and $\rho_{0}(\varepsilon)=D\sum_{m}\delta\left(\varepsilon-\varepsilon_{m}\right)$ is the density of states (DOS) in the pristine case, where the summation is to be done considering the valence and conduction band. In this way we can separate the contribution from both bands, so that $\Omega_{T}=\Omega_{V,T}+\Omega_{C,T}$, where in general $\Omega_{T}=-kT\sum_{\varepsilon}D\ln\left[1+e^{\beta\left(\mu-\varepsilon\right)}\right]$. Now, for $\mu>0$ and low temperature we always have $\beta\left(\mu-\varepsilon\right)\gg1$ for the valence band, so $\Omega_{V,T}\simeq\Omega_{V}\left(T=0\right)$ and $\partial\Omega_{V,T}/\partial B\simeq\partial\Omega_{V}\left(T=0\right)/\partial B$. Hence the valence band magnetization is not affected by the temperature under this conditions. Therefore, we shall omit it for the moment and continue with the conduction band grand potential

$$\Omega_{C,T}=-kT\sum_{m=0}^{\infty}D\ln\left[1+e^{\beta\left(\mu-\varepsilon_{m}\right)}\right]=\sum_{m=0}^{\infty}\Omega_{T}^{m},$$

where $\varepsilon_{m}=\alpha\sqrt{n_{m}B}-s_{m}\mu_{\mathrm{B}}B$ are the conduction energy levels, and we defined $\Omega_{T}^{m}=-kTD\ln\left[1+e^{\beta\left(\mu-\varepsilon_{m}\right)}\right]$. We have $\left(\partial\Omega_{T}^{m}/\partial B\right)_{\mu}=\Omega_{T}^{m}/B+D\left(\partial\varepsilon_{m}/\partial B\right)\left[1+e^{-\beta\left(\mu-\varepsilon_{m}\right)}\right]^{-1}$, so the conduction magnetization $M_{C,T}=-\mathcal{A}^{-1}\left(\partial\Omega_{C,T}/\partial B\right)_{\mu}$ is given by

$$M_{C,T}=\frac{KTD}{\mathcal{A}B}\sum_{m=0}^{\infty}\ln\left[1+e^{\beta\left(\mu-\varepsilon_{m}\right)}\right]-\frac{D}{\mathcal{A}}\sum_{m=0}^{\infty}\frac{\partial\varepsilon_{m}}{\partial B}\frac{1}{1+e^{-\beta\left(\mu-\varepsilon_{m}\right)}}.\label{Mt}$$

We now assume a Fermi energy $\varepsilon_{f}<\mu\leq\varepsilon_{f+1}$, so the conduction magnetization at zero temperature is $M_{C}=-\mathcal{A}^{-1}\left(\partial\Omega_{C}/\partial B\right)_{\mu}=-\mathcal{A}^{-1}\sum_{m=0}^{f}\left[D\left(\varepsilon_{m}-\mu\right)/B+D\left(\partial\varepsilon_{m}/\partial B\right)\right]$. Thus we can rewrite equation (\[Mt\]) as

$$\begin{aligned}
\fl M_{C,T} =  M_{C}+\frac{KTD}{\mathcal{A}B}\sum_{m=0}^{f}\ln\left[1+e^{-\beta\left(\mu-\varepsilon_{m}\right)}\right]+\frac{D}{\mathcal{A}}\sum_{m=0}^{f}\frac{\partial\varepsilon_{m}}{\partial B}\frac{1}{1+e^{\beta\left(\mu-\varepsilon_{m}\right)}}\nonumber\label{Mt2}\\
 +\frac{KTD}{\mathcal{A}B}\sum_{m=f+1}^{\infty}\ln\left[1+e^{\beta\left(\mu-\varepsilon_{m}\right)}\right]-\frac{D}{\mathcal{A}}\sum_{m=f+1}^{\infty}\frac{\partial\varepsilon_{m}}{\partial B}\frac{1}{1+e^{-\beta\left(\mu-\varepsilon_{m}\right)}}.\end{aligned}$$

We can simplify equation (\[Mt2\]) by noticing that the term with the logarithm is in general very small. Indeed, we always have $\ln\left[1+e^{-\beta\left(\mu-\varepsilon_{m}\right)}\right]\leq\ln2<1$ if $m\leq f$ and likewise $\ln\left[1+e^{\beta\left(\mu-\varepsilon_{m}\right)}\right]<\ln2$ if $m>f$. Furthermore, we have $kTD/\mathcal{A}B=2kT/\phi=2kTe/h\sim10^{-7}T\left[\unit{K}\right]\;\textrm{meV/T}{\textrm{\AA}}^2$, which for low $T$ is much smaller than the magnetization at 