I'm reading Kupiainen's notes on the renormalization group and also caught my attention. Actually, this is something that often causes my some confusion. On page 43, in the section about Ginzburg-Landau model, Kupiainen introduces a spin configuration as a function $\phi: \mathbb{Z}^{d} \to \mathbb{R}$. Thus, we can view it as a sequence $\phi = (\phi_{x})_{x\in \mathbb{Z}^{d}} \in \mathbb{R}^{\mathbb{Z}^{d}}$. If $\Lambda \subset \mathbb{Z}^{d}$ is finite, we can consider the restriction $\phi_{\Lambda} = (\phi_{x})_{x \in \Lambda} \in \mathbb{R}^{\Lambda}$, where $\phi_{x}:= \phi(x)$. Because $\Lambda$ is finite, we can consider $\phi_{\Lambda}$ as an usual vector on $\mathbb{R}^{n}$ where $n$ is the cardinality of $\Lambda$. For instance, we can define the Gibbs measure on $\mathbb{R}^{\Lambda}$ as given by: \begin{eqnarray} d\mu_{\Lambda}(\phi) = \frac{1}{Z_{\Lambda}}e^{-H_{\Lambda}(\phi)}\prod_{j=1}^{n}d\phi_{x} \tag{1}\label{1} \end{eqnarray} where $Z_{\Lambda}$ is a normalizing factor, $H_{\Lambda}: \mathbb{R}^{\mathbb{Z}^{d}}\to \mathbb{R}$ is an Hamiltonian with some given boundary conditions and $\prod_{j=1}^{n}d\phi_{x}$ is just the product Borel measure on $\mathbb{R}^{\Lambda}$. On the other hand, on his new set of notes, on page 31 (also about Ginzburg-Landau model) Kupiainen states that "in classical statistical mechanics one considers $\phi(x)$ as a random variable with probability distribution given by (\ref{1})".Now, if I understood it correctly, this means that each $\phi_{x}$ is now a random variable on some underlying probability space. But then, the picture changes a lot, since now instead of simple vectors on $\mathbb{R}^{\Lambda}$, $\phi_{\Lambda}$ is a vector of functions. What does even mean to write $\prod_{j=1}^{n}d\phi_{x}$? Also, if this were the Ising model, we expect $\phi_{\Lambda}$ to be just a vector with entries $\pm 1$ as in the first picture. So, am I missing something here? Why sometimes $\phi_{\Lambda}$ are viewed as vectors and sometimes as vectors of functions? Also, what does $\prod_{j=1}^{n}d\phi_{x}$ mean if each $\phi_{x}$ is a random variable?
$\begingroup$
$\endgroup$
7
-
2$\begingroup$ Not sure, but could the confusion simply be whether one writes $d\mu(\phi )$ or $\mu (\phi ) d\phi $, which are just different ways of saying the same thing? Why are you saying $\phi_{\Lambda } $ is a vector of functions? What functions? $\endgroup$– Michael EngelhardtCommented Apr 15, 2020 at 16:11
-
$\begingroup$ @MichaelEngelhardt as far as I understand, saying that $\phi(x) = \phi_{x}$ is a random variable means that $\phi_{\Lambda} = (\phi_{x})_{x\in \Lambda}$ is a family of random variables indexed by $x \in \Lambda$, and $\phi_{x}: \Omega \to \mathbb{R}$ is a measurable function on some underlying probability space. So each entry of $\phi_{\Lambda}$ is a measurable function. $\endgroup$– JustWannaKnowCommented Apr 15, 2020 at 17:00
-
2$\begingroup$ It seems that Kupiainen is guilty of using terminology that (perhaps implicitly) can lead to confusing, but not that uncommon, notation like $\langle \psi \rangle = \int \psi e^{-\psi^2}\, d\psi$. On the left $\psi$ must be interpreted as a random variable, while on the right $\psi$ is a coordinate on the probability space $\Omega = \mathbb{R}$, with probability measure $d\mu(\psi) = e^{-\psi^2}\, d\psi$. It would be better to write $\langle \Psi \rangle = \int \psi e^{-\psi^2}\, d\psi$ and formally distinguish $\Psi$ and $\psi$, but you can just do that in your head. $\endgroup$– Igor KhavkineCommented Apr 15, 2020 at 17:17
-
2$\begingroup$ @IamWill Just going by your description, I would say Yes. It is not that uncommon a notational/terminological shortcut. $\endgroup$– Igor KhavkineCommented Apr 15, 2020 at 19:13
-
1$\begingroup$ It's the same issue than say in $\mathbb{R}^n$. One may talk of $x_i\in\mathbb{R}$ as the "coordinate" of a specific vector $(x_1,\ldots,x_n)\in\mathbb{R}^n$. But one can also talk about the "coordinate function" $x_i$, which is a map from $\mathbb{R}^n\rightarrow\mathbb{R}$. The random variable is $\phi_x$ seen as a coordinate function. $\endgroup$– Abdelmalek AbdesselamCommented Apr 17, 2020 at 14:20
|
Show 2 more comments