(a) Using 〈1,\frac{1}{2};1,\frac{1}{2}|\frac{3}{2},\frac{3}{2} 〉 1,d^{(1)}_{11} (\beta )=\cos ^{2} (\beta /2) and d^{(1/2)}_{\frac{1}{2}\frac{1}{2}}(\beta )=\cos (\beta /2), we
have
d^{(3/2)}_{\frac{3}{2}\frac{3}{2}}(\beta )= \left\langle 1,\frac{1}{2};1,\frac{1}{2}|\frac{3}{2},\frac{3}{2}\right\rangle \left\langle 1,\frac {1}{2};1,\frac{1}{2}|\frac{3}{2},\frac{3}{2}\right\rangle d^{(1)}_{11} (\beta ) d^{(1/2)}_{\frac{1}{2}\frac{1}{2}}(\beta )=\cos ^{3} \left(\frac {\beta }{2} \right) . (7.424)
Similarly, since 〈1,\frac{1}{2};0,\frac{1}{2}|\frac{3}{2},\frac{1}{2} 〉=\sqrt{2/3} ,〈1,\frac{1}{2};1,-\frac{1}{2}|\frac{3}{2},\frac{1}{2} 〉=1/ \sqrt{3} , and since d^{(1)}_{10} (\beta )=-(1/\sqrt{2} )\sin (\beta ) and d^{(1/2)}_{\frac{1}{2}-\frac{1}{2}}(\beta )=-\sin (\beta /2),we have
d^{(3/2)}_{\frac{3}{2}\frac{1}{2}}(\beta )= \left\langle 1,\frac {1}{2};1,\frac{1}{2}|\frac{3}{2},\frac{3}{2}\right\rangle \left\langle 1,\frac{1}{2};0,\frac{1}{2}|\frac{3}{2},\frac{1}{2}\right\rangle d^{(1)}_{10} (\beta )d^{(1/2)}_{\frac{1}{2}\frac{1}{2}}(\beta )
+\left\langle 1,\frac{1}{2};1,\frac{1}{2}|\frac{3}{2},\frac{3}{2} \right\rangle \left\langle 1,\frac{1}{2};1,-\frac{1}{2}|\frac{3}{2},\frac {1}{2}\right\rangle d^{(1)}_{11} (\beta ) d^{(1/2)}_{\frac{1}{2}-\frac {1}{2}}(\beta )
=-\frac{1}{\sqrt{3} } \sin \beta \cos \left(\frac{\beta }{2} \right) -\frac{1}{\sqrt{3} }\cos ^{2} \left(\frac{\beta }{2} \right)\sin \left (\frac{\beta }{2} \right)
=-\sqrt{3} \sin \left(\frac{\beta }{2} \right) \cos ^{2} \left(\frac {\beta }{2} \right) . (7.425)
To calculate d^{(3/2)}_{\frac{3}{2}-\frac{1}{2}}(\beta ), we need to use the coefficients 〈1,\frac{1}{2};0,-\frac{1}{2}|\frac {3}{2},-\frac{1}{2} 〉=\sqrt{2/3} and 〈1,\frac{1}{2};-1, \frac{1}{2}|\frac{3}{2},-\frac{1}{2} 〉=1/\sqrt{3} along with d^{(1)}_{1,-1} (\beta )=\sin ^{2}(\beta /2):
d^{(3/2)}_{\frac{3}{2}-\frac{1}{2}}(\beta )= \left\langle 1,\frac {1}{2};1,\frac{1}{2}|\frac{3}{2},\frac{3}{2}\right\rangle \left\langle 1, \frac{1}{2};-1,\frac{1}{2}|\frac{3}{2},-\frac{1}{2}\right\rangle d^{(1) }_{1-1} (\beta )d^{(1/2)}_{\frac{1}{2}\frac{1}{2}}(\beta )
+\left\langle 1,\frac{1}{2};1,\frac{1}{2}|\frac{3}{2},\frac{3}{2} \right\rangle \left\langle 1,\frac{1}{2};0,-\frac{1}{2}|\frac{3}{2},-\frac {1}{2}\right\rangle d^{(1)}_{10} (\beta ) d^{(1/2)}_{\frac{1}{2}-\frac {1}{2}}(\beta )
=\frac{1}{\sqrt{3} } \sin ^{2} \left(\frac{\beta }{2} \right) \cos \left(\frac{\beta }{2} \right) +\frac{1}{\sqrt{3} }\sin \beta \sin \left (\frac{\beta }{2} \right)
