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![The eigenvectors in position space (spherical coordinates) of the Hamiltonian operator **H** that represent the hydrogen atom system are very complicated:
\[
\langle r\theta\phi | E_{nlm} \rangle = \sqrt{\left(\frac{2}{na}\right)^3 \frac{(n-l-1)!}{2n(n+l)!}} e^{-r/(na)} \left(\frac{2r}{na}\right)^l \left\{L_{n-l-1}^{2l+1}\left[\frac{2r}{(na)}\right]\right\}Y_l^m (\theta, \phi)
\]
where the integers:
- \( n = 1, 2, \ldots, \infty \)
- \( l = 0, 1, \ldots, n-1 \)
- \( -l \le m \le l \),
The value \( a = 0.529 \times 10^{-10} \) m is known as the Bohr radius. The associated Laguerre polynomial is:
\[
L_q^p(x) = (-1)^p \left(\frac{d}{dx}\right)^p L_{p+q}(x)
\]
where
\[
L_q(x) = \frac{e^x}{q!} \left(\frac{d}{dx}\right)^q \left(e^{-x}x^q\right),
\]
and the spherical harmonic function:
\[
Y_l^m (\theta, \phi) = \varepsilon \sqrt{\frac{(2l+1)(l-|m|)!}{4\pi(l+|m|)!}} e^{im\phi} P_l^m(\cos \theta)
\]
where
\[
\varepsilon = \begin{cases}
(-1)^m, & m \ge 0 \\
1, & m < 0 \\
\end{cases},
\]
\[
P_l^m(x) = (1-x^2)^{|m|/2} \left(\frac{d}{dx}\right)^{|m|} P_l(x),
\]
and
\[
P_l(x) = \frac{1}{2^l l!} \left(\frac{d}{dx}\right)^l (x^2](https://content.bartleby.com/qna-images/question/bbeeb476-a64f-4459-b8af-acf5dbdffec4/6222104c-0ba5-4410-8718-3bb7777a0680/2cpm54j_processed.png)
Transcribed Image Text:The eigenvectors in position space (spherical coordinates) of the Hamiltonian operator **H** that represent the hydrogen atom system are very complicated:
\[
\langle r\theta\phi | E_{nlm} \rangle = \sqrt{\left(\frac{2}{na}\right)^3 \frac{(n-l-1)!}{2n(n+l)!}} e^{-r/(na)} \left(\frac{2r}{na}\right)^l \left\{L_{n-l-1}^{2l+1}\left[\frac{2r}{(na)}\right]\right\}Y_l^m (\theta, \phi)
\]
where the integers:
- \( n = 1, 2, \ldots, \infty \)
- \( l = 0, 1, \ldots, n-1 \)
- \( -l \le m \le l \),
The value \( a = 0.529 \times 10^{-10} \) m is known as the Bohr radius. The associated Laguerre polynomial is:
\[
L_q^p(x) = (-1)^p \left(\frac{d}{dx}\right)^p L_{p+q}(x)
\]
where
\[
L_q(x) = \frac{e^x}{q!} \left(\frac{d}{dx}\right)^q \left(e^{-x}x^q\right),
\]
and the spherical harmonic function:
\[
Y_l^m (\theta, \phi) = \varepsilon \sqrt{\frac{(2l+1)(l-|m|)!}{4\pi(l+|m|)!}} e^{im\phi} P_l^m(\cos \theta)
\]
where
\[
\varepsilon = \begin{cases}
(-1)^m, & m \ge 0 \\
1, & m < 0 \\
\end{cases},
\]
\[
P_l^m(x) = (1-x^2)^{|m|/2} \left(\frac{d}{dx}\right)^{|m|} P_l(x),
\]
and
\[
P_l(x) = \frac{1}{2^l l!} \left(\frac{d}{dx}\right)^l (x^2
![**Instructions:**
1. Determine \(\langle r\theta\phi | E_{3,2,-1} \rangle\) using the above information.
2. Show that your solution is normalized, i.e.,
\[
\iiint \langle r\theta\phi | E_{3,2,-1} \rangle \langle E_{3,2,-1} | r\theta\phi \rangle r^2 \sin\theta \, dr \, d\theta \, d\phi
\]
where \( r \) is integrated from 0 to \(\infty\), \(\theta\) from 0 to \(\pi\), and \(\phi\) from 0 to \(2\pi\).](https://content.bartleby.com/qna-images/question/bbeeb476-a64f-4459-b8af-acf5dbdffec4/6222104c-0ba5-4410-8718-3bb7777a0680/tqiv11s_processed.png)
Transcribed Image Text:**Instructions:**
1. Determine \(\langle r\theta\phi | E_{3,2,-1} \rangle\) using the above information.
2. Show that your solution is normalized, i.e.,
\[
\iiint \langle r\theta\phi | E_{3,2,-1} \rangle \langle E_{3,2,-1} | r\theta\phi \rangle r^2 \sin\theta \, dr \, d\theta \, d\phi
\]
where \( r \) is integrated from 0 to \(\infty\), \(\theta\) from 0 to \(\pi\), and \(\phi\) from 0 to \(2\pi\).
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