Advanced Solid State Physics



Magnetic effects and
Fermi surfaces


Linear response


Crystal Physics



Structural phase

Landau theory
of second order
phase transitions




Exam questions




Course notes

TUG students



Ferrimagnetic materials (e.g. Magnetite) exhibit a crystalstructure with two sublattices A and B with different net magnetic moments. The exchange integrals $J_{AA}$, $J_{BB}$ and $J_{AB}$ are all negative with $|J_{AB}| > |J_{AA}|,|J_{BB}|$. Therefore, the spins of the A site align antiparallel with the spins of the B site. Due to the different net magnetic moments, ferrimagnetic materials have a non vanishing magnetization $M \neq 0$ even without an external field ($B_a = 0$). Similar to the derivation of ferromagnetism, the Hamiltonian of the isotropic Heisenberg model is used for the interacting spins

$$H = - \sum_{i,j} J_{i,j}\vec{S}_i \cdot \vec{S}_j - g \mu_{b} \vec{B_a} \sum_i \vec{S_i}. \nonumber $$

with with $g$ being the Landé g factor and $\mu_B$ the Bohr magneton. For simplification, it is assumed that only neighbouring spins interact with each other and the interaction vanishes for all lattice sites further away. Again, mean field approximation is applied for each of the two sublattices. The main idea is that the interaction of two neighboring spins $\vec{S}_i \cdot \vec{S}_j$ can be replaced by the interaction of a single spin on site $i$ with the mean spin value of its surrounding:

$$ \vec{S}_i \cdot \vec{S}_j \rightarrow \vec{S}_i \cdot \langle \vec{S} \rangle.$$

As there are two different sublattices A and B, the mean field approximation is performed for both lattices. One obtains:

$$ H_{MF,A} = - \sum_{i} \vec{S}_i \cdot \left( \sum_{\delta_B} J_{i,AB} \langle \vec{S}_B \rangle + \sum_{\delta_A} J_{i,AA} \langle \vec{S}_A \rangle + g \mu_{b} \vec{B} \right) .$$ $$ H_{MF,B} = - \sum_{i} \vec{S}_i \cdot \left( \sum_{\delta_A} J_{i,AB} \langle \vec{S}_A \rangle + \sum_{\delta_B} J_{i,BB} \langle \vec{S}_B \rangle + g \mu_{b} \vec{B} \right) .$$

The first sum in the brackets corresponds to the interaction of the spin $i$ with all the neighbouring spins from the other sublatice. The second term is the interaction between spin $i$ with all the nearest spins on the same sublattice.

As the exchange energy is independent of the site $i$, the mean-field Hamiltonians can be rewritten as:

$$ H_{MF,A} = - \sum_{i} \vec{S}_i \cdot \left( z_{AB} J_{AB} \langle \vec{S}_B \rangle + z_{AA} J_{AA} \langle \vec{S}_A \rangle + g \mu_{b} \vec{B} \right) .$$ $$ H_{MF,B} = - \sum_{i} \vec{S}_i \cdot \left( z_{AB} J_{AB} \langle \vec{S}_A \rangle + z_{BB} J_{BB} \langle \vec{S}_B \rangle + g \mu_{b} \vec{B} \right) .$$

Here, the coordination number $z_{i,j}$ was introduced which describes the number of nearest neighbours for each configuration of A and B. It can be seen that the term in the brackets is independent of the site $i$ and acts like a total magnetic field consisting of the external field and the mean fields. This way, one can write the mean field on site A and B as:

$$\vec{B}_{MF,A} = \frac{1}{g \mu_{b}} z_{AB} J_{AB} \langle \vec{S}_B \rangle + \frac{1}{g \mu_{b}} z_{AA} J_{AA} \langle \vec{S}_A \rangle$$ $$\vec{B}_{MF,B} = \frac{1}{g \mu_{b}} z_{AB} J_{AB} \langle \vec{S}_A \rangle + \frac{1}{g \mu_{b}} z_{BB} J_{BB} \langle \vec{S}_B \rangle$$

