# Decimation to a third of spins for a one-dimensional Ising model with H=0

Let us consider a one-dimensional Ising model with nearest-neighbour interaction and periodic boundary conditions, without any external field (${\displaystyle H=0}$). We choose to apply the coarse-graining procedure to our system by grouping spins in blocks of three; this way the ${\displaystyle (i+1)}$-th block (with ${\displaystyle i=0,1,2,\dots }$) will be constituted by the spins ${\displaystyle S_{1+3i}}$, ${\displaystyle S_{2+3i}}$ and ${\displaystyle S_{3+3i}}$ (for example, the first block is ${\displaystyle [S_{1},S_{2},S_{3}]}$, the second one ${\displaystyle [S_{4},S_{5},S_{6}]}$ and so on). In order to define the new block spin we could use the majority rule, but we further simplify the problem requiring that the new block spin ${\displaystyle S_{I}'}$ coincides with the central spin ${\displaystyle S_{2+3i}}$ of the block. In other words, for every block we set:

${\displaystyle P(S_{I}';S_{1+3i},S_{2+3i},S_{3+3i})=\delta _{S_{I}',S_{2+3i}}}$
(for example for the first block we have ${\displaystyle P(S_{1}';S_{1},S_{2},S_{3})=\delta _{S_{1}',S_{2}}}$). Therefore, the coarse-graining procedure consists in summing over the spins at the boundaries of the blocks and leaving untouched the central ones . In the following figure we represent the situation, where the spins over which we sum are indicated by a cross ${\displaystyle \times }$ and the ones leaved untouched by a circle ${\displaystyle \circ }$:

Decimation to ${\displaystyle N/3}$ spins

Now, using the notation introduced in Basic ideas of the Renormalization Group for the general theory, we have:

{\displaystyle {\begin{aligned}e^{-\beta {\mathcal {H}}'}=\operatorname {Tr} _{\lbrace S_{i}\rbrace }P(S_{I}',S_{i})e^{-\beta {\mathcal {H}}}=\operatorname {Tr} _{\lbrace S_{i}\rbrace }\prod _{I}\delta _{S_{I}',S_{2+3i}}\cdot e^{K\sum _{j}S_{j}S_{j+1}}=\\=\sum _{\lbrace S_{i}=\pm 1\rbrace }\delta _{S_{1}',S_{2}}\delta _{S_{2}',S_{5}}\cdots e^{KS_{1}S_{2}}e^{KS_{2}S_{3}}e^{KS_{3}S_{4}}e^{KS_{4}S_{5}}\cdots =\\=\sum _{\lbrace S_{i}=\pm 1\rbrace }e^{KS_{1}S_{1}'}e^{KS_{1}'S_{3}}e^{KS_{3}S_{4}}e^{KS_{4}S_{2}'}\cdots \end{aligned}}}
Let us therefore see how to perform the sum on the first two blocks, ${\displaystyle [S_{1},S_{2},S_{3}]-[S_{4},S_{5},S_{6}]}$:
${\displaystyle \sum _{S_{3},S_{4}=\pm 1}e^{KS_{1}'S_{3}}e^{KS_{3}S_{4}}e^{KS_{4}S_{2}'}}$
From the definitions of ${\displaystyle \cosh }$ and ${\displaystyle \sinh }$ we can write:
${\displaystyle e^{KS_{a}S_{b}}=\cosh K(1+tS_{a}S_{b})\quad \qquad {\text{where}}\quad t=\tanh K}$
so that the sum over ${\displaystyle S_{3}}$ and ${\displaystyle S_{4}}$ becomes:
${\displaystyle \sum _{S_{3},S_{4}=\pm 1}(\cosh K)^{3}(1+tS_{1}'S_{3})(1+tS_{3}S_{4})(1+tS_{4}S_{2}')}$
Expanding the product and keeping in mind that ${\displaystyle S_{i}^{2}=+1}$, we get:
{\displaystyle {\begin{aligned}(1+tS_{1}'S_{3})(1+tS_{3}S_{4})(1+tS_{4}S_{2}')=\\=1+tS_{1}'S_{3}+tS_{3}S_{4}+tS_{4}S_{2}'+t^{2}S_{1}'S_{4}+t^{2}S_{1}'S_{3}S_{4}S_{2}'+t^{2}S_{3}S_{2}'+t^{3}S_{1}'S_{2}'\end{aligned}}}
and clearly all the terms containing ${\displaystyle S_{3}}$ or ${\displaystyle S_{4}}$ (or both) vanish when we perform the sum ${\textstyle \sum _{S_{3},S_{4}=\pm 1}}$. Therefore, the result of the partial sum for the first two blocks is:
${\displaystyle 2^{2}(\cosh K)^{3}\prod _{I}(1+t^{3}S_{1}'S_{2}')}$
(where ${\displaystyle 2^{2}}$ comes from the fact that the constant terms ${\displaystyle 1}$ and ${\displaystyle t^{3}S_{1}'S_{2}'}$ must be summed ${\displaystyle 2^{2}}$ times, two for the possible values of ${\displaystyle S_{3}}$ and two for ${\displaystyle S_{4}}$). Therefore, the partition function of the block spin system will be:
${\displaystyle Z_{N'}[K']=\operatorname {Tr} _{\lbrace S_{I}'\rbrace }2^{2N'}\cosh ^{3N'}K(1+t^{3}S_{I}'S_{I+1}')}$
where ${\displaystyle N'=N/3}$ is the new number of spin variables. However, we know that in general ${\displaystyle Z_{N'}=\operatorname {Tr} _{\lbrace S_{I}'\rbrace }e^{-\beta {\mathcal {H}}'}}$, so let us try to write ${\textstyle Z_{N'}[K']}$in this form. We have:
${\displaystyle 2^{2}(\cosh K)^{3}(1+t^{3}S_{I}'S_{I+1}')=2^{2}(\cosh K)^{3}{\frac {\cosh K'}{\cosh K'}}(1+t^{3}S_{I}'S_{I+1}')}$
and renaming ${\displaystyle t^{3}:=t'}$, so that:
${\displaystyle (\tanh K)^{3}=\tanh K'}$
this term becomes:
${\displaystyle 2^{2}{\frac {(\cosh K)^{3}}{\cosh K'}}\cosh K'(1+t'S_{I}'S_{I+1}')=2^{2}{\frac {(\cosh 1K)^{3}}{\cosh K'}}e^{K'S_{I}'S_{I+1}'}}$
Therefore:
${\displaystyle 2^{2}\cosh ^{3}K(1+t^{3}S_{I}'S_{I+1}')=e^{2\ln 2+\ln {\frac {(\cosh K)^{3}}{\cosh K'}}+K'S_{I}'S_{I+1}'}}$
and we can write:
${\displaystyle -\beta {\mathcal {H}}'(\lbrace S_{I}'\rbrace )=N'g(K,K')+K'\sum _{I}S_{I}'S_{I+1}'}$
where:
${\displaystyle g(K,K')=2\ln 2+\ln {\frac {(\cosh K)^{3}}{\cosh K'}}}$
The new effective Hamiltonian has therefore the same form of the original one with the redefined coupling constant ${\displaystyle K'}$, and exhibits also a new term (${\displaystyle g(K,K')}$) independent of the block spins.

