The magnetic moment associated with the ground electronic state of a molecule is obtained by differentiating the negative of the ground state electronic energy with respect to the external magnetic field (Note that the potential energy of a magnetic moment in an external magnetic field is given by ${\displaystyle E=-\mathbf {m} \cdot \mathbf {H} }$).

${\displaystyle m=-\partial E/\partial H}$

3.1

The bulk magnetisation for 1 mole of the sample is then obtained as I = Nm, where N is the Avogadro number. If excited states are also populated one needs to sum over all the electronic states and weight the contribution of each state by its Boltzman factor,

${\displaystyle I={\frac {-N{\sum (\partial E_{i}/\partial H)e^{-E_{i}/kT}}}{\sum e^{-E_{i}/kT}}}}$

3.2

Recalling the definition of the partition function,

${\displaystyle Z=\sum e^{-E_{i}/kT}}$

3.3

and noting that,

${\displaystyle \partial (logZ)/\partial H=(1/Z)(\partial Z/\partial H)}$

and

${\displaystyle \partial Z/\partial H=-(1/kT)\sum {e^{-E_{i}/kT}\partial E_{i}/\partial H}}$, the above equation for magnetisation(I) may be rewritten as,

${\displaystyle I=NkT\partial (logZ)/\partial H}$

3.4

The molar susceptibility is then obtained by differentiating I with respect to the field,

${\displaystyle \chi _{M}=NkT\partial ^{2}(logZ)/\partial H^{2}}$

3.5

3.4 and 3.5 are the fundamental equations of magnetism.

An important simplification arises when the magnetic response of the ground electronic state of the molecule is purely due to electron spin. If the excited states are sufficiently high in energy such that their contribution may be neglected, one may write,

${\displaystyle Z=\sum _{M_{S}=-S}^{S}e^{-g\beta HM_{S}/kT}}$

3.6

where, S is the total spin quantum number and M_S are the 2S+1 possible z-axis projections. Writing,

${\displaystyle x=g\beta H/kT}$

3.7

${\displaystyle Z=e^{-sx}+e^{(-S+1)x}+....+e^{(S-1)x}+e^{Sx}}$

Summing the above geometric progression, we get

${\displaystyle Z={\frac {e^{Sx}e^{x}-e^{-Sx}}{e^{x}-1}}}$

${\displaystyle ={\frac {e^{x/2}(e^{Sx+x-x/2}-e^{-sx-x/2})}{e^{x/2}(e^{x/2}-e^{-x/2})}}}$

${\displaystyle ={\frac {e^{(2S+1)x/2}-e^{-(2S+1)x/2}}{e^{x/2}-e^{-x/2}}}}$

${\displaystyle ={\frac {Sinh[(2S+1)x/2]}{Sinh(x/2)}}}$

Differentiating ${\displaystyle logZ}$ with respect to H and noting that dx/dH = gβ/kT, one gets the following expression for magnetisation:

${\displaystyle I=(1/2)Ng\beta [(2S+1)Coth(2S+1)(x/2)-Coth(x/2)]}$

Comparing the above expression for the definition of the Brillouin function given below,

${\displaystyle B_{k}(t)={\frac {2k+1}{2k}}Coth[{\frac {2k+1}{2k}}t]-{\frac {1}{2k}}Coth[{\frac {1}{2k}}t]}$

3.8

one may rewrite it as

${\displaystyle I=Ng\beta SB_{S}(y)}$

3.9

where, ${\displaystyle y={\frac {g\beta HS}{kT}}}$

Further simplifications arise when the magnetic field is small as we will see in the next lecture.