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

  • Page ID
    20882
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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 \(-1\).


    This page titled 2.5: Explicit form of the spin-1/2 rotation operator is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Mark E. Tuckerman.

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