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: Final remarks : lecture_8 : The general problem

The simple example revisited

Consider, again, the example of adding to spin-1/2 angular momenta. This time, we will use the Clebsch-Gordan coefficients directly to determine the unitary transformation. Start with the state $\vert 1\;\;1\rangle$. Expanding gives

\begin{displaymath} \vert 1\;\;1\rangle = \sum_{m_1=-1/2}^{1/2}\sum_{m_2=-1/2}^{... ...over 2}\;\;\;m_1;{1 \over 2}\;\;\;m_2\right\vert}1\;\;1\rangle \end{displaymath}

Only one term gives $m_1+m_2=1$, which is clearly $m_1=1/2$ and $m_2=1/2$. Thus,

\begin{displaymath} \vert 1\;\;1\rangle = {\left\vert{1 \over 2}\;\;\;{1 \over 2... ...\over 2};{1 \over 2}\;\;\;{1 \over 2}\right\vert}1\;\;1\rangle \end{displaymath}

The Clebsch-Gordan coefficients, by special case $i$ is just 1, so

\begin{displaymath} \vert 1\;\;1\rangle = {\left\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>} \end{displaymath}

as expected.

The state $\vert 1\;\;0\rangle$ is expanded as

\begin{displaymath} \vert 1\;\;0\rangle = \sum_{m_1=-1/2}^{1/2}\sum_{m_2=-1/2}^{... ...over 2}\;\;\;m_1;{1 \over 2}\;\;\;m_2\right\vert}1\;\;0\rangle \end{displaymath}

This time, since $m_1+m_2=0$, two terms contribute, $m_1=1/2, m_2=-1/2$ and $m_1=-1/2, m_2=1/2$. Hence,

\begin{displaymath} \vert 1\;\;0\rangle = {\left\vert{1 \over 2}\;\;\;{1 \over 2... ...\over 2};{1 \over 2}\;\;\;{1 \over 2}\right\vert}1\;\;0\rangle \end{displaymath}

However, since $j_1+j_2=1$, we can use special case $iii$, and we find
$\displaystyle {\left<{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;-{1 \over 2}\right\vert}1\;\;0\rangle$ $\textstyle =$ $\displaystyle \sqrt{{1! 1! \over 2!}}\sqrt{{1! 1! \over 1! 0! 0! 1!}} = {1 \over \sqrt{2}}$  
$\displaystyle {\left<{1 \over 2}\;\;\;-{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right\vert}1\;\;0\rangle$ $\textstyle =$ $\displaystyle \sqrt{{1! 1! \over 2!}}\sqrt{{1! 1! \over 0! 1! 1! 0!}} = {1 \over \sqrt{2}}$  

so that

\begin{displaymath} \vert 1\;\;0\rangle = {1 \over \sqrt{2}}\left({\left\vert{1 ... ...\;\;\;-{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}\right) \end{displaymath}

It is straightforward to show that $\vert 1\;\;-1\rangle = {\left\vert{1 \over 2}\;\;\;-{1 \over 2};{1 \over 2}\;\;\;-{1 \over 2}\right>}$. The state $\vert\;\;0\rangle$ is expanded to give

\begin{displaymath} \vert\;\;0\rangle = \sum_{m_1=-1/2}^{1/2}\sum_{m_2=-1/2}^{1/... ...over 2}\;\;\;m_1;{1 \over 2}\;\;\;m_2\right\vert}0\;\;0\rangle \end{displaymath}

Again there are two terms that contribute. However, to determine the two Clebsch-Gordan coefficients:

\begin{displaymath} {\left<{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;-{1 \ov... ...\over 2};{1 \over 2}\;\;\;{1 \over 2}\right\vert}0\;\;0\rangle \end{displaymath}

we can use special case $ii$, since $M=J$. In this case,
$\displaystyle {\left<{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;-{1 \over 2}\right\vert}0\;\;0\rangle$ $\textstyle =$ $\displaystyle (-1)^{{1 \over 2}-{1 \over 2}} \sqrt{{1! 1! \over 2! 0! 0!}}\sqrt{{1! 0! \over 0! 1!}} = {1 \over \sqrt{2}}$  
$\displaystyle {\left<{1 \over 2}\;\;\;-{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right\vert}0\;\;0\rangle$ $\textstyle =$ $\displaystyle (-1)^{{1 \over 2}+{1 \over 2}} \sqrt{{1! 1! \over 2! 0! 0!}}\sqrt{{0! 1! \over 1! 0!}} = -{1 \over \sqrt{2}}$  

Hence, the state is given by

\begin{displaymath} \vert\;\;0\rangle = {1 \over \sqrt{2}}\left({\left\vert{1 \o... ...\;\;\;-{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}\right) \end{displaymath}

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: Final remarks : lecture_8 : The general problem
Mark Tuckerman 平成17年2月18日