Canonical ensemble — partition function Z: Worked examples

Before anything, three reminders of the symbols we lean on the whole way down:

Recall What is

$β$ and why do we use it instead of $T$ ? $β$ ::: $β = \frac{1}{k _{B} T}$ . We use it because in $Z = \sum_{i} e^{- β E_{i}}$ the temperature only ever shows up multiplying an energy. Differentiating with respect to $β$ cleanly "pulls down" a factor of $- E_{i}$ , which is exactly what an average energy needs.

Recall What is

$h$ (Planck's constant) and why does it appear in classical $Z$ ? $h$ ::: $h$ is Planck's constant, with units of (energy $\times$ time) = (position $\times$ momentum). In a continuous system we sum over states by integrating over the position–momentum plane; each genuinely distinct state occupies a cell of area $h$ . Dividing by $h$ per degree of freedom turns "area of phase space" into a dimensionless count of states.

Recall The three master formulas we will re-use

$Z = \sum_{i} e^{- β E_{i}}$ ::: partition function (sum of Boltzmann weights) $⟨ E ⟩ = - \frac{\partial ln Z}{\partial β}$ ::: average energy $σ_{E}^{2} = \frac{\partial ^{2} ln Z}{\partial β ^{2}}$ ::: energy fluctuation (the spread of $E$ about $⟨ E ⟩$ )

The scenario matrix

Every problem a partition function can throw at you falls into one of these cells. The examples below are labelled by which cell they hit; together they cover the whole grid.

Cell	What makes it different	Example
A. Cold limit $T \to 0, β \to \infty$	system freezes into lowest state(s)	Ex 1
B. Hot limit $T \to \infty, β \to 0$	all states equally likely	Ex 1
C. Degeneracy $g_{n} > 1$	levels counted with multiplicity	Ex 2
D. Discrete, infinite ladder	geometric series, must converge	Ex 3
E. Continuous / classical	phase-space integral, Gaussians	Ex 4
F. Fluctuations & heat capacity	second derivative of $ln Z$	Ex 5
G. Many independent particles	factorization $Z_{1}^{N}$ , and the $N!$	Ex 6
H. Real-world word problem	pick states from physics, then compute	Ex 7
I. Exam twist / degenerate zero	a level shift or a $g = 0$ -style trap	Ex 8

Ex 1 — Two-level system: both temperature extremes (cells A + B)

Step 1 — Write $Z$ . $Z = e^{- β \cdot 0} + e^{- β ε} = 1 + e^{- β ε}$ Why this step? $Z$ is by definition the sum of Boltzmann weights over all states — here there are exactly two.

Step 2 — Differentiate. $⟨ E ⟩ = - \frac{\partial l n Z}{\partial β} = \frac{ε e ^{- β ε}}{1 + e ^{- β ε}} = \frac{ε}{e ^{β ε} + 1}$ Why this step? The master formula converts "I have $Z$ " into "I have the average energy" with one derivative — no summing $E_{i} P_{i}$ by hand.

Step 3 — Cold limit (cell A). $β \to \infty \Rightarrow e^{β ε} \to \infty \Rightarrow ⟨ E ⟩ \to 0$ . Why this step? Limits are the cheapest correctness test; they must match physics.

Step 4 — Hot limit (cell B). $β \to 0 \Rightarrow e^{β ε} \to 1 \Rightarrow ⟨ E ⟩ \to ε /2$ .

Step 5 — At $k_{B} T = ε$ we have $β ε = 1$ , so $⟨ E ⟩ = \frac{ε}{e + 1}$ .

Recall Numeric value at

$k_{B} T = ε$ $⟨ E ⟩ / ε$ ::: $1/ (e + 1) \approx 0.2689$

Verify: Cold gives $0$ ✓, hot gives $ε /2$ ✓ (both match the forecast). Units: every term in $⟨ E ⟩$ carries a factor $ε$ (an energy), the rest is dimensionless — good.

