Use the Pauli matrices \sigma _{x} =\left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right) , \sigma _{y} =\left(\begin{matrix} 0 & -i \\ i & 0 \end{matrix} \right), and \sigma _{z} =\left(\begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right), to show that
(a) e^{-i\alpha \sigma _{x}} =I\cos \alpha -i\sigma _{x}\sin \alpha , where I is the unit matrix,
(b) e^{i\alpha \sigma _{x}}\sigma _{z}e^{-i\alpha \sigma _{x}} =\sigma _{z}\cos (2\alpha )+\sigma _{y} \sin (2\alpha ).
(a) Using the expansion
e^{-i\alpha \sigma _{x}}=\sum\limits_{n=0}^{\infty } \frac{(-i)^{2n} }{(2n)!} (\alpha )^{2n} \sigma ^{2n}_{x} + \sum \limits_{n=0}^{\infty } \frac{(-i)^{2n+1} }{(2n+1)!} (\alpha )^{2n+1} \sigma ^{2n+1}_{x} , (7.346)
and since \sigma ^{2}_{x} =1, \sigma ^{2n}_{x}=I , and \sigma ^{2n+1}_{x} =\sigma _{x} , where I is the unit matrix, we have
e^{-i\alpha \sigma _{x}}=\left(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) \sum\limits_{n=0}^{\infty } \frac{(-1)^{n} }{(2n)!} (\alpha )^{2n} -i\sigma _{x} \sum\limits_{n=0}^{\infty } \frac{(-1)^{n} }{(2n+1)!} (\alpha )^{2n+1}
=I\cos \alpha -i\sigma _{x}\sin \alpha . (7.347)
(b) From (7.347) we can write
e^{i\alpha \sigma _{x}} \sigma _{z} e^{-i\alpha \sigma _{x}} =(\cos \alpha +i\sigma _{x} \sin \alpha )\sigma _{z} (\cos \alpha -i\sigma _{x} \sin \alpha )
=\sigma _{z} \cos ^{2} \alpha +\sigma _{x} \sigma _{z} \sigma _{x} \sin ^{2} \alpha +i[\sigma _{x}, \sigma _{z}] \sin \alpha \cos \alpha , (7.348)
which, when using the facts that \sigma _{x} \sigma _{z}=-\sigma _{z} \sigma _{x},\sigma ^{2}_{x} =I, and [\sigma _{x}, \sigma _{z}]=-2i\sigma _{y}, reduces to
e^{i\alpha \sigma _{x}} \sigma _{z} e^{-i\alpha \sigma _{x}} =\sigma _{z}\cos ^{2} \alpha -\sigma _{z}\sigma ^{2}_{x}\sin ^{2} \alpha +2\sigma _{y}\sin \alpha \cos \alpha=\sigma _{z}(\cos ^{2} \alpha -\sin ^{2} \alpha )+\sigma _{y}\sin (2\alpha )
=\sigma _{z}\cos (2\alpha )+\sigma _{y}\sin (2\alpha ) . (7.349)