Phy5645/Plane Rotator

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

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

Then, we could get $A={\frac {2}{\sqrt {3\pi }}}$

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

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

Therefore the probability to measure $L_{z}=0$ is ${\frac {2}{3}}$ , the probability to measure $L_{z}=2$ is ${\frac {1}{6}}$ , and the probability to measure $L_{z}=-2$ is </math> is ${\frac {1}{6}}$ . The probability to measure any other value is zero.

c) $<L_{z}>=0$

$<L^{2}>={\frac {4\hbar }{3}}$