View source for Phy5645/Plane Rotator

'''(a)''' <math>A\!</math> can be determined from the normalization condition,

<math>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}.</math>

Therefore, <math>A=\frac{2}{\sqrt{3\pi}}.</math>


'''(b)''' The probability to measure the angular momentum to be <math> \hbar m </math> is

<math> 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}) </math>

Therefore the probability of measuring <math>L_z = 0\!</math> is <math>\tfrac{2}{3},</math> that of measuring <math>L_z = 2\hbar\!</math> is <math>\tfrac{1}{6}</math>, and that of measuring <math> L_z = -2\hbar</math> is also <math>\tfrac {1}{6}.</math>  The probability of measuring any other value is zero.

'''(c)'''

<math>\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 </math>       

<math>\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</math>

Back to [[Orbital Angular Momentum Eigenfunctions#Problems|Orbital Angular Momentum Eigenfunctions]]