2
$\begingroup$

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.

$\endgroup$

1 Answer 1

2
$\begingroup$

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 $g_J$.

$\endgroup$
3
  • $\begingroup$ So if we are asked to find out the magnetic moment of electron what should we do . Method II for sure !! But what is wrong with method I which is precisely my doubt. $\endgroup$
    – Kartheek
    Oct 13, 2014 at 22:28
  • 2
    $\begingroup$ You have to use method 2. As I said method 1 does not work as you measure the expectation value in the $\vec{J}$ direction and not its length squared. This is related to the Wigner-Eckart theorem ( en.wikipedia.org/wiki/Wigner%E2%80%93Eckart_theorem ) which I know too little about to explain more. Maybe someone else can comment on this. $\endgroup$
    – physicus
    Oct 13, 2014 at 22:55
  • 1
    $\begingroup$ Actually i know that is related to Wigner- Eckart theorem , I want some one to explain that theorem in simple terms as it is looking difficult to understand . $\endgroup$
    – Kartheek
    Oct 14, 2014 at 2:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.