Free energy from partition function: Worked examples

Before anything, one reminder of the symbols so you never guess:

Recall The symbols, in plain words

$Z$ ::: the partition function, $Z = \sum_{i} e^{- β E_{i}}$ — one number that sums the "Boltzmann weight" of every microstate.
$β$ ::: shorthand for $1/ (k_{B} T)$ . Big $β$ = cold; small $β$ = hot.
$k_{B}$ ::: Boltzmann's constant $\approx 1.381 \times 1 0^{- 23} J/K$ — the conversion factor between temperature and energy.
$F$ ::: Helmholtz free energy, $F = - k_{B} T ln Z$ .
$U$ ::: average (internal) energy, $U = - \partial_{β} ln Z$ .
$ε$ ::: the energy gap of a two-level system.

Prerequisites live at Boltzmann distribution and microstates and Entropy — Gibbs and Boltzmann definitions; the master result is on the parent topic.

#	Case class	What is special	Example
A	Degenerate input: one accessible state	$Z = 1$ , so $ln Z = 0$	Ex 1
B	Low- $T$ limit ( $β \to \infty$ )	System freezes into ground state	Ex 2
C	High- $T$ limit ( $β \to 0$ )	All states equally likely	Ex 2
D	Finite discrete sum, general $T$	Two-level system, all $T$	Ex 2
E	Continuous integral instead of sum	Classical DOF, Gaussian integral	Ex 3
F	Extensive / many independent parts	$Z = Z_{1}^{N}$ , $F \propto N$	Ex 4
G	Real-world word problem	Numbers, units, an actual answer	Ex 5
H	Exam twist: degeneracy factor	States with multiplicity $g$	Ex 6
I	Sign / limiting sanity of $S, p$	Checking $S = - \partial_{T} F$ everywhere	Ex 7

Worked examples

Worked example Ex 1 — Cell A: a system with only ONE state (degenerate)

A frozen crystal has exactly one accessible configuration, energy $E_{0} = 0$ . Find $Z$ , $F$ , $U$ , $S$ .

Forecast: guess before reading — if there is only one way to be, how messy (entropy) can it be? How much free energy is there?

$Z = e^{- β \cdot 0} = 1$ . Why this step? The sum $\sum_{i} e^{- β E_{i}}$ has a single term; a state at zero energy has Boltzmann weight $e^{0} = 1$ .
$F = - k_{B} T ln Z = - k_{B} T ln 1 = 0$ . Why? $ln 1 = 0$ , so $F$ vanishes. There is no "usefulness" to extract from a system with no choices.
$U = - \partial_{β} ln Z = - \partial_{β} (0) = 0$ . Why? $ln Z$ does not depend on $β$ , so its slope is zero. The one state sits at energy $0$ .
$S = (U - F) / T = (0 - 0) / T = 0$ . Why? Entropy counts choices; one state means zero disorder — this is the third law of thermodynamics in miniature.

Verify: $S = 0$ agrees with the Boltzmann form $S = k_{B} ln Ω = k_{B} ln 1 = 0$ . Every number is zero, which is exactly right for a system that "can only be one way." ✓

Worked example Ex 2 — Cells B, C, D: two-level system at ALL temperatures

A single spin has energies $0$ and $ε > 0$ . Get $Z (T)$ , then examine cold ( $β \to \infty$ ), hot ( $β \to 0$ ), and a numeric mid-point.

Forecast: at very low $T$ the spin should hide in the ground state ( $U \to 0$ ); at very high $T$ both states are equally likely so $U \to ε /2$ . Guess the entropy at each end.

