The 2d Ising model is extremely well studied, nevertheless I have encountered two facts which seem to contradict one another, and I have not been able to find the resolution in the literature. The puzzle is the following.
The critical Ising model is well known to be described by a CFT, and in particular a minimal model. This is described in many places, for example Ginsparg's CFT notes https://arxiv.org/abs/hep-th/9108028. To find the critical temperature, for which the CFT description is valid, perhaps the easiest way is to exploit the Kramers-Wannier duality, which relates the high-temperature/weak-coupling theory to the low-temperature/strong-coupling theory. The critical temperature is then given by the self-dual temperature. This makes it clear that the critical theory is just the usual 2d Ising Hamiltonian, but with the critical value of hte coupling constant $\beta J \equiv K = K_{*}$.
The defining property of the theory at the critical point is that it is invariant under RG flows. In general if $\mathcal{R}$ denotes the RG operation (in any given scheme, for example block-spin RG) and if the Hamiltonian $H$ depends on the coupling constants $\lambda_{1} \lambda_{2}, ..$, then this may be written schematically as
$$ \mathcal{R} H[\lambda_1^*, \lambda_2^* , … ]= H[\lambda_1^*, \lambda_2^* , … ],$$
where $*$ denotes fixed point quantities. Here is where the puzzle arises. Applied to the 2d Ising model with nearest-neighbor (NN) interactions only, the standard block-spin RG generates next-to-nearest-neighbor (NNN) interactions, and even NNNN interactions. See this for a demonstration of this fact. By examining the RG recursion relations, one finds that for no finite $K$ do these new interactions vanish. Therefore, with this RG scheme at least, the 2d Ising model with NN interactions can never be a fixed point of the RG transformation. Any critical theory will necessarily involve additional higher-spin couplings.
So is the critical Ising model $H = – J_* \sum_{<i,j>} s_i s_j$, with only NN interactions, or are there an infinite number of additional higher-spin interactions (which may become negligible in the continuum limit)?
Best Answer
The critical point is not the same thing as the RG fixed point. Let $\mathcal{T}$ denote "theory space" meaning the set of all possible probability measures on real valued fields on the fixed unit lattice $\mathbb{Z}^2$. Block-spin or decimation etc. give you a map $R:\mathcal{T}\rightarrow \mathcal{T}$, namely, a renormalization group transformation. The picture to have in mind is that there is a special point $V_{\ast}\in \mathcal{T}$ such that $R(V_{\ast})=V_{\ast}$. That's the RG fixed point. It is hyperbolic and has a stable manifold $W^s$ (the critical surface) as well as an unstable manifold $W^u$. Now $\beta\mapsto {\rm Ising}(\beta)$ is a special one parameter curve in the big space $\mathcal{T}$. It is the curve made of clean NN Ising Hamiltonians with no extra terms. The critical point is the intersection of this curve with $W^s$ and it corresponds to a special value $\beta_c$ of the inverse temperature. For more about the various ways of seeing a QFT: a set of correlation functions, a probability measure on the space of distributions, a point in $\mathcal{T}$ or a complete orbit of $R$ in $\mathcal{T}$ see the end of Section 4 of my short article QFT, RG, and all that, for mathematicians, in eleven pages.
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