Question 7.P.7: (a) Calculate the expression of 〈2, 0|Y10|1, 0 〉. (b) Use ...

(a) Since

〈2,0|Y_{10}|1,0 〉=\int_{0}^{\pi } \sin \theta d\theta \int_{0} ^{2\pi } Y^{*}_{20} (\theta ,\varphi )Y_{10}(\theta ,\varphi ) Y_{10}(\theta ,\varphi )d\varphi ,           (7.396)

and using the relations Y_{20} (\theta ,\varphi )=\sqrt{5/(16\pi )} (3\cos ^{2} \theta -1) and Y_{10}(\theta ,\varphi )= \sqrt {3/(4\pi )} \cos \theta , we have

=\frac{3}{2} \sqrt{\frac{5}{16\pi } }\int_{0}^{\pi }\cos ^{2} \theta (3\cos ^{2} \theta -1)\sin \theta d\theta .            (7.397)

The change of variables x=\cos \theta leads to

=\frac{3}{2}\sqrt{\frac{5}{16\pi } }\int_{-1}^{1}x^{2} (3x^{2} -1)dx=\frac{1}{\sqrt{5\pi } } .                       (7.398)

(b) Applying the Wigner–Eckart theorem to 〈2,0|Y_{10}|1,0 〉 and using the Clebsch– Gordan coefficient 〈1,1;0,0|2,0 〉=2\sqrt{6} , we have

〈2,0|Y_{10}|1,0 〉=〈1,1;0,0|2,0 〉〈2\parallel Y_{1} \parallel 1〉= \frac{2}{\sqrt{6} } 〈2\parallel Y_{1} \parallel 1〉.               (7.399)

Finally, we may obtain 〈2\parallel Y_{1} \parallel 1〉 from (7.398) and (7.399):

〈2\parallel Y_{1} \parallel 1〉=\sqrt{\frac{3}{10\pi } } .                       (7.400)