# At what densities the many-body approaches are valid?

Suppose we have a n-particle interacting system with a potential $V=a/(r1-r2)$, it is a pseudo-coulomb potential: you can choose it fermion or boson.

Then, at what densities the many-body approaches are valid? in general, is there any relation between the density of the particles and the validity of the many-body approaches, like green functions, DFT, Canonical Transformation, etc.?

• 5 years is long for no upvotes answers, please try copying and pasting into materials.stackexchange.com Commented Jul 23, 2020 at 18:39

In a system with non-interacting particles you may know that the right quantity to look at is $\rho \Lambda^3 \sim \left(\Lambda/l \right)^3$ where $\Lambda$ is the (thermal) de Broglie wavelength and $l = \rho^{-1/3}$ the typical inter-particle distance in the system.

Basically if $\Lambda/l \ll 1$, then your system behaves classically i.e. you do not need to worry about Fermi or Bose statistics and conversely if $\Lambda \gg l$ then you are in the quantum regime and you definitely need to worry about the type of quantum statistics you want to capture.

There are multiple ways to understand/justify how this ratio of lengths comes about:

• You can imagine a representation that is suitable for semi-classical treatments (Wigner representation of the quantum states or Feynman-Kac path integral representation of the partition function). This will give rise to a picture of particles as wave packets of typical size $\Lambda$. You can then argue that one may worry about the fermonic or bosonic nature of the particles when there is sufficient overlap between these wave packets i.e. when $\Lambda / \rho^{-1/3} \ll 1$.

• You can say that quantum statistics arise when it is likely for two particles to be in exactly the same quantum state. If we imagine a system with $N-1$ particles each in a different quantum state, then the probability for the Nth particle to be in exactly the same state as one its peers is roughly

$$p_{same} \sim \frac{N-1}{q_1} \sim \frac{N}{q_1}$$ where $q_1$ is the quantum partition function of a single particle in a box that more or less counts the total number of states available. As it turns out $q_1 = \frac{\Lambda^3}{V}$

so that you have eventually $p_{same} \sim \rho \Lambda^3$. So, if this probability is very small, then you can forget about quantum statistics altogether.

Now, if you have some interaction between the particles, it is slightly more tricky as it adds an additional length scale in the problem. The criterion will still be based on a comparison between the typical length scale between particles and the de Broglie wavelength. However, the density itself may not be imposed but emergent from the interaction and various constraint such as electroneutrality.

• thanks. but i think the question is about the validity of many body approaches. in fact we are not looking to find the bose or fermi statistics about the system. we know the statistics. the problem here is that at what densities the many body approaches are valid, since as far as i know the DFT gives the ground state properties and the ground state has a unique density related to it. now we have 2 questions: 1- are all basic many body approaches corresponds to the ground state 2- what is that density for our system and how to obtain it? Commented Apr 15, 2015 at 17:37
• well I am sorry. I don't even know what you mean by "many-body approach". You seem to refer to DFT which is based on a variational principle and there is indeed a unique (electronic) density field associated to it to the variational problem it wants to solve. Commented Apr 15, 2015 at 17:41
• yeah of course that what you say now and the problem is that how to obtain it in it's simplest form. we just take the first approximation and solve it. ???!!1 Commented Apr 15, 2015 at 17:44