Let us note that ${\textstyle (\tanh K)^{3}=\tanh K'}$ is the recursion relation we are looking for:

${\displaystyle K'=\tanh ^{-1}(\tanh ^{3}K)}$
Rewritten in the form ${\displaystyle t'=t^{3}}$, its fixed points are given by:
${\displaystyle t^{*}={t^{*}}^{3}}$
whose solutions are ${\displaystyle t^{*}=0}$ and ${\displaystyle t^{*}=1}$ (the case ${\displaystyle t^{*}=-1}$ is neglected because ${\displaystyle K>0}$ and so ${\displaystyle \tanh K>0}$). After all, however, ${\displaystyle \tanh K\to 0^{+}}$ if ${\displaystyle K\to 0^{+}}$ (i.e. ${\displaystyle T\to \infty }$) and ${\displaystyle \tanh K\to 1^{-}}$ if ${\displaystyle K\to \infty }$ (i.e. ${\displaystyle T\to 0^{+}}$): in other words, the fixed point ${\displaystyle t^{*}=0}$ corresponds to ${\displaystyle T=\infty }$ while ${\displaystyle t^{*}=1}$ to ${\displaystyle T=0}$. Since ${\displaystyle \tanh K<1\;\forall K\in \mathbb {R} }$, starting from any initial point ${\displaystyle t_{0}<1}$ the recursion relation ${\displaystyle t'=t^{3}}$ makes ${\displaystyle t}$ smaller every time, moving it towards the fixed point ${\displaystyle t^{*}=0}$. We can thus conclude that ${\displaystyle t^{*}=1}$ is an unstable fixed point while ${\displaystyle t^{*}=0}$ is stable, as graphically represented in the following figure:

RG flow for the recursion relation ${\textstyle (\tanh K)^{3}=\tanh K'}$

Note that the fact that the flow converges towards ${\displaystyle T=\infty }$ means that on large spatial scales the system is well described by a Hamiltonian with a high effective temperature, and so the system will always be in the paramagnetic phase (a part when ${\displaystyle T=0}$).

Let us now see how the correlation length transforms. We know that in general, if the decimation reduces the number of spins by a factor ${\displaystyle b}$ (in the case we were considering above, ${\displaystyle b=3}$) we have to rescale distances accordingly, and in particular:

${\displaystyle \xi (t')=\xi (t)/b}$
where in general ${\displaystyle t'=t^{b}}$. Since ${\displaystyle b}$ is in general arbitrary, we can choose ${\displaystyle b={\text{const.}}/\ln t}$ and thus:
${\displaystyle \xi (t')=\xi (t^{b})=\xi (e^{b\ln x})=\xi (e^{\text{const.}})=\left({\frac {\text{const.}}{\ln t}}\right)^{-1}\xi (t)}$
Therefore:
${\displaystyle \xi (t)={\frac {\text{const.}}{\ln t}}\sim {\frac {1}{\ln \tanh K}}}$
which is the exact result we have found at the end of The transfer matrix method.