# Phase space of Ising minimal model + deformations

Consider the Ising field theory, a conformal field theory in 2 dimensions which corresponds to the minimal model $$\mathcal{M}_{4,3}$$ and it's perturbations by the relevant operators $$\epsilon, \sigma$$ corresponding to the energy density (not to be confused with the stress-energy tensor) and the spin respectively, given schematically by the following Hamiltonian

$$H=\mathcal{H}_{\text{Ising}}+m_f\int\epsilon(x) d^2x+h\int\sigma(x)d^2x.$$

Interestingly, the model is integrable in the directions $$m_f=0$$ and $$h=0$$. The operator $$\epsilon$$ is even under fermion $$\mathbb{Z}_2$$ symmetry and $$\sigma$$ is odd. I would like to understand the spontaneous symmetry breaking structure of phase space for small values of the couplings, if it exists. To explain what I mean, assume that initially $$m_f>0$$ and the magnetic field has a very small non-zero value. As I raise the value of the magnetic field is there a point where I can induce a first order transition or is there not? Can I describe it by some sort of Landau type Hamiltonian?

I am aware of the results in the spin lattice version of the Ising model, where there is such a first order transition when crossing the line $$h=0$$ for $$T, however I am not entirely certain that I can use these results to argue for the phase diagram of the continuum theory above (that is described by deformations around the critical point $$T=T_c$$). I know however that the mass the fermion gets for small departures from the critical point is proportional to the distance from the critical point $$m_f\propto|1-T/T_c|$$ so it would seem that in the lattice model, you induce a first order transition the moment you turn on the magnetic field.

What gives? Is the answer trivial and exactly the same as in the lattice model? Is the situation different for a negative mass parameter $$m_f$$? How can I connect the dots together?

The answer is exactly the same as what you get in the lattice model. The field theory is supposed to accurately describe the long-distance, low-energy behavior of the lattice model, so it better reproduces the phase diagram around the critical point. The sign of the mass determines which side of the transition you are in, and once you are on the ferromagnetic side, and with the magnetic field turned on, passing through $$h=0$$ crosses a first-order transition.

Also, once you add the magnetic field term the mapping to fermions no longer works: the $$Z_2$$ symmetry that is necessary for the bosonization/fermionization map to work is broken explicitly. Put it in another way, the $$\sigma$$ operator is a highly non-local operator in the fermionic representation.

• Are you saying that $m_f<0$ represents the antiferromagnetic side of the model? Can you elaborate a bit on what happens in terms of phase transitions on that side of the model? I would appreciate any references on the subject you may have as well. Dec 7, 2021 at 22:40

After a couple days of bibliographic research I finally have answered my question, and posting an answer for posterity here.

As mentioned above, the phase space in the ferromagnetic phase of the Ising model on a square lattice with the same vertical and horizontal couplings at finite temperature is fairly simple: There is one long-range critical point at the Onsager temperature $$T=T_c=\frac{2J}{\ln(1+\sqrt{2})}$$ and the first-order transition happens immediately when the magnetic field is turned on (the magnetization jumps discontinuously to $$+1$$ or $$-1$$ depending on the sign of the magnetic field).

However, the antiferromagnetic model is a little richer. The $$h=0$$ line is the same as the ferromagnetic Ising but the two models differ a little bit when $$h\neq 0$$. In fact the presence of the magnetic field does not destroy the staggered order of the ground state, until it exceeds some critical value $$h_c(T)$$ given by the conjectural relation (but within at most 1% from the exact critical curve):

$$\cosh\frac{h_c(T)}{J}=\sinh^2\frac{2J}{T}$$

This relation also predicts that close to the Onsager point:

$$\left(\frac{h}{J}\right)^2=8\sqrt{2}\frac{T_c-T}{J}$$

which is exactly the result that I had in mind. In fact the same square root law for the critical curve is obtained for a bosonic field theory in 2-D with lagrangian

$$\mathcal{L}=\frac{m^2}{2}\phi^2+\frac{g}{3!}\phi^3+\frac{\lambda}{4!}\phi^4$$

where it can be seen that for small couplings $$g,\lambda<, a first order transition with the VEV of the fundamental field as the order parameter each time one crosses the line $$(g/m^2)^2=3(\lambda/m^2)$$ (interestingly, this happens to be the exact same formula coming from a mean field theory treatment for a coarse-grained field theory with maximum polynomial interaction of degree $$4$$).

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