  Let $\sigma$ be a non-atomic, infinite Radon measure on $\mathbb R^d$, for
example, $d\sigma(x)=z\,dx$ where $z>0$. We consider a system of freely
independent particles $x_1,\dots,x_N$ in a bounded set $\Lambda\subset\mathbb
R^d$, where each particle $x_i$ has distribution
$\frac1{\sigma(\Lambda)}\,\sigma$ on $\Lambda$ and the number of particles,
$N$, is random and has Poisson distribution with parameter $\sigma(\Lambda)$.
If the particles were classically independent rather than freely independent,
this particle system would be the restriction to $\Lambda$ of the Poisson point
process on $\mathbb R^d$ with intensity measure $\sigma$. In the case of free
independence, this particle system is not the restriction of the free Poisson
process on $\mathbb R^d$ with intensity measure $\sigma$. Nevertheless, we
prove that this is true in an approximative sense: if bounded sets
$\Lambda^{(n)}$ ($n\in\mathbb N$) are such that
$\Lambda^{(1)}\subset\Lambda^{(2)}\subset\Lambda^{(3)}\subset\dotsm$ and
$\bigcup_{n=1}^\infty \Lambda^{(n)}=\mathbb R^d$, then the corresponding
particle system in $\Lambda^{(n)}$ converges (as $n\to\infty$) to the free
Poisson process on $\mathbb R^d$ with intensity measure $\sigma$. We also prove
the following $N/V$-limit: Let $N^{(n)}$ be a determinstic sequence of natural
numbers such that $\lim_{n\to\infty}N^{(n)}/\sigma(\Lambda^{(n)})=1$. Then the
system of $N^{(n)}$ freely independent particles in $\Lambda^{(n)}$ converges
(as $n\to\infty$) to the free Poisson process. We finally extend these results
to the case of a free L\'evy white noise (in particular, a free L\'evy process)
without free Gaussian part.