The figure below plots exactly this $⟨ E ⟩ (T)$ . Follow the blue curve rising from $0$ (cold) up toward the red dashed ceiling at $ε /2$ (hot); the yellow dot marks our special point $k_{B} T = ε$ where $⟨ E ⟩ \approx 0.269 ε$ . Notice the curve never overshoots $ε /2$ — energy has nowhere hotter to go than the 50/50 mixture.

Figure — Canonical ensemble — partition function Z

Ex 2 — A degenerate three-level system (cell C)

Step 1 — Build $Z$ with multiplicity. $Z = ground 1 \cdot e^{0} + 3 excited states 3 \cdot e^{- β ε} = 1 + 3 e^{- β ε}$ Why this step? Each microstate contributes its own Boltzmann factor. Three microstates at energy $ε$ contribute $3 e^{- β ε}$ . This is the $Z = \sum_{n} g_{n} e^{- β E_{n}}$ rule in action.

Step 2 — Average energy. $⟨ E ⟩ = - \frac{\partial l n Z}{\partial β} = \frac{3 ε e ^{- β ε}}{1 + 3 e ^{- β ε}} = \frac{3 ε}{e ^{β ε} + 3}$ Why this step? Same master formula; the degeneracy simply rides along inside $Z$ .

Step 3 — Hot limit. $β \to 0 \Rightarrow ⟨ E ⟩ \to \frac{3 ε}{1 + 3} = \frac{3}{4} ε$ . Why this step? Confirms the "all four states equal" prediction.

Verify: Cold limit $β \to \infty$ gives $0$ ✓. Hot limit $\frac{3}{4} ε$ ✓ matches forecast. If we set $g = 1$ we recover Ex 1's $ε / (e^{β ε} + 1)$ ✓ — the formula degrades gracefully.

Ex 3 — Quantum harmonic oscillator: an infinite ladder (cell D)

Step 1 — Write the sum. $Z = \sum_{n = 0}^{\infty} e^{- β ℏ ω (n + 1/2)} = e^{- β ℏ ω /2} \sum_{n = 0}^{\infty} (e^{- β ℏ ω})^{n}$ Why this step? Pulling out the constant $\frac{1}{2} ℏ ω$ exposes a plain geometric series with common ratio $r = e^{- β ℏ ω}$ .

Step 2 — Check the ratio is below 1, then sum. Temperature is always positive, so $β = 1/ k_{B} T > 0$ , and $ℏ ω > 0$ ; therefore the exponent $- β ℏ ω$ is negative, which makes $r = e^{- β ℏ ω}$ a number strictly between $0$ and $1$ . A geometric series with $0 < r < 1$ converges to $\sum_{n \geq 0} r^{n} = \frac{1}{1 - r}$ : $Z = \frac{e ^{- β ℏ ω /2}}{1 - e ^{- β ℏ ω}}$ Why this step? The series converges precisely because $r < 1$ — the fact that energies grow without bound is exactly what forces $r$ below $1$ and guarantees a finite answer.

Step 3 — Rewrite as a $sinh$ . Multiply top and bottom by $e^{+ β ℏ ω /2}$ . The numerator $e^{- β ℏ ω /2} \cdot e^{+ β ℏ ω /2} = 1$ ; the denominator becomes $e^{+ β ℏ ω /2} - e^{- β ℏ ω /2}$ , which is exactly $2 sinh (β ℏ ω /2)$ by the definition $sinh x = \frac{1}{2} (e^{x} - e^{- x})$ : $Z = \frac{1}{e ^{β ℏ ω /2} - e ^{- β ℏ ω /2}} = \frac{1}{2 s i n h ( β ℏ ω /2 )}$ Why this step? The $sinh$ form is compact and its logarithm differentiates cleanly in the next step.