=\sqrt{3} \sin ^{2} \left(\frac{\beta }{2} \right) \cos \left(\frac {\beta }{2} \right) . (7.426)
For d^{(3/2)}_{\frac{3}{2}-\frac{3}{2}}(\beta ) we have
d^{(3/2)}_{\frac{3}{2}-\frac{3}{2}}(\beta )=\left\langle 1,\frac {1}{2};1,\frac{1}{2}|\frac{3}{2},\frac{3}{2}\right\rangle \left\langle 1, \frac{1}{2};-1,-\frac{1}{2}|\frac{3}{2},-\frac{3}{2}\right\rangle d^{(1) }_{1-1} (\beta ) d^{(1/2)}_{\frac{1}{2},-\frac{1}{2}}(\beta )=-\sin ^{3} \left(\frac{\beta }{2} \right), (7.427)
because 〈1,\frac{1}{2};-1,-\frac{1}{2}|\frac{3}{2},-\frac{3}{2} 〉=1,d^{(1)}_{1-1} (\beta )=\sin ^{2} (\beta /2), and d^{(1/2)}_{\frac{1}{2}-\frac{1}{2}} (\beta )=-\sin (\beta /2).
To calculate d^{(3/2)}_{\frac{1}{2}\frac{1}{2}} (\beta ), we need to use the coefficients 〈1,\frac{1}{2};0,\frac{1}{2}|\frac {3}{2},\frac{1}{2} 〉=\sqrt{2/3} and 〈1,\frac{1}{2};1,-\frac{1}{2}|\frac{3}{2},\frac{1}{2} 〉=1/\sqrt{3} along with d^{(1)}_{1-1} (\beta )=\sin ^{2} (\beta /2):
d^{(3/2)}_{\frac{1}{2}\frac{1}{2}} (\beta )=\left\langle 1,\frac {1}{2};1,-\frac{1}{2}|\frac{3}{2},\frac{1}{2}\right\rangle \left\langle 1,\frac{1}{2};1,-\frac{1}{2}|\frac{3}{2},\frac{1}{2}\right\rangle d^{(1) }_{11} (\beta ) d^{(1/2)}_{-\frac{1}{2}-\frac{1}{2}} (\beta )
+\left\langle 1,\frac{1}{2};0,\frac{1}{2}|\frac{3}{2},\frac{1}{2} \right\rangle \left\langle 1,\frac{1}{2};1,-\frac{1}{2}|\frac{3}{2},\frac {1}{2}\right\rangle d^{(1)}_{01} (\beta ) d^{(1/2)}_{\frac{1}{2}-\frac {1}{2}}(\beta )
+\left\langle 1,\frac{1}{2};0,\frac{1}{2}|\frac{3}{2},\frac{1}{2} \right\rangle \left\langle 1,\frac{1}{2};0,\frac{1}{2}|\frac{3}{2},\frac {1}{2}\right\rangle d^{(1)}_{00} (\beta ) d^{(1/2)}_{\frac{1}{2}\frac {1}{2}}(\beta )
+\left\langle 1,\frac{1}{2};1,-\frac{1}{2}|\frac{3}{2},\frac{1}{2} \right\rangle \left\langle 1,\frac{1}{2};0,\frac{1}{2}|\frac{3}{2},\frac {1}{2}\right\rangle d^{(1)}_{10} (\beta ) d^{(1/2)}_{-\frac{1}{2}\frac {1}{2}}(\beta )
=\frac{1}{3} \cos ^{3} \left(\frac{\beta }{2} \right)-\frac{1}{3} \sin (\beta )\sin \left(\frac{\beta }{2} \right)+\frac{2}{3} \cos (\beta )\cos \left(\frac{\beta }{2} \right)-\frac{1}{3}\sin (\beta)\sin \left (\frac{\beta }{2} \right)
=\left[3 \cos ^{2} \left(\frac{\beta }{2} \right)-2\right] \cos \left(\frac{\beta }{2} \right)
=\frac{1}{2}(3\cos \beta -1)\cos \left(\frac{\beta }{2} \right) . (7.428)
Similarly, we have
d^{(3/2)}_{\frac{1}{2}-\frac{1}{2}} (\beta )=\left\langle 1,\frac {1}{2};1,-\frac{1}{2}|\frac{3}{2},\frac{1}{2}\right\rangle \left\langle 1, \frac{1}{2};-1,\frac{1}{2}|\frac{3}{2},-\frac{1}{2}\right\rangle d^{(1) }_{1-1} (\beta ) d^{(1/2)}_{-\frac{1}{2}\frac{1}{2}} (\beta )
