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Two electrons: Singlet $S=0$ and triplet $S=1$

Clock transition: $^{1}S_0 \leftrightarrow {}^3P_0$

1st stage MOT: $^1 S_0 \leftrightarrow {}^1P_{1}$

2nd stage MOT: $^1 S_0 \leftrightarrow {}^3P_{1}$

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$X^2$

${\rm Be}$, ${\rm Mg}$, ${\rm Ca}$, ${\rm Sr}$, ${\rm Ba}$, ${\rm Ra}$

d-block ${\rm Zn}$, ${\rm Cd}$, ${\rm Hg}$

f-block ${\rm Yb}$, ${\rm No}$

M. Inguscio, 2013, Oxford

Intercombination (semi-forbidden) transition ($^1 L_J \leftrightarrow {}^3L'_{J'}$)

Alkaline-earth atoms have two valence electrons. Therefore the total spin angular momentum could only have $S=0$ or $S=1$; the singlet state ($S=0$) and triplet states ($S=1$).

• $\left|S=1, m_S=1\right> = \left|\uparrow\uparrow\right>$ $\left|S=1, m_S=0\right> = \frac{1}{\sqrt{2}}(\left|\uparrow\downarrow\right>+\left|\downarrow\uparrow\right>)$ $\left|S=1, m_S=-1\right> = \left|\downarrow\downarrow\right>$
• $\left|S=0, m_S=0\right> = \frac{1}{\sqrt{2}}(\left|\uparrow\downarrow\right>-\left|\downarrow\uparrow\right>)$

The transition between singlet state ($^1L_J$) and triplet states ($^3L'{J'}$) is forbidden under the perfect LS coupling. However, due to the finite mixing between $^1L'{J'}$ and $^3L'{J'}$ comes from the spin-orbit coupling, **intercombination (semi-forbidden) transition ($^1 L_J \leftrightarrow {}^3L'{J'}$)** happens with narrow linewidth.

• Spin selection rule under the LS coupling: $\Delta S = 0$
• Dressed state, $\left|{}^3L'{J'}\right>{\rm inter}=\cos\theta\left|{}^3L'{J'}\right>+\sin\theta\left|{}^1L'{J'}\right>$
• Narrow linewidth → very cold Doppler temperature

171-Ytterbium: ${}^3P_1$ linewidth: $182{\rm kHz}$, Doppler temperature: $4.4\mu {\rm K}$.

87-Rubidium: $5P_J$ linewidth: $6{\rm MHz}$, Doppler temperature: $146\mu {\rm K}$.

Metastable state ($^3P_0$, $^3P_2$)

The transition between the ground state ($^1S_0$) and two states ($^3P_0$ and $^3P_2$) are also forbidden by angular momentum selection rule.

• $^1S_0\leftrightarrow {}^3P_0$: $J=0 \leftrightarrow 0$