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.


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.

  • 1
    $\begingroup$ Even at $T = 0$, the Heisenberg principle is valid. The lowest energy is defined up this uncertainty : $E = E_{\text{min}} \pm \Delta E$. This is also true for momentum. So even at $T = 0$, the particle occupies a fuzzy region of its classical phase space. That fuzzy region is a single quantum state, while it's a superposition of several classical states. $\endgroup$
    – Cham
    Jul 7 '17 at 18:15
  • $\begingroup$ Also, take note that the temperature is defined for a mixed state only, which isn't the same as a linear superposition (a pure state, by definition). So $T = 0$ even for a pure state defined by a superposition of several momentum states : \begin{equation}| \psi \rangle = \sum_p c(p) \, | p \rangle, \quad \Rightarrow \quad \rho = | \psi \rangle \langle \psi | \quad \Rightarrow \quad S = -\, \mathrm{Tr}(\rho \, \log{\rho}) = 0, \quad \Rightarrow \quad T = 0. \end{equation} $\endgroup$
    – Cham
    Jul 7 '17 at 18:25
  • 1
    $\begingroup$ @Cham In the context we are talking about, equilibrium thermodynamics, this isn't really correct. For a single fermion in thermal equilibrium with a bath that approaches T=0, that fermion will be in its unique ground state (assuming no ground state degeneracy), not just any pure state. For a nonzero bath it would indeed be in a thermal mixed state. $\endgroup$
    – Rococo
    Jul 7 '17 at 18:46
  • $\begingroup$ Even in thermal equilibrium, there's an Heisenberg uncertainty on the energy of the ground state. So even at $T = 0$, there's an uncertainty 0n the momentum. The particle still occupies a fuzzy region in the classical phase space. Don't you agree with that ? $\endgroup$
    – Cham
    Jul 7 '17 at 18:54
  • $\begingroup$ @Cham I don't understand why you talk about the uncertainty principle in this context. A finite $\Delta E$ is only necessary if we want to study processes within a finite time interval. In equilibrium thermodynamics $\Delta t \rightarrow \infty$ so that $\Delta E = 0$: The energies are well defined. $\endgroup$ Aug 6 '21 at 15:41

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