# Renormalization Group and Ising with d=1 and D=1

I have a question about the results of RG on Ising model. I know it's possible to obtain two couple of relations

1. $K'(K)$, $q(K')$
2. $K(K')$, $q(K)$

between the coupling costants. My problem arise when i try to draw the flow diagram of the coupling constants. I know that this model dont allow phase transition except the trivial case $T=0$, but if we reiterate the relation (1) or (2) we increase or decrease the coupling costant. In one case i obtain $$K=0,T=\infty>>>>\cdots>>>>K=\infty,T=0$$ but in the other?

Boolean Ising Model with $d=1$ dimension of lattice, $D=1$ dimension of vector space of the spins on lattice. The energy with zero external field is $$H=-J\sum_{<ij>}S_iS_j$$ note that there are overcounting. Then the partition function can be put in the following form $$Z=\sum_{\{S\}}\prod_ie^{KS_iS_{i+1}}$$ With a partial summation on even spins it became $$Z'=\sum_{S}\prod_ie^{K(S_i+S_{i+1})}+e^{-K(S_i+S_{i+1})}=\sum_{S}\prod_if(K)e^{K'S_iS_{i+1}},$$ where in the last i used the scaling proprieties
$$Z(N,K)=f(K)^{N/2}Z(N/2,K').$$ The relations for $f(K)$ and $K'(K)$ are: $$f(K)=2\cosh^{1/2}2K,$$ $$K'=\frac{1}{2}\log\cosh{2K}.$$ The extensivity of free energy states $-\beta F=\log Z=Nq(K)=\frac{N}{2}\log f(K)+\frac{N}{2}q(K')$.

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I would look through "Elements of Phase Transitions and Critical Phenomena" Ortiz, Nishimori at your library if possible. The 3rd chapter goes over real space renormalization using decimation (renormalize over even numbered spins).

It basically rewrites the partition function for K' and obtains recursion relations between K' and K. One then looks for the fixed points in these recursion relations and these points should be an unstable critical point and two trivial fixed points related to the ordered and disordered state.

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Thanks, i understand, but my question was: if i write the recursion relation $K'(K)$ and it inverse $K(K')$, one tends to decrease the coupling constant, and the other to increase. how to recognize what gives the correct flow of the coupling constant? –  ivax Mar 5 '13 at 8:19

The RG is not a group, it's a semi group so you can only go in one direction, the one that actually renormalize.

Here you can use the relation K'(K) for your flow. but if you use K(K'), you will have something wrong because the RG procedure that give this relation as no physical meaning. it's as if you add spin in your Ising chain.

When you do RG, ,you have to go into te direction that "decrease" the number of site.

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