Exercises — Free energy from partition function

WHAT: apply the master bridge $F=-k_{B}T\ln Z$ directly. WHY: nothing to derive — $F$ is defined as $-k_{B}T\ln Z$ once you have $Z$. $F=-k_{B}T\ln 4=-k_{B}T(1.3863)\approx -1.386k_{B}T.$ The sign is negative because $Z>1$ means $\ln Z>0$, and there is a minus out front. Free energy of a system with many accessible states is low (very negative) — nature likes that.

WHAT: sum Boltzmann weights $e^{-\beta E_i}$ over the two states. WHY: the definition $Z={\sum }_i{e}^{-\beta E_i}$ — each state contributes one exponential. $Z=e^{-\beta \cdot 0}+e^{-\beta \epsilon }=1+e^{-\beta \epsilon }.$ At $\beta \epsilon =1$: $Z=1+e^{-1}=1+0.3679=1.3679$. Picture (Figure 1): the "$1$" is the ground level always counted with full weight; $e^{-\beta \epsilon }$ is the shrunk weight of the excited level — shrunk because it costs energy $\epsilon$.

WHAT: take the $\beta$-derivative of $\ln Z$. WHY tool: $U$ is the average energy ${\sum }_i{p}_i{E}_i$; the parent note showed this weighted sum equals $-{\partial }_{\beta }\ln Z$ — the derivative "pulls down" an $E_i$ from each exponent and the division by $Z$ (hidden inside $\ln$) does the averaging automatically. $\ln Z=\ln (1+e^{-\beta \epsilon }),\frac{\partial \ln Z}{\partial \beta }=\frac{-\epsilon e^{-\beta \epsilon }}{1+e^{-\beta \epsilon }}.$ $U=-\frac{\partial \ln Z}{\partial \beta }=\frac{\epsilon e^{-\beta \epsilon }}{1+e^{-\beta \epsilon }}=\frac{\epsilon }{e^{\beta \epsilon }+1}.$ At $\beta \epsilon =1$: $U/\epsilon =1/(e^1+1)=1/3.7183=0.2689$. Sanity across cases: $\beta \to \infty$ (cold) $\Rightarrow U\to 0$ (ground state). $\beta \to 0$ (hot) $\Rightarrow U\to \epsilon /2$ (both states equally likely). Our value $0.269\epsilon$ sits between, as it must.

WHAT: use the assembled relation from the parent derivation, $S=k_B\beta U+k_B\ln Z$. WHY: it lets us reuse $U$ (already found) and $Z$ (already found) — no new sums.

• $\beta U=\beta \cdot \frac{\epsilon }{e^{\beta \epsilon }+1}=\frac{\beta \epsilon }{e^{\beta \epsilon }+1}=\frac{1}{3.7183}=0.2689$.
• $\ln Z=\ln 1.3679=0.3133$. $S/k_B=0.2689+0.3133=\mathrm{0.5822.}$ Cross-check: the maximum possible entropy for a 2-state system is $\ln 2=0.693$ (both states equally likely). We are below it because at finite $T$ the system still slightly favours the ground state. ✓

WHAT: multiply single-system partition functions, then take the log. WHY tool: independent systems have additive energies in the exponent, and adding in an exponent means multiplying exponentials — so $Z_{\text{tot}}=Z_1^N$. See Boltzmann distribution and microstates for why factorisation holds. $Z_{\text{tot}}=Z_1^N=(1+e^{-\beta \epsilon }{)}^N.$ $F=-k_{B}T\ln Z_1^N=-N{k}_{B}T\ln (1+e^{-\beta \epsilon }).$ At $\beta \epsilon =1$: $F/(N{k}_{B}T)=-\ln (1.3679)=-0.3133$. The lesson (extensivity): $F\propto N$. The logarithm turned a product $Z_1^N$ into a sum $N\ln Z_1$. This is precisely why free energy — not $Z$ itself — is the natural thermodynamic potential. See Helmholtz vs Gibbs free energy.

WHAT: examine $\ln (1+e^{-\beta \epsilon })$ in each temperature extreme. WHY: limits reveal whether the formula behaves physically — a formula that misbehaves at $0$ or $\infty$ is suspect.

