Landé g factor derivation

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Homework Statement



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

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

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


Homework Equations


the second equation is
[tex]\mu=-\frac{e}{2m_{e}} (\vec{J}+\vec{S}) [/tex] using [tex]\vec{L}=\vec{J}-\vec{S}[/tex]

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
[tex]J=\left|L-S\right|[/tex]

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
[tex]\mu=-\frac{e}{2m_{e}} (\vec{L}+2\vec{S})[/tex] to [tex]\mu=-\frac{e}{2m_{e}} (\vec{J}+\vec{S}) [/tex]

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.
 

Answers and Replies

  • #2
vela
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[tex]\vec{J}[/tex], [tex]\vec{L}[/tex], and [tex]\vec{S}[/tex] 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

[tex]\vec{J}^2 = j(j+1)\hbar^2[/tex]

with analogous relationships for [tex]\vec{L}[/tex] and [tex]\vec{S}[/tex].

[tex]\vec{J}[/tex] is the total angular momentum of the electron, which is just the sum of the orbital angular momentum [tex]\vec{L}[/tex] and its spin [tex]\vec{S}[/tex].
 
  • #3
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The vector and corresponding quantum number are related by

[tex]\vec{J}^2 = j(j+1)\hbar^2[/tex]

with analogous relationships for [tex]\vec{L}[/tex] and [tex]\vec{S}[/tex].
Not really. The relationship is actually

[tex]
\vec{J}^2|j,m\rangle=j(j+1)\hbar^2|j,m\rangle
[/tex]

and similarly for [itex]\vec{L}[/itex] and [itex]\vec{S}[/itex]. Recall that they are operators and you need to operate them on something to get the quantum numbers.
 
  • #4
vela
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Not really. The relationship is actually

[tex]
\vec{J}^2|j,m\rangle=j(j+1)\hbar^2|j,m\rangle
[/tex]

and similarly for [itex]\vec{L}[/itex] and [itex]\vec{S}[/itex]. 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.
 
  • #5
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!
 

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