+\left\langle 1,\frac{1}{2};1,-\frac{1}{2}|\frac{3}{2},\frac{1}{2} \right\rangle \left\langle 1,\frac{1}{2};0,-\frac{1}{2}|\frac{3}{2},-\frac {1}{2}\right\rangle d^{(1)}_{10} (\beta ) d^{(1/2)}_{-\frac{1}{2}-\frac {1}{2}}(\beta )
+\left\langle 1,\frac{1}{2};0,\frac{1}{2}|\frac{3}{2},\frac{1}{2} \right\rangle \left\langle 1,\frac{1}{2};-1,\frac{1}{2}|\frac{3}{2},-\frac {1}{2}\right\rangle d^{(1)}_{0-1} (\beta ) d^{(1/2)}_{\frac{1}{2}\frac {1}{2}}(\beta )
+\left\langle 1,\frac{1}{2};0,\frac{1}{2}|\frac{3}{2},\frac{1}{2} \right\rangle \left\langle 1,\frac{1}{2};0,-\frac{1}{2}|\frac{3}{2},-\frac {1}{2}\right\rangle d^{(1)}_{00} (\beta ) d^{(1/2)}_{\frac{1}{2}-\frac {1}{2}}(\beta )
=\frac{1}{3} \sin ^{3} \left(\frac{\beta }{2} \right)-\frac{1}{3} \sin (\beta )\cos \left(\frac{\beta }{2} \right)-\frac{1}{3} \sin (\beta ) \cos \left(\frac{\beta }{2} \right)-\frac{2}{3}\cos (\beta)\sin \left (\frac{\beta }{2} \right)
=-\left[3 \cos ^{2} \left(\frac{\beta }{2} \right)-1\right] \sin \left(\frac{\beta }{2} \right)
=-\frac{1}{2}(3\cos \beta +1)\sin \left(\frac{\beta }{2} \right) . (7.429)
(b) The remaining ten matrix elements of d^{(3/2)} (\beta ) can be inferred from the six elements derived above by making use of the properties of the d-function listed in (7.67).
d^{(j)}_{m^{\prime } m} (\beta )=(-1)^{m^{\prime }-m} d^{(j)}_{mm^{\prime } } (\beta )=(-1)^{m^{\prime }-m}d^{(j)}_{-m^{\prime } -m} (\beta ). (7.67)
For instance, using d^{(j)}_{m^{\prime } m} (\beta )=(-1)^{m^ {\prime }-m}d^{(j)}_{-m^{\prime } -m} (\beta ), we can verify that
d^{(3/2)}_{-\frac{3}{2}-\frac{3}{2}}(\beta )=d^{(3/2)}_{\frac {3}{2}\frac{3}{2}}(\beta ), d^{(3/2)}_{-\frac{1}{2}-\frac{1}{2}}(\beta )=d^{(3/2)}_{\frac{1}{2}\frac{1}{2}}(\beta ), d^{(3/2)}_{-\frac{3}{2}-\frac{1}{2}}(\beta )=-d^{(3/2)}_{\frac{3}{2}\frac{1}{2}}(\beta ) , (7.430)
d^{(3/2)}_{-\frac{3}{2}\frac{1}{2}}(\beta )=d^{(3/2)}_{\frac {3}{2}-\frac{1}{2}}(\beta ), d^{(3/2)}_{-\frac{3}{2}\frac{3}{2}}(\beta )=-d^{(3/2)}_{\frac{3}{2}-\frac{1}{2}}(\beta ), d^{(3/2)}_{-\frac{1}{2}\frac{1}{2}}(\beta )=-d^{(3/2)}_{\frac{3}{2}-\frac{1}{2}}(\beta ) . (7.431)
Similarly, using d^{(j)}_{m^{\prime } m} (\beta )=(-1)^{m^ {\prime }-m} d^{(j)}_{mm^{\prime } } (\beta ) we can obtain the remaining four elements:
d^{(3/2)}_{\frac{1}{2}\frac{3}{2}}(\beta )=-d^{(3/2)}_{\frac {3}{2}\frac{1}{2}}(\beta ), d^{(3/2)}_{-\frac{1}{2}\frac{3}{2}}(\beta )=d^{(3/2)}_{\frac{3}{2}-\frac{1}{2}}(\beta ) , (7.432)
