The ideal Fermi gas

Non-interacting fermions are mostly an idealization. All known fermions (except, to a good approximation, neutrinos) interact with each other rather strongly.

All the same, this model is useful to describe the thermodinamical properties that a uniform assembly of spin-s non-interacting fermions would show at equilibrium. The reason is that Fermi-liquid excitations of a fluid of interacting fermions map one to one on the states of a Fermi gas (see Lifshitz and Pitaevskii - Statistical Physics Part 2 - Pergamon 1980, chap. 1)

We summarize here the elementary Fermi-gas theory, having in mind mainly the conduction electrons of a simple metal such as Na or K. This same model, with appropriate mass and density, finds applications in approximate descriptions of liquid ³He and even of nuclear matter (heavy nuclei, collapsed stars).

The two basic ingredients of the problem are:

1. A single-particle density of states g(E). In the case of electrons in an effective potential representing a solid, g(E) depends on the details of the bandstructure. In the free-electron approximation, the periodic potential energy acting on an electron in the solid is approximated by a constant.
   
   This constant average potential energy can be taken as the 0 of energy, as in the left vertical scale in the figure at the right. [Alternatively one can set the energy zero to that of an electron at rest far away outside the metal. This is natural in photoemission contexts. In this other gauge - right scale in figure - the average potential energy in the metal is a negative quantity, representing "the bottom of the band", i.e. the full depth of the potential-energy well confining electrons within the solid.] In a constant potential, the "bandstructure" is that of a Schroedinger particle moving in free space, a free-electron parabola E(k). The resulting total density of states of a free spin-s fermion of mass m free to move in d dimensions is

   g(E) = Ω(d) g_s/2 [2π(2 m)^½/ℏ]^d V E^d/2-1 for E>0
   g(E) = 0 for E<0,

   where V is the d-dimensional volume of the sample, g_s=(2s+1) is the spin degeneracy and Ω(d) is the (d-1)-dimensional surface of a unitary "sphere" in d dimensions: Ω(1)=2, Ω(2)=2 π, Ω(3)=4 π. Specifically, for electrons g_s=2; in ordinary d=3 metals g(E) = 2^1/2 m_e^3/2 V / (π²ℏ³) E^½. The drastic free-electron approximation yields fair results for IA (alkali) metals and IB metals (Cu, Ag, Au), but makes little sense for most other materials.
   The reasons for using the free-electron approximation are:
   • to describe the metallic features of a generic material in terms of one parameter, namely its electron density n=N/V
   • that most of the calculations can be carried out analytically.
   However, for better quantitative agreement for a specific material, one should really use the actual g(E) given by the bandstructure for that solid. This way, even the electrons of semiconductors and band insulators are described naturally in this independent-particles model (resulting in quite different thermodynamic properties, due to the chemical potential falling in an energy gap!).
2. The Fermi probability distribution f(E) = f_T(E) = 1/(exp[(E-µ)/(k_BT)]+1) is the average number of ideal fermions that occupy a state at energy E when the gas is at equilibrium at temperature = T. µ is called the chemical potential: it is a function of the number N of fermions and of T.

The two basic ingredients are multiplied to yield the overall number of fermions per unit energy: g(E)·f(E). Based on this distribution function, one can compute all the statistical properties of the Fermi gas.

The first problem to be solved is the determination of the chemical potential µ=µ(n,T), as a function of the particle density n=N/V and temperature.

