Classical statistical mechanics predicts that each free electron in a metal should contribute (3/2)k_B to the heat capacity, but measured electronic heat capacities are far smaller — roughly 100 times smaller at room temperature. What is the quantum mechanical explanation?
AElectrons in metals are relativistic, so classical thermodynamics does not apply to them
BThe Pauli exclusion principle blocks most electrons from absorbing thermal energy because all nearby states are occupied; only electrons within ~kT of the Fermi energy have access to empty states and can be thermally excited
CElectron-phonon collisions immediately transfer any absorbed thermal energy to the lattice, so the electrons never actually warm up
DElectron-electron repulsion creates an energy gap just above the Fermi energy, preventing thermal excitation
This is the resolution of the classical heat capacity problem that baffled 19th-century physics. Classical equipartition assigns (3/2)k_B per particle regardless of quantum statistics. The quantum answer: an electron deep in the Fermi sea — say, 2 eV below E_F — cannot absorb a thermal fluctuation of ~0.025 eV (room temperature kT) because every state it might jump into within that energy range is already filled by other electrons (Pauli exclusion). Only electrons within roughly kT of the Fermi surface have accessible empty states nearby. The fraction of thermally active electrons is ~kT/E_F ≈ 0.5% at room temperature for a typical metal, making the electronic heat capacity ~100 times smaller than the classical prediction.
Question 2 Multiple Choice
In a metal at room temperature with E_F ≈ 5 eV and kT ≈ 0.025 eV, approximately what fraction of the conduction electrons are thermally active — capable of absorbing thermal energy?
ANearly all of them — thermal energy at room temperature is sufficient to excite electrons throughout the band
BAbout 50% — electrons above the Fermi level midpoint are thermally accessible
CAbout kT/E_F ≈ 0.5% — only electrons within ~kT of the Fermi surface have access to empty states
DNone — room temperature is far below the quantum threshold for electron excitation in metals
The fraction of thermally active electrons is of order kT/E_F. For typical metals, E_F is a few electron volts while kT at room temperature is ~0.025 eV, giving a ratio of ~0.5%. This tiny fraction is why the electronic heat capacity is so much smaller than the classical prediction: instead of all N electrons contributing (3/2)k_B each, only ~N(kT/E_F) electrons contribute energy of order kT each, giving C_V^{el} ∝ Nk_B(kT/E_F) ∝ T. This linear-T dependence and its small coefficient are direct consequences of the Pauli principle restricting thermal excitations to a thin shell near E_F.
Question 3 True / False
At finite temperature, the Fermi-Dirac distribution becomes a perfect step function with occupation exactly 1 below μ and exactly 0 above μ.
TTrue
FFalse
Answer: False
The perfect step function is only the T = 0 limit. At any finite temperature T > 0, the Fermi-Dirac distribution f(E) = 1/(exp((E−μ)/kT) + 1) smoothly transitions from ~1 to ~0 over an energy width of approximately 4kT around the chemical potential μ. States just below μ have occupation slightly less than 1; states just above μ have occupation slightly more than 0. The step function is an approximation that becomes exact only as T → 0. For metals at room temperature, where kT ≈ 0.025 eV ≪ E_F ≈ 5 eV, the step remains very sharp and the T = 0 approximation is excellent — but it is still not exact.
Question 4 True / False
The linear temperature dependence of electronic heat capacity (C_V ∝ T) is a direct consequence of the Pauli exclusion principle restricting thermal excitations to a thin shell of states near the Fermi energy.
TTrue
FFalse
Answer: True
The argument runs directly: thermal excitations are restricted to electrons within ~kT of E_F (Pauli blocks all deeper electrons). The number of thermally active electrons scales as N(kT/E_F). Each picks up energy of order kT. So the total thermal energy scales as U_el ∝ N(kT/E_F)(kT) = Nk²T²/E_F, and the heat capacity C_V = dU/dT ∝ Nk²T/E_F ∝ T. The linear T dependence follows from the two-step logic: Pauli exclusion limits which electrons participate, and how many participate grows linearly with T as the thermal window expands.
Question 5 Short Answer
Why does the chemical potential μ(T) shift downward from E_F as temperature increases, rather than remaining fixed at E_F?
Think about your answer, then reveal below.
Model answer: The downward shift of μ arises from the asymmetric density of states around E_F. In three dimensions, the density of states g(E) ∝ √E, meaning there are more available states just above E_F than just below it. When temperature smears the Fermi-Dirac step, electrons are promoted from just below E_F to just above it. Because g(E) is larger just above E_F, more electrons are added to the region above E_F than are removed from the region below. To maintain the total electron count fixed (charge neutrality), the chemical potential must shift downward — lowering μ slightly reduces the occupation of high-energy states and increases occupation of low-energy states until the total count is restored. The leading correction is μ(T) ≈ E_F[1 − (π²/12)(kT/E_F)²].
If the density of states were perfectly symmetric around E_F (equal numbers of states just above and just below), the symmetric smearing of the Fermi function would not change the total electron count and μ would remain at E_F. The downward shift is a consequence of the asymmetry: g(E) increases with E, so the asymmetric promotion of electrons requires a compensating downward shift in μ. This effect is tiny in metals at room temperature (the correction is of order (kT/E_F)² ~ 0.003) but becomes important near absolute zero and in systems with sharp features in g(E).