# Static Widom-Rowlinson model

In Elena Pulvirenti's slides she introduced a $$\textbf{static Widom-Rowlinson model of one species}$$. Consider $$\Lambda\subset R^2$$ with periodic boundary conditions, $$\Lambda$$ set of particle configurations with $$\Gamma=\{\gamma\subset \Lambda: N(\gamma)\in \mathbb{N}\}, \,$$ where $$N(\gamma)$$ is the cardinality of $$\gamma$$. The halo of a configuration is $$h(\gamma)=\cup_{x\in \gamma}B_2(x)$$ with radius $$2$$. Let $$H(\gamma)=|h(\gamma)|-N(\gamma)|B_2(0)|$$ be the Hamiltonian. Define the grand-canonical Gibbs measure, $$(1) \mu(d\gamma)=\frac{z^{N(\gamma)}}{\Xi}e^{-\beta H(\gamma)}\mathbb{Q}(d\gamma)$$ where $$\mathbb{Q}$$ is Poisson point process with intensity 1 and $$\Xi$$ is the partition function.

Her result is that the 2-species Widom-Rowlinson model is equivalent to 1-species. $$\textbf{The 2-species WR model}$$ is two types of particles(blue and red) with configurations $$\gamma^B, \gamma^R$$. The grand-canonical Gibbs measure: $$(2) \hat{\mu}(d\gamma^R, d\gamma^B)=\frac{1}{\hat{\Xi}}1_{\{\text{red-blue hard-core}\}}z_R^{N(\gamma^R)}z_B^{N(\gamma^B)}\mathbb{Q}(d\gamma^R)\mathbb{Q}(d\gamma^B)$$ where $$1_{\{\text{red-blue hard-core}\}}$$ means it is $$1$$ if $$d(\gamma^R, \gamma^B)\geq 1$$, otherwise is 0, and $$z_R=e^{\beta\lambda_R}$$ and $$z_B=e^{\beta\lambda_B}$$.

$$\textbf{My question is why 1-species and 2-species are equivalence?}$$ I am confused about that fix the centers of the red discs and integrate over the centers of the blue disc, then: $$\frac{1}{\hat{\Xi}}\int_{\Gamma} 1_{\{\text{red-blue hard-core}\}}z_R^{N(\gamma^R)}z_B^{N(\gamma^B)}\mathbb{Q}(d\gamma^B)=C \frac{z^{N(\gamma^R)}}{\Xi}e^{-\beta H(\gamma^R)}$$ where $$(z_B, z_R)\to (\beta, ze^{\beta V_0})$$ and $$V_0:=|B_2(0)|$$.

So let's think about it this way: if $$A$$ is an event in the two-type model depending only on red then \begin{align*} \mathbb{P}(A) &= \frac{1}{\tilde{\Xi}} \sum_{j,k\geq 0} \frac{z_R^k}{k!} \frac{z_B^j}{j!} \int_{S^k} \int_{S^j} 1_{A} \cdot 1_{RBHC} \,dy\, dx \\ &=\frac{1}{\tilde{\Xi}} \sum_{k\geq0}\frac{z_R^k}{k!} \int_{S^k} 1_A \left(\sum_{j \geq 0} \frac{z_B^j}{j!}\int_{S^j} 1_{RBHC} \,dy \right)\,dx \\ &= \frac{1}{\tilde{\Xi}} \sum_{k\geq0}\frac{z_R^k}{k!} \int_{S^k} 1_A \left(\sum_{j \geq 0} \frac{z_B^j}{j!}\int_{(S \setminus h(\gamma_R))^j} \,dy \right)\,dx \\ &= \frac{1}{\tilde{\Xi}} \sum_{k\geq0}\frac{z_R^k}{k!} \int_{S^k} 1_A \left(\sum_{j \geq 0} \frac{z_B^j}{j!}(|S| - |h(\gamma_R)|)^j \right)\,dx \\ &= \frac{1}{\tilde{\Xi}} \sum_{k\geq0}\frac{z_R^k}{k!} \int_{S^k} 1_A \exp(z_B |S| - z_B |h(\gamma_R)|)\,dx \\ &= \frac{C}{\tilde{\Xi}} \sum_{k\geq0}\frac{z_R^k}{k!} \int_{S^k} 1_A e^{- z_B |h(\gamma_R)|)}\,dx \\ &= \frac{C}{\tilde{\Xi}} \sum_{k\geq0}\frac{(z_R e^{-z_BV_0})^k}{k!} \int_{S^k} 1_A e^{- z_B (|h(\gamma_R)| - k V_0))}\,dx \,. \end{align*}
• So actually neither of those matters in the above computation, they're both just the normalizing constant that makes it a probability measure. My $S$ is the underlying space on which this model is defined (so some finite volume subset of R^d). – Marcus M Jun 28 at 20:10
• If you'd like, you can use the identity $1/\Xi = P(\gamma_R = \emptyset)$, and then the above shows that $\Xi = \tilde{\Xi}/C$. This is equivalent to what I described in my previous comment. – Marcus M Jun 28 at 20:47