Ising model

💼 Case: A lattice composed of spins organized in a periodic structure (classical assumptions)

🧽 Assume:

• Spin $s_i=\pm 1$ so it has to be aligned in one direction (others are energy unfavourable)
• The spins do not change positions just there spin orientation
• Each spin interacts with its nearest neighbours and with the magnetic field $h$ only

• The energy (”classical Hamiltonian” of the system) is

  $$ \boxed{\mathcal H_\text{Is}=-J\sum_{\braket{i,j}}s_i s_j -h\sum_is_i} $$

  where $\sum_{\braket{i,j}}$ means that we sum over all couples of neighbouring lattice sites $i,j$

  The second term is the Zeeman coupling and describes the tendency of the spins to align with $h$

  The first “exchange” term described the interactions between spins. Responsible for cooperative behaviour and phase transitions

  $J$ is the exchange coupling which has units of energy (the $\hbar/2$ term was “absorbed”)

  $h$ has units of energy, we “absorbed” Bohr’s magneton and $g$-factor

🗒️ Notes:

• $J>0$ favours ferromagnetism (parallel alignment of spins)
• $J<0$ favours antiferromagnetism (antiparallel alignment of neighbouring spins)

The Ising model can be solved analytically in $1\rm D$ and $2 \rm D$ (more involved) though not in $\rm 3D$

Mean-field theory for the Ising model

To solve the Ising model in 3D we will use the mean-field approximation.

• To do this we average the previous equation to get the energy of a given spin configuration $\{\overline s_i \}$

  $$ E[\{ \overline s_i\}]=-J\sum_{\braket{i,j}}\overline s_i \overline s_j -h\sum_i \overline s_i $$

⚽ Goal: find the critical temperature where the material changes from paramagnetic (random spins, zero net magnetization) to a ferromagnetic state (aligned spins, finite net magnetization) in the absence of an external magnetic field ($h=0$)

• 🧠 Remember: equation, The magnetization is given as

  $$ M=-\left . \frac{\partial F}{\partial h} \right |_{T,V} $$

  where $F=-k_B T \ln Z(T,V,h)$ and, in the canonical ensemble,

  $$ Z(T,V,h)=\sum_{\{s_i\}} e^{-E[\{s_i \}]/(k_B T)} $$

  where we denote with $\sum_{\{ s_i \}}$ the sum over all spin configurations’

  $$ \sum_{\{ s_i \}}\to \sum_{s_1=\pm 1} \sum_{s_2=\pm 1} \ldots \sum_{s_N=\pm 1} $$

Free spin in the presence of a magnetic field ($J=0$)

• In the case of $J=0$ we can calculate $Z(T,V,h)$ as the Hamiltonian only contains $-h\sum_i s_i$

  $$ \begin{aligned} Z(T,V,h)&=\sum_{s_1=\pm}\sum_{s_2=\pm} \ldots \sum_{S_N=\pm} e^{h(s_1+s_2+\ldots s_N)/(k_B T)}=\left ( \sum_{s=\pm} e^{hs/(k_B T)} \right )^N \\ &=\left ( e^{h/(k_B T)}+e^{-h/(k_B T)} \right )^N=\left [ 2 \cosh \left ( \frac{h}{k_B T} \right ) \right ]^N \\ \Rightarrow \quad F&=-k_B T\ln Z(T,V,h)=-N k_B T \ln \left [2 \cosh \left ( \frac{h}{k_B T} \right ) \right ]

  \end{aligned} $$

  Which applying the chain rule we get

  $$ \boxed{M=-\left . \frac{\partial F}{\partial h} \right |_{T,V}=Nk_B T \frac{2\sinh \left ( \frac{h}{k_B T} \right )}{2\cosh \left ( \frac{h}{k_B T} \right )}\frac{1}{k_B T} =N \tanh \left ( \frac{h}{k_B T} \right )} $$