advanced32 min studyLesson 328 in Physics

Expand about a filled Fermi sea

Derive zero-temperature filling and controlled low-temperature corrections at fixed particle density.

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What you’ll understand

  • Count occupied momentum states with internal degeneracy included.
  • State smoothness and scale conditions for a Sommerfeld expansion.
  • Distinguish fixed-density chemical-potential shifts from fixed-μ calculations.

Fermions cannot all fall into the lowest one-particle state. At zero temperature they fill available states up to a Fermi energy fixed by density and internal degeneracy. At low temperature only a narrow layer around that surface changes occupation substantially. The Sommerfeld expansion turns this picture into a systematic integral approximation. Its most important bookkeeping rule is to specify what is held fixed: a gas with conserved number generally adjusts chemical potential as temperature changes.

Fill a sphere in momentum space

For a uniform nonrelativistic ideal gas in three dimensions, each allowed momentum and internal state can hold at most one identical fermion. At zero temperature every state below chemical potential is filled and every state above it is empty. Counting a sphere of occupied wavevectors relates Fermi wavevector to number density. Its energy then follows from quadratic dispersion. The spin or internal degeneracy must be included once in the count; it does not multiply the energy of each occupied state.

Integrating energy over the filled sphere gives mean energy three fifths of Fermi energy per particle. Differentiating the ground-state energy with respect to volume at fixed number gives a nonzero pressure even at zero temperature. This degeneracy pressure is not caused by a thermal distribution of distinguishable particles and does not vanish merely because temperature does. The ideal formula assumes negligible interactions and nonrelativistic momenta, so applying it to strongly interacting matter or relativistic stars requires further physics.

kF=(6π2ngs)1/3,EF=ℏ2kF22m,TF=EFkB,U(0)N=35EF,P(0)=25nEFk_F=\left(\frac{6\pi^2n}{g_s}\right)^{1/3},\qquad E_F=\frac{\hbar^2k_F^2}{2m},\qquad T_F=\frac{E_F}{k_B},\qquad \frac{U(0)}N=\frac35E_F,\qquad P(0)=\frac25nE_F
n is number density in m⁻³, gs internal degeneracy, kF in m⁻¹, EF in joules, and TF in kelvin. These are zero-temperature thermodynamic-limit formulas for a uniform nonrelativistic ideal Fermi gas.↔ Scroll wide equations horizontally.

Only the surface layer changes appreciably

The Fermi occupation smooths the step over an energy width of order thermal energy. Far below chemical potential, holes are exponentially rare; far above it, particles are exponentially rare. For an integral of a smooth function times the occupation, thermal changes from the two sides mostly cancel at first order in temperature. The leading surviving correction is proportional to temperature squared and the derivative of the smooth function at chemical potential.

This is an asymptotic expansion requiring chemical potential well above the lower spectral edge and the integrand smooth across the thermal window. Gaps, sharp density-of-states singularities, or temperatures comparable with Fermi temperature invalidate the simplest formula. The error notation refers to an appropriate smooth-integrand expansion, not a universal numerical bound for arbitrary spectra. One should inspect the energy dependence near the Fermi surface before treating the first correction as a controlled prediction.

∫0∞ϕ(E)dEe(E−μ)/(kBT)+1=∫0μϕ(E) dE+π26(kBT)2ϕ′(μ)+O((kBT)4)\int_0^\infty\phi(E)\frac{dE}{e^{(E-\mu)/(k_BT)}+1}=\int_0^\mu\phi(E)\,dE+\frac{\pi^2}{6}(k_BT)^2\phi^{\prime}(\mu)+O((k_BT)^4)
φ is a sufficiently smooth temperature-independent weight, μ≫kBT lies away from the lower endpoint, and higher derivatives set the omitted scale. Applying φ=g(E) or Eg(E) gives particle number or energy, respectively.↔ Scroll wide equations horizontally.

