Spontaneous magnetization in Ising model with antiferromagnetic interaction

Question

Consider the Ising model on the square lattice with antiferromagnetic interaction between neighbouring spins.

I am somewhat confused about the spontaneous magnetisation in this model at zero temperature.

My question: which points of the reasoning below are wrong?

1. On the one hand, magnetisation is defined as the average value of one spin, averaged over all configurations (with the Boltzmann distribution) and all spins. This is e.g. eq. (1.7.12) in R. J. Baxter Exactly solved models in statistical mechanics (1989).
2. The ground state in this model on the square lattice, which is bipartite, is: all spins up in one part, all spins down in the other part, in the chessboard pattern.
3. At zero temperature, only the ground state(s) contributes to the magnetisation when we average over states.
4. Averaging over all spins gives the result: the spontaneous magnetisation is equal to zero.
5. On the other hand, according to Baxter's pp. 118-119 (from the chapter on square-lattice Ising model), in a (discrete) translation invariant system, the square of magnetisation can be represented as correlation between spins with the distance between spins sent to infinity. And spontaneous magnetisation is the zero-field limit of that.
6. The square lattice is invariant under any lattice translation, so the expression for magnetisation defined in the previous point is applicable to our model and should give the same result as the definition from the 1st point above.
7. At small fields, when the ground state dominates, if the (Manhattan) distance between spins is even then their correlation is $1$ while if it is odd then it's $-1$, so the large-distance limit of the correlation between spins (which sholud be equal to the square of magnetisation) does not exist.
8. However, on the same pages (and in the note https://arxiv.org/abs/1103.3347 ), Baxter gives the famous Onsager's solution, which is well-defined and nonzero.
9. Onsager's solution should work equally well for ferromagnetic and antiferromagnetic interactions.
10. Onsager's solution should agree with the zero solution obtained in the fist points above.

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For the purpose of improving the automated detection of related questions, I'm noting that I'm interested not only in spontaneous magnetisation but also spontaneous magnetization. :)

Answer 1

Let me discuss the case of positive temperatures, as this is slightly less pathological (but everything remains true, interpreted suitably, at $\beta=\infty$). So, let us consider the planar Ising antiferromagnet at very large $\beta$. There are two extremal Gibbs states, each one corresponding to small finite-temperature perturbations of one of the two chessboard groundstates. One could select one by using a suitable staggered magnetic field, but it is simpler and more elegant to select it using a suitable boundary condition. That is, let us select one of these Gibbs states (say, the one with a positive magnetization at the origin) by taking the thermodynamic limit using as boundary condition the ground state with a $+$ spin at the origin; I'll denote by $\langle\cdot\rangle_\beta^{0+}$. Since the resulting Gibbs measure is extremal, it is mixing (see point 4 of Theorem 6.58 in this book). In particular, $$ \lim_{\|x\|\to\infty} \bigl( \langle\sigma_0\sigma_x\rangle_\beta^{0+} - \langle\sigma_0\rangle_\beta^{0+}\langle\sigma_x\rangle_\beta^{0+} \bigr) = 0, $$ so that $$ \langle\sigma_0\sigma_x\rangle_\beta^{0+} \approx \langle\sigma_0\rangle_\beta^{0+}\langle\sigma_x\rangle_\beta^{0+} $$ when $\|x\|$ is large. In particular, the right-hand side is not the squared magnetization, but $(-1)^{\|x\|_1} (\langle\sigma_0\rangle_\beta^{0+})^2$, as you want (here, $\|x\|_1 = |x_1|+|x_2|$ is the $\ell^1$-norm of the vector $x=(x_1,x_2)$).

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Note finally that, at zero external field, the Ising antiferromagnet can be obtained from the ferromagnet by the following transformation of the microstates: $$ \tilde\sigma_i = (-1)^{\|x\|_1}\sigma_i,\qquad \forall i\in\mathbb{Z}^2. $$ Namely, the configuration $\sigma$ has the same probability under the Gibbs state associated to the Ising ferromagnet as the configuration $\tilde\sigma$ has under the Gibbs state associated to the Ising antiferromagnet. You can thus also understand the above simply by using the standard claim (about the square magnetization) for the ferromagnet and applying the above transformation.