Phy5645/Plane Rotator

(a) $A\!$ can be determined from the normalization condition,

$1=\int _{-\pi }^{\pi }d\phi \,|\psi (\phi )|^{2}=A^{2}\int _{-\pi }^{\pi }d\phi \,\sin ^{4}{\phi }=A^{2}\cdot {\frac {3\pi }{4}}.$

Therefore, $A={\frac {2}{\sqrt {3\pi }}}.$

(b) The probability to measure the angular momentum to be $\hbar m$ is

$P_{m}=|\langle m|\psi \rangle |^{2}=\left|{\frac {1}{\sqrt {2\pi }}}\int _{-\pi }^{\pi }d\phi \,e^{im\phi }\psi (\phi )\right|^{2}={\tfrac {2}{3}}\delta _{m,0}+{\tfrac {1}{6}}(\delta _{m,2}+\delta _{m,-2})$

Therefore the probability of measuring $L_{z}=0\!$ is ${\tfrac {2}{3}},$ that of measuring $L_{z}=2\hbar \!$ is ${\tfrac {1}{6}}$, and that of measuring $L_{z}=-2\hbar$ is also ${\tfrac {1}{6}}.$ The probability of measuring any other value is zero.

(c)

$\langle {\hat {L}}_{z}\rangle ={\frac {4}{3\pi }}\int _{-\pi }^{\pi }d\phi \,\sin ^{2}{\phi }\left(-i\hbar {\frac {d}{d\phi }}\sin ^{2}{\phi }\right)=-{\frac {8}{3\pi }}i\hbar \int _{-\pi }^{\pi }d\phi \,\sin ^{3}{\phi }\cos {\phi }=0$

$\langle {\hat {L}}_{z}^{2}\rangle ={\frac {4}{3\pi }}\int _{-\pi }^{\pi }d\phi \,\sin ^{2}{\phi }\left(-\hbar ^{2}{\frac {d^{2}}{d\phi ^{2}}}\sin ^{2}{\phi }\right)=-{\frac {8}{3\pi }}\hbar ^{2}\int _{-\pi }^{\pi }d\phi \,\sin ^{2}{\phi }(1-2\sin ^{2}{\phi })={\tfrac {4}{3}}\hbar ^{2}$

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