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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

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  • $\begingroup$ Could you give us a reference? $\endgroup$
    – Semoi
    Jun 23, 2020 at 20:53
  • $\begingroup$ 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 $\endgroup$
    – Dari
    Jun 23, 2020 at 20:56
  • $\begingroup$ Semoi meant can you link the source which says what you are saying $\endgroup$
    – SuperCiocia
    Jun 23, 2020 at 21:11
  • $\begingroup$ It is just a text from a script that was given to us in the class. $\endgroup$
    – Dari
    Jun 23, 2020 at 21:40

1 Answer 1

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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.

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  • $\begingroup$ Thank you for the explanation. I will try to prove this for a given electronic configuration. But truth to be told i am totally clueless right now regarding atom physics. But thanks again $\endgroup$
    – Dari
    Jun 23, 2020 at 21:47
  • $\begingroup$ Just work out the total $F$ for your atom and compute $g_F$. $\endgroup$
    – SuperCiocia
    Jun 23, 2020 at 23:36
  • $\begingroup$ @Dari I don't care about my answers specifically, but it's usually good practice to accept answers to your questions. I am saying this because I see that you now have >12 questions and you have never accepted an answer. $\endgroup$
    – SuperCiocia
    Jul 8, 2020 at 4:47

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