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?

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    $\begingroup$ You forgot rescaling which is not raplacing $s(r)$ by $s(r/2)$ but by $2^{\Delta}s(r/2)$ where $\Delta=1/8$ is the scaling dimension of the spin field. $\endgroup$ – Abdelmalek Abdesselam Nov 30 '18 at 15:02
  • $\begingroup$ Thank you for the answer and I agree with it. So the picture for this decimation RG is: after integrating out half spins, the couplings before $n-$body interaction term $s^n$ are not exactly the same as before have a difference of $2^{-n \Delta }$, so by rescaling the $s$ to $2^\Delta s$, the Hamiltonian is restored? And computing the correct $\Delta$ is not a simple task in general. $\endgroup$ – xjtan Dec 1 '18 at 2:28
  • $\begingroup$ And I think it is magic that a simple rescaling of $s$ could fixed all the problem in the coupling. $\endgroup$ – xjtan Dec 1 '18 at 2:38
  • $\begingroup$ I wrote an answer with a bit more detail. $\endgroup$ – Abdelmalek Abdesselam Dec 3 '18 at 15:59

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 $\Delta=d/2$.

  • $\begingroup$ Thanks for the nice answer. I also did some literature search and Kadanoff also wrote on page 295 his book "statistical physics ,static, dynamics and renormalization" that: "The only possibility is that the renormalization scheme that we are using is itself flawed in that in never gets one to the fixed point. That is the right answer.(On the footnote it says that this is based on his private communication with Wilson)". $\endgroup$ – xjtan Dec 4 '18 at 2:27
  • $\begingroup$ Then I was wondering the behavior of critical Hamiltonian as the decimation procedure procede. McGreevy in his note mcgreevy.physics.ucsd.edu/f18/2018F-217-lectures.pdf wrote other two possible situations could be chaos or limited cycle. I think here should be chaos. $\endgroup$ – xjtan Dec 4 '18 at 2:39

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).

  • $\begingroup$ Thanks for the answer, though I do not agree with that their distance on the lattice have not changed. There is one step in RG called rescaling, which rescale the distance between two sites by 1/2 in our context. The reason for the rescaling is for comparing the models before and after the RG. You can check out MacGreevy's note on RG, which I think is fantastic. mcgreevy.physics.ucsd.edu/f18/2018F-217-lectures.pdf $\endgroup$ – xjtan Nov 29 '18 at 12:25
  • $\begingroup$ again, this depends on the procedure. If you want help for your specific procedure, i would recommend outlining the steps you have in mind. Concretely i would suspect that in your case the spin operators get re-defined. $\endgroup$ – Lorenz Mayer Nov 29 '18 at 13:59
  • $\begingroup$ The procedure in my mind is as follows: $\endgroup$ – xjtan Nov 30 '18 at 6:51
  • $\begingroup$ The procedure in my mind is as follows: (1) the square lattice is a bipartite lattice, so we integrated out all the spins on one sub-lattice and get a new Ising model; (2) if we consider the $<S_0 S_2>$ on the old lattice, on the old lattice, the distance is 2 on the unit of lattice constant $a$, in the new lattice ,the distance is 1 in the new lattice constant $2a$; so we rescale the lattice constant back to $a$. $\endgroup$ – xjtan Nov 30 '18 at 7:03

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.

  • $\begingroup$ According to my previous understanding, yes, not fixed point. But adding the operation of rescaling $s$ to $2^\Delta s$ would give a fixed point. $\endgroup$ – xjtan Dec 1 '18 at 2:36

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