---
title: "Fermi-Dirac symmetry about the Fermi level"
subject: "Electrical Engineering Material"
chapter: 1
key: "semester_3/electrical_engineering_material:fermi-symmetry-kt-above-below"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/fermi-symmetry-kt-above-below
site: BCT Engineering
---

# Fermi-Dirac symmetry about the Fermi level

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2082 Kartik (new course), 4 marks
- Video: [Fermi Dirac Distribution in Semiconductor (Basics, Statement, Function, Proof & Case Study)](https://www.youtube.com/watch?v=IEGzFze1y6g) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/fermi-symmetry-kt-above-below
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

For a given fermi energy level EF, show that the probability of emptying energy level KT below EF is equal to probability of occupying level KT above EF.

## Short answer

Using the Fermi-Dirac function f(E)=1/[1+exp((E−EF)/kT)], for E = EF+kT, f=1/(1+e). For E = EF−kT, the empty probability 1−f equals 1/(1+e). Thus the probability of emptiness kT below EF equals the occupation probability kT above EF.

## Answer

The Fermi-Dirac function gives the probability that a state at energy $E$ is occupied:

$$
f(E) = \frac{1}{1 + \exp\left(\dfrac{E - E_F}{kT}\right)}
$$

The probability that a state is **empty** is $1 - f(E)$.

#### Occupation of the level kT above EF

Put $E = E_F + kT$:

$$
f(E_F + kT) = \frac{1}{1 + e^{1}} \quad \cdots (1)
$$

#### Emptying of the level kT below EF

Put $E = E_F - kT$:

$$
\begin{aligned}
1 - f(E_F - kT) &= 1 - \frac{1}{1 + e^{-1}} = \frac{e^{-1}}{1 + e^{-1}}
\end{aligned}
$$

Multiply numerator and denominator by $e$:

$$
1 - f(E_F - kT) = \frac{1}{e + 1} \quad \cdots (2)
$$

#### Comparison

(1) and (2) are identical:

$$
1 - f(E_F - kT) = f(E_F + kT) = \frac{1}{1 + e} = 0.269
$$

Hence the probability of a level $kT$ below $E_F$ being empty equals the probability of a level $kT$ above $E_F$ being occupied. Proved.

#### General result

The same steps work for any $\Delta E$:

$$
1 - f(E_F - \Delta E) = \frac{1}{1 + e^{\Delta E/kT}} = f(E_F + \Delta E)
$$

So $f(E)$ is symmetric about $E_F$ (about the point $f = 1/2$). Physically, the electrons that leave states just below $E_F$ are the same ones that fill states just above $E_F$; the curve drops below 1 under $E_F$ by exactly the amount it rises above 0 over $E_F$.

```text
 f(E)
 1 |-----.
   |      \  <- holes (empty) below EF
 .5|- - - -*
   |        \ <- electrons above EF
 0 |_________'------> E
       EF-kT EF EF+kT
```

## Check yourself

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

1. What is the probability that a state is empty?
   A. f(E)
   B. 1 - f(E)
   C. 1/f(E)
   D. f(E)^2
   Answer: B. Empty and occupied are complementary. (Key point: Empty probability)

2. What is f(E_F + kT)?
   A. 1/(1 + e)
   B. 1/(1 + e^-1)
   C. 1/2
   D. e/(1+e)
   Answer: A. Exponent is +1, giving 1/(1+e) = 0.269. (Key point: Occupation above EF)

3. What is 1 - f(E_F - kT), after simplifying?
   A. 1/(1 + e)
   B. 1/(1 + e^-1)
   C. e/(1 + e)
   D. 0.731
   Answer: A. e^-1/(1 + e^-1) multiplied by e/e is 1/(e + 1), equal to the occupation above. (Key point: Emptying below EF)

4. What general property does this show?
   A. f(E) is symmetric about E_F (the point f = 1/2)
   B. f is zero below EF
   C. f is independent of E
   D. f is linear
   Answer: A. 1 - f(EF - dE) = f(EF + dE) for any dE. (Key point: Symmetry)

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