Problem 1

Problem 2

The d=3 Fermi gas

n = N/V = g_s/2 [(2 m)/ℏ²]^3/2 (k_BT)^3/2 (-1/(4π^3/2)) Plog_3/2(-Exp[µ/k_BT])

1. The T=0 limit (s=½), where the integration is trivial: µ(T=0) = ℏ²(2m)^-1 (3 π² n)^2/3.
   NOTATION: The chemical potential at zero temperature is usually called Fermi energy E_F=µ(T=0). For typical electron densities in metals n~10²⁸ - 10³⁰m^-3, E_F is in the 1 to 10 eV range.
   We give here a few useful expressions for the recurring density of states at the Fermi energy:

   g(E_F) = 3^1/3π^-4/3m_e V ℏ^-2 n^1/3 = 3^1/3π^-4/3m_e N ℏ^-2 n^-2/3 = 3N/(2E_F)

2. The leading corrections in k_BT/µ ("cold" Fermi gas): the best method to derive them is to use the so-called Sommerfeld expansion. This trick is fairly general: one wants a systematic way to find corrections to the an integral of some function H(E)×f(E), as an expansion in powers of T. Consider the primitive function K(E) of H(E), and expand K(E) in a Taylor series around the energy µ(T). [The details of this calculation are detailed, for example, in the Ashcroft-Mermin textbook "Solid State Physics" (Chap 2 and App. C).] By considering the special case H(E)=g(E) one obtains the leading thermal correction to the chemical potential:

   µ = E_F [1- π²/12 (k_BT / E_F)² + O(k_B T / E_F)⁴]

χ= dM/dH = µ_B² g(E_F)/V = (3/2) µ_B² n/E_F = (3^1/3/π^4/3) µ_B² (m_e/ℏ²) n^1/3 = 3^1/3/(4π^4/3) (q_e²/m_e) n^1/3 = 2.20824×10^-9 C²/Kg n^1/3,

From this expression, one gets the susceptibility per electron as VN^-1·χ = VN^-1·µ_B² g(E_F)/V = 3µ_B²/(2E_F). Compare the Fermi-gas Pauli susceptibility to the the Curie susceptibility of free spins (s=½, g=2), amounting to µ_B²/(k_BT). At room temperature, k_BT=0.026 eV << E_F: as a result free spins produce much larger susceptibility (per spin) than those of the electron spins of the metallic band.
[Here we assume that H is measured in the same units as B, like for example in Feynman's books, so that χ is measured in C²/Kg/m. Alternatively, certain authors consider a H' = ε₀c²·H: with this wierd convention M and H' have the same dimensions, and χ'= dM/dH' turns out dimensionless and for the cold electron gas equals 2.77495×10^-15 m n^1/3. The χ' per particle equals of course χ'/n = 2.77495×10^-15 m n^-2/3].
The details of this calculation can be retrieved, for example, in the Ashcroft-Mermin textbook "Solid State Physics" (Chap 31). One can use these basic results for the 3D ideal Fermi gas to solve a broad class of

Problems

1. Compute the total energy U(T=0) and the leading T-correction to the total energy U(T)
   RESULT: U(T=0)=3/5 N E_F
   U(T) = U(T=0) + π²/6 (k_BT)² g(E_F)
2. From the result of the previous problem, deduce the low-temperature expression for the specific heat at constant volume C_V, both the "molar" one and the "per unit volume" one.
   RESULT: C_V^mol = (N_A/N) π²/3 k_B²T g(E_F) = R (π²/2) k_BT/E_F = R 1.1662×10¹⁵ (n^-2/3/m²)(T/K) = 9.69633×10¹⁵ Kg/(s K)²/mol n^-2/3 T
   C_V^volume = (π²/2) k_B n k_BT/E_F = 1.61011×10^-8 Kg/(s K)² n^1/3 T
3. Compute the T=0 pressure P=-dU/dV of the uniform Fermi gas.
   RESULT: P = 2 U / (3 V) = 2 n E_F /5
4. Compute the average momentum |p| of a generic electron in the Fermi gas at T=0, and the finite-T correction.
5. Assume that the conduction electrons in potassium could be regarded as ideal free fermions. K atoms are disposed on a bcc lattice (body centered cubic, i.e. atoms at all vertices and at the center of the cells of a cubic lattice) of cubic side a=5.32 Å, and each atom contributes one electron to the conduction band. For these electrons, compute: µ(T=0), the total kinetic energy density U/V, the magnetic susceptibility.
6. [Advanced!] Compute the leading T-correction to the magnetic susceptibility χ.

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