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For the Ising model with only nearst neighbor interaction on square lattice, if we do the RG by integrating out half degree of freedom, then we would get a new Ising model with many kind of interactions, so Ising model with only nearst neighbor interaction cannot be a fixed point of RG.

In general, the fixed point should include infinite many kind of interactions and we cannot find it exactly.

But for now assume we find it, i.e., we have an Ising model with infinite many kind of interactions and it is a fixed point of RG, and we consider the two point spin-spin correlation,$\langle s(0)s(r)\rangle$. Before the RG, the distance for two spin is $r$, after the RG, the distance becomes $\frac{r}{2}$,but the Hamiltonian remains the same except the number of spins become half .So I think the $\langle s(0)s(r)\rangle=\langle s(0)s(\frac{r}{2})\rangle$. But obviously it is wrong since the spin-spin correlation function should decay as power law. What is wrong with my argument?

Qmechanic
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xjtan
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3 Answers3

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On iteration of the renormalization procedure is the set of transformations:

1) A transformation on space, in particular a rescaling

$$ x \mapsto x' = f(x) \ .$$

2) A transformation of the variables

$$ S(x) \mapsto S'(x') \ . $$

3) A transformation of the state

$$ \langle \cdot \rangle \mapsto \langle \cdot \rangle' \ .$$

To be more explicit, if you consider an Ising model with Hamilton operator

$$ \beta H[S] = \sum_{ x} J^{(1)}(x) S(x) + \sum_{ x,y} J^{(2)}(x,y)S(x) S(y) + \sum_{ x,y,z} J^{(3)}(x,y,z)S(x) S(y) S(z) + \cdots \ ,$$

Then the state $\langle \cdot \rangle$ is simply the Gibbs state of $H[S]$, and $\langle \cdot \rangle'$ is the Gibbs state of a Hamilton operator $H'[S']$, where $H'$ has parameters $J^{'(i)}$.

These have to be chosen s.t. all correlation functions match:

$$\langle S(x_1) S(x_2) \cdots S(x_n) \rangle = \langle S'(x_1') S'(x_2') \cdots S'(x_n') \rangle' \ . $$

Being at a fix point, we have that $\langle \cdot \rangle = \langle \cdot \rangle'$. Your conclusion holds if $S = S'$. But this is not true. In general, the spin operators have some scaling dimension (https://en.wikipedia.org/wiki/Scaling_dimension).

Lorenz Mayer
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You had stopped your consideration and didn't made a final conclusion. The renormalization transformation based on decimations has no fixed point. The accepted answer in this topic Critical 2d Ising Model contains a link to the notes about this matter.

Gec
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The issue is that the decimation procedure does not really allow to multiply by $2^{\Delta}$ in the switch from old spin variables $s(r)$ to $2^{\Delta}s(2r)$. This is the flaw mentioned by Wilson himself on the left column of page 801 of his article "The renormalization group: Critical phenomena and the Kondo problem" in Rev. Mod. Phys. A better transformation is the block spin procedure where the new spins are really new and not just a subset of the old ones. Another problem is that the fixed point should really be a probability measure on ${\mathbb{R}}^{\mathbb{Z}^d}$ rather than $\{-1,1\}^{\mathbb{Z}^d}$ if only to accommodate other real-valued models in the same universality class like the $\phi^4$ model. When doing block spin, the new spins are something like $$ t(r)=2^{\Delta-d}\sum_{u\in 2r+\{0,1\}^d} s(u) $$ For 2D Ising, $\Delta=1/8$ and $d=2$. When repeating the transformation $n$ times, the spacing between values is $2^{n(\Delta-d)}\rightarrow 0$. On the other hand extreme values (e.g. if all $s(r)$ are $+1$'s) go like $2^{n\Delta}\rightarrow \infty$. So just like in the central limit theorem for the binomial distribution, one is approaching the distribution of a real valued random variable with a density. Note that Wilson mentions an intermediate approach due to Kadanoff where the new spins $t$ are still in $\{-1,1\}^{\mathbb{Z}^d}$ but with a coupling parameter $\rho$ to the old spins $s$. Finally, the classical central limit theorem can be understood in the above framework with $\Delta=d/2$.

Abdelmalek Abdesselam
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