# Understanding Ising Model in Statistical Mechanics

A section on the Ising model in the text "Introduction to modern statistical mechanics" by Chandler states the following:

"We consider a system of $$N$$ spins arranged on a lattice. In the presence of a magnetic field, $$H$$, the energy of the system in a particular state, $$\nu$$, is $$E_{\nu} = \sum_{i=1}^{N}H \mu s_i + (\text{energy due to interactions between spins}),$$ where $$s_i = \pm 1$$. A simple model for the interaction energy is $$-J\sum_{ij} ' s_i s_j$$ where $$J$$ is called a coupling constant, and the primed sum extends over nearest-neighbor pairs of spins. The spin system with this interaction energy is called the Ising Model.

Notice that when $$J > 0$$, it is energetically favorable for neighboring spins to be aligned. Hence, we might anticipate that for low enough temperature, this stabilization will lead to a phenomenon called spontaneous magnetization. "

Questions:

• In the last paragraph, by "energetically favorable if $$J > 0$$", is the implication that, assuming $$J > 0$$ and aligned neighbouring spins, it follows that $$-J\sum_{ij}'s_i s_j$$ would be negative and hence the interaction energy reduces the total energy $$E_{\nu}$$ thus moving in the direction of minimum energy, which a stable system in equilibrium tends to after being perturbed. Is this the correct reasoning?

• Lastly, why would we anticipate that for low enough temperature, this stabilization will lead to spontaneous magnetization?

## 2 Answers

Yes, your first statement is the correct reasoning. Or, putting it more briefly, "energetically favourable" is another way of saying "lower energy".

On the second point, you have already established that lower energy goes hand-in-hand with alignment of neighbouring spins $$s_i$$ and $$s_j$$. For a system that is not frustrated in some way, all the spins $$s_i$$ can become aligned with each other. The minimum energy configuration is the same as the completely aligned configuration (in other words, the magnetization $$M=\sum_i s_i$$ will be large and positive or, equally well, large and negative). The Helmholtz free energy is $$F = E - TS$$ and this is minimized for a system in thermal equilibrium at a given temperature $$T$$. At low temperature, $$F$$ is dominated by the energy term $$E$$ (the entropy term $$-TS$$ is small). So the system will tend to a low-energy state, at low temperature.

• Thanks for your response. Could you advise on what expression for free energy you are referring to when you state "At low temperature, the free energy is dominated by the energy term (the entropy term is small) and so the system will tend to a low-energy state, at low temperature"? – John Doe Jan 22 '19 at 12:15
• Yes, no problem. I've edited my answer to make this more explicit. – user197851 Jan 22 '19 at 13:05
1. Yes, for $$J>0$$ we have a ferromagnetic coupling between the spins. Thus the energy is minimized for a parallel alignment of the spins. For $$J<0$$ we have a antiferromagnetic coupling, where the total energy is minimized if the spins are in an antiparallel alignment.

2. If the temperature of the spin system exceeds the Curie temperature, the thermal energy exceeds the exchange energy, and the system is not ferromagnetic anymore. For more information about this I recommend Wikipedia: Curie Temperature.