91影视

The spontaneous emission rate for a charge q, in a starting state$\overline{){\mathbf{蠄}}_{\mathbf{a}}\mathbf{>}}$decaying to the state is given by,

$\mathit{A}\mathbf{=}\frac{{\mathbf{蝇}}_{\mathbf{0}}^{\mathbf{3}}{\left|\mathrm{鈩�}\right|}^{\mathbf{2}}}{\mathbf{3}{\mathbf{鈭�}}_{\mathbf{0}}\mathbf{蟺}\sout{\mathbf{h}}{\mathbf{c}}^{\mathbf{3}}}$鈥� (1)

Here, the matrix element of the diploe moment,$\mathit{鈩�}\mathbf{=}\mathit{q}<{蠄}_b\left|r\right|{蠄}_a>$鈥� (2)

The hydrogen wave function is given by, ${\mathit{蠄}}_{\mathbf{n}\mathbf{l}\mathbf{m}}\mathbf{=}{\mathit{R}}_{\mathbf{n}}\mathbf{\left(}\mathit{r}\mathbf{\right)}{\mathit{Y}}_{\mathbf{l}}^{\mathbf{m}}\left(胃,蠒\right)$鈥� (3) Here,${\mathit{R}}_{\mathbf{n}}$is a radial function and ${\mathit{Y}}_{\mathbf{l}}^{\mathbf{m}}$is a spherical hormonic.

The transitions are allowed from a state $\overline{)nlm>}$only for $l鈫�l卤1$and $m鈫�m,m卤1$for the hydrogen atom (the selection rule). Suppose that the starting state is $\overline{)nlm>}$and finishing state is $\overline{)n^{\text{'}}{l}^{\text{'}}{m}^{\text{'}}>}$.

For $l^{\text{'}}=l+1$, work out the transition rates to all possible final values of $m^{\text{'}}$, for $m^{\text{'}}=m+1$and $m^{\text{'}}=m-1$.

Use the integrals of spherical harmonics to get:

$$\begin{gathered} {鈭�}_0^{2x}{鈭�}_0^{\mathrm{蟺}}{Y}_l^m\left(胃蠒\right)Y_0^0\left(胃蠒\right)Y_l^{-m}\left(胃蠒\right)d胃d蠒=\frac{1}{\sqrt{4\mathrm{蟺}}} \\ {鈭�}_0^{2x}{鈭�}_0^{\mathrm{蟺}}{Y}_l^m\left(胃蠒\right)Y_1^0\left(胃蠒\right)Y_{l+1}^{-m}\left(胃蠒\right)d胃d蠒=\sqrt{\frac{3}{4\mathrm{蟺}}}\sqrt{\frac{(l+m+1)(l-m+1)}{(2l+1)(2l+3)}} \\ {鈭�}_0^{2x}{鈭�}_0^{\mathrm{蟺}}{Y}_l^m\left(胃蠒\right)Y_1^0\left(胃蠒\right)Y_{l-1}^{-(m+1)}\left(胃蠒\right)d胃d蠒=\sqrt{\frac{3}{8\mathrm{蟺}}}\sqrt{\frac{(l+m+1)(l-m+2)}{(2l+1)(2l+3)}} \\ {鈭�}_0^{2x}{鈭�}_0^{\mathrm{蟺}}{Y}_l^m\left(胃蠒\right)Y_1^0\left(胃蠒\right)Y_{l-1}^{-(m+1)}\left(胃蠒\right)d胃d蠒=\sqrt{\frac{3}{8\mathrm{蟺}}}\sqrt{\frac{(l-m)(l-m-1)}{(2l-1)(2l+1)}} \\ And{鈭�}_0^{2x}{鈭�}_0^{\mathrm{蟺}}{Y}_{l_1}^{m_1}\left(胃蠒\right)Y_{l_2}^{m_2}\left(胃蠒\right)Y_{l_2}^{-m_2}\left(胃蠒\right)d胃d蠒=0,m_1+m_2鈮�m_3 \end{gathered}$$

In spherical coordinates, the values are:

$$\begin{gathered} x=r\sin \left(\mathrm{胃}\right)\cos \left(\mathrm{蠒}\right) \\ y=r\sin \left(\mathrm{胃}\right)\sin \left(\mathrm{蠒}\right) \\ z=r\cos \left(\mathrm{胃}\right) \end{gathered}$$

In terms of the spherical harmonics as:

$$\begin{gathered} x+iy=r\sin \left(胃\right)e^{i蠒}=-r\sqrt{\frac{8\mathrm{蟺}}{3}}{Y}_1^1 \\ x-iy=r\sqrt{\frac{8\mathrm{蟺}}{3}}{Y}_1^{-1} \\ z=r\sqrt{\frac{8\mathrm{蟺}}{3}}{Y}_1^0 \end{gathered}$$

The only non-zero matrix elements by using equation (4) are:

localid="1658318624067" 鈥︹赌�(5)

localid="1658376948601"

Here,localid="1658376877539" . And also,.

$l_{l+1}={鈭�}_0^{鈭�}{r}^3{R}_{nl}\left(r\right)R_{n(l+1)}\left(r\right)dr.$

To separate x and y, in (6) and (7), note that:

localid="1658376981160"

As it is zero, add it to (6) and subtracted also from (6), so separate x and y to get:

localid="1658378320569" 鈥︹��. (8)

localid="1658378337745" 鈥︹��. (9)

By adding it to (7) and also subtracted from (7), so separate and to get:

localid="1658378363128" 鈥︹��. (10)

localid="1658378375668" 鈥︹��. (11)

Put all these results together into the square modules of the dipole moment to get:

Substitute from (8), (9), (10) and (11) into this equation to get:

${\left|\mathrm{鈩�}\right|}^2=q^2{l}_{l+1}^2\frac{(l+m+1)(l-m+1)+{\displaystyle \frac{1}{2}}\left[(l+m+1)(l+m+2)+(l-m+1)\right]}{(2l+1)(2l+3)}=q^2{l}_{l+1}^2\frac{l+1}{2l+1}$

Substitute into (1) to get:

role="math" localid="1658382762842" $A=\frac{q^2{l}_{l+1}^2{蝇}_0^3}{3\stackrel{藱}{0_0\mathrm{蟺}\sout{\mathrm{h}}{\mathrm{c}}^3}}\frac{l+1}{2l+1}$

Follow the same method for $l^{\text{'}}=l-1$.

鈥︹��. (12)

role="math" localid="1658381942976"

Separate and to get:

Put all these results to get the dipole moment:

role="math" localid="1658382683044" ${\left|\mathrm{鈩�}\right|}^2=q^2{l}_{l+1}^2\left(\frac{(l+1)(l-m)+{\displaystyle \frac{1}{2}}\left[(l+m-1)(l+m)+(l-m-1)(l-m)\right]}{(2l-1)(2l+1)}\right)=q^2{l}_{l-1}^2\frac{l}{2l-1}$

Substitute into (1) we get:

$A=\frac{q^2{l}_{l-1}^2{蝇}_0^3}{3{\stackrel{藱}{o}}_0\mathrm{蟺}\sout{\mathrm{h}}{\mathrm{c}}^3}\frac{l}{2l-1}$

Hence proved.