Exercise 2.35
Suppose that \(\vec{v}\) is a 3-dimensional unit vector, \(\theta\) is real number, and we can use Taylor expansion to calculate \(\exp (i\theta \vec{v}\cdot\vec{\sigma})\). The Taylor expansion of \(e^{A}\) is given by
(1)\[
e^{A} = \sum_{n=0}^{\infty}\frac{A^n}{n!} = I + A + \frac{A^2}{2!} + \dotsc
\]
where \(A\) is a operator. Meanwhile, the Taylor expansion of \(\sin A\) and \(\cos A\) are given by
(2)\[\begin{split}
\begin{align}
\sin A &= \sum^{\infty}_{n=0}\frac{(-1)^{n}}{(2n+1)!}A^{2n+1} = A - \frac{A^3}{3!} + \frac{A^5}{5!} - \dotsc\\
\cos A &= \sum^{\infty}_{n=0}\frac{(-1)^{n}}{(2n)!}A^{2n} = I - \frac{A^2}{2!} + \frac{A^4}{4!} - \dotsc
\end{align}
\end{split}\]
According to eq. (1), the Taylor expansion for \(\exp (i\theta \vec{v}\cdot\vec{\sigma})\) is given by
(3)\[\begin{split}
\begin{align}
\exp (i\theta \vec{v}\cdot\vec{\sigma}) =& \sum_{n=0}^{\infty}\frac{(i\theta \vec{v}\cdot\vec{\sigma})^n}{n!} = I + i\theta \vec{v}\cdot\vec{\sigma}+\frac{i^2\theta^2(\vec{v}\cdot\vec{\sigma})^2}{2!} \\
&+ \frac{i^3\theta^3(\vec{v}\cdot\vec{\sigma})^3}{3!}
+ \frac{i^4\theta^4(\vec{v}\cdot\vec{\sigma})^4}{4!} + \dotsc \\
=& \sum_{n=0}^{\infty}\frac{(i\theta \vec{v}\cdot\vec{\sigma})^n}{n!} = I + i\theta \vec{v}\cdot\vec{\sigma}-\frac{\theta^2(\vec{v}\cdot\vec{\sigma})^2}{2!} \\
&- \frac{i\theta^3(\vec{v}\cdot\vec{\sigma})^3}{3!}
+ \frac{\theta^4(\vec{v}\cdot\vec{\sigma})^4}{4!} + \dotsc
\end{align}
\end{split}\]
From eq. (3) we can find that,
If \(n\) is even number we have
(4)\[
n\text{ is even: }I -\frac{A^2}{2!} + \frac{A^4}{4!} - \frac{A^6}{6!} + \frac{A^8}{8!} - \frac{A^{10}}{10!}\dotsc
\]
where \(A = \theta(\vec{v}\cdot\vec{\sigma})\)
If \(n\) is odd number we have
(5)\[
n\text{ is odd: }iA -\frac{iA^3}{3!} + \frac{iA^5}{5!} - \frac{iA^7}{7!} + \frac{iA^9}{9!} - \frac{iA^{11}}{11!}\dotsc
\]
where \(A = \theta(\vec{v}\cdot\vec{\sigma})\).
According to eq. (2), eq. (4) and eq. (5), we could rewrite eq. (3) as
(6)\[
\exp (i\theta \vec{v}\cdot\vec{\sigma}) = \cos {[\theta(\vec{v}\cdot\vec{\sigma})]} + i(\vec{v}\cdot\vec{\sigma}) \sin {[\theta(\vec{v}\cdot\vec{\sigma})]}
\]
We could simplify eq. (6) by following properties of \(\vec{v}\cdot\vec{\sigma}\). Note that for \(\vec{v}\cdot\vec{\sigma}\), we have
(7)\[\begin{split}
\begin{align}
\vec{v}\cdot\vec{\sigma} &= v_{x}\sigma_x + v_y\sigma_y + v_z\sigma_z \\
&= \begin{pmatrix}
0 & v_x\\ v_x &0
\end{pmatrix} + i\begin{pmatrix}
0 & -v_y\\ v_y &0
\end{pmatrix}+ \begin{pmatrix}
v_z & 0\\ 0 &-v_z
\end{pmatrix} \\
&= \begin{pmatrix}
v_z & v_x-iv_y\\ v_x+iv_y &-v_z
\end{pmatrix}
\end{align}
\end{split}\]
So for \((\vec{v}\cdot\vec{\sigma})^2\), we have
(8)\[\begin{split}
\begin{align}
(\vec{v}\cdot\vec{\sigma})^2 &= \begin{pmatrix}
v_z & v_x-iv_y\\ v_x+iv_y &-v_z
\end{pmatrix}\begin{pmatrix}
v_z & v_x-iv_y\\ v_x+iv_y &-v_z
\end{pmatrix} \\
&= \begin{pmatrix}
v^2_z+(v_x-iv_y)(v_x+iv_y) & v_z(v_x-iv_y)-v_z(v_x-iv_y)\\
v_z(v_x+iv_y)-v_z(v_x+iv_y) & (v_x-iv_y)(v_x+iv_y) + v^2_z
\end{pmatrix} \\
&= \begin{pmatrix}
v^2_z+v^2_x+v^2_y & 0\\
0 & v^2_x+v^2_y + v^2_z
\end{pmatrix} = \begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix} = I
\end{align}
\end{split}\]
where we use \(v^2_x+v^2_y+v^2_z = 1\) since \(\vec{v}\) is a unit vector. From eq. (7) and eq. (8), we can re-write eq. (4) and eq. (5) as
(9)\[\begin{split}
\begin{align}
n\text{ is even: }& I -I\frac{\theta^2}{2!} + I\frac{\theta^4}{4!} - I\frac{\theta^6}{6!} + I\frac{\theta^8}{8!} - I\frac{\theta^{10}}{10!}\dotsc \\
n\text{ is odd: }&i\theta(\vec{v}\cdot\vec{\sigma}) -(\vec{v}\cdot\vec{\sigma})\frac{i\theta^3}{3!} + (\vec{v}\cdot\vec{\sigma})\frac{i\theta^5}{5!} - (\vec{v}\cdot\vec{\sigma})\frac{i\theta^7}{7!} +\dotsc
\end{align}
\end{split}\]
According to eq. (2) and eq. (9), we could re-write eq. (6) as
(10)\[
\exp (i\theta \vec{v}\cdot\vec{\sigma}) = I\cos {\theta} + i(\vec{v}\cdot\vec{\sigma}) \sin {\theta}
\]