What is the density profile in the context of statistical mechanics?

While doing exercises in Statistical Mechanics I came across the following definition of the Density Profile of a system of $$N$$ non interacting particles

$$\rho(\mathbf{r}) = N\langle \delta(\mathbf{r}-\mathbf{r}_{1})\rangle$$

where $$\delta$$ is the Dirac's delta function and the averaging is carried out over the (canonical) ensemble.

Since the particles are non-interacting, the Hamiltonian of the system is the sum of the single-particle Hamiltonians, each of which reads

$$h_{i} = \frac{\mathbf{p}_{i}^2}{2m} + V(\mathbf{r}_{i})$$

so that the above expression corresponds to

$$\rho(\mathbf{r}) = N \frac{\int{d\mathbf{r}_{1}\delta(\mathbf{r}-\mathbf{r}_{1})e^{-\beta V(\mathbf{r}_{1})}}}{\int{d\mathbf{r}_{1}e^{-\beta V(\mathbf{r}_{1})}}} = N\frac{e^{-\beta V(\mathbf{r})}}{\int{d\mathbf{r}_{1}e^{-\beta V(\mathbf{r}_{1})}}}$$

• What is the physical meaning of the Density Profile as defined like this?

2 Answers

Your first expression would apply to any system of $$N$$ particles, interacting or not, as long as each particle were equivalent to any other. It could be written $$\rho(\mathbf{r}) = \left\langle \sum_{i=1}^N \delta(\mathbf{r}-\mathbf{r}_i) \right\rangle$$ and then (given the equivalence of particles), the sum over $$i$$ can be replaced by a factor $$N$$, and any given particle $$i$$ can be chosen as particle $$1$$.

The interpretation is the natural one of a local number density of particles at position $$\mathbf{r}$$. In other words, if we integrate $$\rho(\mathbf{r})$$ over $$\mathbf{r}$$, allowing the integration to cover any desired volume $$V$$, the result will be the average number of particles within the volume $$V$$. For a given configuration of atoms, the Dirac delta function catches all the ones within the volume, counting $$1$$ for each one, and the angle brackets then give you an average over configurations.

Your final expression (which does only apply in the noninteracting case, as written) has the interpretation of the Boltzmann expression: the probability density of finding a particle at position $$\mathbf{r}$$ is proportional to the Boltzmann factor. The denominator is the normalizing factor. The factor of $$N$$ converts the probability density into a number density.

For uniform systems, the local density $$\rho(\mathbf{r})$$ is equal to the usual density $$\rho=N/V$$ and the $$\mathbf{r}$$-dependence is redundant. But there are many cases where the system is not homogeneous: the density of molecules in the atmosphere, taking account of gravity, is a simple example where the assumptions of your question would apply. More generally, an entire branch of the theory of liquids (density functional theory) is based on the theorem that the free energy is a functional of the single-particle density $$\rho(\mathbf{r})$$. In general, the functional will be unknown, but it can be approximated or modelled in various ways.

It's a generalization of standard (number) density.

First let's consider density as a constant $$\rho=N/V$$. The integral of the density in a region of space of volume $$v$$ gives you the number of particles in that region:

$$\int_v \rho \ d\mathbf r = N \frac v V$$

If $$v=V$$, you get the total number of particles $$N$$.

This is all fine if the system can be considered as homogeneous.

However, let's imagine the following case happens. We will take a 1D example in order to simplify, so I will replace the volume $$V$$ with a length $$L$$.

Let's consider $$N$$ particles on a segment of length $$L$$; the coordinates of the particles are $$x_1, \dots, x_N$$. Now let's imagine that we know that all of the particles are in the segment $$[0,L/2]$$. We then ask someone else (who isn't aware of this) to estimate how many particles are in the segment $$[L/2,L]$$.

Knowing nothing about the system, this person will assume that density is constant and estimate the number of particles in $$[L/2,L]$$ as

$$n_{[L/2,L]}= \int_{L/2}^{L} \rho \ dx$$

where $$\rho= N/L$$. He will then get the result

$$n_{[L/2,L]}= \frac N 2$$

But we know that this is wrong, because there are exactly $$0$$ particles in the segment $$[L/2,L]$$ since all of them are in $$[0,L/2]$$.

On the other hand, if we introduced a microscopic density defined as

$$\rho(x) = \sum_{i=1}^N \delta (x-x_i)$$

we see immediately that, since $$x_k \in [0,l/2]$$ for every $$K$$,

$$\int_{L/2}^{L} \rho(x) \ dx = N$$

which is the correct result.

We also see that if we integrate on the whole interval

$$\int_0^L \rho(x) \ d x = N$$

which is what we expect from density: if we integrate it over all the domain of the system, we get the total number of particles. This of course also generalizes to three dimensions.