Fermions and bosons on an evenly spaced ladder

The model

Take single-particle levels \(0, \epsilon, 2\epsilon, \dots\), nondegenerate and evenly spaced, as for a one-dimensional harmonic oscillator, and an isolated system of \(N\) identical particles of a single spin orientation. Let \(q\) count the energy units above the ground state. By the fundamental assumption, all microstates with the same \(q\) are equally likely.

Fermions (with \(q < N\)). In the ground state the lowest \(N\) levels are filled. Any excited state is reached by lifting the top electron by \(\lambda_1\) levels, the next one by \(\lambda_2 \le \lambda_1\), and so on; the shifts must not increase going down, or two fermions would land on the same level. (Below, levels are labeled by \(\ell\), with \(\ell = 0\) the highest level filled at \(q = 0\).) So a microstate is a list \(\lambda_1 \ge \lambda_2 \ge \cdots\) with \(\sum \lambda_i = q\): a partition of \(q\).

Bosons (with \(N\) large). Each boson can sit at any level. A microstate is fixed by the excitation energies of the bosons that left the ground level, again a list of positive integers summing to \(q\), in any order: again a partition of \(q\).

So both systems have the same number of microstates, the partition number:

\[ \Omega(q) = p(q):\quad 1,\ 1,\ 2,\ 3,\ 5,\ 7,\ 11,\ 15,\ 22,\ 30,\ 42,\ 56,\ 77,\ \dots \]

The occupancy of a level is the number of microstates (or bosons) found there, divided by \(\Omega(q)\). For \(q = 6\), the fermion levels just above the Fermi gap are occupied with probabilities \(5/11, 4/11, 3/11, 2/11, 1/11, 1/11, 0\), mirrored by \(6/11, 7/11, \dots, 1\) just below. The boson levels \(j = 1, 2, 3, \dots\) hold on average \(19/11, 8/11, 4/11, 2/11, 1/11, 1/11\) bosons.

In the thermodynamic limit these occupancies should follow

\[ \bar n_{\rm FD}(\varepsilon) = \frac{1}{e^{(\varepsilon-\mu)/(k_BT)}+1}, \qquad \bar n_{\rm BE}(\varepsilon) = \frac{1}{e^{(\varepsilon-\mu)/(k_BT)}-1}. \]

Even for small \(q\), we can ask which \(\mu\) and \(T\) make the curve pass closest to the exact points. The page does an unweighted least-squares fit (minimizing the sum of squared residuals with the Nelder–Mead method). Move the sliders to try your own guess against the best fit. Energies are in units of \(\epsilon\) and temperatures in units of \(\epsilon/k_B\).

Fitting with Python

Any tool that does nonlinear least squares will do. With SciPy, for \(q = 6\):

import numpy as np
from scipy.optimize import curve_fit

# Fermions: level energies measured from the top level filled at q = 0
e = np.arange(-6, 8)
P = np.array([11, 10, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 1, 0]) / 11
fd = lambda e, mu, T: 1 / (np.exp((e - mu) / T) + 1)
(mu, T), _ = curve_fit(fd, e, P, p0=[0.5, 2])
print(mu, T)          # 0.500  2.245

# Bosons: levels j = 1, ..., 7 (the ground level is left out)
j = np.arange(1, 8)
n = np.array([19, 8, 4, 2, 1, 1, 0]) / 11
be = lambda e, mu, T: 1 / (np.exp((e - mu) / T) - 1)
(mu, T), _ = curve_fit(be, j, n, p0=[-0.1, 2], bounds=([-10, 0.1], [0, 50]))
print(mu, T)          # -0.040  2.284

The fitted values depend somewhat on which levels are included and on how the residuals are weighted. Here the fermion fit uses all levels from \(-q\) to \(q+1\), and the boson fit uses \(j = 1\) to \(q+1\), which reproduces the numbers shown on this page.

Since fermions and bosons have the same \(\Omega(q)\), they also share the entropy \(S/k_B = \ln \Omega(q)\). The temperature follows from the slope,

\[ \frac{1}{T} = \frac{\partial S}{\partial U} \quad\Longrightarrow\quad k_BT \approx \epsilon\,\frac{\Delta q}{\Delta (S/k_B)}. \]

For large \(q\), the Hardy–Ramanujan formula gives \(p(q) \approx e^{\pi\sqrt{2q/3}}/(4\sqrt3\,q)\), so \(S/k_B \approx \pi\sqrt{2q/3}\) and \(k_BT \approx \sqrt{6q}\,\epsilon/\pi\), the result quoted by Arnaud et al.

Arnaud et al. compare three situations for the fermion ladder with the same (average) energy. The isolated system has exactly \(q\) units. In the canonical ensemble the system exchanges energy with a reservoir, and the occupancy of the level at \(\kappa = \ell - \tfrac12 > 0\) above the Fermi gap becomes

\[ \langle N_\kappa\rangle_c = -\sum_{i\ge1} (-1)^i\, x^{\,i(\kappa + i/2)}, \qquad x = e^{-\epsilon/(k_BT)}, \qquad \langle q\rangle = \sum_{n\ge1}\frac{n\,x^n}{1-x^n}. \]

In the grand canonical ensemble it also exchanges particles, and the occupancy is exactly Fermi–Dirac, \(\langle N_\kappa\rangle_{\rm FD} = 1/(x^{-\kappa}+1)\), with \(\langle q\rangle_{\rm FD} = \sum_{n\ \rm odd} n/(x^{-n/2}+1)\). The three agree in the thermodynamic limit, but for small systems they differ, most visibly in the tails.

Remarks

Why the fermion and boson counts agree. The same partition describes both systems, and the correspondence goes deeper. Exchanging particles and holes and reversing the energy scale maps one fermion microstate to another (its conjugate), which corresponds to transposing the partition's Young diagram. Schönhammer and Meden used this kind of correspondence to discuss fermion–boson "transmutation" in one dimension.

How good is Fermi–Dirac for a small isolated system? Near the Fermi level the fit is already reasonable at \(q = 6\), but the exact occupancy drops to zero at \(\kappa = q + \tfrac12\), where Fermi–Dirac still predicts a small positive value. Arnaud et al. show that even at \(q = 6000\) the isolated-system tail differs from Fermi–Dirac by more than 10% at occupancies around \(10^{-4}\).

Heat capacity. In the thermodynamic limit \(U - U_0 \approx \pi^2 (k_BT)^2/6\epsilon\), so \(C = dU/dT \approx \pi^2 k_B^2 T/3\epsilon\): the electronic heat capacity is linear in \(T\), as for electrons in a metal.