$Z = 1 + e^{- β ε}$ . Why? Sum the two Boltzmann weights.
$U = - \partial_{β} ln Z = \frac{ε e ^{- β ε}}{1 + e ^{- β ε}} = \frac{ε}{e ^{β ε} + 1}$ . Why? Differentiate $ln (1 + e^{- β ε})$ with respect to $β$ ; the chain rule brings down $- ε e^{- β ε}$ , and the overall minus flips it positive.
Cell B (cold, $β \to \infty$ ): $e^{β ε} \to \infty$ , so $U \to 0$ and $Z \to 1$ , $F \to 0$ , $S \to 0$ . Why? At $T = 0$ the spin is frozen in its ground state — one state, no entropy.
Cell C (hot, $β \to 0$ ): $e^{β ε} \to 1$ , so $U \to ε /2$ ; and $Z \to 2$ , $S \to k_{B} ln 2$ . Why? Both states equally likely → average energy is the midpoint, and the entropy is " $k_{B}$ times log of two choices."
Cell D (mid, numeric): take $β ε = 1$ . Then $U / ε = 1/ (e^{1} + 1) = 1/ (e + 1) \approx 0.2689$ , and $Z = 1 + e^{- 1} \approx 1.3679$ , so $F / ε = - \frac{1}{1} ln Z \cdot (k_{B} T / ε)$ ... explicitly $F = - k_{B} T ln Z$ ; with $k_{B} T = ε$ (since $β ε = 1$ ), $F = - ε ln (1.3679) \approx - 0.3133 ε$ .

Verify (see figure): $U$ rises smoothly from $0$ to $ε /2$ as $T$ increases — never above $ε /2$ , because in equilibrium you can never make the excited state more likely than the ground state. Units: $U / ε$ is dimensionless, good. ✓

Figure — Free energy from partition function

The figure shows the full curve for cell D and marks the two limiting cells B and C on it.

Worked example Ex 3 — Cell E: continuous integral (1D particle in a box, translational

$Z$ ) A single classical particle of mass $m$ moves freely in 1D over length $L$ . Its energy is purely kinetic, $E = p^{2} /2 m$ . Find $Z$ , then $U$ , using the phase-space integral with measure $d x d p / h$ .

Forecast: there is one quadratic degree of freedom (the momentum), so equipartition predicts $U = \frac{1}{2} k_{B} T$ . Guess what power of $T$ appears in $Z$ .

Set up the integral. Why a sum becomes an integral? Classical position and momentum are continuous, so " $\sum_{i}$ " becomes $\frac{1}{h} \int d x d p$ — the $1/ h$ makes each phase-space cell one state (Planck's constant sets the cell size). $Z = \frac{1}{h} \int_{0}^{L} d x \int_{- \infty}^{\infty} e^{- β p^{2} /2 m} d p .$
Do the position integral: $\int_{0}^{L} d x = L$ . Why? The energy has no $x$ -dependence, so position integrates trivially.
Do the Gaussian momentum integral: $\int_{- \infty}^{\infty} e^{- a p^{2}} d p = π / a$ with $a = β /2 m$ . Why this tool? A bell-shaped $e^{- a p^{2}}$ has no elementary antiderivative, but its area over the whole line is the standard Gaussian result — the only clean way to sum infinitely many continuous states. $\int_{- \infty}^{\infty} e^{- β p^{2} /2 m} d p = \frac{2 π m}{β} .$
Assemble: $Z = \frac{L}{h} \frac{2 π m}{β} \propto β^{- 1/2}$ .
Energy: $ln Z = - \frac{1}{2} ln β + const$ , so $U = - \partial_{β} ln Z = \frac{1}{2 β} = \frac{1}{2} k_{B} T$ . Why? Only the $ln β$ term has a $β$ -slope; its derivative is $- 1/ (2 β)$ , flipped positive.

Verify: $U = \frac{1}{2} k_{B} T$ is exactly equipartition for one quadratic DOF. The $β^{- 1/2}$ scaling of $Z$ from one Gaussian is the fingerprint of a single continuous quadratic mode. Contrast: the oscillator in the parent note had two quadratic DOF ( $x$ and $p$ ), giving $Z \propto β^{- 1}$ and $U = k_{B} T$ . ✓

Worked example Ex 4 — Cell F: extensive many-particle system

Take $N$ independent, distinguishable two-level spins (each with gap $ε$ ). Show $F$ scales with $N$ , and get $U$ and $S$ per spin.