Step 4 — Average energy. $⟨ E ⟩ = - \frac{\partial l n Z}{\partial β} = \frac{ℏ ω}{2} + \frac{ℏ ω}{e ^{β ℏ ω} - 1}$ Why this step? The master derivative again; the first term is zero-point energy, the second is the thermal Planck occupation.

Step 5 — High- $T$ limit. For $β ℏ ω ≪ 1$ , $e^{β ℏ ω} - 1 \approx β ℏ ω$ , so $⟨ E ⟩ \approx \frac{ℏ ω}{β ℏ ω} = k_{B} T$ . Why this step? Must match the classical oscillator result — quantum and classical agree once thermal energy dwarfs the level spacing.

Verify: Low- $T$ limit $β \to \infty$ gives $⟨ E ⟩ \to ℏ ω /2$ (pure zero-point) ✓. High- $T$ gives $k_{B} T$ ✓, matching parent Example 2.

The figure contrasts the two curves. The blue quantum curve flattens onto the green dotted floor $ℏ ω /2$ as it cools (energy freezes at the zero-point value), while the yellow dashed classical line $k_{B} T$ slides straight through the origin. Watch how blue and yellow merge on the right: once $k_{B} T ≫ ℏ ω$ , quantization is invisible.

Ex 4 — Free particle in a box: continuous phase space (cell E)

Step 1 — Set up the phase-space integral. $Z_{1} = \frac{1}{h} \int_{0}^{L} d x \int_{- \infty}^{\infty} d p e^{- β p^{2} /2 m}$ Why this step? For a continuous system we sum over states by integrating over phase space $(x, p)$ , dividing by Planck's constant $h$ (see the recall box above) so each cell of area $h$ counts as one state.

Step 2 — Do the integrals. The $x$ -integral is just $L$ ; the $p$ -integral is a Gaussian $\int e^{- a p^{2}} d p = π / a$ with $a = β /2 m$ : $Z_{1} = \frac{L}{h} \frac{2 π m}{β}$ Why this step? The Gaussian is the tool because $e^{- a p^{2}}$ is exactly the Gaussian shape; its area is a known standard result, so no numerical work is needed.

Step 3 — Average energy. Write $ln Z_{1} = - \frac{1}{2} ln β + const$ : $⟨ E ⟩ = - \frac{\partial l n Z _{1}}{\partial β} = \frac{1}{2 β} = \frac{1}{2} k_{B} T$ Why this step? Only the $β$ -dependence matters for the derivative; the box length $L$ and $h$ both sit in the constant and drop out.

Verify: One quadratic DOF $\Rightarrow \frac{1}{2} k_{B} T$ ✓ (matches forecast and the Equipartition theorem). Units: $⟨ E ⟩$ has units of $k_{B} T$ = energy ✓.

Ex 5 — Two-level fluctuations and heat capacity (cell F)

Before this example we need one new quantity.

Step 1 — Fluctuation from $ln Z$ . With $ln Z = ln (1 + e^{- β ε})$ , $σ_{E}^{2} = \frac{\partial ^{2} l n Z}{\partial β ^{2}} = \frac{ε ^{2} e ^{- β ε}}{( 1 + e ^{- β ε} ) ^{2}}$ Why this step? The second derivative of $ln Z$ is the variance of energy — that identity (from the parent note) turns "compute a spread" into "differentiate twice".

Step 2 — Heat capacity via $σ_{E}^{2} = k_{B} T^{2} C_{V}$ . $C_{V} = \frac{σ _{E}^{2}}{k _{B} T ^{2}} = k_{B} (β ε)^{2} \frac{e ^{- β ε}}{( 1 + e ^{- β ε} ) ^{2}}$ Why this step? The fluctuation–response link lets us get $C_{V}$ without ever differentiating $⟨ E ⟩$ with respect to $T$ directly.

Step 3 — Limits. As $β \to \infty$ , the factor $(β ε)^{2} e^{- β ε} \to 0$ . As $β \to 0$ , the prefactor $(β ε)^{2} \to 0$ . So $C_{V} \to 0$ at both ends. Why this step? Confirms the anomaly's two shoulders.

