Fermi-Dirac statistics — fermions, Fermi energy: Exercises

Everything here rests on two objects from the parent note, Fermi-Dirac statistics:

Level 1 — Recognition

Exercise 1.1

State the value of $f (ε)$ in three limits at fixed $T > 0$ : (a) $ε = μ$ , (b) $ε = μ - 10 k_{B} T$ , (c) $ε = μ + 10 k_{B} T$ .

Recall Solution

The only thing that matters is the sign and size of the exponent $x = (ε - μ) / k_{B} T$ . (a) $x = 0$ : $f = \frac{1}{e ^{0} + 1} = \frac{1}{2}$ . The chemical potential is always the half-filled energy. (b) $x = - 10$ : $e^{- 10} \approx 4.5 \times 1 0^{- 5}$ , so $f = \frac{1}{1.000045} \approx 0.99995 \approx 1$ . Deep states are essentially full. (c) $x = + 10$ : $e^{10} \approx 2.2 \times 1 0^{4}$ , so $f = \frac{1}{22027} \approx 4.5 \times 1 0^{- 5} \approx 0$ . High states are essentially empty.

Exercise 1.2

Copper has electron density $n = 8.5 \times 1 0^{28} m^{- 3}$ . Write down (without deriving) the formula for $k_{F}$ and identify the exponent applied to $n$ in $E_{F}$ .

Recall Solution

$k_{F} = (3 π^{2} n)^{1/3}$ — the one-third power. In $E_{F} = \frac{ℏ ^{2}}{2 m} (3 π^{2} n)^{2/3}$ the density enters through the two-thirds power, because $E_{F} \propto k_{F}^{2}$ and $k_{F} \propto n^{1/3}$ , so $k_{F}^{2} \propto n^{2/3}$ .

Exercise 1.3

Which of these differ between fermions and bosons: (i) the sign in the denominator, (ii) the maximum value of $f$ , (iii) the definition of $μ$ ? Answer yes/no with one sentence each.

Recall Solution

(i) Yes — fermions have $+ 1$ , bosons have $- 1$ . (ii) Yes — fermion $f \leq 1$ (exclusion caps it); boson $f$ can diverge (condensation, see Bose-Einstein Statistics). (iii) No — $μ$ is the chemical potential in both cases; it is the reservoir property that fixes average particle number (see Chemical Potential).

Level 2 — Application

Exercise 2.1

Compute $E_{F}$ (in eV) and the Fermi temperature $T_{F} = E_{F} / k_{B}$ for copper, $n = 8.5 \times 1 0^{28} m^{- 3}$ .

Recall Solution

Step 1 — build the argument. $3 π^{2} n = 3 (9.8696) (8.5 \times 1 0^{28}) = 2.517 \times 1 0^{30} m^{- 3}$ . Why: this is the quantity we raise to $2/3$ . Step 2 — raise to $2/3$ . $(2.517 \times 1 0^{30})^{2/3} = 1.855 \times 1 0^{20} m^{- 2}$ . (This is $k_{F}^{2}$ .) Step 3 — multiply the prefactor. $E_{F} = \frac{( 1.055 \times 1 0 ^{- 34} ) ^{2}}{2 ( 9.11 \times 1 0 ^{- 31} )} \times 1.855 \times 1 0^{20} = 1.13 \times 1 0^{- 18} J .$ Why this step? The bracket we just evaluated is $k_{F}^{2}$ ; the Fermi energy is $ℏ^{2} k_{F}^{2} /2 m$ , so multiplying by the prefactor $ℏ^{2} /2 m$ converts a squared wavenumber into an energy — that is literally the free-particle relation $ε = ℏ^{2} k^{2} /2 m$ evaluated at the top of the Fermi sea. Divide by $1.602 \times 1 0^{- 19}$ : $E_{F} \approx 7.05 eV$ . Step 4 — Fermi temperature. $T_{F} = \frac{1.13 \times 1 0 ^{- 18}}{1.381 \times 1 0 ^{- 23}} \approx 8.2 \times 1 0^{4} K$ . Why divide by $k_{B}$ ? $T_{F}$ is defined so that $k_{B} T_{F} = E_{F}$ — it re-expresses the Fermi energy as the temperature a classical gas would need to reach that same energy scale. Dividing the energy by $k_{B}$ simply converts joules into kelvin. Since room temperature $300 K ≪ T_{F}$ , copper's electron gas is strongly degenerate.

