# Atomic spectroscopy with nuclear spin: hyperfine splitting

In $^{171} Yb$, nuclear spin $1/2$, there is a transition:

$^{2}S_{1/2}(F=0) \rightarrow\; ^{2} F_{7/2} (F=3)$

In this paper on p.3 (top-left) they say that $m_F^{(f)} = 0$ at the final state $^2 F_{7/2}$. At the beginning $m_F^{(i)} = 0$ - the only option. I don't get it: where did one unit of photon polarization go? This transition can't be allowed unless I don't understand the selection rules in this case (which is possible).

Say, we consider LC photon $\lambda = 1$:

1) If there is no nuclear spin, as it is for $^{172} Yb$, there are two transitions: $m_j^{(i)} = 1/2 \rightarrow m_j^{(f)}=3/2$ and $m_j^{(i)} = -1/2 \rightarrow m_j^{(f)}=1/2$.

2) For the case with $^{171} Yb$ shouldn't it be $m_f^{(i)}=0 \rightarrow m_f^{(f)}=1$ the only open transition?

Energy level diagram The transition you are looking at is not the usual electric-dipole transition. In fact, it is a highly forbidden 'octupole' transition (described also on p. 3 of the paper you linked). To see that it is not electric-dipole allowed, we can look at a few selection rules that are broken:

1. The orbital angular momentum changes from $L=0$ (denoted $S$) to $L = 3$ (denoted by $F$). In an electric-dipole transition, this quantum number must change by exactly $\pm 1$.

2. Similarly, the combined electron angular momentum $J$ changes from $1/2$ to $7/2$. In an electric-dipole transition, selection rules allow $J$ to change by 0 or $\pm 1$.

3. The total atom angular momentum $F$ changes from 0 to 3. As with $J$, this value can only change by 0 or $\pm 1$ in electric-dipole transitions.

For these reasons, it is clear that the transition is not electric-dipole allowed. The other property of this transition that you noted is that $m_F =0 \to m_F = 0$. In fact this by itself does not violate selection rules, because this quantity is the projection of the angular momentum onto the quantization axis. For some polarizations of light (ie., parallel to the quantization axis), this can be possible (it's called a $\pi$ transition). For example, a typical Rubidium transition that is electric-dipole allowed would be $|5S_{1/2}, F=1, m_F=0\rangle \to |5P_{3/2}, F=2, m_F =0\rangle$.

In summary, while $m_F = 0 \to m_F =0$ does not break selection rules by itself, the other large changes in angular momentum quantum numbers make the Ytterbium transition highly forbidden.

• Thanks a lot for your response! If I use something like princeton.edu/~romalis/PHYS551/TensorOperators.pdf and extract the selection rules, Clebsch will prohibit $m_F^{(i)} \rightarrow m_F^{(f)}$. So. I think, they were considering some kind of non-collinear configuration... Otherwise, not sure how it could be allowed. – MsTais Jun 8 '18 at 0:27
• If the polarization of the light is parallel to the quantization axis, then $m_F=0 \to m_F =0$ is allowed. In the document you linked, look at the equation between (14) and (15): $\langle 0 0 | \vec{E} \cdot \vec{r} | 1 0 \rangle \ne 0$ – Harry Levine Jun 8 '18 at 2:18