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## Pauli paramagnetism and Landau-Diamagnetism

The calculation of Pauli paramagnetism presented here follows the presentation found in many textbooks. It considers the change in energy due to the Zeeman splitting but neglects the change in energy due to the Landau splitting. There is also a well know calculation that only considers the Landau splitting and neglects the Zeeman splitting resulting in Landau diamagnetism. However since the Zeeman splitting and Landau splitting are nearly equal $g\mu_BB \approx \hbar\omega_c$, it is not clear that it is justified to separate the effects like this. A more rigorous calculation would be to start from the density of states that includes both the Zeeman splitting and the Landau splitting, construct the free energy $F$, and then differentiate this with respect to the magnetic field to determine the magnetization. When this is done then the resulting susceptibility is the sum of Pauli paramagnetism and Landau diamagnetism. In the limit $k_BT > > \hbar\omega$, the Landau diamagnetic susceptibility is $-\frac{1}{3}$ of the Pauli susceptibility. The total susceptibility is $2/3$ the Pauli susceptibility.

In metals there is an additional contribution to the mangetic susceptibility from the free electrons in the conduction band, beside the one from the atomic moments of the to the ion rumps bound electrons. Those non localized electrons do not react independently to external magnetic fields, because of the Pauli principle. With this in mind the according magnetic susceptibility will be derived below.
The energies of free electrons in a external magnetic field $B_{ext}$ parallel the z-axis is,

$$$E=\left( n+ \frac{1}{2}\right) \hbar \omega_c + \frac{\hbar^2}{2m} k_{z}^2 \pm \mu_B * B_{ext}$$$

The first two terms result from the orbital motion of the conduction electrons. Perpendicular to the magnetic field there is a spliting of the orbital motion, which leads to electron states on the so called Landau-cylinders. The spacing between those is $\hbar \omega_c=e\hbar B_{ext}/m$. Through this quantization of the electron states the energy dispersion of the electrons in the fermi sphere is altered compared to the field free states. This leads to the so called Landau diamagnetism of the conduction electrons.

The third term results from the to the electron spin coupled magnetic moment $\mu_s=-g_s\mu_B m_s \approx \mp \mu_B$ $(g_s \approx 2, m_s=\pm \frac{1}{2})$, and is the basis for the Pauli paramagnetism. It can only take those two values in field direction. For the magnetiztion we get therefore,

$$$M=(n_+ - n_-)\mu_B$$$

where $n_+$ is the electron density with $\mu_s$ in field direction and $n_-$ the density with $\mu_s$ against the field direction.

Without the Pauli paramagnetism in the free electron model we get for thermal equilibrium,

$$$n_+ + n_- = n_+ * \frac{n_-}{n_+}=e^{-\frac{\Delta E}{k_BT}}$$$

And for the occupation difference,

$$$\Delta n=n * tanh \left( \frac{\Delta E}{2k_BT} \right)=n * tanh \left( \frac{\mu_B B}{k_BT} \right)$$$

Hence for high temperatures $(\mu_sB \ll k_BT)$ the magnetisation M is,

$$$M= \frac{n \mu_{B}^2 B}{k_BT} \propto \frac{1}{T}$$$

and the susceptibility,

$$$\chi=\mu_0 \frac{n\mu_{B}^2}{k_BT}\propto \frac{1}{T}$$$

However, for metals experiments yield $\chi \propto T^0$. Therefore, the free electron model is not valid for metals (but it still gives good results for paramagnetic ions, Curie law). The Pauli principle needs to be taken into account, as shown in the following schematic.

Source: http://ms.zneb.at/html/exphys5/exse35.htm

The energy splitting generates a surplus of electrons with spin antiparallel to B. The according electron densities can be calculated with the consecutive forumula.

$$$n_{\pm}= \frac{1}{2V} \int\limits_{0}^{\infty} D(E \pm \mu_BB_{ext})f(E) \ dE$$$

With this one can calculate the new magnetization M,

$$$M= (n_+ - n_-)\mu_B = \frac{\mu_B}{2V} \int\limits_{0}^{\infty} \frac{dD}{dE} 2 \mu_B B_{ext} f(E) \ dE$$$ $$$= \frac{\mu_{B}^2 B_{ext}}{V} \int\limits_{0}^{\infty} \frac{dD}{dE} f(E) \ dE$$$ $$$= \frac{\mu_{B}^2 B_{ext}}{V} \left[D(E)f(E)\bigg\rvert_{0}^{\infty} - \int\limits_{0}^{\infty} D(E) \frac{df}{dE} \ dE \right]$$$ $$$= -\frac{\mu_{B}^2 B_{ext}}{V} \int\limits_{0}^{\infty} D(E) \frac{df}{dE} \ dE$$$

For low temperatures $-df/dE$ can be approximated as $\delta(E-E_F)$,

$$$M= \frac{\mu_{B}^2 B_{ext}}{V}D(E_F)$$$

And with the density of states from the free electron gas,

$$$D(E_F) = \frac{V}{2 \pi^2} \left( \frac{2m}{\hbar^2} \right)^{3/2}E_{F}^{1/2}=\frac{3}{2}\frac{nV}{k_B T_F}$$$

the magnetisation is,

$$$M=\frac{3n\mu_{B}^2B_{ext}}{2k_BT_F}$$$

Finally we get the temperature independent Pauli susceptibility:

$$$\chi_P=\mu_0\left(\frac{\partial M}{\partial B_{ext}} \right)_{T,V}=\mu_0 \mu_{B}^2\frac{D(E_F)}{V}=n\frac{3\mu_0\mu_{B}^2}{2k_BT_F}=const.$$$

In general Pauli paramagnetism is much smaller than the paramagnetism due to atomic moments. E.g. for a fermi energy of $1~eV$, a field of $B=17000~T$ is needed to allign all spins.