Total magnetic moment in an atom

Question

I have a doubt regarding the calculation of total angular momentum of electron in an atom. Which is the right way to do it?

Method 1:

Total magnetic moment
$$ \begin{align} \vec{\mu_J} &= \vec{\mu_L}+\vec{\mu_S} \\&= g_L \mu_B\vec{L}+g_s \mu_B\vec{S}. \end{align} $$ Since $g_L = -1$ and $g_S=-2$, $$\begin{align}\vec{\mu_J}& = -\mu_B\vec{L}-2\mu_B\vec{S} \\&= -\mu_B(\vec{L}+2\vec{S}),\end{align}$$ where $$|\mu_J|=\mu_B|\vec{L}+2\vec{S}|$$ and $$|\mu_J|=\mu_B\sqrt{|\vec{L}|^2+4|\vec{S}|^2+4\vec{L}.\vec{S}}.$$

Method 2:

Here we calculate Landé $g$ factor as $$g_J=1+\frac{j(j+1)+s(s+1)-l(l+1)}{2j(j+1)},$$ and then substitute in the equation:
$$|\mu_J| = g \frac{e\hbar}{2m}\sqrt{j(j+1)}.$$

I wanted to know what is wrong with method 1.

Answer 1

The problem is that you do two different things with the two methods. Method 1 gives you the (uniteresting) length of the combined mangetic moment vector while Method 2 gived you its expectation value in the quantization direction which is $\vec{J}=\vec{L}+\vec{S}$. $\vec{\mu}_J$ does obviously not point in the same direction as $\vec{J}$, because of the different g-factors $g_L$ and $g_S$.

If you want to use Method 1 to reproduce the $g_J$ from Method 2 you have to do the following: $$\vec{\mu}_J\cdot\vec{J}=-\mu_B(\vec{L}+2\vec{S})(\vec{L}+\vec{S}).$$ Compute this using $\vec{L}\cdot\vec{S}=\frac{1}{2}(\vec{J}^2-\vec{L}^2-\vec{S}^2)$ and you will reproduce $g_J$.