I attempt to understand the parity and dipole operator from Daniel Steck's notes: Quantum and Atom Optics (page no. 152, section 5.1.1). I have also attached a screenshot at the end of the question.

At the beginning of the section 5.1.1, the author defines the parity operator and derives some basic properties. Based on the fact that the parity operator commutes with a Hamiltonian with an even potential function: $$ \left[\Pi, \frac{p_e^2}{2m_e} - \frac{\alpha}{\left\vert \textbf{r}_e\right\vert}\right] = 0,$$ he eventually states that the diagonal matrix elements of dipole operator $\textbf{d} = -e \textbf{r}_e$ vanish, where $\textbf{r}_e$ is the atomic electron position: $$\langle g | \textbf{d} | g \rangle = \langle e | \textbf{d} | e\rangle = 0.$$

Here, $|g \rangle$ and $|e \rangle$ are eigenstates of a two-level Hamiltonian $H = H_A + H_{AF}$, where $H_A = \hbar\omega_0 | e\rangle \langle e|$ is the free atomic Hamiltonian, and $H_{AF} = -\textbf{d} \cdot \textbf{E}$ is the atom-field interaction Hamiltonian with the assumption that the ground state energy $E_g = 0$.

My Question: I don't see how $\langle g | \textbf{d} | g \rangle = \langle e | \textbf{d} | e\rangle = 0$ is true. Could you show me the details?

The author then uses the results to rewrite the total Hamiltonian $$H = H_A + H_{AF} = \hbar\omega_0 + e \, \textbf{r}_e \cdot \textbf{E}.$$ But I don't see how this Hamiltonian $H$ commutes with the parity operator $\Pi$.

My Question: How is it then justified to apply results to the total Hamiltonian $H$ even if it doesn't commute with the parity operator $\Pi$, whereas the results were derived from a Hamiltonian that commutes with the parity operator?

enter image description here


1 Answer 1


Answer to the First Question

I will show that $\langle a \vert \textbf{r}_e \vert a \rangle = 0$, which also answers my first question.

\begin{eqnarray} \langle a \vert \textbf{r}_e \vert a \rangle &=& \langle a \vert I \textbf{r}_e I \vert a \rangle \\ &=& \langle a \vert \Pi^2 \textbf{r}_e \Pi^2 \vert a \rangle \\ &=& \langle a \vert \Pi \, \Pi \textbf{r}_e \Pi \, \Pi \vert a \rangle \\ &=& \langle a \vert \pi_a \, \Pi \textbf{r}_e \Pi \, \pi_a \vert a \rangle, \end{eqnarray} where the eigenvalue equation for parity operator $\Pi$: $$\Pi \vert a \rangle = \pi_a \vert a \rangle$$ is used. The parity operator $\Pi$ has two eigenvalues $\pm 1$. We assume $\pi_a = 1$.

Then we get, \begin{eqnarray} \langle a \vert \textbf{r}_e \vert a \rangle &=& \langle a \vert \Pi \textbf{r}_e \Pi \vert a \rangle \\ &=& -\langle a \vert \textbf{r}_e \vert a \rangle, \tag{1} \end{eqnarray} where we have used the definition of the parity operator in the notes: $\Pi \textbf{r}_e \Pi = -\textbf{r}_e$.

From Eq. (1) it follows that $\langle a \vert \textbf{r}_e \vert a \rangle = 0$.

Answer to the Second Question

Here author applies the results to $H_A$, not to $H$. Note that $H_A$ has eigenstates $|g\rangle$ and $|e\rangle$ and it commutes with the parity operator $\Pi$.

  • $\begingroup$ For the first question you could also argue that $\langle e \vert \textbf{r}_e \vert e \rangle$ and $\langle g \vert \textbf{r}_e \vert g \rangle$ measure the center of mass in the ground and excited state, respectively. The center of mass (= center of charge) should coincide with the nucleus. $\endgroup$
    – A. P.
    Nov 23, 2022 at 13:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.