Indistinguishable-particle positional measure

Let $\mathrm{\Lambda }\subset {\mathbb{R}}^d$ be a bounded Lebesgue-measurable region in the dimensionless coordinates of Stable regularly tempered classical positional preparation. Use the completed product-Lebesgue sigma-algebra on ${\mathrm{\Lambda }}^N$. For $N\ge 1$, let $D_{\mathrm{\Lambda },N}=\{(x_1,\dots ,x_N)\in {\mathrm{\Lambda }}^N:x_i\mathrm{\ne }{x}_j\text{ whenever }i\mathrm{\ne }j\},{\mathcal{C}}_{\mathrm{\Lambda },N}=D_{\mathrm{\Lambda },N}\mathrm{/}{S}_N,$ where the permutation group $S_N$ acts by relabelling coordinates. Give ${\mathcal{C}}_{\mathrm{\Lambda },N}$ the quotient sigma-algebra: $A\subseteq {\mathcal{C}}_{\mathrm{\Lambda },N}$ is measurable exactly when its preimage under the quotient map $q_N:D_{\mathrm{\Lambda },N}\to {\mathcal{C}}_{\mathrm{\Lambda },N}$ is measurable in the trace of the completed product-Lebesgue sigma-algebra. Identify each orbit with its unordered simple configuration. For $N=0$, set ${\mathcal{C}}_{\mathrm{\Lambda },0}=\{\mathrm{\varnothing }\}$. If ${\lambda }_{\mathrm{\Lambda }}^{\otimes N}$ is product Lebesgue measure restricted to $D_{\mathrm{\Lambda },N}$, define the sector reference measure by ${\mu }_{\mathrm{\Lambda },N}=\frac{1}{N!}(q_N{)}_{\mathrm{\#}}{\lambda }_{\mathrm{\Lambda }}^{\otimes N}(N\ge 1),{\mu }_{\mathrm{\Lambda },0}(\{\mathrm{\varnothing }\})=1.$ Equivalently, for every nonnegative measurable function $f$ on ${\mathcal{C}}_{\mathrm{\Lambda },N}$, $\int f(\omega ){\mu }_{\mathrm{\Lambda },N}(d\omega )=\frac{1}{N!}{\int }_{D_{\mathrm{\Lambda },N}}f(q_N(x_1,\dots ,x_N))d^{dN}x\mathrm{.}$ Collision diagonals have product Lebesgue measure zero, so ${\mu }_{\mathrm{\Lambda },N}({\mathcal{C}}_{\mathrm{\Lambda },N})=\mathrm{\mid }\mathrm{\Lambda }{\mathrm{\mid }}^N\mathrm{/}N!$; in particular, symmetric integrands may use the equivalent integral over all of ${\mathrm{\Lambda }}^N$. Thus ${\mu }_{\mathrm{\Lambda },N}$ is a finite measure, not generally a probability. Two labelled tuples represent the same state exactly when they are in one $S_N$ orbit. The factor $1\mathrm{/}N!$ is an adopted indistinguishable-particle positional convention, not a theorem about classical trajectories or every classical description.

Let ${\mathcal{C}}_{\mathrm{\Lambda }}={⨆}_{N\ge 0}{\mathcal{C}}_{\mathrm{\Lambda },N}$ with the disjoint-union sigma-algebra, and regard each ${\mu }_{\mathrm{\Lambda },N}$ as a measure on this union supported on its $N$-point sector. Define the unit-intensity finite-region Poisson reference by $Q_{\mathrm{\Lambda }}=e^{-\mathrm{\mid }\mathrm{\Lambda }\mathrm{\mid }}{\sum }_{N=0}^{\mathrm{\infty }}{\mu }_{\mathrm{\Lambda },N}\mathrm{.}$ Its total mass is one because ${\mu }_{\mathrm{\Lambda },N}({\mathcal{C}}_{\mathrm{\Lambda },N})=\mathrm{\mid }\mathrm{\Lambda }{\mathrm{\mid }}^N\mathrm{/}N!$ and $e^{-\mathrm{\mid }\mathrm{\Lambda }\mathrm{\mid }}{\sum }_{N\ge 0}\mathrm{\mid }\mathrm{\Lambda }{\mathrm{\mid }}^N\mathrm{/}N!=1$. This is a positional reference measure in the chosen dimensionless coordinates; it supplies no empirical claim about particle trajectories.