Calculate the electric dipole moment p and estimate the transition time for a hydrogen atom electron making an electric dipole transition from the to the $\mathbf{(}\mathbf{n}\mathbf{,}\mathbf{l}\mathbf{,}\mathbf{m}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{+}\mathbf{1}\mathbf{)}$ ground state. Comment on the relationship of the result to that in Example 7.11.

The separation of positive and negative charges in a system is referred to as an electric dipole moment.

Initial state is $\left(n,l,m\right)\mathbf{=}\left(2,1,+1\right)$ and the final state is $\left(1,0,0\right)$.

Where, n is the principal quantum number, l is the azimuthal quantum number, ${\mathbf{m}}_{\mathbf{l}}$ is the magnetic quantum number.

Here, the electric dipole moment is given by,

$\mathbf{p}\mathbf{=}\mathbf{-}\mathbf{eRe}\left(\int \overline{r}{\Psi }_{1,0,0}*\left(\overline{r}\right){\Psi }_{2,1,1}\left(\overline{r}\right)r^2\mathrm{sin\theta drd\theta d\varphi }\right)$ ….. (1)

Where, r is the radius, $\mathbf{\theta }$ is the colatitude, role="math" localid="1659863355058" $\mathbf{\varphi }$ is the azimuth, and $\mathbf{\Psi }$ is the wave function.

Wave functions can be calculated by,

${\mathrm{\Psi }}_{1,0,0}*\left(\overline{\mathrm{r}}\right){\mathrm{\Psi }}_{2,1,1}\left(\overline{\mathrm{r}}\right)=\left(\frac{1}{{\left({\mathrm{a}}_0\right)}^{3/2}}2{\mathrm{e}}^{-\mathrm{r}/\left({\mathrm{a}}_0\right)}\sqrt{\frac{1}{4\mathrm{\pi }}}\right)\left(\frac{1}{{\left(2{\mathrm{a}}_0\right)}^{3/2}}\frac{\mathrm{r}}{\sqrt{3{\mathrm{a}}_0}}{\mathrm{e}}^{-\mathrm{r}/\left(2{\mathrm{a}}_0\right)}\sqrt{\frac{3}{8\mathrm{\pi }}}{\mathrm{e}}^{+\mathrm{i\varphi }}\right)$

Where, ${\mathrm{a}}_0$is the radius of hydrogen atom.

Considering the first integration over $\mathrm{\varphi }$, the second term has ${\mathrm{e}}^{+\mathrm{i\varphi }}$, which will cause the z-component to integrate to zero.

The x-component will have,

$$\begin{gathered} {\int }_0^{2\mathrm{\pi }}\mathrm{cos\varphi }\left(\mathrm{cos\varphi }+\mathrm{isin\varphi }\right)\mathrm{d\varphi }={\int }_0^{2\mathrm{\pi }}{\cos }^2\mathrm{\varphi d\varphi }+\mathrm{i}{\int }_0^{2\mathrm{\pi }}\mathrm{cos\varphi sin\varphi d\varphi } \\ ={\left[\frac{1}{2}\mathrm{\varphi }+\frac{1}{4}\sin 2\mathrm{\varphi }\right]}_0^{2\mathrm{\pi }}+{\left[-\frac{1}{2}{\cos }^2\mathrm{\varphi }\right]}_0^{2\mathrm{\pi }} \\ =\left[\frac{2\mathrm{\pi }}{2}+\frac{1}{4}\sin 4\mathrm{\pi }-0\right]+\mathrm{i}\left[-\frac{1}{2}{\cos }^2\left(2\mathrm{\pi }\right)+\frac{1}{2}{\cos }^2\left(2\mathrm{\pi }\right)\right] \\ =\mathrm{\pi } \end{gathered}$$

You have, y-component as,

$$\begin{gathered} {\int }_0^{2\mathrm{\pi }}\mathrm{cos\varphi }\left(\mathrm{cos\varphi }+\mathrm{isin\varphi }\right)\mathrm{d\varphi }={\int }_0^{2\mathrm{\pi }}{\cos }^2\mathrm{\varphi d\varphi }+\mathrm{i}{\int }_0^{2\mathrm{\pi }}\mathrm{cos\varphi sin\varphi d\varphi } \\ =0+\mathrm{i\pi } \end{gathered}$$

Similarly in x and y terms, the role="math" localid="1659863936500" $\theta$integration is

${\int }_0^{\mathrm{\pi }}{\sin }^3\mathrm{\theta d\theta }=\frac{4}{3}$

Now, by putting everything in eq. 1, you get,

$$\begin{gathered} \mathrm{p}=-\mathrm{eRe}\left({\mathrm{e}}^{+\mathrm{it}∆\mathrm{E}/\mathrm{h}}\frac{\mathrm{\pi }\hat{\mathrm{x}}+\mathrm{i\pi }\hat{\mathrm{y}}}{8{{\mathrm{\pi a}}^4}_0}\frac{4}{3}\int {\mathrm{r}}^2{\mathrm{r}}^2{\mathrm{e}}^{-\mathrm{r}/\left(2{\mathrm{a}}_0\right)}\mathrm{dr}\right) \\ \mathrm{p}=-\mathrm{eRe}\left({\mathrm{e}}^{+\mathrm{it}∆\mathrm{E}/\mathrm{h}}\frac{\hat{\mathrm{x}}+\mathrm{i}\hat{\mathrm{y}}}{8{{\mathrm{\pi a}}^4}_0}\frac{4}{3}\frac{4!}{{\left(3/2{\mathrm{a}}_0\right)}^5}\right) \\ \mathrm{p}=-\mathrm{eRe}\left({\mathrm{e}}^{+\mathrm{it}∆\mathrm{E}/\mathrm{h}}\left(\hat{\mathrm{x}}+\mathrm{i}\hat{\mathrm{y}}\right){\mathrm{a}}_0\frac{2^7}{3^5}\right) \end{gathered}$$

The amplitude of this vector is

${\mathrm{ea}}_0\frac{2^7}{3^5}=0.53{\mathrm{ea}}_0$

Hence,

$$\begin{gathered} \mathrm{p}=0.53\left(1.6\times {10}^{-19}\mathrm{C}\right)\left(0.0529\times {10}^{-9}\mathrm{m}\right) \\ =4.5\times {10}^{-30}\mathrm{Cm} \end{gathered}$$

The frequency is same as in Example 7.11, hence the transition time will be

$$\begin{gathered} \mathrm{Transition}\mathrm{time}\cong \frac{12\left(8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2\right){\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}^3\left(1.055\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)}{{\left(4.5\times {10}^{-30}\mathrm{Cm}\right)}^2{\left(1.55\times {10}^{16}{\mathrm{s}}^{-1}\right)}^3} \\ \cong 4\mathrm{ns} \end{gathered}$$

The character of charge oscillation is different, but the estimated transition time is approximately the same as in the example.