# Relevance of parity argument of dipole selection rules

For context I write once the matrix element relevant for dipole selection rules: $$\left < f|\vec{\epsilon}\cdot\vec{r}|i \right >$$

Reading about the dipole selection rules e.g. here it is often stated that using the Wigner-Eckart Theorem we find that the change of the orbital angular momentum is either $$1$$, $$0$$ or $$-1$$ $$\Delta l=0,\pm1$$. However by parity we must further restrict this result to $$\Delta l=\pm 1$$.

However, let's now consider a specific example. Let's assume $$\vec{\epsilon}=\hat{e_z}$$ such that our operator is proportional to the spherical harmonic $$Y^0_1$$. By the Wigner Eckart theorem I know that: $$\langle l'm'|Y_1^0|lm\rangle \propto \langle10;lm|l'm'\rangle$$ For example, if I plug in explicit wavefunctions of the same parity e.g. $$l=1$$, $$m=1$$, $$l'=1$$ and $$m'=1$$ I have a non-vanishing amplitude! Shouldn't this amplitude vanish by parity? Where is my mistake?

By writing the operator as $$r$$ you are implicitly considering a one electron atom, in which case the total orbital angular momentum $$L$$ and the single electron angular momentum $$l$$ are identical. In this case the parity rule does interact with the angular momentum rule. However in general the rule for total orbital angular momentum $$L$$ is$$\Delta L=0,\pm1$$.

The simplest example of an allowed $$\Delta L=0$$ transition is probably in the carbon spectrum. The carbon ground state is $$1s^22s^22p^2{\,}^3P$$ with even parity. There is a fully allowed transition to $$1s^22s^22p3s{\,}^3P$$ for which $$\Delta l=1$$ ($$p\to s$$) but $$\Delta L=0$$ and $$\Delta S=0$$. This occurs at 1657 nm. (There is fine structure splitting in both triplet terms, but it is quite small.)

With $$Y_1^0$$ you are only considering the linear polarisation term, so the first term in the expansion in the final equation of your link:
$$\propto \langle \ell_n m_n | \ell_i 1 m_i 0 \rangle.$$
From this page, the closest case you can get is: $$\langle L, M| \ell_{1},\ell_{2}, m_{1},m_{2}\rangle =\ \delta _{M,m_{1}+m_{2}} \cdot (\text{bunch of stuff}),$$ amd from your expression $$M = m_n$$, $$m_1 = m_i$$, $$m_2 = 0$$.
Choosing $$m_1 = m_2 = 1$$ as you are trying to do, would give: $$\langle 1, 1| 1, 1, 1,0\rangle =\ \delta _{0,1+0} \cdot (\text{bunch of stuff}) ,$$ so the $$\delta$$ kills the whole thing and you get zero.
• I am not completely sure what you mean. I picked $Y_1^0$ which should correspond to a specific polarized light if I'm not mistaken. Regarding the parity: In the source I have given above they argue with the parity of the spherical functions: "We also know that the $Y_{1m}$ are odd under parity so the other two spherical harmonics must have opposite parity to each other implying that $l_n \neq l_i$" Commented Aug 18, 2021 at 18:30