# Degeneracy split in $n=2$ hydrogen due to uniform electric field. Why is main diagonal zero?

I was checking about the stark effect in the hdyrogen in this site for $n=2$. Now, there are four states to deal with, $\psi_{200}, \psi_{21-1}, \psi_{210}, \psi_{211}$. The perturbation hamiltonian $H_1 = -eEz$. To calculate the degeneracy splitting, we need $\langle2lm|z|2l'm'\rangle$.

This site goes as following: $$0 = \langle2lm|[L_z, z]|2l'm'\rangle = (m-m')\langle2lm|z|2l'm'\rangle$$

Therefore, all terms where $m\neq m'$ are zero. And so, assuming the matrix lines are in that order: $|200\rangle$, $|211\rangle$, $|210\rangle$, $|21-1\rangle$ he claims the matrix is of the form: $$H = \begin{bmatrix} 0 & 0 & \gamma & 0 \\ 0 & 0 & 0 & 0 \\ \gamma & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}, \quad\quad \gamma = \langle200|z|210\rangle$$

However, all the terms in the main diagonal are $\langle2lm|z|2lm\rangle$, and therefore, we have $m=m'$, and thus one shouldn't claim such terms are zero. But, I can't see why should the main diagonal be zero, other than by direct calculation. However, if the main diagonal is not zero, the splitting won't be equidistant. And I guess this is wrong.

Speaking of direct calculation, have a look for instance, in this one: $$\psi_{210} = \frac{1}{\sqrt{24}}a^{-3/2}\frac{r}{a}\exp\left(-\frac{r}{2a}\right)\left(\frac{3}{4\pi}\right)^{1/2}\cos\theta$$

And, using $z=r\cos\theta$, we have for this coefficient in the main diagonal: $$\langle210|z|210\rangle = \int_0^\infty\frac{1}{24}a^{-3}\frac{r^2}{a^2}\exp\left(-\frac{r}{a}\right)\frac{3}{4\pi}r^3dr \int_0^\pi\cos^2\theta\sin\theta d\theta \int_0^{2\pi}d\phi \\ \langle210|z|210\rangle = \frac{1}{24}\frac{1}{a^5}\frac{3}{4\pi}\frac{2}{3}2\pi \int_0^\infty r^5\exp\left(-\frac{r}{a}\right)dr \quad\mbox{ since }\quad \int_0^\pi\cos^2\theta\sin\theta d\theta = \frac{2}{3}$$

However, after some manipulations from the definition of gamma function, we can find: $$\int_0^\infty r^5\exp\left(-\frac{r}{a}\right)dr = 5!a^6$$

Therefore: $$\langle210|z|210\rangle = \frac{5!a}{4\cdot 6} = 5a\neq 0$$

Clearly not zero.

Now, because the parity of the spherical harmonics is given by $P=(-1)^l$ and because the above integral must be zero, we conclude that there is a transition rule: $\Delta l = 1 \neq 0 .$