= Solution
Following the restricted model in the preceding part, its electronic factor for $N$ independent magnetic molecules is $Q_{\mathrm{mag}}=(2\cosh x)^N$. Its field-dependent <Helmholtz free energy> is therefore
$$
\boxed{A_{\mathrm{mag}}(B)=-Nk_{\mathrm B}T\ln\left[2\cosh\frac{\mu B}{k_{\mathrm B}T}\right]+A_{\mathrm{ind}},}
$$
where $A_{\mathrm{ind}}$ denotes terms independent of the field. Equivalently, the change from zero field is $A(B)-A(0)=-Nk_{\mathrm B}T\ln\cosh x$. Retaining the excited-state offset simply adds $N\Delta$ to the restricted free energy.
If $N$ instead counts all molecules in thermal equilibrium, the correct electronic expression is
$$
A_{\mathrm{full}}=-Nk_{\mathrm B}T\ln[2+2e^{-\beta\Delta}\cosh x]+A_{\mathrm{ind}}.
$$
The ordinary translational indistinguishability factor is field independent and does not alter either magnetic derivative.
Back to article page
