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Alternate interpretation of number of degenerate fermions formula in phase space

Physics Asked on September 26, 2021

I’m writing personnal notes on statistical mechanics and I’m tempted to write something that may turn out to be false. So I need a confirmation/infirmation and opinions on the following (I suspect it’s false, but…).

The number of degenerate fermions inside a volume $V$, at temperature $T = 0$, is found by integration on their phase space because of the Pauli exclusion principle (only one fermion per state) :
begin{equation}tag{1}
N = int_0^{p_F} frac{2 cdot 4 pi p^2 , dp}{(2 pi hbar)^3} , V = frac{p_F^3 , V}{3 pi^2 , hbar^3}.
end{equation}
This allows to find the Fermi momentum $p_F$ as a function of the particles density $N/V$. In principle, this integration makes sense only for a large number of particles : $N gg 1$.

Now, I’m tempting to interpret (1) as still valid even for $N = 1$
(only one fermion in the box), since quantum mechanics allows
superpositions of states for a single particle. Doing an integral
over phase space to find just one particle in the box is weird and
shouldn’t be valid (from Pauli’s principle, this summation makes sense
only if $N gg 1)$. But in QM, a single particle could be in a
superposition of states of different momentum, up to the maximal value
allowed (i.e Fermi’s momentum).

Is this interpretation viable ? Or is it completely non-sensical ?

For what is worth, my intuition tells me that this unconventional interpretation may be true after all, since there are some “alternative” interpretations of some calculations in statistical mechanics. But what troubles me is the apparent absence of the Pauli exclusion principle in this interpretation.

In applying the Pauli exclusion principle, we say “only one particle per given state”, but in QM we may say “several states per particle” ! So this seems to suggest that (1) may still be valid in some way, even for $N = 1$.


EDIT : Here’s a tentative justification that the interpretation may be true in some way.

You have a single particle in a rectangular box of sizes $ell$, at temperature $T = 0$. From classical mechanics, its average momentum should be 0. But because of the Heisenberg uncertainty principle, the momentum in the $x$ direction has a minimal uncertainty given by
begin{equation}tag{2}
Delta p_x ge frac{hbar}{2 Delta x} approx frac{hbar}{ell},
end{equation}
since $Delta x approx frac{ell}{2}$ (on average). So the momentum may be anything in this range :
begin{equation}tag{3}
-, Delta p_{x , text{max}} le p_x le Delta p_{x , text{max}}.
end{equation}
Thus, classicaly, the particle occupies a small fuzzy region in its phase space. The volume of that region is given by (1), with $p_F$ replaced by $sim Delta p_x$. The Pauli principle may be satisfied if all other particles are placed alone inside adjacent boxes, so there’s only one particle per quantum state (i.e a fuzzy region in phase space).

The previous argument can be generalized for any number of particles in the same box of volume $V$ : $N > 1$, by splitting the integral (1) in as much parts that there are fermions. For $N = 3$ for example (or any other integer) :
begin{equation}tag{4}
N = int_0^{p_1} dots dp + int_{p_1}^{p_2} dots dp + int_{p_2}^{p_{text{max}}} dots dp,
end{equation}
where $0 < p_1 < p_2$ are completely arbitrary and $p_{text{max}}$ is the maximal value allowed (Fermi’s momentum). This integral splitting is in agreement with Pauli’s exclusion principle. Each particle draws a fuzzy region in its phase space (they can’t be represented by a point, like in classical mechanics). So I now strongly believe that (1) is still valid for any integer $N$, and not just large numbers $N gg 1$.

I need opinions on this.

One Answer

In principle, a single fermion may be in a superposition of many values of $p$. However, when you say that it is at T=0 this is no longer true, because by definition it must be in the lowest available state. That is why this whole summation over states procedure works at all. So I believe your claim is incorrect.

Edit: for completeness, I would also point out that the above formula is only valid for free noninteracting fermions. In the general case, one must find the density of states.

Answered by Rococo on September 26, 2021

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