Exercise 2.2

At what energy (measured from $μ$ , in units of $k_{B} T$ ) is a state exactly 25% occupied?

Recall Solution

Set $f = 0.25$ : $e^{x} + 1 = 4 \Rightarrow e^{x} = 3 \Rightarrow x = ln 3 \approx 1.0986$ . So $ε - μ = (ln 3) k_{B} T \approx + 1.10 k_{B} T$ : a quarter-full state sits just above $μ$ .

Exercise 2.3

The transition region where $f$ falls from $0.9$ to $0.1$ has width $Δ ε$ . Express it in $k_{B} T$ and evaluate for $T = 300 K$ in eV.

Recall Solution

$f = 0.9 \Rightarrow x = ln (1/9) = - 2.197$ ; $f = 0.1 \Rightarrow x = ln 9 = + 2.197$ . Width $= Δ x = 2 ln 9 = 4.394$ , so $Δ ε \approx 4.39 k_{B} T$ . At $300 K$ : $k_{B} T = (1.381 \times 1 0^{- 23}) (300) = 4.14 \times 1 0^{- 21} J = 0.02585 eV$ . $Δ ε \approx 4.39 \times 0.02585 \approx 0.114 eV$ . Compare to $E_{F} \approx 7 eV$ : the smear is $\sim 1.6%$ of $E_{F}$ — a razor-thin fuzzy band.

The figure below makes this concrete. The yellow curve is the ideal $T = 0$ step — full ( $f = 1$ ) up to $μ$ , then a vertical drop to empty ( $f = 0$ ). The blue curve is the finite- $T$ distribution: notice it is no longer a cliff but a smooth slide, and it passes through exactly $f = \frac{1}{2}$ at $ε = μ$ (marked dot). The pink double arrow marks the $0.9 \to 0.1$ band you just computed — its horizontal span is the $4.4 k_{B} T$ width. The point of the picture: all the thermal action lives inside that narrow pink window; deep states stay pinned at $1$ and high states pinned at $0$ , frozen by exclusion.

Figure — Fermi-Dirac statistics — fermions, Fermi energy

Level 3 — Analysis

Exercise 3.1

Show that $f (ε)$ is symmetric about $μ$ in the sense $f (μ + δ) + f (μ - δ) = 1$ for any energy offset $δ$ . Interpret physically.

Recall Solution

Define $δ$ first: let $δ$ be a fixed energy offset from the chemical potential, so that $ε = μ + δ$ sits $δ$ above $μ$ and $ε = μ - δ$ sits $δ$ below. It has units of energy, exactly like $ε$ and $μ$ . Let $x = δ / k_{B} T$ (dimensionless). Then $f (μ + δ) = \frac{1}{e ^{x} + 1}, f (μ - δ) = \frac{1}{e ^{- x} + 1} .$ Multiply the second by $\frac{e ^{x}}{e ^{x}}$ : $f (μ - δ) = \frac{e ^{x}}{1 + e ^{x}}$ . Add: $\frac{1}{e ^{x} + 1} + \frac{e ^{x}}{e ^{x} + 1} = \frac{1 + e ^{x}}{e ^{x} + 1} = 1.$ ∎ Interpretation: the probability a state is empty a distance $δ$ above $μ$ equals the probability a state is occupied a distance $δ$ below $μ$ . Excitations create particle-hole pairs symmetrically — the "hole" left below $μ$ mirrors the "electron" promoted above. This is the electron-hole symmetry that underlies the Sommerfeld expansion.

Exercise 3.2

Evaluate the derivative $- \frac{\partial f}{\partial ε}$ at $ε = μ$ and explain why this quantity acts like a "thermal window."

