Phy5646/CG coeff example1

Find the CG coefficients

$1\rangle :{\dfrac {1}{2}}\otimes 1={\dfrac {3}{2}}\oplus {\dfrac {1}{2}}$

Answer

The addition of $j_{1}=s={\frac {1}{2}}$ and $j_{2}=l=1$ is encountered, for example, in the p-state of an electron. This state is characterised by orbital quantum number $j_{1}=s={\frac {1}{2}}$ and spin quantum number $j_{2}=l=1$. Obviously the possible values of magnetic quantum number for $j_{1}=s={\frac {1}{2}}$ are $m_{s}={\frac {1}{2}},-{\frac {1}{2}}$ and those for $j_{2}=l=1$ are $m_{l}=1,0,-1$. The allowed values of the total angular momentum are between $|j_{1}-j_{2}|\leq j\leq j_{1}+j_{2}$ hence $j={\frac {3}{2}},{\frac {1}{2}}$. To calculate the relevant Clebsch–Gordan coefficients, we have to express the basis vectors $|jmj_{1}j_{2}\rangle$ in terms of $|{\dfrac {1}{2}},1,m_{1}m_{2}\rangle$

Eigenvectors $|jm\rangle$ associated with $j={\dfrac {3}{2}}$:

The state $|{\dfrac {3}{2}},{\dfrac {3}{2}}\rangle$ is given by,

$|{\dfrac {1}{2}},1,{\dfrac {3}{2}},{\dfrac {3}{2}}\rangle$ = $|{\dfrac {1}{2}},1,{\dfrac {1}{2}},1\rangle$

Corresponding CG coefficient, $\langle {\dfrac {1}{2}},1,{\dfrac {1}{2}},{\dfrac {1}{2}}$ $|{\dfrac {1}{2}},1,{\dfrac {3}{2}},{\dfrac {3}{2}}\rangle$ = 1

Now $|{\dfrac {1}{2}},1,{\dfrac {3}{2}},{\dfrac {1}{2}}\rangle$ can be found by

Applying $J_{-}$ to $|{\dfrac {1}{2}},1,{\dfrac {3}{2}},{\dfrac {3}{2}}\rangle$ and

$(J_{1-}+J_{2-})$ to $|{\dfrac {1}{2}},1,{\dfrac {1}{2}},1\rangle$ and the equating the two results,

$J_{-}|{\dfrac {1}{2}},1,{\dfrac {3}{2}},{\dfrac {3}{2}}\rangle$ = $(J_{1-}+J_{2-})|{\dfrac {1}{2}},1,{\dfrac {1}{2}},1\rangle$

$J_{-}|{\dfrac {1}{2}},1,{\dfrac {3}{2}},{\dfrac {3}{2}}\rangle$ = $\hbar {\sqrt {{\frac {3}{2}}\left({\frac {3}{2}}+1\right)-{\frac {3}{2}}\left({\frac {3}{2}}-1\right)}}|{\dfrac {1}{2}},1,{\dfrac {3}{2}},{\dfrac {1}{2}}\rangle$

or $J_{-}|{\dfrac {1}{2}},1,{\dfrac {3}{2}},{\dfrac {3}{2}}\rangle$ = $\hbar {\sqrt {3}}|{\dfrac {1}{2}},1,{\dfrac {3}{2}},{\dfrac {1}{2}}\rangle$