Worked examples — Average energy from partition function

This page is the drill floor for the parent formula $⟨E⟩=-\frac{\partial \ln Z}{\partial \beta },\beta \equiv \frac{1}{k_{B}T}.$ Here $k_B$ is Boltzmann's constant (just a number that converts temperature into energy), and $\beta$ is "coldness" — big $\beta$ means cold, small $\beta$ means hot. Note the definition $\beta \equiv 1/(k_{B}T)$ carefully: it means $1/\beta =k_{B}T$, a fact we lean on whenever a limit produces $1/\beta$. Every symbol used below was earned in the parent note; if you need the derivation, go back there first. Here we only use the tool, across every situation it can face.

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The scenario matrix

Before working anything, let us list every kind of problem this one formula can meet. The formula never changes — but $Z$ (the sum of Boltzmann weights) looks different depending on the situation, and the limits we sanity-check depend on the situation too.

Cell	Case class	What makes it special	Example
A	Finite discrete sum	$Z$ = a handful of terms	Ex. 1 (two-level), Ex. 2 (three-level)
B	Degenerate levels	one energy hit by several states → multiply by count	Ex. 3
C	Infinite discrete sum	need a geometric/known series	Ex. 4 (oscillator)
D	Continuous integral	$Z$ = an integral, use a Gaussian	Ex. 5 (equipartition)
E	Cold limit $T\to 0$ ($\beta \to \infty$)	system freezes into ground state	checks in Ex. 1, 4
F	Hot limit $T\to \infty$ ($\beta \to 0$)	all states equally likely	checks in Ex. 1, 3
G	Shifted zero of energy	add a constant to every $E_i$	Ex. 6
H	Real-world word problem	translate physics → energies → $Z$	Ex. 7
I	Exam twist (given $Z$, find $C_V$)	second derivative / fluctuations	Ex. 8

The rule of the game: six through eight worked examples that together touch every cell. Each one tells you which cell(s) it lands in.

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Example 1 — Two-level system · cells A, E, F

Step 1 — Build $Z$. There are two microstates, so the sum has two terms: $Z=e^{-\beta \cdot 0}+e^{-\beta \epsilon }=1+e^{-\beta \epsilon }.$ Why this step? $Z={\sum }_i{e}^{-\beta E_i}$ is a sum over states, and here "all states" means exactly two.

Step 2 — Take the log and differentiate.

