Questions: Band Structure and Density of States

Question 1

Van Hove singularities in the density of states occur at energies where ∇_k E(k) = 0. Why do these lead to peaks or kinks in g(E)?

• A. At these points electrons cannot move, so they accumulate and increase the density of states
• B. The density of states integral weights each k-point by 1/|∇_k E|; where the gradient vanishes, the integrand diverges, creating a non-analytic feature in g(E)
• C. The Pauli exclusion principle forces more electrons into these energy states
• D. These are points where the crystal potential is strongest

Answer: (B)

The density of states can be written as a surface integral g(E) = integral dS / |∇_k E(k)| over the constant-energy surface E(k) = E. Where ∇_k E → 0, the integrand diverges. In 3D, this produces integrable singularities (kinks or logarithmic divergences) rather than true infinities, but in 1D and 2D the singularities can be stronger. Van Hove showed that these critical points occur at a minimum of four energies in any 3D band (the minimum, maximum, and two types of saddle points).

Question 2

The Fermi surface is the surface in k-space where E_n(k) = E_F. Why is its geometry so important for understanding metallic properties?

• A. The Fermi surface determines the crystal structure of the metal
• B. Only electrons near the Fermi surface participate in low-energy processes (conduction, heat capacity, scattering), so the shape of the Fermi surface controls transport, magnetic oscillations, and response to perturbations
• C. The Fermi surface determines the binding energy of core electrons
• D. The Fermi surface is important only for semiconductors, not metals

Answer: (B)

At typical temperatures, kT << E_F, so the Fermi-Dirac distribution is sharp: states well below E_F are fully occupied and inert; states well above are empty. Only states within ~kT of E_F can be thermally excited or respond to external fields. The Fermi surface is where these active electrons live. Its shape — spherical for free electrons, complex and multi-sheeted for real metals — determines which scattering processes are available, how electrons respond to magnetic fields (de Haas-van Alphen oscillations), and the anisotropy of conductivity.

Question 3

In three dimensions, the free-electron density of states goes as g(E) ∝ √E. How does this change qualitatively in a real crystal?

• T. True
• F. False

Answer: False

This question as stated is not true-false — but to clarify: the free-electron √E density of states is modified significantly in a real crystal. Band gaps create energy ranges where g(E) = 0. Van Hove singularities create peaks and kinks. Flat bands (common in tight-binding models with localized orbitals) produce sharp peaks. The overall shape of g(E) reflects the full band structure and can look nothing like √E. Measuring g(E) experimentally (via photoemission or tunneling spectroscopy) is one of the primary ways to probe electronic structure.

Question 4

Explain why the density of states at the Fermi level, g(E_F), appears in the formulas for so many different physical properties of metals.

Model answer: Low-energy properties of metals are dominated by electrons near the Fermi level, because only these electrons can change their state in response to small perturbations (thermal, magnetic, electric). The number of such electrons available to respond is proportional to g(E_F). Electronic specific heat is C_el = (π²/3)k_B²T g(E_F), reflecting how many states are thermally accessible. Pauli paramagnetic susceptibility is χ = μ_B² g(E_F), since only Fermi-level electrons can flip spin. The BCS superconducting transition temperature depends on g(E_F) through the coupling strength. In each case, g(E_F) measures the 'reservoir' of responsive electrons.

This is why transition metals with their high d-band density of states near E_F have large specific heat coefficients, strong Pauli paramagnetism, and (in some cases) superconductivity — they simply have more electrons available to participate in low-energy physics.