# 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 ($H=0$ ). 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.