What am I missing in the derivation of Landé g factor?

[Otterhoofd]

Homework Statement

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.

Homework 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 $$\mu=-\frac{e}{2m_{e}} (\vec{J}+\vec{S})$$

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.

 

[vela]

$$\vec{J}$$, $$\vec{L}$$, 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 $$\vec{S}$$.

 

[jdwood983]

The vector and corresponding quantum number are related by

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

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

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.

 

[vela]

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.

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

 

[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!