0
$\begingroup$

When considering the Stark Effect, we consider the effect of an external uniform weak electric field which is directed along the positive $z$-axis, $\vec E = E \hat z$, on the ground state of a hydrogen atom. Then using nondegenerate perturbation theory it follows that we can approximate the energy of the ground state by

$$E_{100}=E^{(1)}_{100}-eE\langle100|\hat z|100\rangle+e^2 E^2\sum_{nlm \neq 100}\frac{|\langle nlm|\hat z|100\rangle|^2}{E^{(0)}_{100}−E^{(0)}_{nlm}}$$

Where now $\hat z$ is the position operator. We can show that the second term is zero i.e. $⟨100|\hat z|100⟩=0$. What is the stark effect along $x$-axis?

$\endgroup$

closed as off-topic by G. Smith, Buzz, ZeroTheHero, Jon Custer, user191954 Dec 31 '18 at 13:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – G. Smith, Buzz, ZeroTheHero, Jon Custer, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

1
$\begingroup$

The Hamiltonian of the hydrogen atom has $SO(3)$ symmetry, the choice of using $z$ as quantization axis is a pure convention, so if you change the direction of the electric field, your second order formula becomes

$$E_{100}=E^{(0)}_{100}-eE\langle100|\hat x|100\rangle+e^2 E^2\sum_{nlm \neq 100}\frac{|\langle nlm|\hat x|100\rangle|^2}{E^{(0)}_{100}−E^{(0)}_{nlm}}$$

the second term is still zero, while the second one can be shown to be equal to the one calculated with $\hat z$ in a lot of ways. The simplest one is remembering that choosing $z$ as a quantization axis is convention and using now $x$ as new axis.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.