= -\frac{-\varepsilon e^{-\beta\varepsilon}}{1+e^{-\beta\varepsilon}} = \frac{\varepsilon\,e^{-\beta\varepsilon}}{1+e^{-\beta\varepsilon}}.$$ *Why this step?* $\partial_\beta \ln Z = \frac1Z\partial_\beta Z$, and $\partial_\beta e^{-\beta\varepsilon}=-\varepsilon e^{-\beta\varepsilon}$ — the minus in the exponent supplies the $-$ that cancels the formula's $-$, leaving a positive answer (energies are $\ge0$, so this *must* be positive). **Step 3 — Tidy.** Multiply top and bottom by $e^{\beta\varepsilon}$: $$\boxed{\;\langle E\rangle = \frac{\varepsilon}{e^{\beta\varepsilon}+1}\;}$$ *Why this step?* Multiplying by $e^{\beta\varepsilon}/e^{\beta\varepsilon}=1$ (no change to the value) clears the small exponential from the numerator, giving the tidy "Fermi-like" form $\varepsilon/(e^{\beta\varepsilon}+1)$ whose limits are easier to read by eye. > [!verify] Verify — the two forecasts > - **Cold, $\beta\to\infty$:** $e^{\beta\varepsilon}\to\infty$, so $\langle E\rangle\to 0$. ✓ (frozen in ground state) > - **Hot, $\beta\to 0$:** $e^{\beta\varepsilon}\to 1$, so $\langle E\rangle\to \varepsilon/2$. ✓ (average of the two states) > [!intuition] Reading the figure below > The figure plots $\langle E\rangle/\varepsilon$ (vertical axis, running $0$ to $\varepsilon/2$) against > temperature $k_BT/\varepsilon$ (horizontal axis, cold on the left, hot on the right). The **mint curve** > is our result $1/(e^{\varepsilon/k_BT}+1)$. Notice its two ends: on the far left (cold) the lavender > arrow marks it pinned near $0$ — the spin is frozen in the ground state; on the far right (hot) the > coral dashed line marks the ceiling $\varepsilon/2$ that the curve flattens toward — both states equally > likely. The whole graph is the two forecasts made visible: a smooth climb from $0$ to $\varepsilon/2$. ![[deepdives/dd-physics-2.4.11-d3-s01.png]] --- ## Example 2 — Three-level ladder · cell A > [!example] Statement > Energies $0,\ \varepsilon,\ 2\varepsilon$ (one state each). Find $\langle E\rangle$ at the special temperature > where $\beta\varepsilon = \ln 2$ (i.e. $e^{-\beta\varepsilon}=\tfrac12$). > [!intuition] Forecast > With weights $1,\ \tfrac12,\ \tfrac14$ the top level is least likely, so the average should sit **below** the middle energy $\varepsilon$. Expect something like $0.5\varepsilon$–$0.7\varepsilon$. **Step 1 — Build $Z$.** Three terms, and set $x=e^{-\beta\varepsilon}=\tfrac12$: $$Z = 1 + x + x^2 = 1 + \tfrac12 + \tfrac14 = \tfrac74.$$ *Why this step?* Each level of energy $n\varepsilon$ contributes weight $e^{-\beta n\varepsilon}=x^n$. **Step 2 — Average directly (cross-check of the derivative).** Since the states are $0,\varepsilon,2\varepsilon$ with weights $1,\tfrac12,\tfrac14$: $$\langle E\rangle = \frac{0\cdot 1 + \varepsilon\cdot\tfrac12 + 2\varepsilon\cdot\tfrac14}{7/4} = \frac{\varepsilon}{7/4} = \frac{4\varepsilon}{7}.$$ *Why this step?* $\langle E\rangle=\sum_i p_i E_i$ is always available; here it is quick and confirms the $Z$-method. > [!verify] Verify > $4/7 \approx 0.571$, which is below $\varepsilon$ — matches the forecast. Also below $\varepsilon$ makes sense: the cheap ground state pulls the mean down. --- ## Example 3 — Degenerate excited level + hot limit · cells B, F > [!example] Statement > Ground energy $0$ (one state) and an excited energy $\varepsilon$ that is **three-fold degenerate** > (three different states all share energy $\varepsilon$). Find $\langle E\rangle$ and its hot-limit value. > [!intuition] Forecast > Three doors lead to the excited level but only one to the ground level. So even the hot limit will > **not** split $50$–$50$; the excited level is favoured $3:1$, dragging the hot average toward $\tfrac34\varepsilon$. **Step 1 — Build $Z$ with a multiplicity.** $$Z = 1 + 3e^{-\beta\varepsilon}.$$ *Why this step?* Degeneracy $g$ means $g$ identical terms, so we just multiply that Boltzmann weight by $3$. **Step 2 — Differentiate.** $$\langle E\rangle = -\frac{\partial}{\partial\beta}\ln\!