Question

A hydrogen atom is placed in a (time-dependent) electric $\mathbf{field}\mathbf{}\mathbf{E}\mathbf{=}\mathbf{E}\left(t\right)\stackrel{\mathbf{⏜}}{\mathbf{k}\mathbf{.}}\mathbf{calculate}\mathbf{}\mathbf{all}\mathbf{}\mathbf{four}\mathbf{}\mathbf{matrix}\mathbf{}\mathbf{elements}\mathbf{}{\mathbf{H}}_{\mathbf{ij}}^{\mathbf{,}}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{perturbation}\mathbf{}\stackrel{\mathbf{⏜}}{{\mathbf{H}}^{\mathbf{,}}}\mathbf{=}\mathbf{eEz}$between the ground state (n = 1 ) the (quadruply degenerate) first excited states (n = 2 ) . Also show${\mathbf{thatH}}_{\mathbf{ii}}^{\mathbf{,}}\mathbf{=}\mathbf{0}$ for all five states. Note: There is only one integral to be done here, if you exploit oddness with respect to z; only one of the n = 2 states is “accessible” from the ground state by a perturbation of this form, and therefore the system functions as a two-state configuration—assuming transitions to higher excited states can be ignored.

Short answer

$H_{100,210}^{,}=-0.7449eEa.$

Step-by-step solution

Concept of time varying electric field

Consider a hydrogen atom placed in a time-varying electric field:

$\mathit{E}\mathbf{=}\mathit{E}\left(t\right)\stackrel{\mathbf{⏜}}{\mathbf{z}}$

Where E(t) is some arbitrary function of time. The spatial wave functions formula is given by:

${\mathrm{\psi }}_{\mathrm{n}/\mathrm{m}}={\mathrm{R}}_{\mathrm{nl}}{\mathrm{Y}}_{\mathrm{l}}^{\mathrm{m}}$

Calculating all for matrix elements and showing that Hii,=0

${\psi }_{n/m}=R_{n\left|y\right|m}.\mathrm{From}\mathrm{Tables}4.3\mathrm{and}4.7:$

$$\begin{gathered} {\psi }_{100}=\frac{1}{\sqrt{{\mathrm{\pi a}}^3}}{e}^{-r/a};{\psi }_{200}=\frac{1}{\sqrt{8^{\text{'}}{\mathrm{\pi a}}^3}}\left(1-\frac{r}{2a}\right)e^{-r/2a} \\ {\psi }_{210}=\frac{1}{\sqrt{32{\mathrm{\pi a}}^3}}\frac{r}{a}{e}^{-r/2a}\cos \theta ; \\ {\psi }_{21\pm 1}=\pm \frac{1}{\sqrt{64{\mathrm{\pi a}}^3}}\frac{r}{a}{e}^{-r/2a}\sin \theta e^{\pm i\varphi } \end{gathered}$$

$$\begin{gathered} \mathrm{But}\mathrm{r}\mathrm{cos\theta }=\mathrm{z}\mathrm{and}\mathrm{r}{\mathrm{sin\theta e}}^{\pm \mathrm{i\varphi }}=\mathrm{r}\sin \left(\mathrm{cos\varphi }\pm \mathrm{i}\mathrm{sin\varphi }\right)=\mathrm{r}\mathrm{sin\theta }\mathrm{cos\varphi }\pm \mathrm{ir}\mathrm{sin\theta sin\varphi }.\mathrm{so}{\left|\mathrm{\psi }\right|}^{2\mathrm{is}} \\ \mathrm{an}\mathrm{even}\mathrm{function}\mathrm{of}\mathrm{z}\mathrm{in}\mathrm{all}\mathrm{cases},\mathrm{and}\mathrm{hence}\int \mathrm{z}{\left|\mathrm{\psi }\right|}^2\mathrm{dxdydz},{\mathrm{soH}}_{\mathrm{ii}}^{,}=0\mathrm{moreover},{\mathrm{\psi }}_{100}\mathrm{is} \\ \mathrm{even}\mathrm{in}\mathrm{z},\mathrm{and}\mathrm{so}\mathrm{are}{\mathrm{\psi }}_{200},{\mathrm{\psi }}_{211},\mathrm{and}{\mathrm{\psi }}_{21-1},\mathrm{so}{\mathrm{H}}_{\mathrm{ij}}^{,}=0\mathrm{for}\mathrm{all}\mathrm{except} \\ {\mathrm{H}}_{100,210}^{,}=\mathrm{eE}\frac{1}{\sqrt{{\mathrm{\pi a}}^3}}\frac{1}{\sqrt{32{\mathrm{\pi a}}^3}}\frac{1}{\mathrm{a}}\int {\mathrm{e}}^{-\mathrm{r}/\mathrm{a}}{\mathrm{e}}^{-\mathrm{r}/2\mathrm{a}}{\mathrm{z}}^2{\mathrm{d}}^3\mathrm{r}=\frac{\mathrm{eE}}{4\sqrt{2{\mathrm{\pi a}}^4}}\int {\mathrm{e}}^{-3\mathrm{r}/2\mathrm{a}}{\mathrm{r}}^2{\cos }^2{\mathrm{\theta r}}^2\mathrm{sin\theta drd\theta }. \\ =\frac{\mathrm{eE}}{4\sqrt{2{\mathrm{\pi a}}^4}}{\int }_0^{\infty }{\mathrm{r}}^4{\mathrm{e}}^{-3\mathrm{r}/2\mathrm{a}}\mathrm{dr}{\int }_0^{\mathrm{\pi }}{\cos }^2\mathrm{\theta sin\theta d\theta }{\int }_0^{2\mathrm{\pi }}\mathrm{d\varphi }=\frac{\mathrm{eE}}{4\sqrt{2{\mathrm{\pi a}}^4}}4!{\left(\frac{2\mathrm{a}}{3}\right)}^5\frac{2}{3}2\mathrm{\pi }=\left(\frac{2^8}{3^5\sqrt{2}}\right) \\ \mathrm{or}0:7449\mathrm{eEa}. \end{gathered}$$