Why does a transition between 2 $s$-orbitals work for cesium's clock transition?

Question

I have a small question about the cesium's clock transition. According to the information on the Wiki: https://en.wikipedia.org/wiki/Caesium_standard, the chosen transitions are two hyperfine ground state, F=3 and F=4. Their orbital angular momentums are $s$-orbital but why can they be a pair of transitions coupling by laser field for clock?

I observe the dipole moment term written in the Cesium D Line by Daniel A. Steck:

$$ \langle F\ m_F|e\vec{r}|F'\ m'_{F}\rangle=\langle F||e\vec{r}||F'\rangle(-1)^{F'-1+m_{F}}\sqrt{2F+1} \begin{pmatrix} F'&1&F\\ m'_F&q&-m_F \end{pmatrix} $$

The matrix part is a Wigner $3j$ matrix.

$$ \begin{split} \langle F||e\vec{r}||F'\rangle &\equiv\langle J\ I\ F||e\vec{r}||J'\ I'\ F'\rangle\\ &=\langle J||e\vec{r}||J'\rangle (-1)^{F'+J+1+I}\sqrt{(2F'+1)(2J+1)} \begin{Bmatrix} J&J'&1\\ F'&F&1 \end{Bmatrix} \end{split} $$

$$ \begin{split} \langle J||e\vec{r}||J'\rangle &\equiv\langle L\ S\ J||e\vec{r}||L'\ S'\ J'\rangle\\ &=\langle L||e\vec{r}||L'\rangle (-1)^{J'+L+1+S}\sqrt{(2J'+1)(2L+1)} \begin{Bmatrix} L&L'&1\\ J'&J&S \end{Bmatrix} \end{split} $$

The curly matrix is Wigner $6j$ matrix. The reduced matrix of $\langle L||e\vec{r}||L'\rangle$ will include a radial integration of two orbital angular momentum states.

$$ \langle L||e\vec{r}||L'\rangle\propto\sqrt{2L'+1}\int^{\infty}_0rR_{n1,L}(r)R_{n_2,L'}(r)r^2dr\langle L'0;1,0|L,0\rangle $$

However, in clock transition, two orbital angular momentum is the same($L=L'=0$). This makes the integration vanises(parity issue) and the later CG coefficient vanish ($L=L'=0$ can not make a triangle with another segment 1). So because this dipole element is zero, there should be no Rabi oscillation and this can not be a transition for clock. However, it works. That is what I am very confused about.

Answer 1

Electric Dipole-allowed (E1) transitions are basically never used for making clocks because their linewidths are large. The reason here is because they are allowed to spontaneous decay via dipole transitions (usually many MHz broad). You are correct that the clock transition in Cesium is not electric dipole allowed, which is one reason why its linewidth is very narrow (sub-Hz).

In the case of Cesium's clock transition, microwaves, not lasers, are typically used to drive the transition using a magnetic dipole transition (i.e. using $\mu \cdot B$ rather than $r\cdot E$). These transitions have an even parity selection rule that changes the spin quantum number but doesn't change the orbital numbers (i.e. $\Delta l = \Delta n =0$). This allows for the $S (F=3) \rightarrow S (F=4)$ transition to occur.

You can read more about the magnetic dipole transition at the links below

https://en.m.wikipedia.org/wiki/Selection_rule

https://farside.ph.utexas.edu/teaching/qm/Quantum/node88.html