I am wandering from where the equation for Zeeman effect in weak field $\mathbf{B}$ comes from:

$$ \Delta E=g_j \mu_B M_j, $$ where $g_j$ is giro-magnetic ration (g-factor) defined as: $1+\frac{J(J+1)+S(S+1)-L(L+1)}{2J(J+1)}$ and $\mu_B$ is Bohr's magnetic moment.

I would assume, that this comes from perturbation theory, where we can write:

$$ \Delta E=\langle J M_JLS|H'|J M_JLS\rangle. $$

I know that the $H'=-\mathbf{\hat{\mu} \cdot \hat{B}}$, where $\mathbf{\hat{\mu}}=-\frac{\mu_B}{\hbar}$.

I have trouble calculating the upper expression $\langle J M_JLS|H'|J M_JLS\rangle$. The only idea, that comes to my mind is to rewrite is as wave functions and calculate the whole integral, but that works only for particular set of quantum numbers. I have also trouble to express the dot product $\mathbf{\hat{\mu} \cdot \hat{B}}$ Is there any more general (simpler) approach?

  • 2
    $\begingroup$ are you familiar with the projection theorem? Here is a good explanation of how it is used in order to derive the expression (under the section "Example: Lande $g$-factor") $\endgroup$
    – user275556
    Sep 23, 2021 at 15:36

1 Answer 1


Yes, it comes from perturbation theory.

The Hamiltonian is $H'=-\boldsymbol{\mu} \cdot \mathbf{B}$ and the assumption is that you align your system so that the magnetic field points in the $z$ axis: $\mathbf{B} = B\hat{\mathbf{z}}$.

The magnetic momentum operator $\boldsymbol{\mu}$, for your basis, is usually expressed as: $$ \boldsymbol{\mu} = -\frac{\mu_B g_J \hat{\mathbf{J}}}{\hbar} $$

So now $H'=-\boldsymbol{\mu} \cdot \mathbf{B}$ gives you $\propto \hat{\mathbf{J}} \cdot \hat{\mathbf{z}} = \hat{J_z}$, so: $$ H'=-\boldsymbol{\mu} \cdot \mathbf{B} = \frac{\mu_B\, g_J\, B\, \hat{J_z}}{\hbar}. $$

The action of $\hat{J_z}$ on $|M_J\rangle$ is $\hat{J_z}|M_J\rangle = \hbar M_J|M_J\rangle$, so your perturbation theory term becomes: $$\Delta E=\langle J M_J|H'|J M_J\rangle = \mu_B\, g_J\, B\, M_J $$


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