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

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?

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 $\langle nlm|\,\hat z \,|n\,'l\,'m\,'\rangle\propto \delta_{m,m'}\delta_{l,l\,'\pm1}$.
• Thanks! Now I understood that $\langle g | \hat{z} | g \rangle$ since it is the dot product of two electron states with different parity. I have a new question now (maybe I should start a new topic): shouldn't there be an "overall parity operator" that commutes with $\hat{H}_{el} + \hat{H}_{phot} + \hat{H}_{int}$, since even considering the electromagnetic interaction the overall parity should be conserved. Mar 14, 2014 at 17:03
• @Ralph: There is only one parity operator, and it commutes with the total Hamiltonian. It is trivial for the electron and photon part. For the interaction part, notice that both $\hat d$ and $\hat E$ are vectors and therefore both changes sign under parity.