advanced27 min studyLesson 504 in Physics

Count energy states across the whole Brillouin zone

Derive dimensional density-of-states laws and identify velocity-related singularities.

On this page

What you’ll understand

  • Define density of states with explicit volume and spin conventions.
  • Derive the one-dimensional parabolic-band result.
  • Distinguish a band-path plot from a Brillouin-zone state count.

A band plot assigns energies to selected wavevectors. A density of states answers a different question: how many states lie in a narrow energy interval after every wavevector has been counted? Flat dispersion packs many states into a small energy range, while steep dispersion spreads them out. Deriving this relation requires a wavevector measure and a degeneracy convention. Without them, an apparently correct formula can be wrong by volume factors, spin factors, or powers of two pi.

Declare what is extensive

For a large periodic sample, each independent wavevector occupies a reciprocal-space cell of volume (2π)ᵈ divided by the real-space sample measure. Divide the total density of states by that measure to obtain an intensive quantity. Sum over every band and any genuinely independent spin or valley channels. If spin-split bands are already listed separately, multiplying by an additional spin factor would count them twice. A filled orbital band integrates to one state per primitive cell per spin.

At a nonzero temperature the density of states still describes available one-electron states; occupation is a separate factor supplied by the Fermi function. A sharp peak does not say every state there is occupied, and a zero density at an isolated energy does not necessarily imply a finite insulating gap. Disorder broadening, interactions, and experimental spectral matrix elements can further separate a measured signal from the clean independent-band density defined here.

g(E)Vd=gs∑n∫BZddk(2π)d δ(E−En(k)),Ne=∫g(E)fFD(E) dE\frac{g(E)}{\mathcal V_d}=g_s\sum_n\int_{\rm BZ}\frac{d^dk}{(2\pi)^d}\,\delta(E-E_n(\mathbf k)),\qquad N_e=\int g(E)f_{\rm FD}(E)\,dE
Vd denotes length, area, or volume in dimension d; g is total states per energy and gs an explicitly assumed degeneracy not already included in n. Delta has inverse-energy units; fFD is an occupation probability.↔ Scroll wide equations horizontally.

One-dimensional roots reveal the slope factor

In one dimension, change variables from wavevector to energy separately on each monotonic branch. The delta-function identity contributes the reciprocal absolute energy slope at every root. Both positive and negative roots must be included for a symmetric parabolic band. The absolute value prevents a negative group velocity from producing a negative density of states. Since velocity is energy slope divided by hbar, small velocity naturally enhances the state count per energy.

For a quadratic minimum, the resulting density diverges as inverse square root of energy above the edge. This divergence is integrable: the accumulated count goes to zero as the square root of that energy. In two dimensions a quadratic isotropic band instead gives a constant density above its edge, while in three dimensions it rises as a square root. The dimension changes the number of wavevectors on an energy shell and cannot be repaired by simply multiplying the one-dimensional expression by a constant.

g1D(E)L=gs2π∑ki:E(ki)=E1∣E′(ki)∣=gs2πA(E−Ec),E=Ec+Ak2\frac{g_{1D}(E)}{L}=\frac{g_s}{2\pi}\sum_{k_i:E(k_i)=E}\frac1{|E'(k_i)|}=\frac{g_s}{2\pi\sqrt{A(E-E_c)}},\qquad E=E_c+Ak^2
A>0 has units energy·length², L is sample length, and the parabolic approximation is valid near the isolated minimum Ec. The final expression includes both roots and is zero below the edge.↔ Scroll wide equations horizontally.

A path omits most states

In a three-dimensional crystal a conventional high-symmetry path is a set of lines with zero volume in the Brillouin zone. It is useful for visualizing symmetry and crossings but cannot by itself determine an integral density of states. Use a full-zone integration method and verify its wavevector convergence. Smearing a sparse mesh can hide sampling errors by producing a smooth curve; smoothness is not evidence that the integrated count or a narrow feature has converged.

The function lab isolates the slope of a quadratic. Reset and set coefficient to 1 and position to 1; predict value 1 and slope 2. Reset and set coefficient to 2 and position to 1; predict value 2 and slope 4. For a common physical energy one would also need to change the root position before evaluating a density of states. This adjustable parabola has no state counting, spin degeneracy, or Brillouin-zone integration.

N(E)L=gsπE−EcA,g2D(E)A=gs4πA(E>Ec)\frac{N(E)}{L}=\frac{g_s}{\pi}\sqrt{\frac{E-E_c}{A}},\qquad \frac{g_{2D}(E)}{\mathcal A}=\frac{g_s}{4\pi A}\quad(E>E_c)
N(E) is the number of available states below E in the specified parabolic band, and Acal is sample area. The two-dimensional expression assumes an isotropic quadratic minimum and no additional valley degeneracy.↔ Scroll wide equations horizontally.

A one-dimensional edge count

Use a one-dimensional parabolic band with Ec=0, A=1 eV·nm², and spin degeneracy gs=2. At E=0.25 eV find both wavevector roots, integrated state density below E, and the density of states per length.

  1. The roots satisfy k²=0.25 nm⁻², giving k=±0.5 nm⁻¹. Both roots belong to the local parabolic branch and must be counted; they are not an additional spin degeneracy.

  2. The occupied wavevector interval below this energy has length 1 nm⁻¹. Multiplying by gs/(2π) gives N(E)/L=1/π≈0.3183 nm⁻¹. This is a continuum state density, not a claim of a fractional integer count in a specified tiny ring.

  3. At either root the absolute energy slope is 2A|k|=1 eV·nm. Adding the two reciprocal slopes and multiplying by gs/(2π) gives g(E)/L=2/π≈0.6366 eV⁻¹nm⁻¹, matching the derivative of the integrated count.

The result

The roots are ±0.5 nm⁻¹, the integrated state density is 0.3183 nm⁻¹, and the density of states is 0.6366 eV⁻¹nm⁻¹.

Explore the model

Interactive lab

Interactive lab · function

Read a quadratic’s value and slope

Reset and set a=1, x=1; predict y=1 and slope 2. Reset and set a=2, x=1; predict y=2 and slope 4.

Read a quadratic’s value and slopeThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. 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 forThis is y=ax² with a tangent and signed integral. It illustrates local slopes only and does not count electronic states or integrate a Brillouin zone.

Check your understanding

Why does the one-dimensional state-count formula use an absolute energy slope?

Try it yourself

For a two-dimensional isotropic parabolic band use A=2 eV·nm² and gs=2. Find the number of states per area in an energy window of width 0.40 eV entirely above the band edge.

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 annular wavevector area or use the constant two-dimensional density above the edge.

Compare with the solution

The density of states per area is gs/(4πA)=1/(4π) eV⁻¹nm⁻². Multiplication by the 0.40 eV window gives 0.10/π≈0.03183 nm⁻². The constant result assumes the quadratic dispersion remains valid over the window and no other bands or valleys contribute.

Take it with you

  • Density of states requires an explicit real-space measure and degeneracy count.
  • Every energy root contributes through an absolute slope.
  • A high-symmetry band path cannot replace a full-zone state integral.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.