Landé g factor derivation

1. Jan 20, 2010

Otterhoofd

1. The problem statement, all variables and given/known data

I was looking the calculation of Landé g factor. It starts with

$$\mu=-\frac{e}{2m_{e}} (\vec{L}+2\vec{S})$$ assuming that g of electron =2

The lecture notes then proceed by calculating $$g=1+\frac{J(J+1)+S(S+1)-L(L+1)}{2J(J+1)}$$ using the cosine rule.

2. Relevant equations
the second equation is
$$\mu=-\frac{e}{2m_{e}} (\vec{J}+\vec{S})$$ using $$\vec{L}=\vec{J}-\vec{S}$$

which is, i think, just applying the third hund's rule J=L+S
However, the third Hund's rule also states that for less than half filled
$$J=\left|L-S\right|$$

This then does not give the well known solution posted above. What am i doing wrong? The rest of the calculation is perfectly clear to me, I just don't get the step from
$$\mu=-\frac{e}{2m_{e}} (\vec{L}+2\vec{S})$$ to $$\mu=-\frac{e}{2m_{e}} (\vec{J}+\vec{S})$$

3. The attempt at a solution
Tried various vector equations, but no luck. Please help me, i'm really stuck. I hope and think there is a simple solution! thanks.

2. Jan 20, 2010

vela

Staff Emeritus
$$\vec{J}$$, $$\vec{L}$$, and $$\vec{S}$$ are angular momentum vectors. They're not the same as the quantum numbers j, l, and s. The vector and corresponding quantum number are related by

$$\vec{J}^2 = j(j+1)\hbar^2$$

with analogous relationships for $$\vec{L}$$ and $$\vec{S}$$.

$$\vec{J}$$ is the total angular momentum of the electron, which is just the sum of the orbital angular momentum $$\vec{L}$$ and its spin $$\vec{S}$$.

3. Jan 20, 2010

jdwood983

Not really. The relationship is actually

$$\vec{J}^2|j,m\rangle=j(j+1)\hbar^2|j,m\rangle$$

and similarly for $\vec{L}$ and $\vec{S}$. Recall that they are operators and you need to operate them on something to get the quantum numbers.

4. Jan 20, 2010

vela

Staff Emeritus
D'oh! Yes, you're right of course. I was sloppy.

5. Jan 21, 2010

Otterhoofd

Thanks, i already thought this had to be the case. Explanation in my lecture notes is a bit sloppy I think.

Thanks for your explanation, everything is clear to me again!