Phy5645/HydrogenAtomProblem

(a) To find $N,\!$ we simply take the volume integral of $\psi \psi ^{\ast }.$ Note that $Y_{1}^{-1}\left(\theta ,\phi \right)={\sqrt {\frac {3}{8\pi }}}\sin(\theta )e^{-i\phi },$ and thus the $\phi \!$ dependence in the integral vanishes.

$1={\frac {3}{8\pi }}\int _{\phi =0}^{2\pi }\int _{\theta =0}^{\pi }\int _{r=0}^{\infty }N^{2}r^{2}\sin ^{2}{\theta }e^{-r/a}r^{2}\sin {\theta }\,dr\,d\theta \,d\phi$

$={\tfrac {3}{4}}N^{2}\int _{0}^{\pi }\sin ^{3}{\theta }\,d\theta \int _{0}^{\infty }r^{4}e^{-r/a}\,dr=24a^{5}N^{2}$

Therefore, $N={\frac {1}{\sqrt {24a^{5}}}}.$

(b)

$\psi \psi ^{\ast }(r,\theta ,\phi )={\frac {1}{24a^{5}}}r^{2}\sin ^{2}(\theta )\left({\frac {3}{8\pi }}\right)e^{-r/a}$

$=\left({\frac {1}{24a^{5}}}\right)a^{2}\sin ^{2}\left({\frac {\pi }{4}}\right)\left({\frac {3}{8\pi }}\right)e^{-1}={\frac {\pi e^{-1}}{128a^{3}}}={\frac {0.009}{a^{3}}}$

(c) We simply integrate $\psi \psi ^{\ast }\!$ over the spherical shell given by varying $\phi \!$ and $\theta \!$ with $r=2a.\!$ The spherical harmonics, as we have defined them, are already normalized, so that the probability per unit radial coordinate is

${\frac {dP}{dr}}=\left({\frac {1}{24{a}^{5}}}\right)(2a)^{4}e^{-2}={\frac {2e^{-2}}{3a}}={\frac {0.0902}{a}}.$

(d) We may read the orbital and magnetic quantum numbers directly off of the spherical harmonic, they are $l=1\!$ and $m=-1.\!$ Therefore,

$\langle {\hat {\mathbf {L} }}^{2}\rangle =2\hbar ^{2}$

and

$\langle {\hat {L}}_{z}\rangle =-\hbar .$

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