Hydrogen Atom in a E-field, the Stark Effect.

$$\bgroup\color{black} H_0 = {p^2\over {2\mu}} - {Ze^2\over r} \egroup$$

We choose the axes so that the electric field is in the z direction.

$$\bgroup\color{black} H_1 = e{\cal E} z\egroup$$

This is a small perturbation. For non-degenerate states

$$\bgroup\color{black} E^{(1)}_{nlm}=e{\cal E} \langle\phi_{nlm}\vert z\vert\phi_{nlm}\rangle = 0 \egroup$$

$$\bgroup\color{black} E^{(2)}_{100}=e^2 {\cal E}^2 %script ... ...00}\rangle\right\vert^2 \over {E^{(0)}_1 - E^{(0)}_n}} \egroup$$

$$\bgroup\color{black} \langle\phi_{nlm}\vert z\vert\phi_{100}\... ... \int{d^3rR^*_{nl}(r\cos\theta)R_{10} Y^*_{lm} Y_{00} } \egroup$$

$$\bgroup\color{black} Y_{00}\cos{\theta}={1\over\sqrt{4\pi}} \sqrt{4\pi\over 3} Y_{10} \egroup$$

$$\bgroup\color{black} \langle\phi_{nlm}\vert z\vert\phi_{100}\... ...r^3 dr R^*_{nl} R_{10} \int d\Omega Y^*_{lm} Y_{10} } \egroup$$

$$\bgroup\color{black} = {\delta_{\ell 1} \delta_{m0} \over\sqrt{3} } \int{r^3 R^*_{nl} R_{10} dr } \egroup$$

$$\bgroup\color{black} \left\vert\langle\phi_{nlm}\vert z\vert\... ...ight)^{2n -5} \over {\left(n+1\right)^{2n +5} } }a^2_0 \egroup$$

$$\bgroup\color{black} \equiv f(n)a^2_0 \egroup$$

$$\bgroup\color{black} E^{(2)}_{100} = e^2 {\cal E}^2 \sum^{\in... ...0 \over { {-e^2\over{2a_0}} + {e^2 \over {2a_0n^2}} } } \egroup$$

$$\bgroup\color{black} = a^3_0 {\cal E}^2 %script \sum^{\in... ...%script \sum^{\infty}_{n=2} { n^2 f(n) \over {n^2-1}} \egroup$$

$$\bgroup\color{black} E^{(2)}_{100} = -2 a^3_0{\cal E}^2 (0.74 + 0.10 + \dots ) =-2.25a^3_0{\cal E}^2\egroup$$

$$\bgroup\color{black} \Rightarrow d=-{\partial \Delta E \over {\partial {\cal E}}} %script = 4(1.125)a^3_0 {\cal E} \egroup$$

$$\bgroup\color{black} \alpha {\cal E} \Rightarrow \mbox{induced electric dipole moment} \egroup$$