Average energy from partition function: Exercises

Prerequisites you may want open: Partition function (canonical ensemble), Boltzmann distribution, Helmholtz free energy F = -kT ln Z, Equipartition theorem, Heat capacity and energy fluctuations, Quantum harmonic oscillator — thermal, Two-level system / Schottky anomaly.

Level 1 — Recognition

These check that you can read a $Z$ and reach for the right operation — no long algebra.

Recall Solution L1.1

What: take the logarithm of $Z$ , differentiate with respect to $β$ , flip the sign: $⟨ E ⟩ = - \frac{\partial l n Z}{\partial β} .$ Why each piece:

$ln$ folds the "divide by $Z$ " into the derivative, because $\frac{1}{Z} \partial_{β} Z = \partial_{β} ln Z$ .
$\partial_{β}$ acting on $e^{- β E_{i}}$ manufactures a factor $- E_{i}$ — exactly the energy we want to average.
the minus sign cancels that $- E_{i}$ , leaving a positive energy when all $E_{i} \geq 0$ . Mnemonic: "Log it, slope it, flip it."

Recall Solution L1.2

What: $\frac{\partial ln Z}{\partial β} = - \frac{1}{2 β}$ (the constant $C$ dies under the derivative). So: $⟨ E ⟩ = - \frac{\partial l n Z}{\partial β} = + \frac{1}{2 β} = \frac{1}{2} k_{B} T .$ What it means: any $Z \propto β^{- 1/2}$ is one quadratic degree of freedom — the equipartition result $\frac{1}{2} k_{B} T$ . You never needed the details of the system.

Level 2 — Application

Now grind a real $Z$ through the machine.

Recall Solution L2.1

Setup: $Z = 1 + e^{- β ε}$ , and the standard result is $⟨ E ⟩ = \frac{ε}{e ^{β ε} + 1} .$ Plug $β ε = 2$ : $\frac{⟨ E ⟩}{ε} = \frac{1}{e ^{2} + 1} = \frac{1}{7.389 + 1} = \frac{1}{8.389} = 0.1192.$ So $⟨ E ⟩ \approx 0.119 ε$ . The system sits mostly in the ground state (energy $0$ ) because $ε$ is a bit bigger than the thermal scale $k_{B} T$ .

Recall Solution L2.2

Setup: $Z = \frac{1}{1 - e ^{- β ℏ ω}}$ gives (see Quantum harmonic oscillator — thermal) $⟨ E ⟩ = \frac{ℏ ω}{e ^{β ℏ ω} - 1} .$ Plug $β ℏ ω = 1$ : $\frac{⟨ E ⟩}{ℏ ω} = \frac{1}{e ^{1} - 1} = \frac{1}{1.71828} = 0.5820.$ So $⟨ E ⟩ \approx 0.582 ℏ ω$ . (This uses the convention $E_{0} = 0$ ; with the true ground state $\frac{1}{2} ℏ ω$ you would add $0.5$ .)

Recall Solution L2.3

What: three microstates, so $Z = 1 + e^{- β ε} + e^{- 2 β ε} .$ Differentiate: $\partial_{β} Z = - ε e^{- β ε} - 2 ε e^{- 2 β ε}$ , so

=\frac{\varepsilon e^{-\beta\varepsilon}+2\varepsilon e^{-2\beta\varepsilon}} {1+e^{-\beta\varepsilon}+e^{-2\beta\varepsilon}}.$$ **At $\beta\varepsilon=1$:** with $e^{-1}=0.36788$, $e^{-2}=0.13534$: $$\frac{\langle E\rangle}{\varepsilon}=\frac{0.36788+2(0.13534)}{1+0.36788+0.13534} =\frac{0.63856}{1.50322}=0.4248.$$ So $\langle E\rangle\approx 0.425\,\varepsilon$.

Level 3 — Analysis

Here you reason about limits, signs and shapes, not just numbers.

Recall Solution L3.1

Cold, $T \to 0 \Rightarrow β \to \infty$ : $e^{β ε} \to \infty$ , so $⟨ E ⟩ \to ε /\infty = 0$ . The system is frozen in the ground state — no thermal energy available to climb to $ε$ . Hot, $T \to \infty \Rightarrow β \to 0$ : $e^{β ε} \to e^{0} = 1$ , so $⟨ E ⟩ \to \frac{ε}{1 + 1} = \frac{ε}{2} .$ Both states are now equally likely, and the mean of ${0, ε}$ is $ε /2$ . The full curve rises monotonically from $0$ to $ε /2$ — see the figure.

Figure — Average energy from partition function

Recall Solution L3.2

$C_{V}$ measures how fast $⟨ E ⟩$ changes when you warm the system.

Cold ( $T \to 0$ ): everything sits in the ground state; a tiny warming can't lift anything over the gap $ε$ , so $⟨ E ⟩$ barely moves ⟹ $C_{V} \to 0$ .
Hot ( $T \to \infty$ ): $⟨ E ⟩$ has already saturated at $ε /2$ ; warming further does nothing ⟹ $C_{V} \to 0$ .
In between ( $k_{B} T \sim ε$ ): thermal energy is just enough to start promoting spins across the gap, so $⟨ E ⟩$ changes fastest — a maximum in $C_{V}$ . A function that is $0$ at both ends and positive in the middle must have a peak. This bump is the Schottky anomaly (see Two-level system / Schottky anomaly).

