Various questions on renormalization in lattice systems

Question

Forgive the long, multi questioned-question. The setting of this question is inspired by this answer. Consider some theory on a lattice, for example the 2D $0$-field Ising model $$H=-K\sum_{\langle i,j\rangle} \sigma_i \sigma_j$$ the lattice is $\mathbb{Z}^2$ in this case. We can define the space of all theories $\mathcal T$, i.e. the space of all probability measures of real valued fields defined on $\mathbb{Z}^2$, and define a renormalization map as $R:\mathcal{T}\rightarrow \mathcal{T}$.

• First question: how can we make the idea that renormalization "scales" the system clear? The image of the renormalization map is still on $\mathbb{Z}^2$. How to formalize the idea that the new model has a "larger lattice spacing"? I don't see any notion or measure of the lattice spacing in $H$.

If we accept that there is some notion that the renormalization increases the scale of the system, then the interesting stuff comes when we look at the correlation length $\xi$: if the lattice spacing increases by a factor $b>1$, and the length is measured in units of $b$, then it must be that $\xi$ is mapped to $\xi'=\xi/b$. If we start from a model that has $\xi=\infty$, the image will still have $\xi=\infty$.

Suppose this transformation has a fixed point $V_*$ such that $R(V_*)=V_*$, then by the above argument $V_*$ should have infinite correlation length. Also, since the map reduces the correlation length, the stable manifold of the fixed point, defined as

$$ W^s=\{V\in W^s: \lim_{n\rightarrow \infty}R^n(V)=V_*\}$$ must be composed exclusively of points with $\xi=\infty$. Minor question: is the reciprocal true? Is any point with $\xi=\infty$ on the stable manifold?

• Second question: I've always only seen definitions of the correlation length that are loosely based on an ansatz for the shape of the correlation function of the Ising model $$ \Gamma(r)\sim e^{-r/\xi}$$ in principle the renormalization could map our simple Ising model to a ridiculously complicated distribution, for which the correlation function doesn't have such a simple form. More generally, even if we chose a nice renormalization map, it would still have a fixed point with a stable manifold, and for the argument above to make any sense, all the points on the stable manifold should have a well defined correlation length. What is it?

This all means that really, fixed points and critical points are two separate beasts, and that the critical point of a given model does not correspond to the fixed point of a given renormalization transformation. A model with Hamiltonian $H$ corresponds to a curve in $\mathcal T$ $$ K\rightarrow \mathrm{Ising}(K)$$

and the critical point is the intersection of this curve with the stable manifold of a renormalization procedure, which leads me to the following third question

• Third question: why do we care about fixed points of renormalization? Unless by some miracle the intersection point happens to be the fixed point, finding the fixed point looks useless to me, as finding the stable manifold and where it intersects the curve looks like a daunting problem in general. Is it correct to say that renormalization is not helpful to find the critical temperature of a model? If I understand correctly, it can still be used to derive critical exponents.

Answer 1

Let $\mathbb{Z}^d$ denote the unit square lattice in $d$ dimensions. Let $\Omega=\mathbb{R}^{\mathbb{Z}^d}$ be the Cartesian product of one copy of $\mathbb{R}$ for each lattice site $\mathbf{x}\in\mathbb{Z}^d$. An element $\sigma$ of $\Omega$ is thus a spin configuration $(\sigma_{\mathbf{x}})_{\mathbf{x}\in\mathbb{Z}^d}$. We equip $\Omega$ with the product topology (of a countable product of copies of $\mathbb{R}$) and also with the Borel $\sigma$-algebra $\mathcal{F}$ resulting from this topology. We can now define $\mathcal{T}$ as the set of all probability measures $\mu$ on the measurable space $(\Omega,\mathcal{F})$. Pick some fixed integer $L>1$. For any site $\mathbf{x}\in\mathbb{Z}^d$, define the block $$ B_{\mathbf{x}}=\{\mathbf{y}\in\mathbb{Z}^d\ |\ \mathbf{y}\in L\mathbf{x}+[0,L)^d\} $$ of linear size $L$ near the point $L\mathbf{x}$ (I chose "bottom right corner" but one could also have it at the center). Note that the point $L\mathbf{x}$ belongs to the coarser lattice $(L\mathbb{Z})^d$. We now pick some constant $[\phi]$ and define a map $\Gamma:\Omega\rightarrow\Omega$ as follows. We send the spin configuration $\sigma$ to the new configuration $\Gamma(\sigma)=\tau$ where, for all $\mathbf{x}\in\mathbb{Z}^d$, $$ \tau_{x}=L^{[\phi]-d}\sum_{\mathbf{y}\in B_{\mathbf{x}}} \sigma_{\mathbf{y}}\ . $$ The map $\Gamma$ is continuous and therefore $(\mathcal{F},\mathcal{F})$-measurable. If $\mu$ is a probability measure on $\Omega$, then one can define direct image or push-forward measure $\mu'=\Gamma_{\ast}\mu$. It is the probability distribution of the spin configuration $\Gamma(\sigma)$ if $\sigma$ is sampled according to the probability distribution $\mu$. We thus have a map $R:\mathcal{T}\rightarrow\mathcal{T}, \mu\mapsto\mu'$. This map $R$ is the renormalization group map, in the block spin approach. There are other ways of doing that (decimation, splitting of Gaussian measures as a sum of high and low momentum fields, etc.)

