# Decimation in dimensions higher than one: proliferation of the interactions

As we have already stated, in $d=1$ the recursion relations can be determined without great problems and they don't introduce new interactions. However this is not the case if $d>1$ , and the value of the new coupling constants can't be determined exactly, forcing us to use approximations. Let us see with a generic example how the RG transformation can introduce new interactions in a two-dimensional Ising model with nearest-neighbour interactions.

Suppose we divide our system in blocks containing an odd number of spins and, similarly to what we have seen for $d=1$ in the previous sections, we sum over the spins on the boundary of the block and leave unchanged the one at the center.

Looking at the figure, we see that the spin on the corder of the block 2 is coupled to one spin in bock 1 and one in block 3. When we sum over the spin in block 2 an effective coupling between blocks 1 and 3 will be established: we therefore see that the coarse-graining procedure introduces next-nearest-neighbour interactions between the blocks, so new terms are appearing in the Hamiltonian (which of course, as already stated, respect the symmetries of the original one). We therefore understand that the iteration of the RG will introduce increasingly complicated couplings: this is the so called proliferation of interactions.

Let us now see in detail how to face the problem of the proliferation for a two-dimensional Ising model with nearest-neighbour interaction and $H=0$ . We choose to coarse-grain the system summing over a "chessboard" of spins, as shown in the figure below, which also defines the symbolic notation we are going to use.

We therefore have:

$Z_{N}=\operatorname {Tr} _{\lbrace S_{i}\rbrace }e^{K\sum _{\left\langle i,j\right\rangle }S_{i}S_{j}}=\sum _{\lbrace S_{i}'=\pm 1\rbrace }\sum _{\lbrace \sigma _{i}=\pm 1\rbrace }e^{K\sum _{i}\left(S_{i-x}'+S_{i+x}'+S_{i-y}'+S_{i+y}'\right)\sigma _{i}}$ and performing the sum over $\sigma _{i}$ :
{\begin{aligned}Z_{N}=\sum _{\lbrace S_{i}'=\pm 1\rbrace }\prod _{i}\left[e^{K(S_{i-x}'+S_{i+x}'+S_{i-y}'+S_{i+y}')}+e^{-K(S_{i-x}'+S_{i+x}'+S_{i-y}'+S_{i+y}')}\right]=\\=\sum _{\lbrace S_{i}'=\pm 1\rbrace }\prod _{i}2\cosh \left[K(S_{i-x}'+S_{i+x}'+S_{i-y}'+S_{i+y}')\right]\end{aligned}} The argument of $\cosh$ can assume three possible values, which come from the fact that the $2^{4}=16$ possible spin configurations can be grouped into three categories: all spins aligned, three spins aligned and one not, and two spins aligned and two not. As before, we want to write $Z_{N}[K]=e^{Ng(K,K')}Z_{N'}[K']$ , and from what we have just seen we will have three conditions to satisfy so ${\mathcal {H}}'$ will contain at least three terms: in the coarse-grained Hamiltonian new terms are appearing. We can therefore try for example with the following guess for rewriting $Z_{N}$ :
{\begin{aligned}2\cosh[K(S_{i-x}'+S_{i+x}'+S_{i-y}'+S_{i+y}')]=\\{}\\=z(K)e^{K'(S_{i-x}'S_{i-y}'+S_{i-x}'S_{i-y}'+S_{i+x}'S_{i-y}'+S_{i+x}'S_{i+y}')}\cdot \\\cdot e^{L'(S_{i-x}'S_{i+x}'+S_{i-y}'S_{i+y}')+Q'S_{i-x}'S_{i+x}'S_{i-y}'S_{i+y}'}\end{aligned}} This way, besides nearest-neighbour interactions ($K$ ) we are introducing also next-nearest-neighbour ones ($K'$ and $L'$ ) and also four-spin cluster interactions ($Q'$ ). The situation can be represented as follows:

Note also that the final set of spins resides on a square whose side is ${\sqrt {2}}$ times the original one, so we have $\ell ={\sqrt {2}}$ . Inserting all the possible spin configurations in we get the following equations:

$2\cosh(4K)=z(K)e^{2K'+2L'+Q'}\quad \qquad 2\cosh(4K)=z(K)e^{-Q'}$ $2=z(K)e^{-2L'+Q'}\quad \qquad 2=z(K)e^{-2K'-2L'+Q'}$ Their solutions are given by:
$z(K)=2\cosh ^{1/8}(4K)\cosh ^{1/2}(2K)\quad \qquad K'={\frac {1}{4}}\ln \cosh(4K)$ $L'={\frac {1}{8}}\ln \cosh(4K)\quad \qquad Q'={\frac {1}{8}}\ln \cosh(4K)-{\frac {1}{2}}\ln \cosh(2K)$ If we now reiterate the procedure, more complicated interactions will appear and the problem becomes rapidly intractable. We therefore must do some approximations. We choose to neglect $Q'$ (also because it is the only coupling that can become negative and thus prevent the spins from aligning) and to omit the explicit dependence on $L'$ defining a new constant $K'$ :
$K_{\text{new}}'=K'+L'$ This way the recursion relation involves only $K$ :
$K'={\frac {3}{8}}\ln \cosh(4K)$ Let us therefore see to which conclusions does this lead. The fixed points are given by:
$K^{*}={\frac {3}{8}}\ln \cosh(4K^{*})$ and the non-trivial ($K^{*}\neq 0$ ) numerical solution of this equation is $K_{c}=0.50698\dots$ ; the exact value found by Onsager is $K_{\text{exact}}=0.44069\dots$ , so our approximation is good enough.

If now the initial value $K_{0}$ of $K$ is greater than $K_{c}$ , then the sequence ${\textstyle K{}^{(n)}={\frac {3}{8}}\ln \cosh(4K{}^{(n)})}$ grows indefinitely, while if $K_{0} it tends to zero. Thus, the fixed points $K^{*}=0$ and $K^{*}=\infty$ are stable, while $K^{*}=K_{c}$ is unstable; this can be visually represented as: RG flow of the recursion relation ${\textstyle K^{*}={\frac {3}{8}}\ln \cosh(4K^{*})}$ Let us now linearise the recursion relation near $K_{c}$ and compute a couple of critical exponents. On the base of what we have seen in The origins of scaling and critical behaviour, if we call $\delta K=K-K_{c}$ and $\delta K'=K'-K_{c}$ we have:

$\delta K'=\lambda _{t}\delta K\quad \qquad {\text{where}}\quad \lambda _{t}={\frac {dK'}{dK}}_{|K_{c}}$ Therefore, since also $\ell ={\sqrt {2}}$ , we get:
$y_{t}={\frac {\ln \lambda _{t}}{\ln \ell }}={\frac {1}{\ln {\sqrt {2}}}}\ln \left({\frac {dK'}{dK}}_{|K_{c}}\right)={\frac {\ln \left[{\frac {3}{2}}\tanh(4K_{c})\right]}{\ln 2/2}}=1.070\dots$ and so:
$\nu ={\frac {1}{y_{t}}}=0.9345\dots$ and from the hyperscaling relation:
$\alpha =2-{\frac {d}{y_{t}}}=0.131\dots$ Onsager's exact result, as we know, gives $\alpha =0$ (since the specific heat diverges logarithmically) and thus $\nu =1$ . We therefore see that our approximation is sufficiently accurate (even if still improvable).