• $T\to 0$ ($\beta \to \infty$): $e^{-\beta \epsilon }\to 0$, so $\ln (1+0)=0$, giving $F\to 0$. Physically the system is frozen in the ground state (energy $0$, entropy $0$), so $F=U-TS\to 0$. ✓
• $T\to \infty$ ($\beta \to 0$): $e^{-\beta \epsilon }\to 1$, so $\ln 2$, giving $F\to -k_{B}T\ln 2\to -\infty$. The entropy term $-TS$ dominates ($S\to k_B\ln 2$), and $F$ dives to $-\infty$. ✓ Figure 2 plots $F/\epsilon$, $U/\epsilon$, and $TS/\epsilon$ vs $k_{B}T/\epsilon$ so you can see $F=U-TS$ pulling downward as $T$ grows.

WHAT: do the two Gaussian integrals, then differentiate $\ln Z$. WHY the Gaussian tool: the Boltzmann weight $e^{-\beta (\dots )}$ of a quadratic Hamiltonian is a Gaussian in $p$ and in $x$; the standard result ${\int }_{-\infty }^{\infty }{e}^{-a{x}^2}dx=\sqrt{\pi /a}$ answers "what is the total weight over all positions and momenta?" $Z=\frac{1}{h}\int e^{-\beta p^2/2m}dp\int e^{-\beta m{\omega }^2{x}^2/2}dx=\frac{1}{h}\sqrt{\frac{2\pi m}{\beta }}\sqrt{\frac{2\pi }{\beta m{\omega }^2}}=\frac{2\pi }{h\beta \omega }=\frac{1}{\hslash \beta \omega }=\frac{k_{B}T}{\hslash \omega }.$ (Used $h=2\pi \hslash$.) Energy: $\ln Z=-\ln \beta +\text{const}$, so $U=-\frac{\partial \ln Z}{\partial \beta }=-\left(-\frac{1}{\beta }\right)=\frac{1}{\beta }=k_{B}T.$ This is equipartition: two quadratic degrees of freedom ($p$ and $x$), each contributing $\frac{1}{2}{k}_{B}T$. Heat capacity: $C_V=\frac{\partial U}{\partial T}=\frac{\partial (k_{B}T)}{\partial T}=k_B$. Free energy: $F=-k_{B}T\ln (k_{B}T/\hslash \omega )$.

WHAT: use $S=-\partial F/\partial T$. WHY: this is the thermodynamic identity $dF=-SdT-pdV+\mu dN$ read at fixed $V,N$; see Legendre transforms in thermodynamics. Let $x=k_{B}T/\hslash \omega$, so $F=-k_{B}T\ln x$. Then $S=-\frac{\partial F}{\partial T}=k_B\ln \left(\frac{k_{B}T}{\hslash \omega }\right)+k_{B}T\cdot \frac{1}{T}=k_B\ln \left(\frac{k_{B}T}{\hslash \omega }\right)+k_B.$ Check $F=U-TS$: with $U=k_{B}T$, $U-TS=k_{B}T-T[k_B\ln x+k_B]=k_{B}T-k_{B}T\ln x-k_{B}T=-k_{B}T\ln x=F.✓$ The $k_{B}T$ from $U$ cancels the $+k_{B}T$ inside $-TS$, leaving pure $-k_{B}T\ln Z$ — the same "$U$ cancels" magic as the parent derivation.

