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



Paramagnetic materials have atoms with magnetic moments which can be aligned with an applied field so that the induced field will add to the applied field. This results in a larger $B$ -field inside the material than outside and a positive magnetic susceptibility. Paramagnetism is similar to diamagnetism in the sense that the materials will only show magnetization when a magnetic field is applied. Once the applied field is switched off, the moments will randomize again and the magnetization decay exponentially to zero.

The magnetization of a paramagnet is the sum of the aligned magnetic moments per unit volume. The magnetic moment of an atom depends on the total angular momentum quantum number $J$. Since the magnetic quantum numbers are restricted to the values $m_J = (-J,-J+1,\cdots , J-1, J)$, the $z$-component of the magnetic moment can can take the values,

$$\mu_{m_J}=m_J g_J \mu_B,$$

where $g_J$ is the Landé g factor, $\mu_B$ is the Bohr magneton. For a magnetic field applied in the $z$-direction, the energies of the magnetic states will be,

$$E_{m_J} = -\mu_{m_j}B_z = -m_J g_J \mu_B B_z.$$

The occupation probabilty of state $m_J$ is given by a Boltzmann distribution,

$$p_{m_J} = \frac{\exp\left(\frac{-E_{m_J}}{k_{B}T}\right)}{\sum\limits_{m_J = -J}^{m_J=J} \exp\left(\frac{-E_{m_J}}{k_{B}T}\right)}.$$

The $J=\frac{1}{2}$ case
For a spin $\frac{1}{2}$ system where $J = \frac{1}{2}$, there are two states spin-up aligned parallel to the $B$ field and spin-down aligned antiparallel to the $B$ field. The spin up state has a lower energy and will have a higher occupation. Let $N_{\uparrow}$ be the number of spin-up states, $N_{\downarrow}$ be the number of spin-down states, and $N=N_{\uparrow}+N_{\downarrow}$ be the total number of states. The occupations of spin-up and spin-down are given by,

\[ \begin{equation} \frac{N_{\uparrow}}{N}=\frac{\exp\left(\frac{\mu B}{k_{B}T}\right)}{\exp\left(\frac{\mu B}{k_{B}T}\right)+\exp\left(\frac{-\mu B}{k_{B}T}\right)} \end{equation} \] \[ \begin{equation} \frac{N_{\downarrow}}{N}=\frac{\exp\left(\frac{-\mu B}{k_{B}T}\right)}{\exp\left(\frac{\mu B}{k_{B}T}\right)+\exp\left(\frac{-\mu B}{k_{B}T}\right)} \end{equation} \]

Here $\mu =\frac{1}{2} g_{\frac{1}{2}} \mu_B$. The magnetization is magentic moment per unit volume $V$,

\[ \begin{equation} M=\mu\frac{N_{\uparrow} - N_{\downarrow}}{V}. \end{equation} \]

The occupation probabilities can be used to write this as,

\[ \begin{equation} M=n\mu \tanh\left(\frac{\mu B}{k_B T}\right), \end{equation} \]

where $n=\frac{N}{V}$ is the density of the spins. A plot of the magentization shows that there are two interesting limits, $\mu B >> k_B T$ and $\mu B << k_B T$.

$\large \frac{M}{M_s}$

$\large \frac{\mu B}{k_BT}$

For $\mu B >> k_B T$, all of the spins align and the magnetization approaches the saturation magnetization $M_s = n\mu$. For small arguments, $\tanh(x)\approx x$ so in the limit $\mu B << k_B T$, the relation between the magnetization an the $B$ field is linear,

\[ \begin{equation} M \approx \frac{n \mu^2 B}{k_B T} = \frac{CB}{T}. \end{equation} \]

The linear magnetic susceptibility $\chi_m =\frac{dM}{dH} = \frac{n\mu_0 g_{1/2}^2 \mu_B^2}{4k_BT}$ describes how the magnetization changes with applied magnetic field near zero field. This has the form $\chi_m= \frac{C}{T}$ which is known as the Curie law where $C$ is called the Curie constant.

$\large \frac{\chi_m}{C}$


General $J$
For a general value of the total angular momentum quantum number, the magnetization can be written in terms of the average value of $m_J$,

\[ \begin{equation} M=ng_J \mu_B \langle m_J \rangle \end{equation} \]

The average value of $m_J$ is,

$$\langle m_J \rangle = \sum\limits_{m_J = -J}^{m_J=J}m_Jp_{m_J} = \frac{\sum\limits_{m_J = -J}^{m_J=J}m_j\exp\left(\frac{m_J g_J \mu_B B}{k_{B}T}\right)}{\sum\limits_{m_J = -J}^{m_J=J} \exp\left(\frac{m_J g_J \mu_B B}{k_{B}T}\right)}.$$

Notice that the numerator is the derivative of the denominator so this can be written,

$$\langle m_J \rangle = \frac{1}{Z}\frac{dZ}{dx},$$

Where $Z= \sum\limits_{m_J = -J}^{m_J=J} \exp\left(m_J x\right)$ and $x = \frac{ g_J \mu_B B}{k_{B}T}$. The sum from $-J$ to $J$ can be written as the difference of two infinite sums,

$$Z= \sum\limits_{m_J = -\infty}^{m_J=J} \exp\left(m_J x\right) - \sum\limits_{m_J = -\infty}^{m_J=-J-1} \exp\left(m_J x\right).$$

This is convenient because both of the infinite sums are geometric series that can be summed,

$$Z= \left(e^{Jx} -e^{-(J+1) x}\right)\left(1+e^{-x}+e^{-x^2}++e^{-x^3}+\cdots\right)= \frac{e^{Jx} -e^{-(J+1) x}}{1-e^{-x}}.$$

This can be rearranged to be expressed as, $$Z= \frac{\text{sinh}\left(\left(J+\frac{1}{2}\right)x\right)}{\text{sinh}\left(\frac{x}{2}\right)}.$$ $$ \langle m_{J} \rangle = \frac{1}{Z} \frac{dZ}{dx} = \left ( J + \frac{1}{2} \right ) \coth \left ( \left ( J + \frac{1}{2} \right ) x \right ) - \frac{1}{2} \coth \left ( \frac{1}{2} x \right ).$$
$\langle m_J \rangle = JB_J(Jx)$

$\large x=\frac{g_J\mu_BB}{k_BT}$

Where $B_J(x)$ is the Brillouin function,

$$B_J(x)= \frac{2J+1}{2J}\text{coth}\left(\frac{2J+1}{2J}x\right)-\frac{1}{2J}\text{coth}\left(\frac{x}{2J}\right).$$

The magnetization can then be expressed as,

\[ \begin{equation} M=ng_J \mu_B J \left(\frac{2J+1}{2J} \coth\left( \frac{2J+1}{2J} \frac{g_J\mu_BJB}{k_BT}\right)-\frac{1}{2J} \coth\left(\frac{1}{2J} \frac{g_J\mu_B JB}{k_B T}\right)\right) \end{equation} \]

In the high field limit, $\frac{\mu_BB}{k_BT} >> 1$, the magnetization saturates at $M_s = ng_J\mu_BJ$. In the low field limit, $\frac{\mu_BB}{k_BT} < < 1$, the susceptibility has the form of a Curie law, $$ \chi_m =\frac{dM}{dH} \approx \frac{n\mu_0 g_{J}^2 J(J+1)\mu_B^2}{3k_BT}.$$