Phase transition in Ising Model

Question

Consider the NN Ising model as \begin{equation} H = -J \sum_{<ij>} \sigma_{i} \sigma_{j} - h \sum_{i} \sigma_{i} \end{equation} This model has a global $\mathbb{Z_{2}}$ symmetry in the absence of $h$. We know that phase transition occurs when $d>1$.

We can write the continuum field theory for that model as - \begin{equation} F[m] = \int d^dx \Big[\frac{t}{2}m^2(x) + u m^4(x) + \dfrac{K}{2}(\nabla m)^2+.......-h \cdot m\Big] \end{equation} It has a symmetry of $m \to -m$ in the absence of $h$.

My question is: Any phase transition that breaks the $\mathbb{Z_{2}}$ symmetry is always a continuous (2nd Oreder) transition or not (1st order)? How do we prove that statement rigorously?

Answer 1

No, there is no reason for such a phase transition to be necessarily of second order. In fact, in the one-dimensional Ising model with formal Hamiltonian $$ -\sum_{i,j\in\mathbb{Z}} J_{|j-i|} \sigma_i\sigma_j $$ with $J_r \sim 1/r^2$, it is known that the phase transition is first order (in the sense that the spontaneous magnetization is discontinuous at $\beta_c$).

On the other hand, for the nearest-neighbor Ising model on $\mathbb{Z}^d$, $d\geq 2$, it has been proved that the phase transition is always continuous (note: when $d \geq 4$, this result had been known for decades). Their argument also applies to the one-dimensional model when $J_r\sim r^{-\alpha}$, $1<\alpha<2$ (moreover, it is well known that there is no phase transition when $\alpha >2$).

The rigorous proofs are rather technical and I shall not even attempt to explain them here. If you are interested, the papers cited above contain all the relevant information.