# Linear Stark effect in Hydrogen: Why do states $\langle 2,0,0|H_1|2,0,0\rangle$ and $\langle 2,1,0|H_1|2,1,0\rangle$ vanish?

I was reading sec. 10.4, example (4), about the 'Linear Stark effect in Hydrogen' in the book 'Quantum Mechanics: A Modern Development' by Leslie E. Ballentine (Also related to example (ii) in ch. 13.1). We consider the first excited state of the Hydrogen atom and denote the degenerate stationary states as $$|n,l,m\rangle$$, where $$n$$ is the principal quantum number, and $$l$$ and $$m$$ are the angular momentum quantum numbers. The first excited state is four-fold degenerate with the degenerate states being $$|2,0,0\rangle, \; |2,1,1\rangle, \; |2,1,0\rangle, \; |2,1,-1\rangle$$. So we can write the degenerate state vector belonging to the first excited state energy, say $$E_1$$, as:

$$|\Psi^{(0)}\rangle=c_1|2,0,0\rangle+c_2|2,1,1\rangle+c_3|2,1,0\rangle+c_4|2,1,-1\rangle$$

The coefficients are determined by diagonalizing the matrix of the perturbation, $$H_1=e\textbf{E}\cdot\textbf{r}=e|\textbf{E}|r\cos{\theta}$$ in the four-dimensional subspace spanned by the four degenerate basis vectors.

The matrix element $$\langle n,l,m|H_1|n',l',m'\rangle$$ vanishes unless $$m=m'$$, and the text states that, with this argument, the only non-vanishing elements in the $$4\times 4$$ matrix are $$\langle 2,1,0|H_1|2,0,0\rangle=\langle 2,0,0|H_1|2,1,0\rangle^*$$. This conclusion confuses me a lot because what about the states $$\langle 2,0,0|H_1|2,0,0\rangle$$ and $$\langle 2,1,0|H_1|2,1,0\rangle$$? These states also fulfill the requirement that $$m=m'$$, so why are these not considered non-vanishing elements of the four-dimensional matrix?

• Hint: parity Commented Aug 6, 2023 at 11:14

Consider any state $$|n,l,m\rangle$$. It has the wave function $$\psi_{nlm}(\vec{r})$$. Then the diagonal matrix element is \begin{align} \langle n,l,m|e\vec{E}\cdot\vec{r}|n,l,m\rangle &= \int d^3r\ \psi_{nlm}^*(\vec{r})\ e\vec{E}\cdot\vec{r}\ \psi_{nlm}(\vec{r}) \\ &= \int d^3r\ \underbrace{e\vec{E}\cdot\vec{r}\ |\psi_{nlm}(\vec{r})|^2}_{f(\vec{r})} \end{align}
The integrand has odd parity, meaning $$f(-\vec{r})=-f(\vec{r})$$. Therefore the contributions from $$\vec{r}$$ and $$-\vec{r}$$ to the integral cancel each other. And hence the integral is $$0$$.