d^{(3/2)}_{\frac{1}{2}-\frac{3}{2}}(\beta )=d^{(3/2)}_{-\frac {3}{2}\frac{1}{2}}(\beta ), d^{(3/2)}_{-\frac{1}{2}-\frac{3}{2}}(\beta )=-d^{(3/2)}_{-\frac{3}{2}-\frac{1}{2}}(\beta ) . (7.433)
Collecting the six matrix elements calculated in (a) along with the ten elements inferred above, we obtain the matrix of d^{(3/2)} (\beta ):
\left(\begin{matrix} \cos ^{3}\left(\frac{\beta }{2} \right) & -\sqrt{3}\sin \left(\frac{\beta }{2} \right)\cos ^{2}\left(\frac{\beta }{2} \right) & \sqrt{3}\sin ^{2}\cos \left(\frac{\beta }{2} \right) & -\sin ^{3}\left(\frac{\beta }{2} \right) \\ \sqrt{3}\sin \left(\frac{\beta }{2} \right)\cos ^{2}\left(\frac{\beta }{2} \right) & \frac{1}{2}(3\cos \beta -1)\cos \left(\frac{\beta }{2} \right) & -\frac{1}{2}(3\cos \beta +1)\sin \left(\frac{\beta }{2} \right) &\sqrt{3} \sin ^{2}\left(\frac{\beta }{2} \right)\cos \left(\frac{\beta }{2} \right) \\ \sqrt{3}\sin ^{2}\left(\frac {\beta }{2} \right)\cos \left(\frac{\beta }{2} \right) & \frac{1}{2}(3\cos \beta +1)\sin \left(\frac{\beta }{2} \right) & \frac{1}{2}(3\cos \beta -1)\cos \left(\frac{\beta }{2} \right) & -\sqrt{3}\sin \left(\frac{\beta }{2} \right)\cos ^{2}\left(\frac{\beta }{2} \right) \\ \sin ^{3}\left(\frac {\beta }{2} \right) & \sqrt{3}\sin ^{2}\left(\frac{\beta }{2} \right)\cos \left(\frac{\beta }{2} \right) &\sqrt{3}\sin \left(\frac{\beta }{2} \right) \cos ^{2}\left(\frac{\beta }{2} \right) & \cos ^{3}\left(\frac{\beta }{2} \right) \end{matrix} \right) , (7.434)
which can be reduced to
d^{(3/2)} (\beta )=\frac{\sin \beta }{2} \left(\begin{matrix} \frac{\cos ^{2} (\beta /2)}{\sin (\beta /2)} & -\sqrt{3}\cos \left(\frac {\beta }{2} \right) & \sqrt{3}\sin \left(\frac{\beta }{2} \right) & -\frac {\sin ^{2} (\beta /2)}{\cos (\beta /2)} \\ \sqrt{3}\cos \left(\frac {\beta }{2} \right) & \frac{3\cos \beta -1}{2\sin (\beta /2)} & -\frac{3\cos \beta +1}{2\sin (\beta /2)} & \sqrt{3}\sin \left(\frac{\beta }{2} \right) \\ \sqrt{3}\sin \left(\frac{\beta }{2} \right) & \frac{3\cos \beta +1}{\cos (\beta /2)} & \frac{3\cos \beta -1}{2\sin (\beta /2)} & -\sqrt{3} \cos \left(\frac{\beta }{2} \right) \\ \frac{\sin ^{2} (\beta /2)}{\cos (\beta /2)} & \sqrt{3}\sin \left(\frac{\beta }{2} \right) & \sqrt{3}\cos \left(\frac{\beta }{2} \right) & \frac{\cos ^{2} (\beta /2)}{\sin (\beta /2)} \end{matrix} \right) . (7.435)
Following the method outlined in this problem, we can in principle find the matrix of any d-function. For instance, using the matrices of d^{(1)} and d^{(1/2)} along with the Clebsch–Gordan coefficients resulting from the addition of j_{1} =1 and j_{2} =1, we can find the matrix of d^{(2)} (\beta ).