Density depletion for Fermions In my recent advanced statistical physics class, I read about the density depletion of Fermions, which are "defending" a given volume around them against other Fermions, while the exchange hole shrinks like $\lambda n^3$ for $T \gg T_F$.
What is the intuitive interpretation or derivation of the density depletion defined by as
$$n  \int d^3r \: ( g(\vec{r}) - 1) ) \, ,$$
where $g(\vec{r})$ is the two particle correlation function?
I hope I can get a flavor of this definition, since I can't understand what exactly is calculated.
My own interpretation (which might be totally wrong) is the following:
we calculate the density times the "probability to insert another Fermion at any distance (hence the integral)" which then gives us the "average" density.
However, this appears even to myself pretty hand-wavy.
 A: This answer might make things a bit clearer, but still misses my wish for a derivation:
Assume that there is a Fermion at position $\vec{r} = 0$. Then $g(\vec{r})$ is the probability to find another Fermion at a distance $\vec{r}$ from this Fermion. The integral over space of $g(\vec{r}) - 1$ then gives us a relative quantity of how much Fermions have removed from the vicinity of the Fermion at $\vec{r}=0$. Multiplying it with n, the density, gives the right units.
Additionally, think of it in the way that the density depletion is $0$ if the probability for another Fermion in the vicinity is 1. (1-1)=0, hence the density stays the same. But for a smaller probability of finding another Fermion in the vicinity, the density would decrease. The interpretation of a negative density depletion is that another fermion would be (still in probabilities!), that a fermion at $\vec{r}$ could expel that percentage of another fermion. For $T=0$ the density depletion is $-1$ meaning the Fermion is defending a given volume by expelling one other Fermion.
