Close the “loophole” in Equation 9.78 by showing that if${\mathit{l}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{l}\mathbf{=}\mathbf{0}$then$\mathbf{⟨}{\mathbf{n}}^{\mathbf{\text{'}}}{\mathbf{l}}^{\mathbf{\text{'}}}{\mathbf{m}}^{\mathbf{\text{'}}}\mathbf{\left|}\mathbf{r}\mathbf{\right|}\mathbf{n}\mathbf{l}\mathbf{m}\mathbf{⟩}\mathbf{=}\mathbf{0}$

No transitions occurs unless.$\Delta l=\pm 1$. . 9.78

$l^{\text{'}}=l=0$

The selection rules for states, with a spherically symmetric potential, the starting and ending states had to be related by:

$[{(l^{\text{'}}+l+1)}^2-1][{(l^{\text{'}}-l)}^2-1]=0$

One solution of this equation leads to equation 9.78, which states that no transitions occurs unless$\Delta l=\pm 1.$also the solution$l^{\text{'}}=l=0$seems to be correct and acceptable.

Then the two states where the transition occur between them are$⟨n^{\text{'}}00\left|and\right|n00⟩$, the matrix elements of the transition rate formula have the form of:

$⟨{\psi }_b\left|r\right|{\psi }_a⟩=⟨n^{\text{'}}00\left|r\right|n00⟩$

For$l=m=0$ the angular part of the wave function is:

$Y_0^0=\frac{1}{\sqrt{4\pi }}$

So the wave function is constant. In the above equation we have three integrals, and they depend on the values of $x,y$and $z$in spherical coordinates.

For$x=rsin\left(\theta \right)cos\left(\theta \right)$we get:

$⟨n^{\text{'}}00\left|x\right|n00⟩~\underset{0}{\overset{\pi }{\int }}\underset{0}{\overset{2\pi }{\int }}{\sin }^2\left(\theta \right)cos\left(\varphi \right)$

For$y=r\sin \left(\theta \right)\sin \left(\varphi \right)$, we get:

$⟨n^{\text{'}}00\left|y\right|n00⟩~\underset{0}{\overset{\pi }{\int }}\underset{0}{\overset{2\pi }{\int }}{\sin }^2\left(\theta \right)\sin \left(\varphi \right)$

And for,$z=rcos\left(\theta \right)$ we get:

$⟨n^{\text{'}}00\left|z\right|n00⟩~\underset{0}{\overset{\pi }{\int }}\underset{0}{\overset{2\pi }{\int }}\sin \left(\theta \right)\cos \left(\theta \right)d\theta$

Thus:$⟨n^{\text{'}}00\left|r\right|n00⟩=0$