Forecast: independent parts → their $Z$ 's multiply. What does the $ln$ then do to $F$ ?

$Z_{tot} = Z_{1}^{N} = (1 + e^{- β ε})^{N}$ . Why? For independent systems $\sum e^{- β (E_{a} + E_{b} + \dots)}$ factorizes into a product; see the parent note's mistake #4. (Distinguishable, so no $1/ N!$ .)
$F = - k_{B} T ln Z_{1}^{N} = - N k_{B} T ln (1 + e^{- β ε})$ . Why? $ln (a^{N}) = N ln a$ — the logarithm turns the product into a sum, making $F$ extensive ( $\propto N$ ).
$U_{tot} = N \frac{ε}{e ^{β ε} + 1}$ . Why? $ln Z_{tot} = N ln Z_{1}$ , and $\partial_{β}$ is linear, so total energy is $N$ times the single-spin energy from Ex 2.
Numeric check, $β ε = 1$ , $N = 1000$ : $U_{tot} / ε = 1000/ (e + 1) \approx 268.9$ .

Verify: doubling $N$ exactly doubles $F$ and $U$ — the hallmark of an extensive quantity. Had we (wrongly) written $F = - k_{B} T Z$ , $F$ would scale like $(1 + e^{- β ε})^{N}$ , an exponential in $N$ — clearly not extensive. The $ln$ is what saves us. ✓ (See Helmholtz vs Gibbs free energy for why extensivity matters.)

Worked example Ex 5 — Cell G: real-world word problem (units and a number)

A colour centre in a crystal has a ground state and one excited state $ε = 0.10 eV$ above it. At room temperature $T = 300 K$ , what fraction of these centres are excited, and what is the free energy per centre in eV?

Forecast: $0.10$ eV versus $k_{B} T \approx 0.0259$ eV at $300$ K, so $β ε \approx 3.9$ — the gap is a few $k_{B} T$ , meaning the excited state is fairly rare. Guess "a few percent."

Compute $β ε$ . Why? This dimensionless ratio controls everything. $k_{B} T = 8.617 \times 1 0^{- 5} eV/K \times 300 K = 0.02585 eV$ , so $β ε = 0.10/0.02585 = 3.868$ .
Excited fraction $p_{exc} = \frac{e ^{- β ε}}{1 + e ^{- β ε}} = \frac{1}{e ^{β ε} + 1}$ . Why? Boltzmann probability of the upper state, normalized by $Z$ . $p_{exc} = \frac{1}{e ^{3.868} + 1} = \frac{1}{47.83 + 1} = 0.02048 \approx 2.0%.$
Free energy per centre: $F = - k_{B} T ln (1 + e^{- β ε}) = - 0.02585 eV \times ln (1 + 0.02090)$ . $F = - 0.02585 \times ln (1.02090) = - 0.02585 \times 0.020684 = - 5.35 \times 1 0^{- 4} eV .$

Verify: $\approx 2%$ excited matches our forecast ("a few percent"). Units: $k_{B} T$ in eV times a dimensionless log gives eV, good. $F$ is small and negative, as expected when the excited state barely contributes. ✓

Worked example Ex 6 — Cell H: exam twist with a degeneracy factor

A two-level system where the excited level is 3-fold degenerate: ground energy $0$ (1 state), excited energy $ε$ (3 states). Find $Z$ , and the excited-population fraction; note how degeneracy shifts the high- $T$ limit.

Forecast: with three excited states versus one ground state, at very high $T$ the system spends more time excited — the population should tilt toward $3/4$ , not $1/2$ .