Expressing the average spin in terms of the overall magnetization density,

$$ \vec{M} = \underbrace{\frac{N}{V}}_{n} g \mu_{b} \langle \vec{S} \rangle $$

where $n$ represents the spin density and introducing $$\gamma = \frac{1}{\mu_0}\frac{z J}{n g^2 \mu_b^2},$$ one obtains:

$$\vec{B}_{MF,A} = \mu_0 \gamma_{AB} \vec{M}_B + \mu_0 \gamma_{AA} \vec{M}_A $$ $$\vec{B}_{MF,B} = \mu_0 \gamma_{AB} \vec{M}_A + \mu_0 \gamma_{BB} \vec{M}_B $$

where $\mu_0$ is the vacuum permeability and $\gamma_{AB} = \gamma_{BA}$. As defined on the paramagnetism page, the energies of the magnetic states for a field in the $z$-direction are given by:

$$E_{m_J} =-\vec{\mu}\cdot\vec{B}_{total} = -m_J g\mu_B B_{total},$$

where $m_J = \pm\frac{1}{2}$. Because of the two different sublattices A and B, the spins can take on the following energies:

$$ E_A = \pm \frac{1}{2} g \mu_b \left( B_{MF,A} + B_a \right), E_B = \pm \frac{1}{2} g \mu_b \left( B_{MF,B} + B_a \right) $$

The derivation of the magnetization follows a similar scheme as on the paramagnetism page. The two possible spin states up and down are differently occupied. $N_{\uparrow}$ is the number of spin-up states, $N_{\downarrow}$ is the number of spin-down states and $N=N_{\uparrow}+N_{\downarrow}$ is total number of spins. The population probability of both spin states can then be calculated using Boltzmann factors:

$$\frac{N_{\uparrow}}{N}=\frac{\exp\left(\frac{\mu B}{k_{B}T}\right)}{\exp\left(\frac{\mu B}{k_{B}T}\right)+\exp\left(\frac{-\mu B}{k_{B}T}\right)},$$ $$\frac{N_{\downarrow}}{N}=\frac{\exp\left(\frac{-\mu B}{k_{B}T}\right)}{\exp\left(\frac{\mu B}{k_{B}T}\right)+\exp\left(\frac{-\mu B}{k_{B}T}\right)}. $$

In these equations, $\mu=\frac{1}{2} g \mu_B$ and $B=B_{total}$ is the total magnetic field on site A or B. The magnetization per unit volume is given by

$$M = \mu \frac{N_{\uparrow}-N_{\downarrow}}{V}.$$

By inserting the occupation probabilities the equations for the sublattice magnetizations become,

$$M_A = n_A \mu \tanh\left(\frac{\mu (B_{MF,A} + B_a)}{k_B T}\right) = M_{s,A} \tanh\left(\frac{\mu (\mu_0 \gamma_{AB} M_B + \mu_0 \gamma_{AA} M_A + B_a)}{k_B T}\right), $$ $$M_B = n_B \mu \tanh\left(\frac{\mu (B_{MF,B} + B_a)}{k_B T}\right) = M_{s,B} \tanh\left(\frac{\mu (\mu_0 \gamma_{AB} M_A + \mu_0 \gamma_{BB} M_B + B_a)}{k_B T}\right).$$

In general, the magnetic susceptibility is defined as

$$\chi_m = \left.\frac{\partial M}{\partial H_a}\right\vert_{H_a = 0}.$$

For spin models, the relative permeability $\mu_r=1$ because the spins are in vacuum. Therefore $H_a=B_a/\mu_0$ and the equation for the magnetic susceptibility can be written as

$$\chi_m = \left. \mu_0 \frac{\partial M}{\partial B_a}\right\vert_{B_a = 0}.$$

The starting point for the calculation of the susceptibility are the expressions for the magnetization of the A and B lattice.

$$M_A = \frac{1}{2} n_A g \mu_b \tanh\left(\frac{ g \mu_b (\mu_0 \gamma_{AB} M_B + \mu_0 \gamma_{AA} M_A + B_a)}{2k_B T}\right) $$ $$M_B = \frac{1}{2} n_B g \mu_b \tanh\left(\frac{ g \mu_b (\mu_0 \gamma_{AB} M_A + \mu_0 \gamma_{BB} M_B + B_a)}{2k_B T}\right) $$

