I'm trying to rederive the expressions for the square of the sublattice magnetization of the $120^\circ$ Neel antiferromagnetic (NAFM) and the stripe phase on the triangular lattice as shown in equations (2) and (3) of (https://arxiv.org/abs/1802.00935) or equation (2) of (https://arxiv.org/abs/1309.6309). For a lattice with $N$ sites
$m_{Neel}^2 = \frac{1}{3} \displaystyle \sum_{\alpha = 1}^3 \left[ \frac{1}{\frac{N}{6} \big(\frac{N}{6} + 1 \big)} \left\langle \left(
\sum_{i \in \alpha} \boldsymbol{S}_i \right)^2 \right\rangle \right]$
where $\alpha = 1,2,3$ represent the three sublattices of the $120^\circ$ order and the factor $\frac{N}{6}\big(\frac{N}{6} + 1\big)$ is used to normalize the value of $m_{Neel}^2$ so that $m_{Neel}^2 \in (0,1)$.
$m_{stripe}^2 = \frac{1}{2} \displaystyle \sum_{\beta = 1}^2 \left[ \frac{1}{\frac{N}{4} \big(\frac{N}{4} + 1 \big)} \left\langle \left(
\sum_{i \in \beta} \boldsymbol{S}_i \right)^2 \right\rangle \right]$
where $\beta = 1,2$ represent the two sublattices of the stripe order.
The corresponding pictures of the two phases are shown below
Edit: I found a definition of magnetic order parameters in $\S 2.3.1$ of Quantum Magnetism (2004) edited by Schollowock et al. For completeness I've summarized it below (I've added the derivations, a picture and changed the notations slightly). Any errors that remain are my own.
Suppose we have a Hamiltonian $H$ and suppose the system spontaneously breaks some continuous symmetry (like rotation) to settle into a classical ground state. To break the symmetry we add a field to the Hamiltonian
$$
H^{new} = H – h \sum_i \boldsymbol{e}_i^{z'} \cdot \boldsymbol{S}_i
$$
where $\boldsymbol{e}_i^{z'}$ is the local $z$-direction at site-$i$ as shown in the figure below.
Our goal is define an order parameter to to measure the magnetization of the state. From the two possible degrees of freedom, $S^2$ and $m_z$, we chose $m_z$. First we define the order-parameter-operator as
$$
\hat{m}_z \equiv \textstyle \frac{1}{N} \displaystyle\sum_i S_i^{z'}
= \textstyle \frac{1}{N} \displaystyle \sum_i \boldsymbol{e}_i^{z'} \cdot \boldsymbol{S}_i
$$
so that the order parameter is the expectation value of $\hat{m}_z$ in the ground state
$$
m_z \equiv \lim_{h \to 0} \lim_{N \to \infty} \langle \hat{m}_z \rangle.
$$
Note: we assume the spins are classical vectors and the ground state of the system has planer ordering (the spins lie in a plane). We can then pick the plane of the spins to be the $z$–$x$ plane and relate the components spin $\boldsymbol{S}_i$ (the spin at site-$i$ in the global frame) with the spin in the local frame, $\boldsymbol{S}_i'$, by the rotation matrix as follows
$$
\boldsymbol{S}_i' = R_y (\phi_i) \boldsymbol{S}_i =
\begin{bmatrix}
S_i^x \cos{\phi_i} – S_i^z \sin{\phi_i} \\
S^y \\
S_i^x \sin{\phi_i} + S_i^z \cos{\phi_i}
\end{bmatrix}
$$
Note: $\phi_i$ is the angle between the local axis $\boldsymbol{e}_i^{z'}$ and the global axis $\boldsymbol{e}^{z}$ at site $i$. Hence, the $z$-component of the spin at site $i$ in the local frame is
$$
\boxed{ S_i^{z'} = S_i^x \sin{\phi_i} + S_i^z\cos{\phi_i}.}
$$
Example 1: Ferromagnetic (FM) order on any 2D lattice. For a FM order for all $i$ in the system $\phi_i = 0$ thus
$$
\begin{align}
m_z = \lim_{h \to 0} \lim_{N \to \infty} \langle \hat{m}_z \rangle \: & = \: \lim_{h \to 0} \lim_{N \to \infty} \textstyle \frac{1}{N} \displaystyle \big\langle \sum_i S_i^{z'} \big\rangle. \\
& = \lim_{h \to 0} \lim_{N \to \infty} \textstyle \frac{1}{N} \displaystyle \big\langle \sum_i S_i^x \sin{\phi_i} + S_i^z \cos{\phi_i} \big\rangle. \\
& = \lim_{h \to 0} \lim_{N \to \infty} \textstyle \frac{1}{N} \displaystyle \big\langle \sum_i S_i^{z} \big\rangle.
