# 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

• Could you give us a reference? Jun 23, 2020 at 20:53
• Reference? I simply cannot (from the many things) understand how can i for a given Term Symbol after calculating Lande g-factor and Zeeman energy (for orbital spin coupling) in a weak magnetic field say that no splitting takes place. Like, in which cases Splitting does not take place
– Dari
Jun 23, 2020 at 20:56
• Semoi meant can you link the source which says what you are saying Jun 23, 2020 at 21:11
• It is just a text from a script that was given to us in the class.
– Dari
Jun 23, 2020 at 21:40

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.
• Just work out the total $F$ for your atom and compute $g_F$. Jun 23, 2020 at 23:36