# Decimation to a half of spins for a one-dimensional Ising model with H not 0

Let us now see a different decimation procedure for the same one-dimensional Ising model, when $H\neq 0$ . This time the idea of the procedure is to sum over the spins that are on even sites and leaving unaltered those on odd sites: Decimation to $N/2$ spins, with the same notation as before

We write the partition function as:

{\begin{aligned}Z_{N}(K,h)=\operatorname {Tr} e^{-\beta {\mathcal {H}}}=\sum _{\lbrace S_{i}=\pm 1\rbrace }e^{K\sum _{i}S_{i}S_{i+1}+h\sum _{i}S_{i}}=\\=\sum _{S_{1}=\pm 1}\cdots \sum _{S_{N}=\pm 1}e^{K(S_{1}S_{2}+S_{2}S_{3})+hS_{2}+{\frac {h}{2}}(S_{1}+S_{3})}\cdot e^{K(S_{3}S_{4}+S_{4}S_{5})+hS_{4}+{\frac {h}{2}}(S_{3}+S_{5})}\cdots \end{aligned}} Indicating with $S_{i}'$ the spins that are kept untouched and summing over the $N/2$ even spins:
{\begin{aligned}Z_{N}(K,h)=\\=\sum _{\lbrace S_{2i+1}'=\pm 1\rbrace }\sum _{\lbrace S_{2i}=\pm 1\rbrace }e^{S_{2}\left[K(S_{1}'+S_{3}')+h\right]+{\frac {h}{2}}(S_{1}'+S_{3}')}e^{S_{4}\left[K(S_{3}'+S_{5}')+h\right]+{\frac {h}{2}}(S_{3}'+S_{5}')}\cdots =\\=\sum _{\lbrace S_{2i+1}'=\pm 1\rbrace }\left[e^{\left(K+{\frac {h}{2}}\right)(S_{1}'+S_{3}')+h}+e^{\left(-K+{\frac {h}{2}}\right)(S_{1}'+S_{3}')-h}\right]\cdot \\\cdot \left[e^{\left(K+{\frac {h}{2}}\right)(S_{3}'+S_{5}')+h}+e^{\left(-K+{\frac {h}{2}}\right)(S_{3}'+S_{5}')-h}\right]\cdots =\\=\sum _{\lbrace S_{2i+1}'=\pm 1\rbrace }\prod _{i=0}^{N-1}\left[e^{\left(K+{\frac {h}{2}}\right)(S_{2i+1}'+S_{2i+3}')+h}+e^{\left(-K+{\frac {h}{2}}\right)(S_{2i+1}'+S_{2i+3}')-h}\right]\end{aligned}} Now, since $Z$ must not change after the RG transformation we can write:
$Z_{N}(K,h)=e^{Ng(K,h)}Z_{N'}(K',h')=e^{Ng(K,h)}\operatorname {Tr} _{\lbrace S'\rbrace }e^{-\beta {\mathcal {H}}'}$ where:
$-\beta {\mathcal {H}}'(K',h')=K'\sum _{i=0}^{N'-1}S_{2i+1}'S_{2i+3}'+h'\sum _{i=0}^{N'-1}S_{2i+1}'$ Since $N'=N/2\Rightarrow N=2N'$ , this means that:
{\begin{aligned}Z_{N}(K,h)=e^{2N'g}\sum _{\lbrace S_{2i+1}'=\pm 1\rbrace }e^{K'\sum _{i}S_{2i+1}'S_{2i+3}'+h'\sum _{i}S_{2i+1}'}=\\=\sum _{\lbrace S_{2i+1}'=\pm 1\rbrace }\prod _{i=0}^{N'-1}e^{K'S_{2i+1}'S_{2i+3}'+{\frac {h'}{2}}(S_{2i+1}'+S_{2i+3}')+2g}\end{aligned}} Therefore, we must have:
$e^{\left(K+{\frac {h}{2}}\right)(S_{2i+1}'+S_{2i+3}')+h}+e^{\left(-K+{\frac {h}{2}}\right)(S_{2i+1}'+S_{2i+3}')-h}=e^{K'S_{2i+1}'S_{2i+3}'+{\frac {h'}{2}}(S_{2i+1}'+S_{2i+3}')+2g}$ and this equality must hold for all the possible values of $S_{2i+1}'$ and $S_{2i+3}'$ . In particular:

The solutions of these equations are:

$K'={\frac {1}{4}}\ln {\frac {\cosh(2K+h)\cosh(2K-h)}{\cosh ^{2}h}}\quad \qquad h'=h+{\frac {1}{2}}\ln {\frac {\cosh(2K+h)}{\cosh(2K-h)}}$ $g={\frac {1}{8}}\ln \left[16\cosh(2K+h)\cosh(2K+h)\cosh ^{2}h\right]$ which are the recursion relations for this decimation procedure. Defining:
$x=e^{-4K}\quad \qquad y=e^{-2h}\quad \qquad z=e^{-8g}$ (where of course $0\leq x,y,z\leq 1$ ) the recursion relations can be more easily written as:
$x'=x{\frac {(1+y)^{2}}{(x+y)(1+xy)}}\quad \qquad y'=y{\frac {x+y}{1+xy}}\quad \qquad z'={\frac {z^{2}xy^{2}}{(x+y)(1+xy)(1+y)^{2}}}$ Note that $x'$ and $y'$ do not depend on $z$ : this means that the constant $g$ is not involved in the singular behaviour of the free energy density. In fact, from we have:
$f_{N}(K,h)={\frac {1}{2}}f_{N'}(K',h')-k_{B}Tg(K,h)$ and since $g$ does not influence the RG flow of the variables $x$ and $y$ (i.e. $K$ and $h$ ), the critical properties of the system are not altered by $g$ ; since as we know these are determined by the behaviour of the singular part of $f$ , $g$ is part of the regular one.