When considering the Stark Effect, we consider the effect of an external uniform weak electric field which is directed along the positive $z$-axis, $\vec{\varepsilon} = \varepsilon \vec{k}$, 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 $$E_{100} = E_{100}^{(1)} + \epsilon \varepsilon \langle 100| \hat{Z}| 100 \rangle + e^2 \varepsilon^2 \sum_{nlm \neq 100}\frac{|\langle nlm| \hat{Z}| 100 \rangle|^2}{E_{100}^{(0)}-E_{nlm}^{(0)}}.$$ We can show that the second term is zero i.e. $\langle 100| \hat{Z}| 100 \rangle = 0$.
How does it follow from this that the following conclusion can be made "The underlying physics behind this is that when the hydrogen atom is in the ground state, it has no permanent electric dipole moment"?
Thanks for any assistance.
Best Answer
First of all, it should be $z$, not $\hat{z}$. You're taking the mean value of the $z$-coordinate in the ground state.
Now, the ground state has spherical symmetry. This means that whatever $\langle z \rangle$ is, it better be equal to $\langle x \rangle$ and $\langle y \rangle$, since in the ground state $|100\rangle$ there is nothing that makes the $z$ direction special.
The dipole moment operator for the electron is $q \mathbf{r}$, with $q$ its charge and $\mathbf{r}$ the position operator. So the mean value of the dipole moment in the ground state is $q \langle 100 | \mathbf{r} | 100 \rangle$. But $\mathbf{r} = x \mathbf{\hat{x}} + y \mathbf{\hat{y}} + z \mathbf{\hat{z}}$, and since we know that the mean value of the coordinates is zero, we get that the mean value of the dipole moment is zero too.