\big(1+3e^{-\beta\varepsilon}\big) = \frac{3\varepsilon\,e^{-\beta\varepsilon}}{1+3e^{-\beta\varepsilon}} = \frac{3\varepsilon}{e^{\beta\varepsilon}+3}.$$ *Why this step?* Same log-slope-flip; the $3$ rides along untouched by the derivative. > [!verify] Verify — hot limit > $\beta\to 0$: $e^{\beta\varepsilon}\to1$, so $\langle E\rangle\to \dfrac{3\varepsilon}{1+3}=\dfrac34\varepsilon$. ✓ matches the $3:1$ forecast (three excited doors vs one ground door). > Cold limit $\beta\to\infty$: $\langle E\rangle\to 0$ ✓ (ground state still wins when it is the only cheap option). See the [[Two-level system / Schottky anomaly]] for how degeneracy reshapes the heat-capacity bump. --- ## Example 4 — Quantum oscillator: infinite sum + cold freeze · cells C, E > [!example] Statement > A 1D quantum oscillator has levels $E_n = n\hbar\omega$, $n=0,1,2,\dots$ (infinitely many). > Find $\langle E\rangle$ and its behaviour as $T\to 0$. > [!intuition] Forecast > Infinitely many levels, but the high ones are exponentially unlikely. As $T\to0$ everything should > collapse to the ground level $n=0$, giving $\langle E\rangle\to 0$. **Step 1 — Sum the geometric series.** With $r=e^{-\beta\hbar\omega}$ and $|r|<1$: $$Z = \sum_{n=0}^{\infty} r^n = \frac{1}{1-r} = \frac{1}{1-e^{-\beta\hbar\omega}}.$$ *Why this step?* This is the only place we need a **known infinite sum**: $\sum r^n = 1/(1-r)$. It converges because $r<1$ (positive energies, positive $\beta$). **Step 2 — Log and differentiate.** $\ln Z = -\ln(1-e^{-\beta\hbar\omega})$, so $$\langle E\rangle = -\frac{\partial}{\partial\beta}\Big[-\ln(1-e^{-\beta\hbar\omega})\Big] = \frac{\hbar\omega\, e^{-\beta\hbar\omega}}{1-e^{-\beta\hbar\omega}} = \frac{\hbar\omega}{e^{\beta\hbar\omega}-1}.$$ *Why this step?* Chain rule on the log delivers the numerator $\hbar\omega e^{-\beta\hbar\omega}$. > [!verify] Verify > - **Cold, $\beta\to\infty$:** denominator $e^{\beta\hbar\omega}-1\to\infty$, so $\langle E\rangle\to 0$. ✓ > - **Hot, $\beta\to 0$:** $e^{\beta\hbar\omega}-1\approx\beta\hbar\omega$, so $\langle E\rangle\to \hbar\omega/(\beta\hbar\omega)=1/\beta=k_BT$. ✓ (classical [[Equipartition theorem]] value for an oscillator; deeper treatment in [[Quantum harmonic oscillator — thermal]]). --- ## Example 5 — Continuous quadratic mode (equipartition) · cell D > [!example] Statement > A degree of freedom with energy $E=\tfrac12 a x^2$, $x\in(-\infty,\infty)$ (a "spring" coordinate). > Find $\langle E\rangle$. > [!intuition] Forecast > This is the classic case that should give exactly $\tfrac12 k_BT$ per quadratic term — the equipartition result. **Step 1 — $Z$ is now an integral.** $$Z = \int_{-\infty}^{\infty} e^{-\beta a x^2/2}\,dx.$$ *Why this step?* Continuous states → the sum over microstates becomes an integral over the coordinate $x$. **Step 2 — Do the Gaussian integral by substitution.** Start from the master fact $\int_{-\infty}^{\infty} e^{-u^2}\,du=\sqrt{\pi}$. Our exponent is $-\tfrac12\beta a\,x^2$, so let $u = \sqrt{\tfrac{\beta a}{2}}\,x$, which gives $du = \sqrt{\tfrac{\beta a}{2}}\,dx$, i.e. $dx = \sqrt{\tfrac{2}{\beta a}}\,du$. Then $$Z = \int_{-\infty}^{\infty} e^{-u^2}\,\sqrt{\tfrac{2}{\beta a}}\,du = \sqrt{\tfrac{2}{\beta a}}\int_{-\infty}^{\infty} e^{-u^2}\,du = \sqrt{\tfrac{2}{\beta a}}\,\sqrt{\pi} = \sqrt{\frac{2\pi}{\beta a}}\ \propto\ \beta^{-1/2}.$$ *Why this step?* The substitution turns our integral into the standard one; the prefactor $\sqrt{2/(\beta a)}$ is exactly the $dx/du$ bookkeeping, which is why the answer is $\sqrt{2\pi/(\beta a)}$ and not just $\sqrt{\pi}$. **Step 3 — Only the $\beta$-dependence matters.** $\ln Z = -\tfrac12\ln\beta + \text{const}$, so $$\langle E\rangle = -\frac{\partial \ln Z}{\partial\beta} = -\left(-\frac{1}{2\beta}\right)=\frac{1}{2\beta}=\tfrac12 k_BT.$$ *Why this step?