# What is an order parameter? Thermal average or spatial integral of a thermal average?

Sometimes the the order parameter is defined as the thermal average of a spatially varying field $\textbf{m}(\textbf{x})$ i.e., $\langle\textbf{m}(\textbf{x})\rangle$.

Sometimes the order parameter (density) is defined as the spatial integral $\frac{\textbf{M}}{V}=\frac{1}{V}\int\textbf{m}(\textbf{x}) d^3\textbf{x}$ where $\textbf{m}(\textbf{x})$ is supposed to be some coarse-grained microscopic variable such as spin.

Which one is the correct and most general definition of the order parameter?

A note: Chaikin and Lubensky's book says that when $\textbf{m}(\textbf{x})$ is independent of $\textbf{x}$, the order parameter is given by $$\langle\textbf{m}(\textbf{x})\rangle=\frac{\textbf{M}}{V}$$ where the left hand side is thermal average and right hand side is defined by the the spatial integral expression. The definitions will be consistent with each other if in the integral expression $\textbf{m}(\textbf{x})$ is actually $\langle\textbf{m}(\textbf{x})\rangle$.

• I think what you are looking for is the Ergodic hypothesis – noah Aug 17 '17 at 20:49
• Just a note: To talk about any thermodynamic quantity you will need to, at some point, take a spatial ensemble average (which is basically a normalized spatial integral). – honeste_vivere Aug 21 '17 at 14:43

In the magnetic system, most of the time, people just calculate the thermal average of the magnetization at an arbitrary spatial point, say, $\langle m(\mathbf{x}) \rangle$. If you want to get the $M$ (the total magnetization), you just need to do a spatial integration over that.
In your problem, I think the authors are discussing the ferromagnetic phase, in that case, there is a translational invariance about the order parameter $\langle m(\mathbf{x}) \rangle$, which means at every spatial point $\mathbf{x}$, the value of the thermal average of $m(\mathbf{x})$ is the same, so you can integrate over the whole space (which is simply equivalent to time the volume of the system) and divide that by the volume of the system.
In some other cases, like spin density wave (or antiferromagnetic), when we calculate the $\langle m(\mathbf{x}) \rangle$, we would get some spatial dependence of the magnetization, e.g. $$m(\mathbf{x}) \propto \cos(\mathbf{Q} \cdot \mathbf{x}+\theta )$$
• You said, "If you want to get the MM (the total magnetization), you just need to do a spatial integration over that." Do you mean $\textbf{M}=\int \langle \textbf{m}(\textbf{x})\rangle d^3\textbf{x}$ instead of $\textbf{M}=\int\textbf{m}(\textbf{x})d^3\textbf{x}$? @ChuaNChen – SRS Aug 19 '17 at 3:16
• @SRS Yes, that's my understanding, I think the two processes "commute" with each other. It's like the thermal average over a sum of individual $m(x)$ is equivalent to the sum over the thermal average over each individual $m(x)$. – Chuan Chen Aug 19 '17 at 3:44