91影视

No transitions occurs unless.$螖l=卤1$. . 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:

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

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

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

$鉄�{蠄}_b\left|r\right|{蠄}_a鉄�=鉄�n^{\text{'}}00\left|r\right|n00鉄�$

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

$Y_0^0=\frac{1}{\sqrt{4蟺}}$

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

$鉄�n^{\text{'}}00\left|x\right|n00鉄�~\underset{0}{\overset{蟺}{鈭�}}\underset{0}{\overset{2蟺}{鈭�}}{\sin }^2\left(胃\right)cos\left(蠒\right)$

For$y=r\sin \left(胃\right)\sin \left(蠒\right)$, we get:

$鉄�n^{\text{'}}00\left|y\right|n00鉄�~\underset{0}{\overset{蟺}{鈭�}}\underset{0}{\overset{2蟺}{鈭�}}{\sin }^2\left(胃\right)\sin \left(蠒\right)$

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

$鉄�n^{\text{'}}00\left|z\right|n00鉄�~\underset{0}{\overset{蟺}{鈭�}}\underset{0}{\overset{2蟺}{鈭�}}\sin \left(胃\right)\cos \left(胃\right)d胃$

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