# The stark effect on ground state [closed]

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?

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

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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.