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

$P(S_{I}';S_{1+3i},S_{2+3i},S_{3+3i})=\delta _{S_{I}',S_{2+3i}}$

(for example for the first block we have

$P(S_{1}';S_{1},S_{2},S_{3})=\delta _{S_{1}',S_{2}}$).
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

$\times$ and the ones leaved untouched by a circle

$\circ$:

Decimation to

$N/3$ spins

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

${\begin{aligned}e^{-\beta {\mathcal {H}}'}=\operatorname {Tr} _{\lbrace S_{i}\rbrace }P(S_{I}',S_{i})e^{-\beta {\mathcal {H}}}=\operatorname {Tr} _{\lbrace S_{i}\rbrace }\prod _{I}\delta _{S_{I}',S_{2+3i}}\cdot e^{K\sum _{j}S_{j}S_{j+1}}=\\=\sum _{\lbrace S_{i}=\pm 1\rbrace }\delta _{S_{1}',S_{2}}\delta _{S_{2}',S_{5}}\cdots e^{KS_{1}S_{2}}e^{KS_{2}S_{3}}e^{KS_{3}S_{4}}e^{KS_{4}S_{5}}\cdots =\\=\sum _{\lbrace S_{i}=\pm 1\rbrace }e^{KS_{1}S_{1}'}e^{KS_{1}'S_{3}}e^{KS_{3}S_{4}}e^{KS_{4}S_{2}'}\cdots \end{aligned}}$

Let us therefore see how to perform the sum on the first two blocks,

$[S_{1},S_{2},S_{3}]-[S_{4},S_{5},S_{6}]$:

$\sum _{S_{3},S_{4}=\pm 1}e^{KS_{1}'S_{3}}e^{KS_{3}S_{4}}e^{KS_{4}S_{2}'}$

From the definitions of

$\cosh$ and

$\sinh$ we can write:

$e^{KS_{a}S_{b}}=\cosh K(1+tS_{a}S_{b})\quad \qquad {\text{where}}\quad t=\tanh K$

so that the sum over

$S_{3}$ and

$S_{4}$ becomes:

$\sum _{S_{3},S_{4}=\pm 1}(\cosh K)^{3}(1+tS_{1}'S_{3})(1+tS_{3}S_{4})(1+tS_{4}S_{2}')$

Expanding the product and keeping in mind that

$S_{i}^{2}=+1$, we get:

${\begin{aligned}(1+tS_{1}'S_{3})(1+tS_{3}S_{4})(1+tS_{4}S_{2}')=\\=1+tS_{1}'S_{3}+tS_{3}S_{4}+tS_{4}S_{2}'+t^{2}S_{1}'S_{4}+t^{2}S_{1}'S_{3}S_{4}S_{2}'+t^{2}S_{3}S_{2}'+t^{3}S_{1}'S_{2}'\end{aligned}}$

and clearly all the terms containing

$S_{3}$ or

$S_{4}$ (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:

$2^{2}(\cosh K)^{3}\prod _{I}(1+t^{3}S_{1}'S_{2}')$

(where

$2^{2}$ comes from the fact that the constant terms

$1$ and

$t^{3}S_{1}'S_{2}'$ must be summed

$2^{2}$ times, two for the possible values of

$S_{3}$ and two for

$S_{4}$).
Therefore, the partition function of the block spin system will be:

$Z_{N'}[K']=\operatorname {Tr} _{\lbrace S_{I}'\rbrace }2^{2N'}\cosh ^{3N'}K(1+t^{3}S_{I}'S_{I+1}')$

where

$N'=N/3$ is the new number of spin variables.
However, we know that in general

$Z_{N'}=\operatorname {Tr} _{\lbrace S_{I}'\rbrace }e^{-\beta {\mathcal {H}}'}$, so let us try to write

${\textstyle Z_{N'}[K']}$in this form. We have:

$2^{2}(\cosh K)^{3}(1+t^{3}S_{I}'S_{I+1}')=2^{2}(\cosh K)^{3}{\frac {\cosh K'}{\cosh K'}}(1+t^{3}S_{I}'S_{I+1}')$

and renaming

$t^{3}:=t'$, so that:

$(\tanh K)^{3}=\tanh K'$

this term becomes:

$2^{2}{\frac {(\cosh K)^{3}}{\cosh K'}}\cosh K'(1+t'S_{I}'S_{I+1}')=2^{2}{\frac {(\cosh 1K)^{3}}{\cosh K'}}e^{K'S_{I}'S_{I+1}'}$

Therefore:

$2^{2}\cosh ^{3}K(1+t^{3}S_{I}'S_{I+1}')=e^{2\ln 2+\ln {\frac {(\cosh K)^{3}}{\cosh K'}}+K'S_{I}'S_{I+1}'}$

and we can write:

$-\beta {\mathcal {H}}'(\lbrace S_{I}'\rbrace )=N'g(K,K')+K'\sum _{I}S_{I}'S_{I+1}'$

where:

$g(K,K')=2\ln 2+\ln {\frac {(\cosh K)^{3}}{\cosh K'}}$

The new effective Hamiltonian has therefore the same form of the original one with the redefined coupling constant

$K'$, and exhibits also a new term (

$g(K,K')$) independent of the block spins.

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

$K'=\tanh ^{-1}(\tanh ^{3}K)$

Rewritten in the form

$t'=t^{3}$, its fixed points are given by:

$t^{*}={t^{*}}^{3}$

whose solutions are

$t^{*}=0$ and

$t^{*}=1$ (the case

$t^{*}=-1$ is neglected because

$K>0$ and so

$\tanh K>0$). After all, however,

$\tanh K\to 0^{+}$ if

$K\to 0^{+}$ (i.e.

$T\to \infty$) and

$\tanh K\to 1^{-}$ if

$K\to \infty$ (i.e.

$T\to 0^{+}$): in other words, the fixed point

$t^{*}=0$ corresponds to

$T=\infty$ while

$t^{*}=1$ to

$T=0$.
Since

$\tanh K<1\;\forall K\in \mathbb {R}$, starting from any initial point

$t_{0}<1$ the recursion relation

$t'=t^{3}$ makes

$t$ smaller every time, moving it towards the fixed point

$t^{*}=0$. We can thus conclude that

$t^{*}=1$ is an unstable fixed point while

$t^{*}=0$ 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 $T=\infty$ 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 $T=0$).

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 $b$ (in the case we were considering above, $b=3$) we have to rescale distances accordingly, and in particular:

$\xi (t')=\xi (t)/b$

where in general

$t'=t^{b}$. Since

$b$ is in general arbitrary, we can choose

$b={\text{const.}}/\ln t$ and thus:

$\xi (t')=\xi (t^{b})=\xi (e^{b\ln x})=\xi (e^{\text{const.}})=\left({\frac {\text{const.}}{\ln t}}\right)^{-1}\xi (t)$

Therefore:

$\xi (t)={\frac {\text{const.}}{\ln t}}\sim {\frac {1}{\ln \tanh K}}$

which is the exact result we have found at the end of

The transfer matrix method.