4
$\begingroup$

Let's consider a bounded (maybe compact) set $\Lambda \subset \mathbb{R}^{d}$ with particles interacting on it. Suppose, for each $N \in \mathbb{N}$, $U_{N}: (\mathbb{R}^{d})^{N} \to \mathbb{R}\cup \{+\infty\}$ is function which represents the potential energy of a system consisting in $N$ interacting particles. Now, the partition function of such a system in the grand-canonical ensemble is given by: \begin{eqnarray} \Xi_{\Lambda}(\beta, z) := \sum_{N=0}^{\infty}\frac{z^{N}}{N!} \int_{\Lambda}d\mu(x_{1})\cdots \int_{\Lambda}d\mu(x_{N}) e^{-\beta U_{N}(x_{1},...,x_{N})} \tag{1}\label{1} \end{eqnarray}

My Question is: What is the $\sigma$-algebra in which the product measures $\mu(x_{1})\times\cdots\times \mu(x_{N})$ are defined? I mean, for each $N$, the $N$-th term of the sum (\ref{1}) involves a product of $N$ measures and I'm having trouble understanding what is happening here. It seems to me that this is just a weak-limit in the sense that: \begin{eqnarray} \int d\mu_{n} f \to \int d\mu f \tag{2}\label{2} \end{eqnarray} but this implies that each $\mu_{n} = \mu(x_{1})\times \cdots \times \mu(x_{N}) $ is defined, for every $N$, on a "bigger" $\sigma$-algebra. What is this $\sigma$-algebra?

$\endgroup$
  • 4
    $\begingroup$ woh you know it's serious when capital xi comes out :-) $\endgroup$ – nomen Feb 7 at 22:56
6
$\begingroup$

Looking at your formula (1), it appears that $\mu$ must be a measure defined on a $\sigma$-algebra $\mathscr F$ over the finite set $\Lambda$. The natural $\sigma$-algebra over the finite set $\Lambda$ is the largest $\sigma$-algebra over $\Lambda$, which is the (power) set $2^\Lambda$ of all subsets of $\Lambda$. By definition, the product measure $\mu^{\otimes N}$ is defined on the product $\sigma$-algebra $\mathscr F^{\otimes N}$. If $\mathscr F=2^\Lambda$, then $\mathscr F^{\otimes N}=(2^\Lambda)^{\otimes N}=2^{\Lambda^N}$, the set of all subsets of $\Lambda^N$.

Your formula (1) can then be rewritten simply as $$ \Xi_{\Lambda}(\beta, z) := \sum_{N=0}^{\infty}\frac{z^{N}}{N!} \int_{\Lambda^N}e^{-\beta U_{N}(x_{1},...,x_{N})}\,\mu^{\otimes N}\Big(\prod_{j=1}^N dx_j\Big) \\ =\sum_{N=0}^{\infty}\frac{z^{N}}{N!} \sum_{(x_{1},...,x_{N})\in\Lambda^N}e^{-\beta U_{N}(x_{1},...,x_{N})}\,\prod_{j=1}^N \mu(\{x_j\}). $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I just realized I mistyped my question. Actually, for continuous systems, $\Lambda$ must be a bounded set, not finite. I'll edit it. How does your answer change in this case? $\endgroup$ – IamWill Feb 7 at 14:57
  • $\begingroup$ The case $\Lambda$ finite is assumed when we're dealing with $\mathbb{Z}^{d}$ rather than $\mathbb{R}^{d}$. Your answer seems to address the $\mathbb{Z}^{d}$ case and it is still very useful to me because I get in troble understanding the problem in $\mathbb{Z}^{d}$ a well. But my original post should have considered $\Lambda$ bounded and I didn't notice my mistake. $\endgroup$ – IamWill Feb 7 at 14:59
  • $\begingroup$ @Willy.K: Iosif's answer applies also in $\mathbb{R}^d$. Looks to me you are trying to follow the lectures by Brydges which are rather advanced stuff without knowing measure theory and in particular the definition of product $\sigma$-algebras and product of measures, etc. You need to review that first. To do it quickly, you can for example read the first chapter of the book "Probability: Theory and Examples" by Rick Durrett. $\endgroup$ – Abdelmalek Abdesselam Feb 7 at 15:05
  • $\begingroup$ @AbdelmalekAbdesselam this time I'm not following Brydges notes. But I would expect $\mu(x_{1})\times \cdots \times \mu(x_{N})$ to be a measure on, I don't know, $\mathbb{R}^{\infty}$ and $\int_{\Lambda}$ to be just the integral restriced to $\Lambda$. Adapting the answer would lead to a $\sigma$-algebra consisting on $2^{\Lambda}$, that is, every subset of $\Lambda$ is measurable, right? $\endgroup$ – IamWill Feb 7 at 15:11
  • $\begingroup$ In some references, the $\sigma$-algebra is assumed to be $\cup_{N}\Gamma_{N}(\Lambda)$ where $\Gamma_{N}(\Lambda) := \{(x_{1},...,x_{N})\in (\mathbb{R}^{d})^{\Lambda}, x_{i}\in \Lambda\}$ but I don't know if this is the case. $\endgroup$ – IamWill Feb 7 at 15:17
4
$\begingroup$

The configuration space is the disjoint union of $\Lambda^N$ for each nonnegative integer $N$. You can take the Borel $\sigma$-algebra on each of these (or Lebesgue if you prefer, but you're unlikely to encounter non-Borel sets in real life).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for your answer! Just to clarify: I set $\Omega = \cup_{N}\Lambda^{N}$ to be my configuration space and I can take the Borel $\sigma$-algebra on each $\Lambda^{N} \subset \mathbb{R}^{dN}$, right? With what $\sigma$-algebra should I equip $\Omega$? $\endgroup$ – IamWill Feb 7 at 19:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.