What are "parity considerations" in deciding the form of the Hamiltonian?

Question

In "introductory Quantum Optics", by Gerry and Knight, the Jeynes model is considered. In this model of electron-EM field interaction the electron is approximated by a two state system ($\lvert g\rangle$ and $\lvert e\rangle$), and the form of the dipole operator $\hat{d}$ is said to be constrained by parity consideration not to have on-diagonal terms:

Only the off-diagonal elements of the dipole operator are nonzero, since by parity consideration $\langle e\rvert\hat{d}\lvert e\rangle=0=\langle g\rvert\hat{d}\lvert g\rangle$.

Why? What does parity have to do with it?

Answer 1

This comes from the microscopic origin of the model. For example, in the case of the hydrogen atom, the dipole operator is given by (up to some signs) $\hat d=e \hat z E$ where I have assumed that the electric field is in the direction $z$, and $\hat z$ is the position operator of the electron (of charge $-e$).

Let's now have a look at the effects of the parity operator $\hat \Pi$. We have $[\hat H,\hat \Pi]=0$, meaning that the eigenstates such that $\hat\Pi\,|g/e\rangle=\pm|g/e\rangle$ and we also have $\hat \Pi\, \hat z\,\hat \Pi=-\hat z$. It is thus easy to show that $\langle g/e|\,\hat z\,|e/g\rangle=0$ by symmetry, which answers the question.

Microscopically, one can show that the selection rule of the of the matrix elements of $\hat z$ between the eigenstates $|nlm\rangle$ of the hydrogen atom are such that $\langle nlm|\,\hat z \,|n\,'l\,'m\,'\rangle\propto \delta_{m,m'}\delta_{l,l\,'\pm1}$.