# Neel ordering on the square lattice vs mean-field AFM Heisenberg model

Question: It seems like the Neel order of the AFM Heisenberg model on the square lattice is actually stronger than the (bipartite) fully-connected case. This seems counterintuitive. Am I simply wrong about the above statement, or can this actually happen and be understood intuitively?

Details:

Consider the antiferromagnetic Heisenberg model for spin-1/2:

$$$$\hat{H}=\sum_{\langle i,j\rangle} (X_i X_j +Y_i Y_j + Z_iZ_j)$$$$

where the $$X,Y,Z$$ are 2x2 Pauli operators (thus I'm using the convention where each spin has "norm" 1).

The model obviously behaves quite differently for different connectivity specified by $$\langle i,j\rangle$$; here, I would focus mainly on two cases: the square lattice and the complete-bipartite graph case (also known as Lieb-Mattis model).

It is well-understood that for both cases, the ground state is in the Neel phase, where the order parameter $$$$M=\frac{1}{N}(\sum_{i\in A} Z_i - \sum_{j\in B}Z_j)$$$$ known as staggered magnetization has a non-zero expectation value in the thermodynamic limit. i.e, $$\langle M ^2\rangle$$ converges to a non-zero value when $$N\rightarrow \infty$$.

In my understanding, we can say that the "dense enough" connectivity of the system induces this Neel order in the ground state, since if the graph is just a 1-dimensional graph (e.g. a cycle of $$N$$ sites), then long range Neel order will not even appear. This makes me want to think that the complete-bipartite case has the "maximal" Neel order, because it has the largest connectivity without violating bipartiteness (which is obviously essential for the Neel ordering).

However, when you actually calculate the Neel order for the Lieb-Mattis model, you get $$$$\langle M^2\rangle = \frac{1}{3} + \frac{1}{3N}$$$$ while on the other hand for the square lattice, it seems pretty well established that the Neel order parameter is around $$\langle M^2\rangle = 0.376996 > 1/3$$ in the thermodynamic limit. (cf. https://arxiv.org/abs/0705.2746?context=cond-mat and https://journals.aps.org/prb/abstract/10.1103/PhysRevB.56.11678 be careful with the difference in convention here)

This goes against my intuition I mention above. Is it the case that higher connectivity in the Lieb-Mattis model actually weakens the Neel order compared to the square lattice? If that is true, I guess checking the Neel ordering for the honeycomb (hexagonal) lattice would be a good consistency check, but I couldn't find reference for that. If anyone knows such reference, or anything about the Neel order in general that sorts this puzzling situation, those information would be greatly appreciated!

(Added) After I posted this, I found several other papers that have numerical data on the Neel order for square lattices and honeycomb lattices. However, the square lattice Neel order is still stronger than the complete bipartite graph, and furthermore, the honeycomb lattice order parameter seems to be just a bit smaller than the square lattice case, reconfirming my "intuition" that the more connectivity the graph has, the stronger the Neel order should become (which turns out to be false when thinking about the complete bipartite graph and hence my question)

e.g., In the inset of fig. 2 (a) from https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.113.027201 you can see that $$\langle M^2\rangle$$ is a little above 0.09x4=0.36 for the square lattice $$J_2 =0$$ case (which corresponds to the model I'm considering now). In fig. 2 (e) of https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.110.127203 you can see that $$\langle M^2\rangle$$ is around 0.081x4=0.325 for the hexagonal lattice $$J_2/J_2 =0$$ case.

I found a stupid "mistake" in my definition. The thing I write above only takes into account of the $$z$$ component, so it is most natural to take the other two directions into account and make the order parameter a vector. This is what people are calculating in the above references, so obviously, I should compare it with value 1 and not 1/3!