Zeeman Effect and Lande $g$-factor I am studying the Zeeman effect for spin-orbital coupling and there is a section which i do not fully understand:
In case of a weak magnetic field we can show that no splitting occurs by calculating the Lande $g$-factor and the zeeman energy for the spin-orbital coupling.
I know both formulas regarding zeeman energy and $g$-factor for spin-orbital coupling, but what i do not understand is how for a given electronic configuration one can say that there is no splitting judging from the result one get for zeeman energy and $g$-factor
 A: It would be nice if you could link the source from which you are quoting that paragraph, so that we can read the context that statement is made within.
To first-order, the Zeeman energy $\Delta E$ arising from the interation of the atomic magnetic dipole momentum $\boldsymbol{\mu}$ with an external magnetic field $\mathbf{B}$ is given by:
$$ \Delta E =  -\langle \boldsymbol{\mu}\cdot \mathbf{B}\rangle = \\ g_F\, m_F\, \mu_{\mathrm{B}}\,B \quad \mathrm{or} \quad (g_J\, m_J+g_I\,m_I)\, \mu_{\mathrm{B}}\,B  \quad \mathrm{or} \quad (g_S\, m_S+g_L\,m_L + m_I\,g_I)\, \mu_{\mathrm{B}}\,B, $$
where $\mu_{\mathrm{B}}$ is the Borh magneton, and I have chosen the $\mathbf{B}$ field to lie along the $z$ axis. Each of the expressions is using a different basis (and it is hence valid for different strengths of the magnetic fields): $|F, m_F\rangle$, $|m_J, m_I\rangle,$ and $|m_S, m_L, m_I\rangle$ respectively.
For weak fields, you'd use the $|F, m_F\rangle$ basis and hence see that the Zeeman (first-order) correction vanishes for $g_F=0$. $g_F$ is calculated from the quantum numbers $J^2, F^2$ etc. so there might be some combination of those which makes it zero.
