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Landé g factor derivation

  1. Jan 20, 2010 #1
    1. The problem statement, all variables and given/known data

    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.


    2. Relevant 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]

    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. jcsd
  3. Jan 20, 2010 #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].
     
  4. Jan 20, 2010 #3
    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.
     
  5. Jan 20, 2010 #4

    vela

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    D'oh! Yes, you're right of course. I was sloppy.
     
  6. Jan 21, 2010 #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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