Step 4 — Locate the peak numerically. Maximizing $C_{V}$ over $x = β ε$ gives the peak at $x \approx 2.399$ , i.e. $k_{B} T \approx ε /2.399 \approx 0.417 ε$ .

Recall Peak of the Schottky heat capacity

$k_{B} T_{peak} / ε$ ::: $\approx 0.417$ (where $x = β ε \approx 2.399$ )

Verify: $C_{V} \to 0$ both limits ✓. Peak sits near $k_{B} T \sim ε$ ✓ as forecast.

In the figure the red curve is $C_{V} / k_{B}$ : it starts at zero on the cold left, rises to a single hump, then decays back to zero on the hot right. The yellow dot marks the peak at $k_{B} T \approx 0.417 ε$ . This lone bump — not a step, not a plateau — is the fingerprint of a two-level system in real heat-capacity data.

Ex 6 — $N$ independent atoms: factorization and the $N!$ trap (cell G)

Step 1 — Factorize. Total energy $E = \sum_{k = 1}^{N} ϵ_{k}$ , so $Z_{N} = \sum_{all configs} e^{- β \sum_{k} ϵ_{k}} = \prod_{k = 1}^{N} (\sum_{states of k} e^{- β ϵ_{k}}) = Z_{1}^{N}$ Why this step? $e^{sum} = \prod e^{terms}$ , and a sum over independent labels splits into a product of independent sums. This is why $Z$ is so powerful for large systems.

Step 2 — Total energy. $ln Z_{N} = N ln Z_{1}$ , so $⟨ E ⟩_{N} = - \frac{\partial l n Z _{N}}{\partial β} = N ⟨ E ⟩_{1} = \frac{N ε}{e ^{β ε} + 1}$ Why this step? The logarithm turns the product into a sum, and energy is extensive — it scales linearly with $N$ .

Step 3 — The indistinguishable case. If instead these were free identical gas particles, we divide: $Z_{N} = Z_{1}^{N} / N!$ . Why this step? Swapping two identical particles is not a new microstate; the product overcounts by $N!$ . Dropping this factor breaks the extensivity of entropy — the Gibbs paradox.

Verify: For distinguishable atoms energies simply add, so $⟨ E ⟩_{N} = N ⟨ E ⟩_{1}$ ✓. The $1/ N!$ multiplies $Z$ by a constant (independent of $β$ ), so it changes the free energy and entropy (introduced in Ex 8 and in Helmholtz free energy F) but not $⟨ E ⟩ = - \partial_{β} ln Z$ — a useful sanity check.

Ex 7 — Real-world word problem: paramagnet in a field (cell H)

Step 1 — Choose states and energies from the physics. Two states: aligned with energy $E_{+} = - μ B$ and anti-aligned with energy $E_{-} = + μ B$ . So $Z = e^{- β (- μ B)} + e^{- β (+ μ B)} = e^{+ β μ B} + e^{- β μ B} = 2 cosh (β μ B)$ Why this step? The word problem hands us no formula — we must first translate "aligned/anti-aligned" into concrete energies before any machinery applies. The two exponentials combine into $2 cosh$ by the definition $cosh x = \frac{1}{2} (e^{x} + e^{- x})$ .

Step 2 — Average the observable $m$ . Magnetization is $⟨ m ⟩ = \sum_{i} m_{i} P_{i}$ , with $P_{i} = e^{- β E_{i}} / Z$ : $⟨ m ⟩ = \frac{( + μ ) e ^{+ β μ B} + ( - μ ) e ^{- β μ B}}{2 c o s h ( β μ B )} = μ \frac{e ^{β μ B} - e ^{- β μ B}}{e ^{β μ B} + e ^{- β μ B}} = μ tanh (β μ B)$ Why this step? The Boltzmann distribution weights any observable, not just energy. The aligned state (larger weight, since it has lower energy) carries $+ μ$ ; the ratio of the $sinh$ -like top to the $cosh$ -like bottom is exactly $tanh$ .