* Any $\beta$-independent constant (the $\ln(2\pi/a)/2$ part) vanishes under $\partial_\beta$ — we only ever needed the power of $\beta$. The last equality uses the definition $\beta\equiv 1/(k_BT)$, so $1/\beta = k_BT$ exactly (restated from the top of this page), turning $1/(2\beta)$ into $\tfrac12 k_BT$. > [!verify] Verify > At $T$ where $k_BT=4$ (in some energy unit), $\langle E\rangle = 2$. Units: $[k_BT]=$ energy ✓. Each quadratic term = $\tfrac12 k_BT$ — the [[Equipartition theorem]] in two lines. --- ## Example 6 — Shift the zero of energy · cell G > [!example] Statement > Take Example 1's two-level system but measure energies from a shifted origin: states have energies > $E_0$ and $E_0+\varepsilon$ (both raised by a constant $E_0$). Show $\langle E\rangle$ = (old answer) $+\,E_0$. > [!intuition] Forecast > Sliding every energy up by $E_0$ should slide the average up by exactly $E_0$ — nothing physical changed, only the ruler's zero. **Step 1 — New partition function.** $$Z' = e^{-\beta E_0} + e^{-\beta(E_0+\varepsilon)} = e^{-\beta E_0}\big(1 + e^{-\beta\varepsilon}\big) = e^{-\beta E_0} Z,$$ where $Z=1+e^{-\beta\varepsilon}$ is the old one. *Why this step?* A common factor $e^{-\beta E_0}$ pulls out of every term. **Step 2 — Logs turn products into sums.** $$\ln Z' = -\beta E_0 + \ln Z.$$ *Why this step?* This is exactly why the $\ln$ form is so powerful — the constant shift becomes an additive $-\beta E_0$ term. **Step 3 — Differentiate.** $$\langle E\rangle' = -\frac{\partial}{\partial\beta}\big(-\beta E_0 + \ln Z\big) = E_0 - \frac{\partial \ln Z}{\partial\beta} = E_0 + \frac{\varepsilon}{e^{\beta\varepsilon}+1}.$$ *Why this step?* $\partial_\beta(-\beta E_0)=-E_0$, and the formula's leading minus flips it to $+E_0$. > [!verify] Verify > With $E_0=5$, $\varepsilon$ such that $e^{\beta\varepsilon}=1$ (hot), old answer $=\varepsilon/2$; new $=5+\varepsilon/2$. The shift is exactly $E_0$. ✓ Also note $C_V$ is **unchanged** by a constant shift (the second derivative kills $-\beta E_0$). --- ## Example 7 — Word problem: two conformations of a molecule · cell H > [!example] Statement > A protein hinge has a **folded** state (energy $0$) and an **open** state (energy $\varepsilon = 2\,k_B T_{\text{room}}$, > i.e. $\beta\varepsilon=2$ at room temperature). What fraction of energy $\langle E\rangle/\varepsilon$ is stored at room $T$? > [!intuition] Forecast > The open state costs $2k_BT$ — noticeably more than thermal energy — so it should be fairly rare. Expect $\langle E\rangle/\varepsilon$ to be small, maybe around $0.1$. **Step 1 — Translate to energies.** Two states, $0$ and $\varepsilon$ — same shape as Example 1. *Why this step?* Word problems reduce to "list the states and their energies." **Step 2 — Use the Example 1 result** with $\beta\varepsilon = 2$: $$\frac{\langle E\rangle}{\varepsilon} = \frac{1}{e^{\beta\varepsilon}+1} = \frac{1}{e^{2}+1}.$$ *Why this step?* We already derived $\langle E\rangle=\varepsilon/(e^{\beta\varepsilon}+1)$; just plug in the number. **Step 3 — Evaluate.** $e^2\approx 7.389$, so $\dfrac{1}{7.389+1}=\dfrac{1}{8.389}\approx 0.119$. *Why this step?* We turn the symbolic fraction into a number so we can compare it to our forecast and to a measurable probability — a formula only becomes a *prediction* once it is a number. We need $e^2$ because $\beta\varepsilon=2$ sits in the exponent, and $e^2$ is just the fixed constant $\approx 7.389$. > [!verify] Verify > $\langle E\rangle/\varepsilon\approx 0.119$ — about $12\%$, small as forecast. Cross-check against the open-state Boltzmann probability: $p_{\text{open}}=\dfrac{e^{-\beta\varepsilon}}{1+e^{-\beta\varepsilon}}=\dfrac{e^{-2}}{1+e^{-2}}\approx 0.119$. These match because **only the open state carries energy** ($E_{\text{folded}}=0$), so $\langle E\rangle=\varepsilon\,p_{\text{open}}$, hence $\langle E\rangle/\varepsilon=p_{\text{open}}$. ✓ The prediction–verification cycle closes cleanly: the energy fraction *is* the open-state occupation probability. --- ## Example 8 — Exam twist: given $Z$, find the heat capacity · cell I > [!example] Statement > For the two-level system ($0,\varepsilon$), compute the heat capacity $C_V = \partial\langle E\rangle/\partial