Now suppose that the original measure is such that the two-point function satisfies $$ \langle\sigma_{\mathbf{x}_1}\sigma_{\mathbf{x}_2}\rangle_{\mu} \sim e^{- \frac{|\mathbf{x}_1-\mathbf{x}_2|}{\xi}} $$ at large distance. Note that, to avoid confusion, I put as a subscript the probability measure with respect to which the expectation $\langle\cdot\rangle$ is taken. Also note that the $\sim$ is rather vague. It could mean the LHS is roughly equal to the RHS times a constant or even a power law decay in the distance $|\mathbf{x}_1-\mathbf{x}_2|$.

Let us do the computation for the new measure $\mu'=R(\mu)$. Pretty much by definition of the direct image measure, $$ \langle\sigma_{\mathbf{x}_1}\sigma_{\mathbf{x}_2}\rangle_{\mu'}= \langle(\Gamma(\sigma))_{\mathbf{x}_1}(\Gamma(\sigma))_{\mathbf{x}_2}\rangle_{\mu} $$ $$ =L^{2[\phi]-2d}\sum_{\mathbf{y}_1\in B_{\mathbf{x}_1},\mathbf{y}_2\in B_{\mathbf{x}_2}} \langle\sigma_{\mathbf{y}_1}\sigma_{\mathbf{y}_2}\rangle_{\mu} $$ $$ \simeq L^{2[\phi]} \langle\sigma_{L\mathbf{x}_1}\sigma_{\mathbf{Lx}_2}\rangle_{\mu} $$ from the approximation that the two-point function of $\mu$ does not change much if the points roam around the $L$ blocks near $L\mathbf{x}_1$ and $L\mathbf{x}_2$. So the result is $$ \sim e^{-\frac{|L\mathbf{x}_1-L\mathbf{x}_2|}{\xi}}=e^{-\frac{|\mathbf{x}_1-\mathbf{x}_2|}{\xi'}} $$ with $\xi'=\frac{\xi}{L}$. So you see that the correlation length has shrunk by a factor of $L$.

The above regime is for noncritical measures. In that case the best choice of $[\phi]$ is $\frac{d}{2}$, in order to converge to a well defined fixed point. Indeed, take $\mu_{\rm triv}$ to be the measure where all the $\sigma_{\mathbf{x}}$ are iid $N(0,1)$ random variables. One then has $R(\mu_{\rm triv})=\mu_{\rm triv}$ just from undergraduate probability.

For 2D Ising and for the critical measure $\mu$, one expects that the choice $[\phi]=\frac{1}{8}$ in the definition of $R$ entails convergence to a fixed point which is nontrivial. You can redo a similar two-point function calculation as above in this case, them you will see that, because $\langle\sigma_{\mathbf{x}_1}\sigma_{\mathbf{x}_2}\rangle_{\mu}$ decays like $1/|\mathbf{x}_1-\mathbf{x}_2|^{1/4}$, the fixed point condition $\mu=\mu'$ is only consistent with $1/8$ as a pick for the scaling dimension $[\phi]$.