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1. Determine (rop|E3,2-1) using the above information. 2. Show that your solution is normalized, i.e., I| (rep|E3,2,-1) (E3,2,-1|rO4) r² sin® dr d@ dø where r is integrated from 0 to o, 0 from 0 to TT, and ø from 0 to 27n.

1. Determine (rop|E3,2-1) using the above information. 2. Show that your solution is normalized, i.e., I| (rep|E3,2,-1) (E3,2,-1|rO4) r² sin® dr d@ dø where r is integrated from 0 to o, 0 from 0 to TT, and ø from 0 to 27n.

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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
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 \(2\pi\).
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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