Question 7.P.9: Find the rotation matrix d^(1) corresponding to j = 1.

To find the matrix of d^{(1)} (\beta )=e^{-i\beta \hat{J} _{y}/ \hbar } for j = 1, we need first to find the matrix representation of \hat{J} _{y} within the joint eigenstates \left\{|j,m 〉\right\} of \hat{\vec{J} }^{2} and \hat{J} _{z}. Since the basis of j = 1 consists of three states |1,-1 〉,|1 ,0 〉,|1,1 〉, the matrix representing \hat{J} _{y} within this basis is given by

=\frac{i\hbar }{\sqrt{2} } \left(\begin{matrix} 0 & -1 & 0 \\ 1 & 0 & -1 \\ 0 & 1 & 0 \end{matrix} \right) .                              (7.415)

We can easily verify that J^{3}_{y} =J_{y} :

J^{2}_{y} =\frac{\hbar ^{2} }{2} \left(\begin{matrix} 1 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 1 \end{matrix} \right) ,          J^{3}_{y} =\frac{i \hbar ^{3}}{\sqrt{2} } \left(\begin{matrix} 0 & -1 & 0 \\ 1 & 0 & -1 \\ 0 & 1 & 0 \end{matrix} \right)=\hbar ^{2} J_{y}.                (7.416)

We can thus infer

J^{2n}_{y} =\hbar ^{2n-2} J^{2}_{y}          (n>0),           J^{2n+1}_{y} =\hbar ^{2n}J_{y}.                            (7.417)

Combining these two relations with

=\sum\limits_{n=0}^{\infty } \frac{1}{(2n)!} \left(-\frac{i\beta }{\hbar } \right) ^{2n} J^{2n}_{y} + \sum\limits_{n=0}^{\infty } \frac{1}{(2n+1)!} \left(-\frac{i\beta }{\hbar } \right) ^{2n+1} J^{2n+1}_{y} ,         (7.418)

we obtain

=\hat{I} +\left(\frac{\hat{J} _{y}}{\hbar } \right) ^{2}\left [\sum\limits_{n=0}^{\infty }\frac{(-1)^{n} }{(2n)!}(\beta )^{2n}-1 \right] -i\frac{\hat{J} _{y}}{\hbar } \sum\limits_{n=0}^{\infty }\frac{(-1)^{n} }{(2n+1)!} \beta ^{2n+1},          (7.419)

where \hat{I} is the 3\times 3 unit matrix. Using the relations \sum\limits_{n=0}^{\infty } [(-1)^{n}/(2n) !](\beta )^{2n}=\cos \beta and \sum\limits_{n=0} ^{\infty } [(-1)^{n}/(2n+1)!](\beta )^{2n+1}=\sin \beta , we may write

e^{-i\beta \hat{J} _{y}/\hbar } =\hat{I} +\left(\frac{\hat{J} _{y}}{\hbar } \right) ^{2} [\cos \beta -1]-i \frac{\hat{J} _{y}}{\hbar } \sin \beta .                   (7.420)

Inserting now the matrix expressions for J_{y} and J^{2}_{y} as listed in (7.415) and (7.416), we obtain

e^{-i\beta \hat{J} _{y}/\hbar } =\hat{I} +\frac{1}{2} \left (\begin{matrix} 1 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 1 \end{matrix} \right) (\cos \beta -1)-i\frac{i}{\sqrt{2} } \left(\begin{matrix} 0 & -1 & 0 \\ 1 & 0 & -1 \\ 0 & 1 & 0 \end{matrix} \right) \sin \beta                  (7.421)

or

d^{(1)} (\beta )=\left(\begin{matrix} d^{(1)}_{11} & d^{(1)}_{1,0} & d^{(1)}_{1-1} \\ d^{(1)}_{01} & d^{(1)}_{00} & d^{(1)}_{0-1} \\ d^{(1)}_{-11} & d^{(1)}_{-10} & d^{(1)}_{-1-1} \end {matrix} \right) =\left(\begin{matrix} \frac{1}{2}(1+\cos \beta ) & -\frac{1}{\sqrt{2} }\sin \beta & \frac{1}{2}(1-\cos \beta ) \\ \frac{1}{\sqrt{2} }\sin \beta & \cos \beta & -\frac{1}{\sqrt{2} }\sin \beta \\ \frac{1}{2}(1-\cos \beta ) & \frac{1}{\sqrt{2} }\sin \beta & \frac{1}{2}(1+\cos \beta ) \end{matrix} \right) .        (7.422)

Since \frac{1}{2}(1+\cos \beta )=\cos ^{2} (\beta /2) and \frac{1}{2}(1-\cos \beta )=\sin ^{2} (\beta /2) , we have

d^{(1)} (\beta )=e^{-i\beta \hat{J} _{y}/\hbar }=\left(\begin {matrix} \cos ^{2} (\beta /2) & -\frac{1}{\sqrt{2} }\sin (\beta ) & \sin ^{2} (\beta /2) \\ \frac{1}{\sqrt{2} }\sin (\beta ) & \cos (\beta ) & -\frac{1}{\sqrt{2} }\sin (\beta ) \\ \sin ^{2} (\beta /2) & \frac{1}{\sqrt{2} }\sin (\beta ) & \cos ^{2} (\beta /2) \end{matrix} \right) .                (7.423)

This method becomes quite intractable when attempting to derive the matrix of d^{(j)} (\beta ) for large values of j. In Problem 7.10 we are going to present a simplermethod for deriving d^{(j)} (\beta ) for larger values of j; this method is based on the addition of angular momenta.