Question

Show that the spontaneous emission rate (Equation 9.56) for a transition from $\mathit{n}\mathbf{,}\mathit{l}$to${\mathit{n}}^{\mathbf{\text{'}}}\mathbf{,}{\mathit{l}}^{\mathbf{\text{'}}}$ in hydrogen is

$\frac{{\mathbf{e}}^{\mathbf{2}}{\mathbf{\omega }}^{\mathbf{2}}{\mathbf{l}}^{\mathbf{2}}}{\mathbf{3}\mathbf{\pi }\dot{{\mathbf{o}}_{\mathbf{0}}{\mathbf{hc}}^{\mathbf{3}}}}\mathbf{\times }\mathbf{\{}\begin{array}{ll}\frac{\mathbf{l}\mathbf{+}\mathbf{1}}{\mathbf{2}\mathbf{l}\mathbf{+}\mathbf{1}}& \mathbf{if}\mathbf{}{\mathbf{l}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{l}\mathbf{+}\mathbf{1}\\ \frac{\mathbf{l}}{\mathbf{2}\mathbf{l}\mathbf{-}\mathbf{1}}& \mathbf{if}\mathbf{}{\mathbf{l}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{l}\mathbf{-}\mathbf{1}\end{array}$

where

$\mathit{l}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}{\mathit{r}}^{\mathbf{3}}{\mathit{R}}_{\mathbf{n}\mathbf{l}}\mathbf{\left(}\mathit{r}\mathbf{\right)}{{\mathit{R}}_{\mathbf{n}}}^{\mathbf{\text{'}}}{\mathit{J}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathit{d}\mathit{r}$

(The atom starts out with a specific value of m, and it goes to$\mathit{a}{\mathit{m}}_{\mathbf{y}}$of the state’s mconsistent with the selection rules:${\mathit{m}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{m}\mathbf{+}\mathbf{1}\mathbf{,}\mathit{m}$ or m -1 . Notice that the answer is independent of m .) Hint: First calculate all the nonzero matrix elements of x,y,and z between role="math" localid="1658313179553" $\mathbf{\left|}\mathbf{n}\mathbf{\right|}\mathit{m}\mathbf{>}$and$\overline{){\mathbf{n}}^{\mathbf{\text{'}}}{\mathbf{l}}^{\mathbf{\text{'}}}{\mathbf{m}}^{\mathbf{\text{'}}}\mathbf{>}}$for the case . From these, determine the quantity

Then do the same for${\mathit{l}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{l}\mathbf{-}\mathbf{1}$.

Short answer

Hence showed that $A=\frac{q^2{l}_{l-1}^2{\omega }_0^3}{3\dot{o_0\mathrm{\pi }\sout{\mathrm{h}}{\mathrm{c}}^3}}\frac{l+1}{2l+1}$and $A=\frac{q^2{l}_{l-1}^2{\omega }_0^3}{3\dot{o_0\mathrm{\pi }\sout{\mathrm{h}}{\mathrm{c}}^3}}\frac{l+1}{2l-1}$.

Step-by-step solution

Define Spontaneous emission rate.

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

$\mathit{A}\mathbf{=}\frac{{\mathbf{\omega }}_{\mathbf{0}}^{\mathbf{3}}{\left|\wp \right|}^{\mathbf{2}}}{\mathbf{3}{\mathbf{\in }}_{\mathbf{0}}\mathbf{\pi }\sout{\mathbf{h}}{\mathbf{c}}^{\mathbf{3}}}$… (1)

Here, the matrix element of the diploe moment,$\mathit{\wp }\mathbf{=}\mathit{q}<{\psi }_b\left|r\right|{\psi }_a>$… (2)

The hydrogen wave function is given by, ${\mathit{\psi }}_{\mathbf{n}\mathbf{l}\mathbf{m}}\mathbf{=}{\mathit{R}}_{\mathbf{n}}\mathbf{\left(}\mathit{r}\mathbf{\right)}{\mathit{Y}}_{\mathbf{l}}^{\mathbf{m}}\left(\theta ,\varphi \right)$… (3) Here,${\mathit{R}}_{\mathbf{n}}$is a radial function and ${\mathit{Y}}_{\mathbf{l}}^{\mathbf{m}}$is a spherical hormonic.

Obtain the transition using the integrals of spherical harmonics.

The transitions are allowed from a state $\overline{)nlm>}$only for $l\to l\pm 1$and $m\to m,m\pm 1$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:

$$\begin{gathered} {\int }_0^{2x}{\int }_0^{\mathrm{\pi }}{Y}_l^m\left(\theta \varphi \right)Y_0^0\left(\theta \varphi \right)Y_l^{-m}\left(\theta \varphi \right)d\theta d\varphi =\frac{1}{\sqrt{4\mathrm{\pi }}} \\ {\int }_0^{2x}{\int }_0^{\mathrm{\pi }}{Y}_l^m\left(\theta \varphi \right)Y_1^0\left(\theta \varphi \right)Y_{l+1}^{-m}\left(\theta \varphi \right)d\theta d\varphi =\sqrt{\frac{3}{4\mathrm{\pi }}}\sqrt{\frac{(l+m+1)(l-m+1)}{(2l+1)(2l+3)}} \\ {\int }_0^{2x}{\int }_0^{\mathrm{\pi }}{Y}_l^m\left(\theta \varphi \right)Y_1^0\left(\theta \varphi \right)Y_{l-1}^{-(m+1)}\left(\theta \varphi \right)d\theta d\varphi =\sqrt{\frac{3}{8\mathrm{\pi }}}\sqrt{\frac{(l+m+1)(l-m+2)}{(2l+1)(2l+3)}} \\ {\int }_0^{2x}{\int }_0^{\mathrm{\pi }}{Y}_l^m\left(\theta \varphi \right)Y_1^0\left(\theta \varphi \right)Y_{l-1}^{-(m+1)}\left(\theta \varphi \right)d\theta d\varphi =\sqrt{\frac{3}{8\mathrm{\pi }}}\sqrt{\frac{(l-m)(l-m-1)}{(2l-1)(2l+1)}} \\ And{\int }_0^{2x}{\int }_0^{\mathrm{\pi }}{Y}_{l_1}^{m_1}\left(\theta \varphi \right)Y_{l_2}^{m_2}\left(\theta \varphi \right)Y_{l_2}^{-m_2}\left(\theta \varphi \right)d\theta d\varphi =0,m_1+m_2\ne m_3 \end{gathered}$$

Find <ψbxψa>,<ψbyψa>and <ψb|z|ψa> .

In spherical coordinates, the values are:

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

In terms of the spherical harmonics as:

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

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

localid="1658318624067" ……(5)

localid="1658376948601"

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

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

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" ……. (8)

localid="1658378337745" ……. (9)

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

localid="1658378363128" ……. (10)

localid="1658378375668" ……. (11)

Find the value of A.

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:

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

Substitute into (1) to get:

role="math" localid="1658382762842" $A=\frac{q^2{l}_{l+1}^2{\omega }_0^3}{3\dot{0_0\mathrm{\pi }\sout{\mathrm{h}}{\mathrm{c}}^3}}\frac{l+1}{2l+1}$

Find the non-zero matrix element.

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

……. (12)

role="math" localid="1658381942976"

Separate and to get:

Put all these results to get the dipole moment:

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

Substitute into (1) we get:

$A=\frac{q^2{l}_{l-1}^2{\omega }_0^3}{3{\dot{o}}_0\mathrm{\pi }\sout{\mathrm{h}}{\mathrm{c}}^3}\frac{l}{2l-1}$

Hence proved.