Recall Solution

Why this tool (the derivative)? Thermal and electrical response of a metal comes from states that change their occupation as you nudge energy or temperature. The slope $- \partial f / \partial ε$ measures exactly how sharply $f$ changes — where it is large, states are "active." With $x = (ε - μ) / k_{B} T$ : $- \frac{\partial f}{\partial ε} = \frac{1}{k _{B} T} \cdot \frac{e ^{x}}{( e ^{x} + 1 ) ^{2}} .$ At $ε = μ$ , $x = 0$ : $\frac{e ^{0}}{( 1 + 1 ) ^{2}} = \frac{1}{4}$ , so $- \frac{\partial f}{\partial ε}_{μ} = \frac{1}{4 k _{B} T} .$ This is a bump peaked at $μ$ with height $\sim 1/ (4 k_{B} T)$ and width $\sim k_{B} T$ — a "window" of width $k_{B} T$ that selects only the electrons near the Fermi surface. Everything else is frozen by exclusion.

Exercise 3.3

The Fermi velocity is $v_{F} = ℏ k_{F} / m$ . Compute $v_{F}$ for copper and compare it to the thermal speed $k_{B} T / m$ at $300 K$ . What does the comparison tell you?

Recall Solution

Step 1 — $k_{F}$ . $k_{F} = (3 π^{2} n)^{1/3} = (2.517 \times 1 0^{30})^{1/3} = 1.362 \times 1 0^{10} m^{- 1}$ . Step 2 — $v_{F}$ . $v_{F} = \frac{( 1.055 \times 1 0 ^{- 34} ) ( 1.362 \times 1 0 ^{10} )}{9.11 \times 1 0 ^{- 31}} = 1.58 \times 1 0^{6} m/s$ . Step 3 — thermal speed. $k_{B} T / m = \frac{4.14 \times 1 0 ^{- 21}}{9.11 \times 1 0 ^{- 31}} = 6.74 \times 1 0^{4} m/s$ . Ratio: $v_{F} / v_{thermal} \approx 23$ . Even at $300 K$ the electrons at the Fermi surface move $\sim 23 \times$ faster than a classical gas would predict — because Pauli exclusion, not temperature, is doing the pushing. This is the microscopic origin of degeneracy pressure.

Level 4 — Synthesis

Exercise 4.1

Using the density of states $g (ε) = \frac{V}{2 π ^{2}} (\frac{2 m}{ℏ ^{2}})^{3/2} ε$ , show that at $T = 0$ the total number of electrons gives back the parent note's $E_{F} = \frac{ℏ ^{2}}{2 m} (3 π^{2} n)^{2/3}$ .

Recall Solution

Why integrate? At $T = 0$ , $f = 1$ below $E_{F}$ and $0$ above, so summing occupied states means integrating $g (ε)$ from $0$ to $E_{F}$ . $N = \int_{0}^{E_{F}} g (ε) d ε = \frac{V}{2 π ^{2}} (\frac{2 m}{ℏ ^{2}})^{3/2} \int_{0}^{E_{F}} ε d ε .$ The integral gives $\frac{2}{3} E_{F}^{3/2}$ : $N = \frac{V}{3 π ^{2}} (\frac{2 m}{ℏ ^{2}})^{3/2} E_{F}^{3/2} .$ Solve for $E_{F}$ . Write $n = N / V$ : $E_{F}^{3/2} = 3 π^{2} n (\frac{ℏ ^{2}}{2 m})^{3/2} \Rightarrow E_{F} = (3 π^{2} n)^{2/3} \frac{ℏ ^{2}}{2 m} .$ Exactly the parent formula. ∎

Exercise 4.2

Show that the average energy per electron at $T = 0$ is $\frac{3}{5} E_{F}$ , and hence write the total ground-state energy.