Recall Solution L3.3

What tool and why: when $ℏ ω ≪ k_{B} T$ , the quantity $x \equiv β ℏ ω$ is small, so we expand the exponential with its Taylor series $e^{x} = 1 + x + \frac{x ^{2}}{2} + \dots$ — the right tool because it turns a hard function into a leading power we can read off. Do it: $e^{x} - 1 = x + \frac{x ^{2}}{2} + \dots \approx x$ for small $x$ . Hence $⟨ E ⟩ = \frac{ℏ ω}{e ^{β ℏ ω} - 1} \approx \frac{ℏ ω}{β ℏ ω} = \frac{1}{β} = k_{B} T .$ What it means: the discrete quantum ladder blurs into a continuum and you recover the classical oscillator's $k_{B} T$ — that's two quadratic modes (kinetic + potential), each contributing $\frac{1}{2} k_{B} T$ , exactly equipartition.

Level 4 — Synthesis

Combine the average-energy machine with fluctuations, free energy and multiple modes.

Recall Solution L4.1

Direct route (cross-check): with probabilities $p_{0} = \frac{1}{1 + e ^{- β ε}}$ and $p_{1} = \frac{e ^{- β ε}}{1 + e ^{- β ε}}$ , and energies $0, ε$ : $⟨ E ⟩ = ε p_{1}, ⟨ E^{2} ⟩ = ε^{2} p_{1},$ $σ_{E}^{2} = ε^{2} p_{1} - ε^{2} p_{1}^{2} = ε^{2} p_{1} (1 - p_{1}) = ε^{2} p_{0} p_{1} .$ Derivative route: $ln Z = ln (1 + e^{- β ε})$ gives the same $σ_{E}^{2} = ε^{2} \frac{e ^{- β ε}}{( 1 + e ^{- β ε} ) ^{2}} = ε^{2} p_{0} p_{1}$ . They agree — as the fluctuation identity guarantees. At $β ε = 1$ : $p_{1} = \frac{e ^{- 1}}{1 + e ^{- 1}} = \frac{0.36788}{1.36788} = 0.26894$ , $p_{0} = 0.73106$ , so $\frac{σ _{E}^{2}}{ε ^{2}} = p_{0} p_{1} = 0.73106 \times 0.26894 = 0.19661.$

Recall Solution L4.2

What: $C_{V} = k_{B} β^{2} σ_{E}^{2}$ . With $σ_{E}^{2} = ε^{2} p_{0} p_{1}$ and $β ε = 1$ (so $β^{2} ε^{2} = 1$ ): $\frac{C _{V}}{k _{B}} = β^{2} ε^{2} p_{0} p_{1} = 1 \times 0.19661 = 0.19661.$ So $C_{V} \approx 0.197 k_{B}$ per spin at this temperature — on the rising flank of the Schottky bump of L3.2.

Recall Solution L4.3

Key fact: $ln$ of a product is a sum: $ln Z = ln Z_{spin} + ln Z_{mode}$ . Apply the machine: the derivative is linear, so

=\langle E\rangle_{\text{spin}}+\langle E\rangle_{\text{mode}}.$$ **Result:** $\langle E\rangle=\dfrac{\varepsilon}{e^{\beta\varepsilon}+1}+\dfrac{1}{2}k_BT$. Independent subsystems ⟹ multiply $Z$ ⟹ add energies. This is *why* the log form is so powerful: factorization becomes addition.

Level 5 — Mastery

Open-ended reasoning and a degenerate/limiting case.

Recall Solution L5.1

What: each of the $g$ excited microstates carries the same weight $e^{- β ε}$ , so $Z = 1 + g e^{- β ε} .$ Differentiate: $\partial_{β} Z = - g ε e^{- β ε}$ , hence

=\frac{\varepsilon}{\tfrac1g e^{\beta\varepsilon}+1}.$$ **High-$T$ limit ($\beta\to0$):** $e^{\beta\varepsilon}\to1$, so $$\langle E\rangle\to\frac{g\varepsilon}{1+g}.$$ **Interpretation:** at infinite $T$ all $1+g$ microstates are equally likely; the fraction in the excited band is $g/(1+g)$, and each of those carries energy $\varepsilon$. Setting $g=1$ recovers the ordinary two-level answer $\varepsilon/2$. Degeneracy tilts the high-$T$ average toward $\varepsilon$.

Recall Solution L5.2

The reasoning: $⟨ E ⟩ = \sum_{i} p_{i} E_{i}$ is a weighted average with weights $p_{i} \geq 0$ summing to $1$ . A weighted average of numbers can never be smaller than the smallest of them nor larger than the largest: $E_{m i n} = \sum_{i} p_{i} E_{m i n} \leq \sum_{i} p_{i} E_{i} \leq \sum_{i} p_{i} E_{m a x} = E_{m a x} .$ Consequences: as $T \to 0$ all weight moves to $E_{m i n}$ so $⟨ E ⟩ \to E_{m i n}$ ; as $T \to \infty$ all $p_{i}$ become equal so $⟨ E ⟩ \to$ the plain arithmetic mean of the levels (weighted by degeneracy). The derivative machine must always land inside this window — a useful sanity check on any answer.

Recall Master checklist

Take $ln Z$ ::: turns division by $Z$ into a clean derivative and factorization into addition. Slope in $β$ ::: $\partial_{β}$ pulls a factor $- E_{i}$ out of every weight $e^{- β E_{i}}$ . Flip the sign ::: cancels the $- E_{i}$ , giving positive energy for $E_{i} \geq 0$ . Second slope ::: $\partial_{β}^{2} ln Z = σ_{E}^{2}$ , and $C_{V} = k_{B} β^{2} σ_{E}^{2}$ . Always sanity-check ::: $E_{m i n} \leq ⟨ E ⟩ \leq E_{m a x}$ for finite spectra.