After a very nice discussion in my previous question, I decided to move on and try to formulate the variational principle for the grand canonical ensemble. I tried following the references cited in the answer to my post, but I don't seem to be getting anywhere, probably because of different approaches and notations.

My setting is the following. The phase space now is $\Omega_{N} := (\Lambda \times \mathbb{R}^{d})^{N}$ for some $N \ge 1$ and $\Lambda \subset \mathbb{R}^{d}$ nonempty and compact. Equip $\Omega_{N}$ with its Borel $\sigma$-algebra $\mathbb{B}_{N}$. Let $\mu_{\Lambda,\beta,z}^{(N)}$ be the measure on $(\Omega_{N},\mathbb{B}_{N})$ with has the following density with respect to the Lebesgue measure: $$d\mu_{\Lambda,\beta,z}^{(N)} := \frac{1}{N!}z^{N}e^{-\beta H_{\Lambda,\beta}^{(N)}(x)}dx$$ where $z \in \mathbb{R}$ is a parameter e $H_{\Lambda,\beta}^{(N)}$ an $N$-body Hamiltonian. Here, I am assuming $\mu_{\Lambda,\beta,z}^{(0)}(\{\emptyset\}) = 1$.

Set $\Omega := \bigcup_{N=0}^{\infty}(\Omega_{N}\times \{N\})$, where I am assuming $\Omega_{0} = \{\emptyset\}$ for simplicity. For each $E \in \mathbb{B}_{N}$, let $\pi_{N}(E) := \{x \in \Omega_{N}: (x,N) \in E\}$. We equip $\Omega$ with the $\sigma$-algebra $\mathcal{F}$, defined by: $$\mathcal{F}:= \{E \subset \Omega: \mbox{$\pi_{N}(E) \in \mathbb{B}_{N}$ for every $N \in \mathbb{N}$}\}$$ We introduce a probability measure $\mu_{\Lambda,\beta,z}$ on $(\Omega,\mathcal{F})$ by setting: $$\mu_{\Lambda,\beta,z} := \frac{1}{Z_{\Lambda,\beta,z}}\sum_{N=0}^{\infty}\mu_{\Lambda,\beta,z}^{(N)}$$ where the partition function $Z_{\Lambda,\beta,z}$ only normalizes the measure.

This is the tricky part for me. The measure $\mu_{\Lambda,\beta,z}$ is supposed to be the Gibbs measure that minimizes a given functional, which coincides with the free energy at its minimum. What is this functional? In principle, it is supposed to be defined over all measures $\mu$ on $(\Omega,\mathcal{F})$. However, we only defined our objects (e.g. Hamiltonian) on $\Omega_{N}$. In other words, what would something like $\mathbb{E}_{\mu}[H_{\Lambda,\beta}]$ mean? Should I, instead, define the variational principle in terms of all measures $\mu$ on $\Omega$ which have the form $\mu = \sum_{N=0}^{\infty}\mu_{N}$, where $\mu_{N}$ is a measure on $\Omega_{N}$?

  • $\begingroup$ I'll provide an answer if I find time. In the meantime, you might have a look at Section 1.3 of this paper, which should be readable. The ideas are exactly the same as for the canonical ensemble (but the reference measure is a Poisson point process now rather than Lebesgue). $\endgroup$ Aug 15 at 16:37
  • $\begingroup$ @YvanVelenik thank you for this new reference. I am not very familiar with Poisson point processes, but I tried to learn something from it. I sketched an answer of my own question below. I would highly appreciate your comments and/or corrections to it. $\endgroup$
    – IamWill
    Aug 16 at 8:49

1 Answer 1


Attempted Proof: Suppose we have a family $\{f_{N}\}_{N\in \mathbb{N}}$ of functions $f_{N}: \Omega_{N} \to \mathbb{R}$. It induces a function $f: \Omega \to \mathbb{R}$ by setting $$f(x,N) := f_{N}(x). \tag{1}\label{1}$$ Hence, we can define the Hamiltonian $H_{\Lambda,\beta}: \Omega \to \mathbb{R}$ by $H_{\Lambda,\beta}(x,N) := H_{\Lambda,\beta}^{(N)}(x)$.

Let $\delta$ be a counting measure on $\mathbb{N}$, that is, $\delta(\{N\}) = 1$. Let $\mu$ be a measure on $(\Omega, \mathcal{F})$ with density $$d\mu = \rho(x,N) d\delta dx. \tag{2}\label{2}$$ If $f$ is defined as in (\ref{1}), then expectation with respect to $\mu$ leads to: $$\mathbb{E}_{\mu}[f] := \int_{\Omega} f(x,N)\rho(x,N)d\delta dx = \sum_{N=0}^{\infty}\int_{\Omega_{N}} f_{N}(x)\rho(x,N)dx \tag{3}\label{3}$$ We denote the set of all measures $\mu$ of the form (\ref{2}) by $\mathcal{M}(\Omega)$.

The entropy functional is now defined by: $$\mathcal{M}(\Omega) \ni \mu \mapsto S(\mu):= \mathbb{E}_{\mu}[\log\rho] \equiv \sum_{N=0}^{\infty}\int_{\Omega_{N}}\rho(x,N)\log(\rho(x,N))dx \tag{4}\label{4}$$ Thus, the measure $\mu_{\Lambda,\beta,z}$ defined in my post is the solution of: $$\inf_{\mu \in \mathcal{M}(\Omega)}(\mathbb{E}_{\mu}[H_{\Lambda,\beta}] - TS(\mu) -\mu \mathbb{E}_{\mu}[\mathcal{N}]) \tag{5}\label{5}$$ where $\mathcal{N}(x,N) := N$ is the number function.

  • $\begingroup$ At first glance, this seems fine. One of the $\mu$ in (5) should be a $kT\log(z)$, right? Of course, $\mu$ is the standard notation for the chemical potential, but you're already using this symbol for the probability measure... $\endgroup$ Aug 16 at 9:28
  • $\begingroup$ @YvanVelenik yes, $\mu$ is a bad notation for the chemical potential in my case. Thanks for your comments. Following the canonical case, it is probably also a good idea to define $S(\mu) = -\infty$ when $\mu$ has not the form of (\ref{2}), right? $\endgroup$
    – IamWill
    Aug 16 at 10:43
  • $\begingroup$ Yes. The measure has to be absolutely continuous wrt to the underlying Poisson point process. $\endgroup$ Aug 16 at 10:54
  • $\begingroup$ @YvanVelenik I did the calculations and my reasoning seems to lead to the right result. The only point is that there is a factor $1/N!$ missing. Should I simply define the counting measure $\delta(\{N\}) = 1/N!$ instead? This solves the problem, but it seems a bit artificial to me. $\endgroup$
    – IamWill
    Aug 16 at 18:08
  • $\begingroup$ It’s not artificial : the points are indistinguishable, we number them for convenience, but the numbering is meaningless. The "1/N!" is implicit in the definition of the Poisson point process, which is why you don’t see it in the paper I linked. $\endgroup$ Aug 16 at 19:37

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