---
title: "Number of states and density of states"
subject: "Electrical Engineering Material"
chapter: 1
key: "semester_3/electrical_engineering_material:density-of-states-derivation"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/density-of-states-derivation
site: BCT Engineering
---

# Number of states and density of states

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2081 Baisakh, 8 marks; 2073 Shrawan, 8 marks; 2068 Baisakh, 6 marks (asked 3 times)
- Video: [Density of States Derivation Part 1](https://www.youtube.com/watch?v=z7YGS67GETo) — Jordan Edmunds Chetty
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/density-of-states-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

What do you understand by number of states and density of states in quantum mechanics? Derive appropriate expressions for them.

## Short answer

For free electrons in a 3D box, the number of states per unit volume up to energy E is S_v(E) = (π/3)(8m/h²)^(3/2) E^(3/2). The density of states is g(E) = dS_v/dE = (π/2)(8m/h²)^(3/2) E^(1/2) = (8√2 π m^(3/2)/h³) √E, proportional to √E.

## Answer

- The **number of states** $S(E)$ is the total number of allowed electron states (including spin) with energy from 0 up to $E$ (usually per unit volume).
- The **density of states** $g(E)$ is the number of states per unit energy per unit volume at energy $E$: $g(E) = dS_v/dE$. So $g(E)\,dE$ is the number of states between $E$ and $E + dE$ per unit volume.

#### Electron in a three-dimensional box
Treat the free electrons of a metal as particles in a cubic box of side $L$ (infinite well in 3D). The energies are

$$
E = \frac{h^2}{8mL^2}\left(n_1^2 + n_2^2 + n_3^2\right), \qquad n_1, n_2, n_3 = 1, 2, 3, \ldots
$$

Each set $(n_1, n_2, n_3)$ is one orbital state, i.e. one point in "n-space" with unit volume per point. Define

$$
n^2 = n_1^2 + n_2^2 + n_3^2 = \frac{8mL^2E}{h^2}
$$

All states with energy $\le E$ lie inside a sphere of radius $n$. Since $n_1, n_2, n_3$ are positive, only **one octant** (1/8) of the sphere counts.

```text
   n2
   ^
   | . . . .
   | . . . . .    one octant of a sphere
   | . . . . . .  of radius n; each dot
   | . . . . . .  = one orbital state
   +--------------> n1
```

#### Number of states
Each orbital state holds 2 electrons (spin), so

$$
\begin{aligned}
S(E) &= 2\times\frac{1}{8}\times\frac{4}{3}\pi n^3 = \frac{\pi}{3}n^3 \\
&= \frac{\pi}{3}\left(\frac{8mL^2E}{h^2}\right)^{3/2}
\end{aligned}
$$

Dividing by the volume $L^3$:

$$
S_v(E) = \frac{\pi}{3}\left(\frac{8m}{h^2}\right)^{3/2}E^{3/2}
$$

#### Density of states

$$
\begin{aligned}
g(E) &= \frac{dS_v}{dE} = \frac{\pi}{3}\left(\frac{8m}{h^2}\right)^{3/2}\cdot\frac{3}{2}E^{1/2} \\
&= \frac{\pi}{2}\left(\frac{8m}{h^2}\right)^{3/2}E^{1/2} = \frac{8\sqrt{2}\,\pi\,m^{3/2}}{h^3}E^{1/2}
\end{aligned}
$$

So $g(E) \propto \sqrt{E}$.

```text
 g(E)
  ^                 __..--
  |           _.--''
  |       _.-'      g(E) ~ sqrt(E)
  |    .-'
  |  /
  | /
  +-------------------------> E
```

#### Use
Multiplying $g(E)$ by the Fermi–Dirac function $f(E)$ gives the number of electrons per unit energy, $n_E = g(E)f(E)$. Integrating up to $E_F$ at 0 K gives the Fermi energy $E_F = \frac{h^2}{8m}\left(\frac{3n}{\pi}\right)^{2/3}$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What does g(E) dE represent?
   A. Number of electrons between E and E + dE
   B. Number of states per unit volume between E and E + dE
   C. Total states up to E
   D. Probability of occupation
   Answer: B. g(E) = dS_v/dE, the states per unit energy per unit volume, so g dE counts states in the interval. (Key point: Definition of g(E))

2. Why is only one octant (1/8) of the n-sphere counted?
   A. Only energy-positive states exist
   B. Spin restricts it to 1/8
   C. The box has 8 faces
   D. n1, n2, n3 are all positive integers
   Answer: D. Since n1, n2, n3 are positive, only the positive octant of the sphere of radius n contains valid states. (Key point: Octant of sphere)

3. Why is the factor 2 included in S(E) = 2 x (1/8)(4/3) pi n^3?
   A. Each orbital state holds two electrons of opposite spin
   B. There are two directions of motion
   C. Two octants are counted
   D. Energy is doubled
   Answer: A. Each (n1, n2, n3) orbital state can hold 2 electrons (spin up and down). (Key point: Factor of 2)

4. What is the dependence of g(E) on E in 3D?
   A. g(E) proportional to E^(3/2)
   B. g(E) proportional to sqrt(E)
   C. g(E) proportional to 1/E
   D. g(E) constant
   Answer: B. S_v is proportional to E^(3/2), so its derivative g(E) is proportional to E^(1/2). (Key point: Energy dependence)

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