Georgi Lie Algebras Chapter 2 Solutions

==Problem 2A==
Find all components of matrix $e^{i\alpha A}$ where $A$ is $$\begin{equation} A = \begin{pmatrix} 0&&0&&1 \\ 0&&0&&0 \\ 1&&0&&0 \end{pmatrix} \end{equation}$$

==Solution==
Simply exp will yield, $$\begin{equation} \left(\begin{array}{rrr} \frac{1}{2} \, {\left(e^{\left(2 i \, y\right)} + 1\right)} e^{\left(-i \, y\right)} & 0 & \frac{1}{2} \, {\left(e^{\left(2 i \, y\right)} - 1\right)} e^{\left(-i \, y\right)} \\ 0 & 1 & 0 \\ \frac{1}{2} \, {\left(e^{\left(2 i \, y\right)} - 1\right)} e^{\left(-i \, y\right)} & 0 & \frac{1}{2} \, {\left(e^{\left(2 i \, y\right)} + 1\right)} e^{\left(-i \, y\right)} \end{array}\right) \end{equation}$$ Then applying Demovire's theorem, one gets, $$\begin{equation} \left(\begin{array}{rrr} \frac{1}{2} \, {\left(\cos\left(2 \, y\right) + i \, \sin\left(2 \, y\right) + 1\right)} {\left(\cos\left(y\right) - i \, \sin\left(y\right)\right)} & 0 & \frac{1}{2} \, {\left(\cos\left(2 \, y\right) + i \, \sin\left(2 \, y\right) - 1\right)} {\left(\cos\left(y\right) - i \, \sin\left(y\right)\right)} \\ 0 & 1 & 0 \\ \frac{1}{2} \, {\left(\cos\left(2 \, y\right) + i \, \sin\left(2 \, y\right) - 1\right)} {\left(\cos\left(y\right) - i \, \sin\left(y\right)\right)} & 0 & \frac{1}{2} \, {\left(\cos\left(2 \, y\right) + i \, \sin\left(2 \, y\right) + 1\right)} {\left(\cos\left(y\right) - i \, \sin\left(y\right)\right)} \end{array}\right) \end{equation}$$

==Problem 2B==
If $[A,B]=B$, calculate $$\begin{equation} e^{i\alpha A}B e^{-i\alpha A} \end{equation}$$ Using equation $2.44$, and setting $Y=-Z$, we get $$\begin{align*} RHS=X-i[-Z,X]-\frac{1}{2}[-Z,[-Z,X]]+\cdots \\ &&=&& X-i[X,Z]-1/2[[X,Z],Z]+\cdots \end{align*}$$ Applying this to equation in our problem, $$\begin{align*} e^{i\alpha A} B e^{-i \alpha A} = B - i\alpha[B,A]-\frac{1}{2}\alpha^2[[B,A],A]+\cdots \\ = B + i\alpha B - \frac{1}{2}\alpha^2[B,A]+\cdots \\ = B + i \alpha B - \frac{1}{2} \alpha^2B + \cdots \\ = B e^{-i \alpha} \end{align*}$$