6
$\begingroup$

The spin-only formula

$$\mu_\mathrm{so} = \mu_\mathrm{B} \sqrt{n(n+2)} = \mu_\mathrm{B}\cdot 2\sqrt{S(S+1)}$$

is usually a good first approximation to calculate the magnetic moment of transition metal complexes. (I am aware of its limitations.) The usual derivation of this formula involves the Zeeman splitting of the ground state into its separate $M_S$ states under the influence of a magnetic field, followed by evaluating the average magnetic moment using a Boltzmann distribution.

However, if I were to write that $\vec{\mu} = \gamma_\mathrm{e} g_\mathrm{e} \vec{S}$, then I could arrive at the simpler "derivation"

$$\begin{align} |\vec{\mu}| &= |\gamma_\mathrm{e}|\cdot g_\mathrm{e} \cdot |\vec{S}| \\ &= \frac{e}{2m_\mathrm{e}} \cdot 2 \cdot \sqrt{S(S+1)}\hbar \\ &= \mu_\mathrm{B}\cdot 2\sqrt{S(S+1)} \end{align}$$

Is this a coincidental result? I'm assuming there is something wrong with the second "derivation", or even worse, it is just an entirely unphysical interpretation, since I am using $S$ instead of $S_z$. What is the mistake?

$\endgroup$
  • 1
    $\begingroup$ The first equation has units of Bohr Magneton as does your last one. As I understand it the 'derivation' applies to the spin of an isolated electron, i.e it relates its spin magnetic moment to its spin angular momentum, so I think that the similarity is accidental. $\endgroup$ – porphyrin Mar 25 '17 at 17:59
  • $\begingroup$ @porphyrin Which part of it would not be applicable to a multi-electron system (after coupling of $s_1$, $s_2$, .. to give $S$)? $\endgroup$ – orthocresol Mar 25 '17 at 18:14
  • $\begingroup$ With more than one spin then you will have to work out how to properly add the angular momentum, so probably end up with the full derivation anyway. $\endgroup$ – porphyrin Mar 26 '17 at 9:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.