/2)\cos(\Delta t/2)(\vec{\psi}+\hat{z}).\vec{\sigma}}\\\color{red}{-2\sin^2(\Delta t/2)\vec{\psi}.\hat{z}I}\color{red}{-2i\sin^2(\Delta t/2)(\vec{\psi}\times\hat{z}).\vec{\sigma}}-I+\vec{\psi}.\hat{z}I+(\vec{\psi}\times\hat{z}).\vec{\sigma}-i\sin(\Delta t)I+\cos(\Delta t)(\vec{\psi}+\hat{z}).\vec{\sigma}\\-i\sin(\Delta t)(\vec{\psi}.\hat{z})I+\sin(\Delta t)(\vec{\psi}\times\hat{z}).\vec{\sigma}\bigg]\\ $$
The red coloured terms constitute that of the required expression, but how do I deal with the rest of the terms ?
The problem is with the calculation of $\exp(-i|x\rangle\langle x|\Delta t)$.
$$ (|x\rangle\langle x|)^{2m}=|x\rangle\langle x|=(|x\rangle\langle x|)^{2m+1}\\ $$ \begin{align} &e^{-i|x\rangle\langle x|\Delta t}=\sum_{k=0}^\infty\frac{1}{k!}(-i\Delta t)^k(|x\rangle\langle x|)^k\\ &=\sum_{m=0}^\infty\frac{1}{(2m)!}(-i\Delta t)^{2m}(|x\rangle\langle x|)^{2m}+\sum_{k=0}^\infty\frac{1}{(2m+1)!}(-i\Delta t)^{2m+1}(|x\rangle\langle x|)^{2m+1}\\ &=I+\sum_{m=1}^\infty\frac{(-i)^{2m}}{(2m)!}(\Delta t)^{2m}.|x\rangle\langle x|-\sum_{k=0}^\infty\frac{i(-i)^{2m}}{(2m+1)!}(\Delta t)^{2m+1}.|x\rangle\langle x|\\ &=I+|x\rangle\langle x|(\sum_{m=1}^\infty\frac{(-i)^{2m}}{(2m)!}(\Delta t)^{2m}+1-1)-|x\rangle\langle x|\sum_{k=0}^\infty\frac{i(-i)^{2m}}{(2m+1)!}(\Delta t)^{2m+1}\\ &=I-|x\rangle\langle x|+|x\rangle\langle x|\bigg(\sum_{m=0}^\infty\frac{(-1)^m}{(2m)!}(\Delta t)^{2m}-i\sum_{k=0}^\infty\frac{(-1)^m}{(2m+1)!}(\Delta t)^{2m+1}\bigg)\\ &=I+|x\rangle\langle x|\Big(\cos(\Delta t)-i\sin(\Delta t)-1\Big)=I+(e^{-i\Delta t}-1)|x\rangle\langle x| \end{align}
Therefore,
$\boxed{e^{-i|x\rangle\langle x|\Delta t}=I+(e^{-i\Delta t}-1)|x\rangle\langle x|\quad\&\quad e^{-i|\psi\rangle\langle\psi|\Delta t}=I+(e^{-i\Delta t}-1)|\psi\rangle\langle\psi|}$
\begin{align} &U(\Delta t)=\exp(-i|\psi\rangle\langle\psi|\Delta t).\exp(-i|x\rangle\langle x|\Delta t)\\ &=(I+(e^{-i\Delta t}-1)|\psi\rangle\langle \psi|)(I+(e^{-i\Delta t}-1)|x\rangle\langle x|)\\ &=I+(e^{-i\Delta t}-1)(|\psi\rangle\langle \psi|+|x\rangle\langle x|)+(e^{-i\Delta t}-1)^2|\psi\rangle\langle \psi|x\rangle\langle x|\\ &=I-(1-e^{-i\Delta t})(|\psi\rangle\langle \psi|+|x\rangle\langle x|)+(1-e^{-i\Delta t})^2|\psi\rangle\langle \psi|x\rangle\langle x|\\ &=I-(2\sin^2\Delta /2+i2\sin\Delta t/2\cos\Delta t/2)(|\psi\rangle\langle \psi|+|x\rangle\langle x|)+(2\sin^2\Delta 2+i2\sin\Delta t/2\cos\Delta t/2)^2|\psi\rangle\langle\psi|x\rangle\langle x|\\ &=I-{2i}\sin\Delta t/2(\cos\Delta t/2-i\sin\Delta /2)(|\psi\rangle\langle \psi|+|x\rangle\langle x|)\\ &\quad -4\sin^2\Delta t/2(\cos\Delta t/2-i\sin\Delta /2)^2|\psi\rangle\langle \psi|x\rangle\langle x|\\ &=I-2i\sin\Delta t/2(\cos\Delta t/2-i\sin\Delta t/2)(I+\frac{(\vec{\psi}+\hat{z}).\vec{\sigma})}{2}\\ &\quad -4\sin^2\Delta t/2(\cos\Delta t/2-i\sin\Delta t/2)^2\frac{1}{4}(I+(\vec{\psi}+\hat{z}).\vec{\sigma}+(\vec{\psi}.\vec{\sigma})(\hat{z}.\vec{\sigma}))\\\\ &\text{Making use of the identity} (\vec{a}.\vec{\sigma})(\vec{b}.\vec{\sigma})=(\vec{a}.\vec{b})I+i(\vec{a}\times\vec{b}).