2.5: Explicit form of the spin-1/2 rotation operator

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For spin-1/2, the rotation operator

$\begin{displaymath} {R_{\alpha}^{(s)}({\bf n})}= \exp\left(-i{\alpha \over 2}\stackrel{\rightarrow}{\sigma}\cdot{\hat{\bf n}}\right) \end{displaymath}$

can be written as an explicit 2$\times$ 2 matrix. This is accomplished by expanding the exponential into a Taylor series:

$\begin{displaymath} \exp\left(-i{\alpha \over 2}\stackrel{\rightarrow}{\sigma}\c... ...4(\stackrel{\rightarrow}{\sigma}\cdot{\hat{\bf n}})^4 - \cdots \end{displaymath}$

Note that

$\begin{displaymath} (\stackrel{\rightarrow}{\sigma}\cdot{\hat{\bf n}})^2 = (\sta... ...ot{\hat{\bf n}}+ i\sigma({\hat{\bf n}}\times{\hat{\bf n}}) = 1 \end{displaymath}$

Thus, the Taylor series becomes:

$\displaystyle \exp\left(-i{\alpha \over 2}\stackrel{\rightarrow}{\sigma}\cdot{\hat{\bf n}}\right)$	$\textstyle =$	$\displaystyle 1-{i\alpha \over 2}\stackrel{\rightarrow}{\sigma}\cdot{\hat{\bf n... ...gma}\cdot{\hat{\bf n}}) + {1 \over 4!}\left({i\alpha \over 2}\right)^4 - \cdots$
	$\textstyle =$	$\displaystyle \left[1-{1 \over 2!}\left({\alpha \over 2}\right)^2 + {1 \over 4!... ... \over 2}\right) - {1 \over 3!}\left({\alpha \over 2}\right)^3 + \cdots \right]$
	$\textstyle =$	$\displaystyle \cos\left({\alpha \over 2}\right) - i\stackrel{\rightarrow}{\sigma}\cdot{\hat{\bf n}} \sin\left({\alpha \over 2}\right)$

Thus,

$\begin{displaymath} {R_{\alpha}^{(s)}({\bf n})}= \cos\left({\alpha \over 2}\righ... ...w}{\sigma}\cdot{\hat{\bf n}} \sin\left({\alpha \over 2}\right) \end{displaymath}$

As a 2$\times$2 matrix,

$\begin{displaymath} \stackrel{\rightarrow}{\sigma}\cdot{\hat{\bf n}}= \sigma_x n... ... = \left(\matrix{n_z & n_x-in_y \cr n_x + in_y & -n_z}\right) \end{displaymath}$

so that the rotation operator becomes

$\begin{displaymath} {R_{\alpha}^{(s)}({\bf n})}= \left(\matrix{\cos\left({\alph... ... \over 2}\right)+in_z\sin\left({\alpha \over 2}\right)}\right) \end{displaymath}$

Now consider the example of $\alpha = 2\pi$. In this case, it is easy to see that the rotation operator reduces to

$\begin{displaymath} R_{2\pi}^{(s)}({\hat{\bf n}}) = \left(\matrix{-1 & 0 \cr 0 & -1}\right) = -{\rm I} \end{displaymath}$

Interestingly, a rotation through an angle $2\pi$ of a spin state returns the state to its original value but causes it to pick up an overall phase factor

$\begin{displaymath} -1 = e^{i\pi} \end{displaymath}$

While this phase factor cannot affect any physical property, it is, nevertheless observable in the experiment depicted below:

$\vert\chi\rangle$, is split by a partially reflecting material into two beams. One of these is sent through a magnetic field region tuned to generate a rotation by $\alpha = 2\pi$ of the spin state, so that the new state is $\vert\chi'\rangle$. The beams are then brought back together and allowed to interfere. The overlap, $\langle \chi'\vert\chi\rangle=-1$ is measured, which will yield the over phase factor $-1$.

\(\displaystyle \exp\left(-i{\alpha \over 2}\stackrel{\rightarrow}{\sigma}\cdot{\hat{\bf n}}\right)\)	\(\textstyle =\)	$\displaystyle 1-{i\alpha \over 2}\stackrel{\rightarrow}{\sigma}\cdot{\hat{\bf n... ...gma}\cdot{\hat{\bf n}}) + {1 \over 4!}\left({i\alpha \over 2}\right)^4 - \cdots$
	\(\textstyle =\)	$\displaystyle \left[1-{1 \over 2!}\left({\alpha \over 2}\right)^2 + {1 \over 4!... ... \over 2}\right) - {1 \over 3!}\left({\alpha \over 2}\right)^3 + \cdots \right]$
	\(\textstyle =\)	$\displaystyle \cos\left({\alpha \over 2}\right) - i\stackrel{\rightarrow}{\sigma}\cdot{\hat{\bf n}} \sin\left({\alpha \over 2}\right)$

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