# Why is it that when the atom has inversion symmetry, its dipole moment vanishes when the atom is in an eigenstate?

I am now reading my lecture notes on dipole moment and there's a point which confused me. It says that:

Let us consider the $$|1s\rangle$$ and $$|2p_x\rangle$$ states of a hydrogen atom. The atom has inversion symmetry, so all the eigenstates have either even (such as $$|1s\rangle$$ ) or odd (such as $$|2p_x\rangle$$). Therefore, when the atom is in an eigenstate, its dipole moment vanishes, i.e.

$$\langle 1s | d |1s \rangle = \langle 2_x | d |2p_x\rangle = 0$$

Can someone please further explain this line?

Suppose $$\psi$$ is even, like $$|1s\rangle$$, then $$\hat d\psi$$ is odd and therefore $$\psi^* \hat d \psi$$ is odd. When we integrate an odd function from $$-\infty$$ to $$\infty$$ the result is zero.
Now suppose $$\psi$$ is odd, like $$|2p_x\rangle$$, then $$\hat d\psi$$ is even and therefore $$\psi^* \hat d \psi$$ is odd. And as before when we integrate from $$-\infty$$ to $$\infty$$ we get zero.
So the result of $$\langle\psi|\hat d|\psi\rangle$$ is zero if $$\psi$$ is either an even function or an odd function.