# Stark effect: find ground state

1. Jun 4, 2017

### JulienB

1. The problem statement, all variables and given/known data

Hi everybody! I have a problem related to first-order perturbation theory, and I'm not sure I'm tackling the problem correctly. Here is the problem:

Consider a hydrogen atom in an externally applied electric field $\vec{F}$. Use first-order perturbation theory to find the perturbed ground state wavefunction. (Take $\vec{F}=F\hat{z}$ and, just to make things easier, include only the $n = 2$ states.)

2. Relevant equations

(I use Griffiths as a source)

Correction of wave function: $\psi_n^1 = \sum_{m\neq n} \frac{\langle \psi_m^0 | H' | \psi_n^0 \rangle}{E_n^0 - E_m^0}$

3. The attempt at a solution

I think that the perturbation due to $\vec{F}$ is $H'=eFz$. My main difficulty is to interpret the equation given in Griffiths for a wave function in spherical coordinates. Here is my attempt:

$\psi_{100}^1= \sum_{nlm\neq 100}^{\infty} \frac{|nlm \rangle \langle nlm | H' | 100 \rangle}{E_1^0 - E_n^0}$

Is that the correct way to deal with this equation? Then I calculated $\langle nlm | H' | 100 \rangle$ (generally), and using $z=r \cos \theta$, $Y_0^0 \cos \theta = \frac{1}{\sqrt{3}} Y_1^0$ and the orthogonality of spherical harmonics I get:

$\langle nlm | H' | 100 \rangle = \frac{eF}{\sqrt{3}} \delta_{l1} \delta_{m0} \frac{2}{\sqrt{a^3}} \int dr\ r^3 R_{nl}^* e^{-r/a}$

which means that $\langle nlm | H' | 100 \rangle=0$ for $l\neq 1$, $m\neq 0$. Then the exercise asks to consider only $n=2$ states, so the only $nlm$ triplet for which the correction is not $0$ is $|210\rangle$. So now I can integrate the radial function and get:

$\langle 210 | H' 100 \rangle = \frac{128 \sqrt{2}}{243} eFa$

which takes me to a correction:

$\psi_0^1=\frac{|210\rangle \langle 210 | H' | 100 \rangle}{E_1^0 - E_2^0}$
$= - \frac{256 eF}{729 \sqrt{3}} \frac{1}{Ry} \frac{r}{\sqrt{a^3}} e^{-r/2a}$

(The $-$ comes from the energy difference: $E_1^0 - E_2^0 = -Ry + Ry/4 = -3Ry/4$)

Finally I can add the correction to the unperturbated ground state wave function and get:

$|100\rangle_\text{perturbated} = \frac{1}{\sqrt{\pi a^3}} e^{-r/a} + \psi_0^1$

Does that make sense? The result is kind of ugly, but from what I read on the internet and in Griffiths it would be surprising if that wasn't the case. I've also only very recently started to use the Dirac notation, hopefully I didn't get confused along the way.

Julien.

2. Jun 9, 2017

### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

3. Jun 16, 2017

### kuruman

I did not check your numbers but the method looks right. Note that the perturbed wavefunction you found is not normalized. It's not pretty but that's PT for you.