WHAT: differentiate $U$ with respect to $T$, using $\beta =1/k_{B}T$. WHY: $C=\partial U/\partial T$ measures how much energy the system soaks up per degree — the fingerprint of a two-level ("Schottky") system. Write $U=\epsilon (e^{\beta \epsilon }+1{)}^{-1}$. Chain rule via $\beta$: $\frac{d\beta }{dT}=-\frac{1}{k_B{T}^2}=-k_B{\beta }^2$. $\frac{\partial U}{\partial \beta }=\epsilon \cdot \left(-\frac{\epsilon e^{\beta \epsilon }}{(e^{\beta \epsilon }+1{)}^2}\right)=-\frac{{\epsilon }^2{e}^{\beta \epsilon }}{(e^{\beta \epsilon }+1{)}^2}.$ $C=\frac{\partial U}{\partial T}=\frac{\partial U}{\partial \beta }\frac{d\beta }{dT}=\left(-\frac{{\epsilon }^2{e}^{\beta \epsilon }}{(e^{\beta \epsilon }+1{)}^2}\right)(-k_B{\beta }^2)=k_B(\beta \epsilon {)}^2\frac{e^{\beta \epsilon }}{(e^{\beta \epsilon }+1{)}^2}.$ At $\beta \epsilon =2$: $e^2=7.389$, denominator $(8.389{)}^2=70.38$. $C/k_B=4\cdot \frac{7.389}{70.38}=4\cdot 0.10498=\mathrm{0.4199.}$ Physics: $C\to 0$ at both $T\to 0$ (frozen) and $T\to \infty$ (no room to absorb more), peaking in between — the Schottky anomaly. See Entropy — Gibbs and Boltzmann definitions.

WHAT: weight each state by $e^{-\beta (E-\mu N)}$ instead of $e^{-\beta E}$. WHY the new tool: now particle number can change, so we let the reservoir "pay" $\mu$ per particle. This is the grand canonical recipe. Step 1 — write $\mathcal{Z}$. The two states are empty $(N=0,E=0)$ and occupied $(N=1,E=\epsilon )$: $\mathcal{Z}=e^{-\beta (0-\mu \cdot 0)}+e^{-\beta (\epsilon -\mu \cdot 1)}=1+e^{-\beta (\epsilon -\mu )}.$ Step 2 — grand potential. Apply the definition directly: $\Phi =-k_{B}T\ln (1+e^{-\beta (\epsilon -\mu )}).$ Step 3 — average occupation. The trick $⟨n⟩=\frac{1}{\beta }\frac{\partial \ln \mathcal{Z}}{\partial \mu }$ works because differentiating with respect to $\mu$ pulls down a factor $\beta N$ from each exponent, exactly building the number-weighted average: $⟨n⟩=\frac{1}{\beta }\frac{\partial \ln \mathcal{Z}}{\partial \mu }=\frac{e^{-\beta (\epsilon -\mu )}}{1+e^{-\beta (\epsilon -\mu )}}=\frac{1}{e^{\beta (\epsilon -\mu )}+1}.$ Interpretation. This is the Fermi–Dirac distribution — the same algebraic shape as the two-level energy, but now it counts particles. Setting $\mu =\epsilon$ (at any $T$) gives $⟨n⟩=\frac{1}{e^0+1}=\frac{1}{2}$: the chemical potential is exactly the energy at which a level is half-filled.

WHAT: use $p=-\partial F/\partial V$ at fixed $T,N$; see Ideal gas from the partition function. WHY: only $Z_1$'s $V$-dependence matters for pressure, and $Z_1\propto V$, so $\ln Z_1=\ln V+(\text{V-independent})$. $F=-N{k}_{B}T\ln Z_1=-N{k}_{B}T(\ln V-3\ln \lambda ).$ $p=-\frac{\partial F}{\partial V}=N{k}_{B}T\frac{\partial \ln V}{\partial V}=N{k}_{B}T\cdot \frac{1}{V}=\frac{N{k}_{B}T}{V}.$ Multiplying both sides by the volume $V$ gives the ideal-gas law: $pV=N{k}_{B}T.✓$ The whole ideal-gas law is just the derivative of $\ln V$. (The $1/N!$ Gibbs correction shifts $F$ by a $V$-independent amount, so it leaves $p$ untouched — it fixes entropy extensivity, not pressure.)

L1 plug into $F=-k_{B}T\ln Z$ ::: L2 differentiate $\ln Z$ for $U,S$ ::: L3 multiply $Z$'s, log makes $F$ extensive ::: L4 Gaussian integrals + equipartition for the oscillator ::: L5 heat-capacity peak, grand ensemble, ideal-gas law — all from one $\ln Z$.