(a) Let us use a lighter notation for |\psi〉 :|\psi〉 =\frac{1}{\sqrt{5} }|1,-1〉 +\sqrt{\frac{3}{5} }|1,0〉 +\frac{1}{\sqrt{5} }|1,1〉.
From (5.56)
\hat{J}_{\pm }|j,m〉 =\hbar \sqrt{(j\mp m)(j\pm m+1)} |j,m\pm 1〉.
we can write \hat{L}_{+}|l,m 〉= \hbar \sqrt{l(l+1)-m(m+1)} |l,m+1 〉; hence the only terms that survive in 〈\psi |\hat{L}_{+}| \psi〉 are
〈\psi |\hat{L}_{+}| \psi〉=\frac{\sqrt{3} }{5} 〈1,0|\hat{L}_{+} |1,-1〉 +\frac{\sqrt{3} }{5} 〈1,1|\hat{L}_{+}|1,0〉 =\frac{2\sqrt{6} }{5} \hbar, (5.275)
since 〈1,0|\hat{L}_{+}|1,-1〉 = 〈1,1|\hat{L}_{+}|1,0〉 =\sqrt{2} \hbar .
(b) If \hat{L}_{z} were measured, we will find three values l_{z} =-\hbar ,0, and \hbar. The probability of finding the value l_{z} =-\hbar is
P_{-1} =\left| 〈1,-1|\psi〉 \right| ^{2} =\left|\frac{1}{\sqrt{5} } 〈1,-1|1,-1〉 +\sqrt{\frac{3}{5} } 〈1,-1|1,0〉 +\frac{1}{\sqrt{5} } 〈1,-1|1,1〉 \right| ^{2}
=\frac{1}{5}, (5.276)
since 〈1,-1|1,0〉 = 〈1,-1|1,1 〉=0 and 〈1,-1|1,-1〉 =1 . Similarly, we can verify that the probabilities of measuring l_{z}=0 and \hbar are respectively given by
P_{0} =\left| 〈1,0|\psi〉 \right| ^{2} =\left|\sqrt{\frac{3}{5} } 〈1,0|1,0〉\right| ^{2} =\frac{3}{5}, (5.277)
P_{1} =\left| 〈1,1|\psi〉 \right| ^{2} =\left|\sqrt{\frac{1}{5} } 〈1,1|1,1〉\right| ^{2} =\frac{1}{5}. (5.278)
(c) After measuring l_{z} =-\hbar the system will be in the eigenstate |lm〉=|1,-1〉, that is, \psi (\theta ,\varphi )=Y_{1} ,-1(\theta ,\varphi ). We need first to calculate the expectation values of \hat{L}_{x},\hat{L}_{y}, \hat {L}^{2}_{x}, and \hat{L}^{2}_{y} using |1,-1〉. Symmetry requires that 〈1,-1|\hat{L}_{x}|1,-1 〉=〈1,-1|\hat{L}_{y}|1,-1 〉=0.
The expectation values of \hat{L}^{2}_{x} and \hat{L}^{2}_{y} are equal, as shown in (5.60);
〈\hat{J}^{2}_{x} 〉=〈\hat{J}^{2}_{y} 〉=\frac{1}{2}\left [〈j,m|\hat{\vec{J}}^{2}|j,m 〉 -〈j,m|\hat{J}^{2}_{z}|j,m 〉\right] =\frac{\hbar ^{2} }{2} \left[j(j+1)-m^{2} \right].
they are given by
〈\hat{L}^{2}_{x} 〉=〈\hat{L}^{2}_{y} 〉=\frac{1}{2}[〈\hat{\vec {L}}^{2}-\hat{L}^{2}_{z}〉]=\frac{\hbar ^{2} }{2}\left[l(l+1)-m^{2} \right]=\frac{\hbar ^{2} }{2}; (5.279)
in this relation, we have used the fact that l = 1 and m = -1. Hence
\Delta L_{x}=\sqrt{〈\hat{L}^{2}_{x} 〉} =\frac{\hbar }{\sqrt{2} } =\Delta L_{y}, (5.280)
and the uncertainties product \Delta L_{x} \Delta L_{y} is given by
\Delta L_{x} \Delta L_{y}=\sqrt{〈\hat{L}^{2}_{x} 〉〈\hat {L}^{2}_{y} 〉} =\frac{\hbar ^{2} }{2}. (5.281)