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| This problem taken from ''Quantum Mechanics: Concepts and
| | '''(a)''' To find <math>N,\!</math> we simply take the volume integral of <math>\psi\psi^\ast.</math> Note that <math>Y_1^{-1}\left(\theta, \phi \right) = \sqrt{\frac{3}{8\pi}}\sin(\theta)e^{-i\phi},</math> and thus the <math>\phi\!</math> dependence in the integral vanishes. |
| Applications'' by Nouredine Zettili: Exercise 6.3
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| '''An electron in a hydrogen atom is in the energy eigenstate''' <math>
| | <math>1=\frac{3}{8\pi}\int_{\phi= 0}^{2\pi} \int_{\theta = 0}^{\pi} \int_{r=0}^{\infty} |
| \psi_{2,1,1} \left(r, \theta, \phi \right) = Nre^{-\frac{r} | | N^{2}r^{2}\sin^{2}{\theta}e^{-r/a}r^{2}\sin{\theta}\,dr\,d\theta\,d\phi</math> |
| {2a_o}}Y_{1,-1}\left(\theta, \phi \right)</math>. '''(a) Find the normalization constant, N.''' | |
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| Take the volume integral of <math>\psi\psi*</math>. <math>Y_{1,-1}\left(\theta, \phi
| | <math>=\tfrac{3}{4}N^2 \int_{0}^{\pi} \sin^{3}{\theta}\,d\theta \int_{0}^{\infty}r^{4}e^{-r/a}\,dr=24a^5N^2</math> |
| \right) = \sqrt{\frac{3}{8\pi}}sin(\theta)e^{-i\phi} </math> and as such the
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| phi dependence in the integral vanishes :
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| <math>\int _{\phi= 0}^{2\pi} \int_{\theta = 0}^{\pi} \int_{r=0}^{\infty} | | Therefore, <math>N = \frac{1}{\sqrt{24a^5}}.</math> |
| N^{2}r^{2}sin^{2}(\theta) \frac{3}{8\pi}e^{-\frac{r}
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| {a_o}}r^{2}sin(\theta)drd\theta d\phi</math>
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| <math>\Longrightarrow \frac{3N^2}{8\pi} \int_{0}^{\pi} sin^{3}(\theta)d
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| \theta \int_{0}^{2\pi} e^{-2i\phi}d\phi \int_{ 0}^{\infty}r^{4}e^{-
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| \frac{r}{a_o}}dr = 1 </math>
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| <math>\Longrightarrow \frac{3N^2}{8\pi} \frac{4}{3}(2\pi)(24a^5) = 1</math> | | <math>\psi\psi^\ast(r,\theta,\phi) = \frac{1}{24a^5}r^{2}\sin^{2}(\theta)\left(\frac{3}{8\pi}\right)e^{-r/a}</math> |
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| Therefore <math>N^{2}\left(24a^{5}\right) = 1 </math> so <math>N = \sqrt\frac{1}{24a^5}</math>
| | <math>=\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}</math> |
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| '''(b)What is the probability per unit volume of finding the electron at''' <math> r = a_{o}, \theta = 45^{\circ}, \phi = 60^{\circ}? </math> | | '''(c)''' We simply integrate <math>\psi\psi^\ast\!</math> over the spherical shell given by varying <math>\phi\!</math> and <math>\theta\!</math> with <math>r = 2a.\!</math> The spherical harmonics, as we have defined them, are already normalized, so that the probability per unit radial coordinate is |
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| <math>\psi\psi* = N^{2}r^{2}sin^{2}(\theta)\left(\frac{3}{8\pi}\right)e^{- | | <math>\frac{dP}{dr}=\left(\frac{1}{24{a}^5}\right)(2a)^{4}e^{-2} = \frac{2e^{-2}}{3a} = \frac{0.0902}{a}.</math> |
| \frac{r}{a_o}}</math> | |
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| <math>\Longrightarrow \left(\frac{1}{24{a_o}^5}\right) {a_o}^2 | | '''(d)''' We may read the orbital and magnetic quantum numbers directly off of the spherical harmonic, they are <math>l=1\!</math> and <math>m=-1.\!</math> Therefore, |
| sin^{2}\left(\frac{\pi}{4}\right)\left(\frac{3}{8\pi}\right)e^{-1} =
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| \left(\frac{\pi e^{-1}}{128{a_o}^3}\right) = \frac{0.009}{{a_o}^3}</math>
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| | <math>\langle\hat{\mathbf{L}}^2\rangle=2\hbar^{2}</math> |
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| '''(c) What is the probability per unit radial interval (dr) of finding the electron at''' <math> a =a_{o} ? </math>
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| Average over <math>\phi</math> and <math>\theta</math> at <math>r = 2a_{o}</math>
| | <math>\langle\hat{L}_z\rangle=-\hbar.</math> |
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| <math>\left| Y_{1,-1}\right|^2 = \left(\frac{3}{8\pi}\right)sin^{2}(\theta)
| | Back to [[Hydrogen Atom#Problems|Hydrogen Atom]] |
| = \left(\frac{3}{16\pi}\right)</math>
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| <math>\psi\psi* = N^{2}r^{2}e^{-\frac{r}{a}}s \left| Y_{1,-1}\right|^{2} </math>
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| <math>\Longrightarrow \left(\frac{1}{24{a}^5}\right)(2a)^{2}e^{-\frac{2a}
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| {a}}\left(\frac{3}{16\pi}\right) = \left(\frac{1}{32\pi a}
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| \right)e^{-2} = \frac{0.0013}{a}</math>
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| '''(d) If''' <math> \mathbf{ \hat L^2}</math> '''and''' <math>\mathbf{ \hat L_{z}}</math> '''are made, what will the results be?'''
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| l=1, m = -1 are the l and m of the eigenstate <math>Y_{1,-1}(\theta, \phi)</math>
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| <math>\hat L^2 = \hbar^{2} l (l +1) = 2\hbar^{2} </math>
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| <math>\hat L_z = \hbar m = -\hbar </math>
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(a) To find 
we simply take the volume integral of 
Note that 
and thus the 
dependence in the integral vanishes.
Therefore, 
(b)
(c) We simply integrate 
over the spherical shell given by varying and 
with 
The spherical harmonics, as we have defined them, are already normalized, so that the probability per unit radial coordinate is
(d) We may read the orbital and magnetic quantum numbers directly off of the spherical harmonic, they are 
and 
Therefore,
and
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