(a) Since Y_{20}(x,y,z)=\sqrt{5/16\pi } (3z^{2}-r^{2})/r^{2} and Y_{2,\pm 1}(x,y,z)=\mp \sqrt{15/8\pi } (x\pm iy)z/r^{2}, we can write
\frac{2z^{2}-x^{2}-y^{2} }{r^{2} }=\frac{3z^{2}-r^{2}}{r^{2}} =\sqrt{\frac{16\pi }{5} } Y_{20} and \frac{xz}{r^{2}} =\sqrt{\frac{2\pi }{15} } (Y_{2,-1}-Y_{21}); (5.301)
hence
\psi (x,y,z)=\frac{1}{4\sqrt{\pi } }\sqrt{\frac{16\pi }{5} }Y_{20}+\sqrt{\frac{3}{\pi } } \sqrt{\frac{2\pi }{15} } (Y_{2,-1}-Y_{21})=\frac{1}{\sqrt{5} } Y_{20}+\sqrt{\frac{2}{5} } (Y_{2,-1}-Y_{21}). (5.302)
Having expressed ψ in terms of the spherical harmonics, we can now easily write
\hat{\vec{L}}^{2} \psi (x,y,z)=\frac{1}{\sqrt{5} }\hat{\vec {L}}^{2}Y_{20}+\sqrt{\frac{2}{5} }\hat{\vec{L}}^{2}(Y_{2,-1}-Y_{21})=6\hbar ^{2} \psi (x,y,z). (5.303)
and
\hat{L}_{z} \psi (x,y,z)=\frac{1}{\sqrt{5} }\hat{L}_{z} Y_{20}+\sqrt{\frac{2}{5} }\hat{L}_{z}(Y_{2,-1}-Y_{21})=-\hbar \sqrt{\frac{2}{5} }\hat{L}_{z}(Y_{2,-1}+Y_{21}). (5.304)
This shows that \psi (x,y,z) is an eigenstate of \hat{\vec{L}}^{2} with eigenvalue 6\hbar ^{2} ;\psi (x,y,z) is, however, not an eigenstate of \hat{L}_{z}. Thus the total angular momentum of the particle is
\sqrt{〈\psi |\hat{\vec{L}}^{2}|\psi 〉} =\sqrt{6} \hbar . (5.305)
(b) Using the relation \hat{L}_{+}Y_{lm} =\hbar \sqrt{l(l+1)-m(m+1)} Y_{l m+1}, we have
\hat{L}_{+}\psi (x,y,z)=\frac{1}{\sqrt{5} }\hat{L}_{+}Y_{20} +\sqrt{\frac{2}{5} }\hat{L}_{+}(Y_{2,-1}-Y_{21})=\hbar \sqrt{\frac {6}{5} } Y_{21}+\hbar \sqrt{\frac{2}{5} }\left(\sqrt{6} Y_{20} -2Y_{22} \right); (5.306)
hence
〈\psi |\hat{L}_{+}|\psi 〉=\left[\frac{1}{\sqrt{5} }〈2,0|+\sqrt {\frac{2}{5} }(〈2,-1|-〈2,1|)\right]
\times \left[\hbar \sqrt{\frac{6}{5} }Y_{21}+\hbar \sqrt{\frac {2}{5} }\left(\sqrt{6}Y_{20}-2Y_{22} \right) \right]
=0. (5.307)
(c) Since |\psi 〉=(1/\sqrt{5} )Y_{20}+\sqrt{2/5} (Y_{2,-1}-Y_{21}), a calculation of 〈\psi |\hat{L}_{z}|\psi 〉 yields
〈\psi |\hat{L}_{z}|\psi 〉=0, with probability P_{0} =\frac{1}{5}, (5.308)
〈\psi |\hat{L}_{z}|\psi 〉=-\hbar, with probability P_{-1} =\frac{2}{5}, (5.309)
〈\psi |\hat{L}_{z}|\psi 〉=\hbar, with probability P_{1} =\frac{2}{5}. (5.310)
(d) Since \psi (x,y,z)=(1/4\sqrt{\pi } )(2z^{2}-x^{2}-y^{2}) /r^{2}+\sqrt{3/\pi } xz/r^{2} can be written in terms of the spherical coordinates as
\psi (\theta ,\varphi )=\frac{1}{4\sqrt{\pi } }(3\cos \theta ^{2}-1 )+\sqrt{\frac{3}{\pi } }\sin \theta \cos \theta \cos \varphi, (5.311)
the probability of finding the particle at the position θ and φ is
P(\theta ,\varphi)=\left|\psi (\theta ,\varphi )\right| ^{2} \sin \theta d\theta d\varphi =\left[\frac{1}{4\sqrt{\pi } }(3\cos ^{2}\theta -1 )+\sqrt{\frac{3}{\pi } }\sin \theta \cos \theta \cos \varphi \right]^{2} \sin \theta d\theta d\varphi ; (5.312)
hence
P\left(\frac{\pi }{3},\frac{\pi }{2} \right) =\left[\frac{1}{4\sqrt {\pi } }\left(3\cos ^{2}\frac{\pi }{3}-1 \right)+0 \right] ^{2} (0.03) ^{2} \sin \frac{\pi }{3}=9.7\times 10^{-7}. (5.313)