# Formulating the variational principle in grand canonical ensemble

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 $$\Omega_{N}$$?

• 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). Aug 15 at 16:37
• @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. Aug 16 at 8:49

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.
• 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... Aug 16 at 9:28
• @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? Aug 16 at 10:43
• @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. Aug 16 at 18:08