Step 3 — Cold limit (cell A flavour). $T \to 0 \Rightarrow β \to \infty \Rightarrow tanh (β μ B) \to 1$ , so $⟨ m ⟩ \to μ$ : every spin locks onto the field. Why this step? At zero temperature there is no energy to borrow, so the system sits in its lowest state — fully aligned.

Step 4 — Hot limit (cell B flavour). $T \to \infty \Rightarrow β \to 0 \Rightarrow tanh (β μ B) \to 0$ , so $⟨ m ⟩ \to 0$ : the two orientations become equally likely and cancel. Why this step? Infinite temperature erases the energy bias; up and down are equiprobable.

Verify: $⟨ m ⟩ = μ tanh (β μ B) > 0$ for $B > 0$ ✓ (aligned with the field, matching the forecast). Reversing the field $B \to - B$ flips the sign of $⟨ m ⟩$ (the function is odd in $B$ ) ✓ — physically the magnetization follows the field direction. Units: $⟨ m ⟩$ carries the moment $μ$ and $tanh$ is dimensionless ✓.

Ex 8 — Exam twist: shifting the energy zero (cell I)

First, the two thermodynamic quantities this example checks:

Step 1 — New partition function. $Z^{'} = \sum_{i} e^{- β (E_{i} + E_{0})} = e^{- β E_{0}} \sum_{i} e^{- β E_{i}} = e^{- β E_{0}} Z$ Why this step? The common factor $e^{- β E_{0}}$ pulls straight out of the sum, so $Z^{'}$ is just $Z$ times a constant weight.

Step 2 — Probabilities are invariant. $P_{i}^{'} = \frac{e ^{- β (E_{i} + E_{0})}}{Z ^{'}} = \frac{e ^{- β E_{0}} e ^{- β E_{i}}}{e ^{- β E_{0}} Z} = \frac{e ^{- β E_{i}}}{Z} = P_{i}$ . Why this step? The shift cancels top and bottom — this is why only energy differences are ever physical.

Step 3 — Energy and free energy shift by $E_{0}$ . Using $ln Z^{'} = ln Z - β E_{0}$ : $⟨ E ⟩^{'} = - \frac{\partial l n Z ^{'}}{\partial β} = - \frac{\partial}{\partial β} (ln Z - β E_{0}) = ⟨ E ⟩ + E_{0}$ $F^{'} = - k_{B} T ln Z^{'} = - k_{B} T (ln Z - β E_{0}) = F + E_{0}$ Why this step? The $β E_{0}$ term differentiates to $E_{0}$ for the energy, and $- k_{B} T \cdot (- β E_{0}) = E_{0}$ for $F$ (since $k_{B} T β = 1$ ). Both shift by the sea-level change, as they must.

Step 4 — Entropy is unchanged. $S^{'} = \frac{⟨ E ⟩ ^{'} - F ^{'}}{T} = \frac{(⟨ E ⟩ + E _{0} ) - ( F + E _{0} )}{T} = \frac{⟨ E ⟩ - F}{T} = S$ . Why this step? $E_{0}$ cancels in the difference $⟨ E ⟩ - F$ , so entropy — a count of accessible states — cannot depend on where we put zero energy.

Step 5 — The $g = 0$ trap. A level with degeneracy $g = 0$ contributes $0 \cdot e^{- β E} = 0$ : it simply doesn't exist and adds nothing to $Z$ . No divergence, no special case — a "missing" level is invisible to the sum. Why this step? Exam questions love to insert a phantom level; the sum handles it automatically.

Verify: Probabilities unchanged ✓ (a constant energy shift is unphysical). $⟨ E ⟩$ and $F$ both shift by $E_{0}$ ✓; entropy $S$ unchanged ✓ — all matching the forecast.