Zeeman effect-derivation

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?

• 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")
– user275556
Commented Sep 23, 2021 at 15:36

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: $$\Delta E=\langle J M_J|H'|J M_J\rangle = \mu_B\, g_J\, B\, M_J$$