Question

Question: Show that the angular normalization constant in Table 7.3 for the case $\mathbf{(}\mathbf{l}\mathbf{,}{\mathbf{m}}_{\mathbf{l}}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}$ is correct.

Short answer

Answer

It has been proved that the normalization for the case $\left(l,m_l\right)\mathbf{=}\left(1,0\right)$is correct.

Step-by-step solution

Given data

The wave function corresponding to $\left(l,m_l\right)=\left(1,0\right)$ is,

${\Theta }_{l,m_l}\left(\theta \right){\Phi }_{m_l}\left(\varphi \right)=\sqrt{\frac{3}{4\pi }}\cos \theta$

Normalization

$$\begin{gathered} \text{TheangularpartoftheHydrogenatomwavefunctionshouldsatisfytheflowingcondition,} \\ \underset{0}{\overset{\pi }{\int }}{\left\{{\Theta }_{l,m_l}\left(\theta \right)\right\}}^{2}2\pi sin\theta d\theta =1.....\left(I\right) \end{gathered}$$

Determining whether the given normalization constant is correct

In the given wave function,

${\Theta }_{l,m_l}\left(\theta \right)=\sqrt{\frac{3}{4\pi }}\cos \theta$

Check equation (I) as,

$$\begin{gathered} =\frac{3}{4\pi }\times 2\pi \underset{0}{\overset{\pi }{\int }}{\cos }^2\theta \sin \theta d\theta \\ =\frac{3}{2}\underset{0}{\overset{\pi }{\int }}{\cos }^2\theta \sin \theta d\theta \end{gathered}$$

Let us assume,

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

Then the integral becomes,

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

Thus the normalization is correct.