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?


1 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 $\langle nlm|\,\hat z \,|n\,'l\,'m\,'\rangle\propto \delta_{m,m'}\delta_{l,l\,'\pm1}$.

  • $\begingroup$ 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. $\endgroup$
    – Ralph
    Mar 14, 2014 at 17:03
  • $\begingroup$ @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. $\endgroup$
    – Adam
    Mar 15, 2014 at 0:39

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