Recall Solution

Why: the mean energy weights each occupied state by $ε$ , i.e. $⟨ ε ⟩ = \frac{\int _{0}^{E_{F}} ε g d ε}{\int _{0}^{E_{F}} g d ε}$ . Since $g \propto ε$ : $⟨ ε ⟩ = \frac{\int _{0}^{E_{F}} ε ^{3/2} d ε}{\int _{0}^{E_{F}} ε ^{1/2} d ε} = \frac{\frac{2}{5} E _{F}^{5/2}}{\frac{2}{3} E _{F}^{3/2}} = \frac{2/5}{2/3} E_{F} = \frac{3}{5} E_{F} .$ So total energy $U_{0} = N ⟨ ε ⟩ = \frac{3}{5} N E_{F}$ . Meaning: even at absolute zero the gas carries enormous kinetic energy — for copper, $\frac{3}{5} (7 eV) = 4.2 eV$ per electron. That stored energy is what resists compression as degeneracy pressure.

Exercise 4.3

From $U_{0} = \frac{3}{5} N E_{F}$ and $E_{F} \propto n^{2/3} \propto V^{- 2/3}$ , derive the degeneracy pressure $P = - \partial U_{0} / \partial V$ and show $P = \frac{2}{5} n E_{F}$ .

Recall Solution

First, why $E_{F} \propto V^{- 2/3}$ ? We hold the electron number $N$ fixed and shrink the box. Since $n = N / V$ , the density scales as $n \propto V^{- 1}$ ; feeding this into $E_{F} \propto n^{2/3}$ gives $E_{F} \propto (V^{- 1})^{2/3} = V^{- 2/3}$ . Squeezing the box raises every electron's Fermi energy — that rising energy is the pressure. Write $E_{F} = C V^{- 2/3}$ where $C = \frac{ℏ ^{2}}{2 m} (3 π^{2} N)^{2/3}$ (holding $N$ fixed). Then $U_{0} = \frac{3}{5} N C V^{- 2/3}$ . $P = - \frac{\partial U _{0}}{\partial V} = - \frac{3}{5} N C (- \frac{2}{3}) V^{- 5/3} = \frac{2}{5} \cdot \frac{N C}{V} V^{- 2/3} = \frac{2}{5} \frac{N}{V} E_{F} .$ So $P = \frac{2}{5} n E_{F}$ . For copper: $P = \frac{2}{5} (8.5 \times 1 0^{28}) (1.13 \times 1 0^{- 18}) \approx 3.8 \times 1 0^{10} Pa \approx 3.8 \times 1 0^{5} atm$ . This vast internal pressure — from exclusion alone — is what supports white dwarf stars.

Level 5 — Mastery

Exercise 5.1

Using the finite- $T$ chemical potential from the parent note, $μ (T) = E_{F} [1 - \frac{π ^{2}}{12} (\frac{k _{B} T}{E _{F}})^{2}],$ compute the fractional shift $\frac{E _{F} - μ}{E _{F}}$ for copper at $T = 300 K$ . Is the common shorthand " $μ = E_{F}$ " safe here?

Recall Solution

$\frac{E _{F} - μ}{E _{F}} = \frac{π ^{2}}{12} (\frac{k _{B} T}{E _{F}})^{2} = \frac{π ^{2}}{12} (\frac{1}{T _{F} / T})^{2}$ . With $T = 300 K$ , $T_{F} = 8.2 \times 1 0^{4} K$ : $\frac{T}{T _{F}} = 3.66 \times 1 0^{- 3}$ . $\frac{π ^{2}}{12} (3.66 \times 1 0^{- 3})^{2} = 0.8225 \times (1.34 \times 1 0^{- 5}) = 1.10 \times 1 0^{- 5}$ . So $μ$ sits about 11 parts per million below $E_{F}$ . For copper at room temperature the shorthand $μ \approx E_{F}$ is extremely safe — but it is a $(T / T_{F})^{2}$ effect that grows and cannot be ignored in hot or low-density systems.

Exercise 5.2

A white-dwarf-like electron gas has $n = 1 0^{36} m^{- 3}$ . Compute $E_{F}$ in joules and in eV, then in units of the electron rest energy $m_{e} c^{2} = 8.19 \times 1 0^{- 14} J$ . Comment on whether the non-relativistic formula is trustworthy.