\end{align}
$$
Therefore, for a FM order the local coordinate system doesn't matter.
Example 2: Neel antiferromagnet order on any two-sublattice system. Call the sublattices $A$ and $B$. Without loss of generality if we let $\phi_{i \in A} = 0$, then $\phi_{i \in B} = \pi$ so that
$$
\begin{align}
m_z & = \lim_{h \to 0} \lim_{N \to \infty} \textstyle \frac{1}{N} \displaystyle \big\langle \sum_{i \in A} S_i^{z'} + \sum_{i \in B} S_i^{z'} \big\rangle. \\
& = \lim_{h \to 0} \lim_{N \to \infty} \textstyle \frac{1}{N} \displaystyle \big\langle \sum_{i \in A} S_i^z \cos{0} + \sum_{i \in B} S_i^z \cos{\pi} \big\rangle. \\
& = \lim_{h \to 0} \lim_{N \to \infty} \textstyle \frac{1}{N} \displaystyle \big\langle \sum_i \epsilon_i S_i^{z} \big\rangle
\end{align}
$$
where $\epsilon_i = +1$ for $i \in A$ and $\epsilon_i = -1$ for $i \in B$. If we introduce the short hand notation $S_{\alpha}^\mu \equiv \displaystyle \sum_{i \in \alpha} S_i^\mu$ for $\mu = x, y, z$ and $\alpha = A, B, C$, then we can rewrite the above expression as
$$
m_z = \lim_{h \to 0} \lim_{N \to \infty} \textstyle \frac{1}{N} \displaystyle \big\langle S_A^z – S_B^{z} \big\rangle.
$$
Example 3: $120^\circ$ Neel antiferromagnetic order on a triangular lattice. In this case we have three sublattices $A, B, C$ where we set $\phi_{i \in A} = 0, \: \phi_{i \in B} = 2 \pi/3$, and $\phi_{i \in C} = 4 \pi/3$ so that
$$
\begin{align}
m_z & = \lim_{h \to 0} \lim_{N \to \infty} \textstyle \frac{1}{N} \displaystyle \big\langle \sum_{i \in A} S_i^{z'} + \sum_{i \in B} S_i^{z'} + \sum_{i \in C} S_i^{z'} \big\rangle. \\
& = \lim_{h \to 0} \lim_{N \to \infty} \textstyle \frac{1}{N} \displaystyle \big\langle \displaystyle
\sum_{i \in A} S_i^z
+ \sum_{i \in B} \textstyle \big(S_i^x \sin{\frac{2 \pi}{3}} + S_i^z \cos{\frac{2 \pi}{3}} \big)
\displaystyle
+ \sum_{i \in C} \textstyle \big(S_i^x \sin{\frac{4 \pi}{3}} + S_i^z \cos{\frac{4 \pi}{3}} \big)\big\rangle. \\
& = \lim_{h \to 0} \lim_{N \to \infty} \textstyle \frac{1}{N} \displaystyle \big\langle S_A^z + \textstyle \frac{\sqrt{3}}{2} S_B^x – \frac{1}{2} S_B^z – \frac{\sqrt{3}}{2} S_C^x – \frac{1}{2} S_C^z \big\rangle
\end{align}
$$
Finite lattices. The ground state of a finite system can not spontaneously break symmetry (SSB). This implies $(\hat{m}_z)^2$ is used instead of $\hat{m}_z$. Moreover, the ground state of finite antiferromagnetic systems with even sites is rotationally invariant so $\langle S_i^x S_i^x \rangle = \langle S_i^y S_i^y \rangle = \langle S_i^z S_i^z \rangle.$ Using these two facts we define a new finite-system order-parameter as follows
$$
\begin{align}
\overline{m} = \sqrt{ \Big\langle \textstyle \big( \frac{1}{N} \displaystyle \sum_i \boldsymbol{S}_i' \big)^2 \Big\rangle }
& = \sqrt{ \textstyle \frac{1}{N^2} \displaystyle \sum_i \langle \boldsymbol{S}_i' \cdot \boldsymbol{S}_i' \rangle} \\
& = \sqrt{ \textstyle \frac{1}{N^2} \displaystyle \sum_i \big\langle \big( S_i^{x'} S_i^{x'} \rangle + \langle S_i^{y'} S_i^{y'} \rangle + \langle S_i^{z'} S_i^{z'} \big) \big\rangle } \\
& = \sqrt{ 3 \Big\langle \big(\textstyle \frac{1}{N} \displaystyle \sum_i S_i^{z'} \big) \big(\textstyle \frac{1}{N} \displaystyle \sum_i S_i^{z'} \big) \Big\rangle } \\
& = \sqrt{ 3 \big\langle (\hat{m}_z)^2 \big\rangle }.