T$ > and find the temperature (in terms of $\varepsilon/k_B$) where it peaks — the **Schottky** peak. > [!intuition] Forecast > Heat capacity measures how much the average energy responds to a nudge in $T$. It must vanish at both > extremes (frozen at cold, saturated at hot) and therefore **peak somewhere in the middle**, at a > temperature of order $\varepsilon/k_B$. > [!definition] What is $\sigma_E^2$? > $\sigma_E^2$ is the **energy variance**: how much the energy jitters around its average. In symbols > $\sigma_E^2 \equiv \langle E^2\rangle - \langle E\rangle^2$, where $\langle E^2\rangle=\sum_i p_i E_i^2$ > is the average of the *squared* energy. It is always $\ge 0$ (a spread cannot be negative). > The parent note showed that the **second** $\beta$-derivative of $\ln Z$ manufactures exactly this > spread: differentiating twice pulls out an $E_i^2$ and an $(\sum E_i)^2$ piece, giving > $$\partial^2_\beta \ln Z = \langle E^2\rangle - \langle E\rangle^2 = \sigma_E^2.$$ > Feeding this into $C_V=\partial\langle E\rangle/\partial T$ and using $\beta=1/k_BT$ (so $\partial_T = -k_B\beta^2\partial_\beta$) gives the fluctuation identity > $$C_V = k_B\beta^2\,\sigma_E^2.$$ **Step 1 — Get $C_V$.** From Example 1, $\langle E\rangle=\varepsilon/(e^{\beta\varepsilon}+1)$. Differentiating in $T$ (or equivalently using $C_V=k_B\beta^2\sigma_E^2$ from the box) gives $$C_V = k_B\,(\beta\varepsilon)^2\,\frac{e^{\beta\varepsilon}}{(e^{\beta\varepsilon}+1)^2}.$$ *Why this step?* $\langle E\rangle$ depends on $T$ only through the combination $\beta\varepsilon$; the chain rule brings down the $(\beta\varepsilon)^2$ prefactor (this is the [[Heat capacity and energy fluctuations]] identity made concrete). **Step 2 — Find the peak.** Let $u=\beta\varepsilon$, so $C_V(u)=k_B\,u^2 e^{u}/(e^{u}+1)^2$. To locate the maximum we set $dC_V/du=0$. Carrying out that derivative (product + quotient rule) and dividing off the common positive factor $u\,e^{u}/(e^{u}+1)^2$ leaves the tidy condition $$u = 2\,\frac{e^{u}-1}{e^{u}+1} = 2\tanh(u/2).$$ *Why this step?* A peak is where the slope is zero, so we solve $dC_V/du=0$. The messy derivative factorises: the $\tfrac{e^u-1}{e^u+1}$ combination is exactly $\tanh(u/2)$ (definition of $\tanh$), which is why the condition collapses to $u=2\tanh(u/2)$. Solving numerically: $u^\ast \approx 2.399$, i.e. $k_BT^\ast \approx \varepsilon/2.399 \approx 0.417\,\varepsilon$. > [!verify] Verify > Peak at $k_BT^\ast\approx 0.417\,\varepsilon$ — of order $\varepsilon/k_B$ and firmly between the two zero limits, as forecast. Plugging $u^\ast=2.399$ back gives $C_V^\ast\approx 0.439\,k_B$. Both cold ($u\to\infty$) and hot ($u\to0$) limits give $C_V\to0$ ✓ — the hallmark [[Two-level system / Schottky anomaly]] bump. ✓ --- > [!recall]- Which cell did each example fill? > Ex.1 ::: A, E, F (finite discrete + both limits) > Ex.2 ::: A (three-level finite sum) > Ex.3 ::: B, F (degeneracy + hot limit) > Ex.4 ::: C, E (infinite geometric sum + cold freeze) > Ex.5 ::: D (continuous Gaussian integral, equipartition) > Ex.6 ::: G (shifted energy zero) > Ex.7 ::: H (real-world word problem) > Ex.8 ::: I (given $Z$ → heat capacity, Schottky peak) > [!mnemonic] The universal recipe > **"List → weight → sum → log → slope → flip."** List states, weight by $e^{-\beta E}$, sum into $Z$, > take $\ln$, take the $\beta$-slope, flip the sign: same six moves for every cell above. ## Active recall > [!recall] Self-test > quick-1 ::: Why does a three-fold degeneracy push the hot-limit energy above $\varepsilon/2$? Because the excited level has three doors vs one for the ground, so the hot average tends to $\tfrac34\varepsilon$. > quick-2 ::: Why does shifting all energies by $E_0$ leave $C_V$ unchanged? The shift adds $-\beta E_0$ to $\ln Z$, whose second $\beta$-derivative is zero. > quick-3 ::: In the oscillator, why does the hot limit give $k_BT$ not $\tfrac12 k_BT$? A vibrational mode has *two* quadratic terms (kinetic + potential), each $\tfrac12 k_BT$.