I am trying to understand the renormalization group approach to phase transitions. A central quantity is the correlation length $\xi$ given by $G(r) = \langle\phi(r)\phi(0)\rangle \sim e^{-r/\xi}$ for $r \to \infty$. A phase transitions occurs when $\xi$ diverges.

The renormalization group approach is based on the idea that only slow modes $\phi_k = \int dx \phi(x) e^{-ikx}$, where $k \ll \Lambda$, are responsible for the phase transition (i.e. only long range behaviour). Therefore one can integrate out some fast modes $\Lambda' < k < \Lambda$ without changing the physics. My question regards why the correlation length only depends on slow modes?

If $G(r) \sim e^{-r/\xi}$ then its Fourier transform $\tilde{G}(p)$ around $p \approx 0$ is something like $\tilde{G}(p) \sim \frac{1}{p^2+1/\xi^2} + ...$ If $\xi$ diverges then $\tilde{G}(0)$ also diverges. In this sense it is true that the long range behaviour of $G(r)$ is determined only by the 'slow' $p \approx 0$ behaviour of $\tilde{G}(p)$.

$\tilde{G}(p)$ can be written in terms of the modes $\phi_k$ as: $$\tilde{G}(p) = \int dr \langle\phi(r)\phi(0)\rangle e^{-ipr} = \int dr \int dk_1 \int dk_2 e^{-ipr+ik_1r}\langle\phi_{k_1}\phi_{k_2}\rangle = 2\pi \int dk_2 \langle\phi_{p}\phi_{k_2}\rangle$$

Here one can see that $\tilde{G}(p)$ depends on arbitrary fast modes $\phi_{k_2}$ even for $p \approx 0$. Therefore it should not be a valid step to integrate out fast modes because they still influence $\tilde{G}(p)$ and therefore also the correlation length $\xi$. Can anyone resolve this?

Edit: This is a response to the answer by bbrink and concerns the Ising model $H = -\sum_{ij} J_{ij}\sigma_i\sigma_j$. As bbrink pointed out in real space the RG is quite simple to understand. If we consider only observables consisting of even spins, e.g. $\langle \sigma_{2i} \sigma_{2j} \rangle = \frac{1}{Z} \sum_{all\;spins} \sigma_{2i} \sigma_{2j} e^{-\beta H}$ then we can first sum over all the odd spins and obtain an effective Hamiltonian $H'$ (and effective temperature $\beta'$) for the even spins: $e^{-\beta' H'} = \sum_{odd\;spins} e^{-\beta H}$. The expectation value for even spins is the same for the initial and primed Hamiltonian. Then the RG uses the fact that $H'$ is again a Hamiltonian on a lattice (and might even look similar to the Ising model). This self similarity is then exploited to obtain some information about the system.

Now we would like to do the same thing in momentum space. Momentum space means that instead of writing down the Hamiltonian using $\sigma_i$ we write it down using their Fourier transform $\tilde{\sigma}_k = \sum_n \sigma_n e^{-ikn}$ where $k$ is an integer multiples of $2\pi/L$. The Hamiltonian then reads $H = - \sum_{kl} \tilde{J}_{kl}\tilde{\sigma}_k^\dagger\tilde{\sigma}_l$. The reasons for this is because in many systems $\tilde{J}$ has a much nicer form than $J$.

Now we would like to do the same step as in real space, namely summing over every second spin. It is usually said that intuitively summing over every second spin (i.e. short ranges) is like summing over the upper half of the spectrum of $k$ (i.e. high frequencies).

But this does not work for the correlation length. In particular:

$$\langle \sigma_{2n} \sigma_{2m} \rangle = \sum_{k,l} \langle \tilde{\sigma}_k\tilde{\sigma}_l\rangle e^{2i(kn+lm)}$$

which depends on all values of $k$ and $l$. Therefore it is not possible to write down an effective Hamiltonian in momentum space for the correlation length.

But still what is usually done in the literature for field theoretic models is to do the renormalization step in momentum space (i.e. summing over some high frequency modes) and deriving the RG equations. Using these RG equations one then derives for example the behavior of the correlation length near the critical point. But this should not give the right result.


1 Answer 1


All degrees of freedom contribute to correlations. Coarse-graining does not change the correlation functions and other statistical moments of a systems---it changes the Hamiltonian/action, introducing new interaction terms such that the statistical moments are preserved.

Moreover, if the correlation length is infinite after renormalizing a model, then the correlation length was also infinite before renormalizing the model (i.e., the original system had to be tuned to the critical point in the first place, before any renormalization takes place.).