\end{align}
$$
The order parameters defined above has limitations: (i) It supposes that the classical and quantum ground states of $H$ have the same type of ordering. (ii) It doesn't deal with systems where the classical ground state has a large number of non-trivial degeneracy. (iii) It assumes we know the form of the classical ground state in the first place. To avoid these issues they define a universal order parameter by
$$
m^+ = \sqrt{ \textstyle \frac{1}{N^2} \displaystyle \sum_{i, j}^N \big| \langle \boldsymbol{S}_i \boldsymbol{S}_j \rangle \big| }.
$$
For spin systems with non-collinear GS (like the $120^\circ$ AFM) the definitions $\overline{m}$ ad $m^+$ are not identical. Moreover, the universal order parameter can not distinguish between different types of ordering.
- Why does no SSB in finite systems imply we have to look at $(\hat{m}_z)^2$ instead of $\hat{m}_z$?
- Is it correct to say they're deriving/defining the universal order parameter by first using dimensional analysis to get $\langle \boldsymbol{S}_i \boldsymbol{S}_j \rangle$ and then dividing by $N^2$ to get an intensive quantity?
- How are the normalization constants in $m_{Neel}^2$ and $m_{stripe}^2$ determined starting either from $\overline{m}$ or $m^+$?
Best Answer
This is a common point to get confused about SSB. To really understand it, you would actually need to get back to the foundations of statistical mechanics which is a monster itself, so here I'll just explain the gist of it.
In a finite system, the ground state of a quantum system would just be the equal superposition of the different symmetry broken states, so the expectation value of $\hat m_z$ will always be strictly 0 because of cancelation. This is why you would want a way to measure your order parameter that doesn't always cancel out. Squaring it is the simplest and mathematically the nicest way to do so. Typically, you would get a nonzero value for $\langle \hat m_z^2\rangle$ even in the disordered phase, but that will tend to 0 in the thermodynamic limit $N\rightarrow\infty$.
One important point I'd like to say is that this "universal order parameter" is not truly universal, since it will fail to capture not only exotic states like topological order (cf. e.g. toric code), but also fairly common non-magnetic orders such as valence-bond solids.
With that in mind, I would say this is just like one possible attempt to try to generalize the simplest kind of order parameter so that it captures a slightly broad range of magnetic orders. And to do that, they first sum up the simplest quantity that probes the system -- the two-point correlations -- and then normalize it to get intensivity.
I don't think the term "universal order parameter" is commonly used. A quick google search will reveal that different people propose different quantities claiming it to be a somewhat universal order parameter.
Honestly, it doesn't really matter as long as they give you an intensive quantity. The differences in the details of those constants will only result in a constant factor in the resulting order parameter. Since the central role for the order parameter is to see whether it is 0 or nonzero (i.e. distinguish different phases), such constant factors don't play an important role.
In practice though, of course it's nicer to have quantities where you can calculate that the maximum (minimum) possible value it can take is $\pm 1$, because then you have some sense of "how strong" the order is by measuring them. These $N/6$ and $N/4$ factors do exactly that, and you can calculate to see that indeed their maximum value is $1$ assuming $s=1/2$.