Question

Question: Show that the normalization constant $\sqrt{\mathbf{15}\mathbf{/}\mathbf{32}\mathbf{\pi }}$ given in Table 7.3 for the angular parts of the $\mathit{l}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{}{\mathit{m}}_{\mathbf{l}}\mathbf{=}\mathbf{\pm }\mathbf{2}$ wave function is correct.

Short answer

Answer

It has been proved that the normalization for the case $\mathit{l}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{}{\mathit{m}}_{\mathbf{l}}\mathbf{=}\mathbf{\pm }\mathbf{2}$is correct.

Step-by-step solution

Given data

$$\begin{gathered} \text{Thewavefunctioncorrespondingto}\left(l,m_l\right)=\left(2,\pm 2\right)\text{is,} \\ {\Theta }_{l,m_l}\left(\theta \right){\Phi }_{m_l}\left(\varphi \right)=\sqrt{\frac{15}{32\pi }}{\sin }^2\theta e^{\pm 2i\varphi } \end{gathered}$$

Normalization

$$\begin{gathered} \mathbf{\text{TheangularpartoftheHydrogenatomwavefunctionshouldsatisfytheconditionas:}} \\ \underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}{\left\{{\Theta }_{l,m_l}\left(\theta \right)\right\}}^{\mathbf{2}}\mathbf{2}\mathit{\pi }\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{d}\mathit{\theta }\mathbf{=}\mathbf{1}\mathbf{}\mathbf{} \end{gathered}$$

Determining whether the given normalization constant is correct

In the given wave function,

${\Theta }_{l,m_l}\left(\theta \right)=\sqrt{\frac{15}{32\pi }}{\sin }^2\theta$

Check equation (I) as,

$$\begin{gathered} =\frac{15}{32\pi }\times 2\pi \underset{0}{\overset{\pi }{\int }}{\sin }^5\theta d\theta \\ =\frac{15}{16}\underset{0}{\overset{\pi }{\int }}{\sin }^4\theta \sin \theta d\theta \\ =\frac{15}{16}\underset{0}{\overset{\pi }{\int }}{\left({\sin }^2\theta \right)}^2\sin \theta d\theta \\ =\frac{15}{16}\underset{0}{\overset{\pi }{\int }}{\left(1-{\cos }^2\theta \right)}^2\sin \theta d\theta \end{gathered}$$

Let us assume,

$$\begin{gathered} \cos \theta =z \\ -\sin \theta d\theta =dz \end{gathered}$$

Then the integral becomes,

$$\begin{gathered} =\frac{15}{16}\underset{1}{\overset{-1}{\int }}{\left(1-z^2\right)}^2\left(-dz\right) \\ =\frac{15}{16}\underset{-1}{\overset{1}{\int }}\left(1+z^4-2z^2\right)dz \\ =\frac{15}{16}{\left[z+\frac{z^5}{5}-2\frac{z^3}{3}\right]}_{-1}^1 \\ =\frac{15}{16}\left(2+\frac{2}{5}-\frac{4}{3}\right) \\ =1 \end{gathered}$$

Thus the normalization is correct.