To demonstrate my claims, it's easier to consider the Ising model. Suppose our spins are on a square lattice (1 or 2d) with lattice spacing $a$. The correlation function (really just the second moment since I won't assume the means are zero) is given by $$\langle \sigma(\mathbf{r}_k) \sigma(\mathbf{r}_\ell) \rangle = \frac{1}{Z} \sum_{\vec{\sigma}} \sigma(\mathbf{r}_k) \sigma(\mathbf{r}_\ell) e^{-\mathcal H(\vec{\sigma})}$$ where $\sigma(\mathbf{r}_i) = \pm 1$ is the state of spin at site $i$ (which has some spatial position $\mathbf{r}_i$), $Z$ is the partition function, and $\mathcal H$ is the Hamiltonian, which I will write as $$\mathcal H = -\sum_{ij} J(|\mathbf{r}_i-\mathbf{r}_j|) \sigma(\mathbf{r}_i) \sigma(\mathbf{r}_j) - \sum_i h_i \sigma(\mathbf{r}_i),$$ with $J(|\mathbf{r}_i-\mathbf{r}_j|)$ the exchange coupling between spins at positions $\mathbf{r}_i$ and $\mathbf{r}_j$ and $h(\mathbf{r}_i)$ is the field bias. Temperature has been absorbed into these terms.

Now, suppose we were to coarse grain this model in real space by averaging out every other spin. The coarse-grained Hamiltonian is $$e^{-\mathcal H'} = \sum_{{\rm every~other~spin}} e^{-\mathcal H}$$ (note that I haven't rescaled the lattice yet).

Two things to notice:

  1. the partition function is not changed by this operation: $$Z' = \sum_{\rm remaining~spins} e^{-\mathcal H'} = \sum_{\rm remaining~spins} \sum_{{\rm every~other~spin}} e^{-\mathcal H} = \sum_{\rm all~spins} e^{-\mathcal H} = Z.$$

  2. This also holds true for the statistical moments, as claimed earlier. Suppose spins $k$ and $\ell$ are not integrated out. Then the second moment is $$\langle \sigma_k \sigma_\ell \rangle = \frac{1}{Z} \sum_{\rm remaining~spins} \sigma_k \sigma_\ell e^{-\mathcal H'} = \frac{1}{Z} \sum_{\rm remaining~spins} \sigma_k \sigma_\ell \sum_{{\rm every~other~spin}} e^{-\mathcal H} = \frac{1}{Z}\sum_{\rm all~spins}\sigma_k \sigma_\ell e^{-\mathcal H};$$ i.e., coarse-graining does not change correlation functions, as claimed. This holds true even in field theoretic models like the $\phi^4$ theory.

However, so far I have only done the coarse-graining step of the renormalization process. There is a second step, the rescaling step. In the case of the lattice model we would rescale the lattice, determining some new lattice spacing, e.g. in 1d $a' = 2a$, and we would rescale the positions $\mathbf{r}'_i$ as well ($\mathbf{r}'_i = \mathbf{r}_i/2$ in 1d). Therefore, if we have assigned the spins spatial positions and we found that $$\langle \sigma(\mathbf{r}_k)\sigma(\mathbf{r}_\ell)\rangle \sim \exp(-|\mathbf{r}_k-\mathbf{r}_\ell|/\xi),$$ for some $\xi < \infty$. Since this is the same quantity before and after coarse-graining we can just write down the correlation function in our new rescaled coordinates by setting $\mathbf{r}_i = 2\mathbf{r}'_i$ (in 1d, for other dimensions the $2$ will be different). So, in terms of our new coordinates the correlation function would be: $$\langle \sigma(\mathbf{r}'_k)\sigma(\mathbf{r}'_\ell)\rangle \sim \exp(-2|\mathbf{r}'_k-\mathbf{r}'_\ell|/\xi),$$ from which we could define a "new" correlation length $\xi' = \xi/2$. It looks like the correlation length got smaller after renormalization, and so superficially it seems like the correlations changed, but that's not really true. All that's happening here is that we changed the units that we were measuring distance with. And of course, if $\xi = \infty$, then $\xi' = \infty$.

This is similarly true in field theoretic models, except you also generally have the field renormalization factors which further make it look like the correlation functions change after coarse-graining. But again, what's really happening is that you are making a change of variables $\phi(\mathbf{x}) = \mathcal Z \phi'(\mathbf{x}')$ and measuring correlations between $\phi'$'s instead of $\phi$'s, i.e., $\langle \phi(\mathbf{x}) \phi(\mathbf{y}) \rangle \sim C \langle \phi'(\mathbf{x}') \phi(\mathbf{y}')\rangle$, for some rescaling factor $C$.

  • $\begingroup$ Thanks for your answer. What you do is exactly how the RG group should work. In real space this is absolutely clear to me: If we consider only observables consisting of even spins (e.g. $<\sigma_{2i} \sigma_{2j}>$), then we can sum over all odd spins first and obtain an effective Hamiltonian for the even spins. My problem is that the same reasoning does not work in momentum space. Summing over every second spin in position space is not the same as summing over the upper half of the spectrum in momentum space. I will add a calculation to my question, maybe you can have a look. $\endgroup$
    – toaster
    Commented Nov 6, 2020 at 14:58

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