Recall Solution

Step 1 — argument. $3 π^{2} n = (29.608) (1 0^{36}) = 2.961 \times 1 0^{37} m^{- 3}$ . Step 2 — $2/3$ power. $(2.961 \times 1 0^{37})^{2/3}$ . Write the base as $29.608 \times 1 0^{36}$ ; then $(1 0^{36})^{2/3} = 1 0^{24}$ and $(29.608)^{2/3} = 9.63$ , so the result is $9.63 \times 1 0^{24} m^{- 2}$ . (This is $k_{F}^{2}$ .) Step 3 — energy. $E_{F} = \frac{( 1.055 \times 1 0 ^{- 34} ) ^{2}}{2 ( 9.11 \times 1 0 ^{- 31} )} \times 9.63 \times 1 0^{24} = 6.11 \times 1 0^{- 8} \times 9.63 \times 1 0^{24} \times \dots$ More cleanly, the prefactor $\frac{ℏ ^{2}}{2 m} = \frac{( 1.055 \times 1 0 ^{- 34} ) ^{2}}{2 ( 9.11 \times 1 0 ^{- 31} )} = 6.11 \times 1 0^{- 39} J\cdotm^{2}$ , so $E_{F} = 6.11 \times 1 0^{- 39} \times 9.63 \times 1 0^{24} = 5.88 \times 1 0^{- 14} J .$ In eV: $5.88 \times 1 0^{- 14} /1.602 \times 1 0^{- 19} = 3.67 \times 1 0^{5} eV \approx 0.37 MeV$ . Step 4 — compare to rest energy. $\frac{E _{F}}{m _{e} c ^{2}} = \frac{5.88 \times 1 0 ^{- 14}}{8.19 \times 1 0 ^{- 14}} = 0.72$ . $E_{F}$ is about $72%$ of the electron rest energy — the Fermi-surface electrons are firmly relativistic. The non-relativistic dispersion $ε = ℏ^{2} k^{2} /2 m$ is no longer trustworthy here; you must use the relativistic energy–momentum relation, which softens the degeneracy pressure and ultimately produces the Chandrasekhar mass limit for white dwarfs.

Exercise 5.3

Take the limit $μ \to - \infty$ (equivalently $e^{- μ / k_{B} T} \to \infty$ ) in the Fermi-Dirac distribution and show it reduces to the Maxwell-Boltzmann form. Explain the physical regime where this happens.

Recall Solution

Why this limit? When states are very rarely occupied ( $f ≪ 1$ ), no two particles ever compete for the same state, so exclusion becomes irrelevant and quantum statistics should collapse to the classical answer. If $μ \to - \infty$ then for any finite $ε$ , $(ε - μ) / k_{B} T \to + \infty$ , so $e^{(ε - μ) / k_{B} T} ≫ 1$ and the $+ 1$ is negligible: $f (ε) = \frac{1}{e ^{(ε - μ) / k_{B} T} + 1} ⟶ e^{- (ε - μ) / k_{B} T} = e^{μ / k_{B} T} e^{- ε / k_{B} T} .$ This is $f \propto e^{- ε / k_{B} T}$ — the Maxwell-Boltzmann factor. When: high temperature and/or low density, i.e. $T ≫ T_{F}$ , where occupation is sparse and the " $+ 1$ " (the fermion signature) stops mattering. See the figure comparing the two.

Rapid self-test

Copper

E_{F}

in eV?

\approx 7.0 eV

Fermi temperature of copper?

\approx 8.2 \times 1 0^{4} K

Value of

- \partial f / \partial ε

ε = μ

1/ (4 k_{B} T)

Average electron energy at

T = 0

\frac{3}{5} E_{F}

Degeneracy pressure in terms of

n, E_{F}

P = \frac{2}{5} n E_{F}

Fermi-Dirac reduces to which classical law as

μ \to - \infty

Maxwell-Boltzmann,

f \propto e^{- ε / k_{B} T}