Fix number before reporting the chemical potential

At fixed density, insert the square-root state density into the number integral and solve for the chemical potential shift that keeps number unchanged. Chemical potential falls slightly below its zero-temperature Fermi energy. Substituting that shift into the energy integral gives a quadratic increase in total energy and a heat capacity linear in temperature. Holding chemical potential at Fermi energy instead changes number and yields different intermediate coefficients; that is another ensemble path, not the fixed-density result.

The linear heat capacity reflects the small fraction of particles within the thermal surface layer, each changing energy by roughly thermal energy. The probability lab can illustrate exclusion-limited occupation fluctuations for equal-energy fermionic states at a prescribed occupation probability. It cannot display a Fermi surface or carry out the Sommerfeld integral. Its probability must be supplied from the Fermi function, and its finite equal-energy count must not be mistaken for the entire dispersive gas.

μ(T)EF=1−π212(TTF)2+⋯ ,U(T)N=35EF[1+5π212(TTF)2+⋯ ],CVNkB≃π22TTF\frac{\mu(T)}{E_F}=1-\frac{\pi^2}{12}\left(\frac{T}{T_F}\right)^2+\cdots,\qquad \frac{U(T)}N=\frac35E_F\left[1+\frac{5\pi^2}{12}\left(\frac{T}{T_F}\right)^2+\cdots\right],\quad \frac{C_V}{Nk_B}\simeq\frac{\pi^2}{2}\frac{T}{T_F}
These are low-temperature fixed-N,V expansions for the three-dimensional nonrelativistic ideal gas, with T/TF≪1. μ and EF are in joules and CV in J/K; gs is already encoded in EF.↔ Scroll wide equations horizontally.

Evaluate the first low-temperature corrections

An ideal Fermi gas has TF=1000 K and temperature 100 K at fixed density. Using π²≈9.87, estimate μ/EF and CV/(NkB). State why the two corrections have different temperature powers.

  1. The small parameter is T/TF=0.10. Its square is 0.01, so the chemical-potential fractional decrease is (9.87/12)×0.01=0.008225.

  2. Thus μ/EF≈0.991775. The leading heat-capacity ratio is (9.87/2)×0.10=0.4935.

  3. Energy rises quadratically with temperature in the expansion, so differentiating gives heat capacity linear in temperature. The chemical-potential shift itself remains quadratic because it is determined by the fixed-number condition.

The result

μ is about 0.9918 EF and CV about 0.4935 NkB. These are leading low-temperature estimates, not exact values at all temperatures.

Explore the model

Interactive lab

Interactive lab · probability

Compare fluctuations of fermionic states

For each fresh run Reset and set BOTH probability p=0.5, trials n=20; then Reset and set p=0.1, n=20. Predict count variances 5 and 1.8. These can represent independent equal-energy states at two chosen Fermi occupations. The model does not vary temperature, chemical potential, or a momentum distribution.

Compare fluctuations of fermionic statesThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forOccupation fluctuations are greatest near half filling and suppressed near empty or full states.

Check your understanding

Why must μ(T) be adjusted when deriving low-temperature corrections for a conserved-density ideal Fermi gas?

Try it yourself

At zero temperature, increase the density of a uniform nonrelativistic ideal Fermi gas by eight while keeping mass and internal degeneracy fixed. Find the changes in kF, EF, mean energy per particle, and pressure.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Count the powers of n in kF, then use EF proportional to kF² and P proportional to nEF.

Compare with the solution

kF doubles and EF increases fourfold. Mean energy per particle, 3EF/5, also quadruples. Pressure rises by eight times four, a factor of 32. The new momenta must still remain nonrelativistic and interactions negligible for these ideal-gas scalings to apply.

Take it with you

  • Pauli filling sets a density-dependent Fermi energy and nonzero zero-temperature pressure.
  • The Sommerfeld expansion requires a smooth spectrum near μ and T≪TF.
  • Fixed number determines μ(T); the leading three-dimensional heat capacity is linear in T.

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