$Z = 1 + 3 e^{- β ε}$ . Why? Each of the 3 excited microstates has weight $e^{- β ε}$ ; degeneracy multiplies, it does not add a new exponent. This is the general rule $Z = \sum_{levels} g_{ℓ} e^{- β E_{ℓ}}$ with degeneracy $g_{ℓ}$ .
Excited fraction $p_{exc} = \frac{3 e ^{- β ε}}{1 + 3 e ^{- β ε}}$ . Why? Total excited weight over $Z$ .
High- $T$ limit ( $β \to 0$ ): $p_{exc} \to \frac{3}{1 + 3} = \frac{3}{4} = 0.75$ . Why? When Boltzmann weights are all $\approx 1$ , states are counted purely by how many there are: 3 excited vs 1 ground.
Numeric, $β ε = 1$ : $p_{exc} = \frac{3 e ^{- 1}}{1 + 3 e ^{- 1}} = \frac{1.1036}{2.1036} = 0.5246$ .

Verify: compare to the non-degenerate Ex 2 at $β ε = 1$ , where $p_{exc} = 1/ (e + 1) = 0.2689$ . Tripling the excited degeneracy roughly doubled the population — sensible. High- $T$ limit $3/4$ ≠ $1/2$ , exactly the twist. ✓

Worked example Ex 7 — Cell I: sign & limiting sanity of

$S$ Using the non-degenerate two-level system (Ex 2), verify $S = - \partial F / \partial T$ gives a non-negative entropy that peaks at $k_{B} ln 2$ , checking the notorious sign.

Forecast: entropy must be $\geq 0$ everywhere, be $0$ at $T = 0$ , and saturate at $k_{B} ln 2$ (two equally likely states) as $T \to \infty$ .

Write $F (T) = - k_{B} T ln (1 + e^{- ε / k_{B} T})$ and differentiate with the correct sign $S = - \partial F / \partial T$ . Why the minus? Because $d F = - S d T - p d V + μ d N$ ; the minus is built into the differential (parent note, mistake #3).
Result: $S = k_{B} ln (1 + e^{- β ε}) + \frac{ε / T}{e ^{β ε} + 1}$ . Why two terms? First term is $k_{B} ln Z$ ; second is $U / T$ , exactly $S = U / T + k_{B} ln Z$ .
Numeric, $β ε = 1$ (so $k_{B} T = ε$ ): first term $k_{B} ln (1.3679) = 0.3133 k_{B}$ ; second term $= \frac{1}{e + 1} k_{B} = 0.2689 k_{B}$ . Sum $S = 0.5822 k_{B}$ .
Limits: $T \to 0 \Rightarrow S \to 0$ ; $T \to \infty \Rightarrow S \to k_{B} ln 2 = 0.6931 k_{B}$ . Why? Both computed above in Ex 2, and $0.5822 < 0.6931$ , so the mid-value sits correctly below the ceiling.

Verify: $S > 0$ throughout, monotone rising to $k_{B} ln 2$ , and if we had used the wrong sign $S = + \partial F / \partial T$ we'd get a negative entropy — an instant contradiction. ✓

Recall One-line map of every cell

Which examples cover: degenerate, cold, hot, general discrete, continuous, extensive, word-problem, degeneracy, sign? ::: Ex1 (A), Ex2 (B,C,D), Ex3 (E), Ex4 (F), Ex5 (G), Ex6 (H), Ex7 (I) — all nine cells covered.

Active recall

Two-level system, excited fraction at $β ε = 1$ ? ::: $1/ (e + 1) \approx 0.269$ .
1D free particle $U$ from Gaussian $Z$ ? ::: $\frac{1}{2} k_{B} T$ (one quadratic DOF, equipartition).
Why does $F$ scale like $N$ for independent parts? ::: $Z = Z_{1}^{N}$ and $ln (Z_{1}^{N}) = N ln Z_{1}$ — the log makes $F$ extensive.
Excited level with degeneracy 3 — high- $T$ population? ::: $3/4$ , since states are counted by multiplicity when weights $\to 1$ .
Correct sign of entropy from $F$ ? ::: $S = - \partial F / \partial T$ (minus built into $d F$ ).

Worked examples — Free energy from partition function

The scenario matrix

Worked examples

Active recall