\vec{\sigma}\\\\ &=I-2i\sin\Delta t/2(\cos\Delta t/2-i\sin\Delta t/2)(I+\frac{(\vec{\psi}+\hat{z}).\vec{\sigma}}{2})\\ &\quad -\sin^2\Delta t/2(\cos\Delta t/2-i\sin\Delta t/2)^2(I+(\vec{\psi}+\hat{z}).\vec{\sigma}+(\vec{\psi}.\hat{z})I+i(\vec{\psi}\times\hat{z}).\vec{\sigma})\\ &=I-2i\sin\Delta t/2.e^{-i\Delta t/2}(I+\frac{(\vec{\psi}+\hat{z}).\vec{\sigma}}{2})-\sin^2\Delta t/2.e^{-i\Delta t}(I+(\vec{\psi}+\hat{z}).\vec{\sigma}\\ &\quad +(\vec{\psi}.\hat{z})I+i(\vec{\psi}\times\hat{z}).\vec{\sigma})\\ &=e^{-i\Delta t}[e^{i \Delta t}I-2i\sin\Delta t/2.e^{i\Delta t/2}(I+\frac{(\vec{\psi}+\hat{z}).\vec{\sigma}}{2})-\sin^2\Delta t/2(I+(\vec{\psi}+\hat{z}).\vec{\sigma}\\ &\quad+(\vec{\psi}.\hat{z})I+i(\vec{\psi}\times\hat{z}).\vec{\sigma})]\\ &=e^{-i\Delta t}[(\cos\Delta t/2+i\sin\Delta t/2)e^{i\Delta t/2}I-2i\sin\Delta t/2.e^{i\Delta t/2}I-i\sin\Delta t/2.e^{i\Delta t/2}(\vec{\psi}+\hat{z}).\vec{\sigma}\\ &\quad -\sin^2\Delta t/2(I+(\vec{\psi}+\hat{z}).\vec{\sigma}+(\vec{\psi}.\hat{z})I+i(\vec{\psi}\times\hat{z}).\vec{\sigma})]\\ &=e^{-i\Delta t}[\cos\Delta t/2.e^{i\Delta t/2}I-i\sin\Delta t/2.e^{i\Delta t/2}I-i\sin\Delta t/2.e^{i\Delta t/2}(\vec{\psi}+\hat{z}).\vec{\sigma}\\ &\quad -\sin^2\Delta t/2(I+(\vec{\psi}+\hat{z}).\vec{\sigma}+(\vec{\psi}.\hat{z})I+i(\vec{\psi}\times\hat{z}).\vec{\sigma})]\\ &=e^{-i\Delta t}[e^{i\Delta t/2}.e^{-i\Delta t/2}I-i\sin\Delta t/2.e^{i\Delta t/2}(\vec{\psi}+\hat{z}).\vec{\sigma}-\sin^2\Delta t/2I\\ &\quad -\sin^2\Delta t/2(\vec{\psi}+\hat{z}).\vec{\sigma}-\sin^2\Delta t/2(\vec{\psi}.\hat{z})I-i\sin^2\Delta t/2(\vec{\psi}\times\hat{z}).\vec{\sigma}]\\ \\ &\text{Substituting } \sin^2\Delta t/2=1-\cos^2\Delta t/2\text{,}\\ \\ &=e^{-i\Delta t}[I-i\sin\Delta t/2.(\cos\Delta t/2+i\sin\Delta t/2)(\vec{\psi}+\hat{z}).\vec{\sigma}-(1-\cos^2\Delta t/2)I\\ &\quad -\sin^2\Delta t/2(\vec{\psi}+\hat{z}).\vec{\sigma}-\sin^2\Delta t/2(\vec{\psi}.\hat{z})I-i\sin^2\Delta t/2(\vec{\psi}\times\hat{z}).\vec{\sigma}]\\ &=e^{-i\Delta t}[I-i\sin(\Delta t/2).\cos(\Delta t/2)(\vec{\psi}+\hat{z}).\vec{\sigma}\color{blue}{+\sin^2(\Delta t/2)(\vec{\psi}+\hat{z}).\vec{\sigma}}-I+\cos^2(\Delta t/2)I\\ &\quad \color{blue}{-\sin^2(\Delta t/2)(\vec{\psi}+\hat{z}).\vec{\sigma}}-\sin^2(\Delta t/2)(\vec{\psi}.\hat{z})I-i\sin^2(\Delta t/2)(\vec{\psi}\times\hat{z}).\vec{\sigma}]\\ &=e^{-i\Delta t}[-i\sin(\Delta t/2).\cos(\Delta t/2)(\vec{\psi}+\hat{z}).\vec{\sigma}+\cos^2(\Delta t/2)I-\sin^2(\Delta t/2)(\vec{\psi}.\hat{z})I\\ &\quad -i\sin^2(\Delta t/2)(\vec{\psi}\times\hat{z}).\vec{\sigma}]\\\\ &\text{Rearranging the terms,}\\\\ &=e^{-i\Delta t}[\cos^2(\Delta t/2)I-\sin^2(\Delta t/2)(\vec{\psi}.\hat{z})I-i\sin(\Delta t/2).\cos(\Delta t/2)(\vec{\psi}+\hat{z}).\vec{\sigma}\\ &\quad -i\sin^2(\Delta t/2)(\vec{\psi}\times\hat{z}).\vec{\sigma}]\\ &\color{red}{=e^{-i\Delta t}\bigg[\Big(\cos^2(\Delta t/2)-\sin^2(\Delta t/2)(\vec{\psi}.\hat{z})\Big)I}\\ &\quad \quad \quad \quad \color{red}{-2i\sin(\Delta t/2)\Big(\cos(\Delta t/2)\frac{(\vec{\psi}+\hat{z})}{2}+\sin(\Delta t/2)\frac{(\vec{\psi}\times\hat{z})}{2}\Big).\vec{\sigma}\bigg]}\\ \end{align}