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

<br /> \vec{J}^2|j,m\rangle=j(j+1)\hbar^2|j,m\rangle<br />

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

<br /> \vec{J}^2|j,m\rangle=j(j+1)\hbar^2|j,m\rangle<br />

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!