Above $T_c$, the hyperbolic tangent can be expanded with $\tanh(x) \approx x$ (for $x \ll 1$). By applying this expansion to the formulas and neglecting the interactionof the spins on the same sublattice ($\gamma_{AA} = \gamma_{BB} = 0) $ one gets:

$$M_A \approx \frac{n_A g^2 \mu_b^2}{4 k_B T} \left( \mu_0 \gamma_{AB} M_B + B_a \right) = \frac{C_A}{\mu_0 T} \left( \mu_0 \gamma_{AB} M_B + B_a \right)$$ $$M_B \approx \frac{n_B g^2 \mu_b^2}{4 k_B T} \left( \mu_0 \gamma_{AB} M_A + B_a \right) = \frac{C_B}{\mu_0 T} \left( \mu_0 \gamma_{AB} M_A + B_a \right)$$

For $B_a = 0$ this system of equations has a non trivial solution if

$$ \begin{vmatrix} T & -\gamma_{AB}C_A \\ -\gamma_{AB} C_B & T \end{vmatrix} = 0 $$

This leads to the ferrimagnetic Curie Temperature $T_c = \sqrt{C_A C_B} |\gamma_{AB}|$.

In order to calculate the susceptibility above $T_c$, $M_A$ and $M_B$ are calculated explicitly by solving the linear system of equations with $B_a \neq 0$. The susceptibility then results in:

$$\chi_m = \mu_0 \frac{\partial (M_A + M_B)}{\partial B_a} = \frac{(C_A + C_B)T - 2 |\gamma_{AB}| C_A C_B}{T^2 - T_c^2}$$

In an antiferromagnetic material the antisymmetric spin orientation is favorable as in the case of ferrimagnetic material. The only difference is that the magnetic moments on the two sublattices are the same, therefore $C_A = C_B = C$ and $\gamma_{AA} = \gamma_{BB}$. We again start of with the expressions for the magnetizations, expand the hyperbolic tangent, and use $\vec{M}_A = -\vec{M}_B$. This results in:

$$M_A \approx \frac{C}{\mu_0 T} \left( \mu_0 (\gamma_{AB} - \gamma_{AA}) M_B + B_a \right),$$ $$M_B \approx \frac{C}{\mu_0 T} \left( \mu_0 (\gamma_{AB} - \gamma_{AA}) M_A + B_a \right).$$

For $B_a = 0$ this system of equations has a non trivial solution if

$$ \begin{vmatrix} T & -(\gamma_{AB} - \gamma_{AA}) C \\ -(\gamma_{AB} - \gamma_{AA}) C & T \end{vmatrix} = 0 $$

This leads to the Neél temperature $T_N = |\gamma_{AB}-\gamma_{AA}|C$

For the susceptibility above $T_N$, the material is paramagnetic and $\vec{M_A} = \vec{M_B} = \frac{1}{2}\vec{M}$. With these considerations, expanding the hyperbolic tangent leads to:

$$ M = 2 M_A \approx \frac{2C}{\mu_0 T}\left( \mu_0 (\gamma_{AB} + \gamma_{AA}) \frac{M}{2} + B_a \right) $$

Solving this equation explicitly leads to.

$$ M = \frac{1}{\mu_0} \frac{2C}{T - C \left(\gamma_{AB} + \gamma_{AA} \right)} B_a$$

The paramagnetic susceptibility is then calculated as before:

$$\chi_m = \mu_0 \frac{\partial M}{\partial B_a} = \frac{2C}{T - C \left(\gamma_{AB} + \gamma_{AA} \right)} = \frac{2C}{T+\Theta}$$

Here, $\Theta$ is the paramagnetic Neél temperature. A factor of 2 appears in the Curie-Weiss form because $C$ represents the Curie constant of the sublattice, not the whole lattice.


Written by Kelvin Walenta, February 2021.