The Stark Effect for n=2 Hydrogen.

The Stark effect for the n=2 states of hydrogen requires the use of degenerate state perturbation theory since there are four states with (nearly) the same energies. For our first calculation, we will ignore the hydrogen fine structure and assume that the four states are exactly degenerate, each with unperturbed energy of $E_0$. That is $H_0\phi_{2\ell m}=E_0\phi_{2\ell m}$. The degenerate states $\phi_{200}$, $\phi_{211}$, $\phi_{210}$, and $\phi_{21(-1)}$.

The perturbation due to an electric field in the z direction is $H_1 = +e{\cal E}z$. So our first order degenerate state perturbation theory equation is

$$\bgroup\color{black}\sum_i\alpha_i\left<\phi^{(j)}_n\left\ver... ...right\vert\phi^{(i)}\right> = (E_0+E^{(1)}_n) \alpha_j.\egroup$$

Because of the exact degeneracy assumed, $H_0$ and $E_0$ can be eliminated from the equation.

$$\bgroup\color{black}\sum_i\alpha_i(E_0\delta_{ij}+\left<\phi^... ...ight\vert\phi^{(i)}\right>) = (E_0+E^{(1)}_n) \alpha_j.\egroup$$

$$\bgroup\color{black}E_0\alpha_j + \sum_i\alpha_i\left<\phi^{(... ...ert\phi^{(i)}\right> = E_0\alpha_j +E^{(1)}_n \alpha_j.\egroup$$

$$\bgroup\color{black}\sum_i\alpha_i\left<\phi^{(j)}_n\left\ver... ...l E}z\right\vert\phi^{(i)}\right> = E^{(1)}_n \alpha_j.\egroup$$

This is just the eigenvalue equation for $H_1$ which we can write out in (pseudo)matrix form

$$\bgroup\color{black}\left(\matrix {\cr H_1 \cr\cr}\right) \l... ...x {\alpha_1 \cr\alpha_2 \cr\alpha_3 \cr\alpha_4}\right) \egroup$$

Now, in fact, most of the matrix elements of $H_1$ are zero. We will show that because $[L_z,z]=0$, that all the matrix elements between states of unequal $m$ are zero. Another way of saying this is that the operator z doesn't ``change'' $m$. Here is a little proof.

$$\bgroup\color{black}\left< Y_{lm}\left\vert[L_z,z]\right\vert... ...') \left<Y_{lm}\left\vert z \right\vert Y_{l'm'}\right> \egroup$$

This implies that $\left<Y_{lm}\left\vert z \right\vert Y_{l'm'}\right> = 0$ unless $m=m'$.

Lets define the one remaining nonzero (real) matrix element to be $\gamma$.

$$\bgroup\color{black} \gamma = e{\cal E} \left<\phi_{200}\left\vert z\right\vert\phi_{210}\right> \egroup$$

The equation (labeled with the basis states to define the order) is

$$\bgroup\color{black} \matrix {\phi_{200} \cr \phi_{211} \cr \... ...x {\alpha_1 \cr\alpha_2 \cr\alpha_3 \cr\alpha_4}\right) \egroup$$

We can see by inspection that the eigenfunctions of this operator are $\phi_{211}$, $\phi_{21-1}$, and ${1\over\sqrt{2}}\left( \phi_{200} \pm\phi_{210} \right)$ with eigenvalues (of $H_1$) of 0, 0, and $\pm\gamma$.

What remains is to compute $\gamma$. Recall $Y_{00}={1\over\sqrt{4\pi}}$ and $Y_{10}=\sqrt{3\over 4\pi}\cos\theta$.

$$\bgroup\color{black} \gamma = e{\cal E} \int{ \left(2a_0\righ... ...}} \left({r\over{a_0}}\right) e^{-r/2a_0} Y_{10} d^3r }\egroup$$

$$\bgroup\color{black} = 2e{\cal E} \left(2a_0\right)^{-3}{1\ov... ...\int{ {1\over\sqrt{4\pi}} \cos{\theta} Y_{10} d\Omega } \egroup$$

$$\bgroup\color{black} = 2eE(2)^{-3}{1\over\sqrt{3}}{1\over\sqr... ...er{a^4_0}} - {r^5\over{2a^5_0}} \right)} e^{-r/a_0} dr \egroup$$

$$\bgroup\color{black} = {a_0 e {\cal E} \over{12}} \left[ \in... ...e^{-x}dx} - {1\over 2}\int^\infty_0 {x^5e^{-x}dx}\right]\egroup$$

$$\bgroup\color{black} = {a_0 e {\cal E} \over{12}} \left[4\cdo... ... 1 \over 2} \right] = {a_0 e {\cal E} \over{12}} (-36) \egroup$$

$$\bgroup\color{black} = -3e{\cal E}a_0 \qquad\Rightarrow\qquad E^{(1)}_n = \mp 3e{\cal E}a_0 \egroup$$

If the states are not exactly degenerate, we have to leave in the diagonal terms of $H_0$. Assume that the energies of the two (mixed) states are $E_0\pm\Delta$, where $\Delta$ comes from some other perturbation, like the hydrogen fine structure. (The $\phi_{211}$ and $\phi_{21(-1)}$ are still not mixed by the electric field.)

$$\bgroup\color{black}\left( \matrix { {E_0 - \Delta} & \gamma ... ...t) = E \left( \matrix { \alpha_1 \cr \alpha_2 }\right) \egroup$$

$$\bgroup\color{black} E = E_0 \pm \sqrt{ \gamma^2 + \Delta^2 } \egroup$$

This is OK in both limits, $\Delta\gg\gamma$, and $\gamma\gg\Delta$. It is also correct when the two corrections are of the same order.