---
title: "Degenerate state, Fermi energy and density of states"
subject: "Electrical Engineering Material"
chapter: 1
key: "semester_3/electrical_engineering_material:degenerate-state-density-of-states"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/degenerate-state-density-of-states
site: BCT Engineering
---

# Degenerate state, Fermi energy and density of states

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2078 Kartik, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/degenerate-state-density-of-states
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

Define Degenerate state and Fermi energy. Derive an expression showing the relationship between density of states and energy.

## Short answer

Degenerate states have equal energy but different quantum numbers. Fermi energy is the highest occupied level at 0 K. Density of states g(E) = (8√2 π m^(3/2)/h³) √E, derived by counting states in k-space (octant of sphere) per unit volume.

## Answer

#### Degenerate state
Two or more different quantum states (different sets of quantum numbers, i.e. different wave functions) that have the **same energy** are called **degenerate states**. For an electron in a cubic box of side $L$,

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

The states $(n_1, n_2, n_3) = (1,1,2), (1,2,1), (2,1,1)$ all have $n_1^2 + n_2^2 + n_3^2 = 6$, so they have the same energy $6h^2/8mL^2$. This level is **three-fold degenerate**. With spin, each state is also two-fold spin degenerate.

#### Fermi energy
The **Fermi energy** $E_F$ is the energy of the highest occupied level in a metal at absolute zero (0 K). At 0 K all states below $E_F$ are filled and all above are empty. At any temperature it is the energy at which the probability of occupation is exactly 1/2, $f(E_F) = 1/2$. For copper $E_F \approx 7$ eV.

#### Density of states versus energy
**Density of states** $g(E)$ is the number of states per unit energy per unit volume at energy $E$.

Let $n^2 = n_1^2 + n_2^2 + n_3^2 = \dfrac{8mL^2E}{h^2}$. Each state is a point with unit volume in n-space. All states with energy up to $E$ lie inside a sphere of radius $n$, but only the positive octant is allowed ($n_i > 0$).

```text
   n3
   |
   |  . . .
   |  . . . .      states with energy <= E
   |  . . . . .    fill 1/8 of a sphere
   +-----------n2  of radius n
  /
 n1
```

Number of orbital states $= \frac{1}{8}\cdot\frac{4}{3}\pi n^3$. With 2 spin states each:

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

Per unit volume (divide by $L^3$):

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

Differentiating:

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

So $g(E) \propto \sqrt{E}$: the number of available states per unit energy increases as the square root of energy.

```text
 g(E)
  ^                      ___..
  |                __..--
  |  filled   _.-''|
  |  at 0 K .'     |  empty
  |       /        |
  |     /          |
  +----------------+--------> E
                  E_F
```

#### Fermi energy from $g(E)$
At 0 K all states up to $E_F$ are full, so the electron concentration is

$$
n = \int_0^{E_F} g(E)\,dE = \frac{\pi}{3}\left(\frac{8m}{h^2}\right)^{3/2}E_F^{3/2} \quad\Rightarrow\quad E_F = \frac{h^2}{8m}\left(\frac{3n}{\pi}\right)^{2/3}
$$

## Check yourself

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

1. Degenerate states are:
   A. Different quantum states having the same energy
   B. The same quantum state occupied by two electrons
   C. States with zero energy
   D. States above the Fermi level
   Answer: A. Different wave functions (quantum numbers) with equal energy are degenerate. (Key point: Degenerate state)

2. The Fermi energy at 0 K is:
   A. The energy where all states are empty
   B. The average energy of all electrons
   C. The energy of the highest occupied level
   D. The lowest energy in the band
   Answer: C. At 0 K, states below E_F are full and those above are empty. (Key point: Fermi energy)

3. Why is only 1/8 of the sphere in n-space used to count states?
   A. Because n1, n2, n3 must all be positive
   B. Because of electron spin
   C. Because energy is symmetric in time
   D. Because only one of eight bands is filled
   Answer: A. Quantum numbers are positive integers, so only the positive octant is allowed. (Key point: n-space counting)

4. How does the density of states g(E) of free electrons depend on energy?
   A. g is proportional to E^2
   B. g is proportional to E^(3/2)
   C. g is constant
   D. g is proportional to sqrt(E)
   Answer: D. Differentiating S_v proportional to E^(3/2) gives g(E) proportional to E^(1/2). (Key point: g(E) dependence)

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