Free energy from partition function: Question bank

Before we start, a one-line reminder of the symbols so nothing is used unexplained:

Definition Symbol refresher

$k_{B}$ — Boltzmann's constant, the fixed conversion factor ( $\approx 1.38 \times 1 0^{- 23} J/K$ ) that turns temperature into an energy per microstate.
$T$ — the absolute temperature (in kelvin) of the heat bath the system sits in; higher $T$ means the bath pushes more energy into the system.
$V$ — the volume the system occupies (held fixed in the canonical setup).
$N$ — the number of particles in the system (also held fixed in the canonical setup).
$Z = \sum_{i} e^{- β E_{i}}$ — the partition function, a single number that sums the Boltzmann weight $e^{- β E_{i}}$ of every microstate $i$ .
$β = 1/ (k_{B} T)$ — "inverse temperature"; large $β$ = cold, small $β$ = hot.
$F = - k_{B} T ln Z$ — the Helmholtz free energy (natural variables $T, V, N$ ).
$U = - \frac{\partial ( ln Z )}{\partial β}$ — internal (average) energy, i.e. the derivative of $ln Z$ with respect to $β$ .
$S = - \frac{\partial F}{\partial T}$ — entropy.

Two pictures anchor everything below. The first shows why multiplying $Z$ 's becomes adding $F$ 's (extensivity). The second shows the shape of $F (T)$ and why its slope is $- S$ .

Figure — Free energy from partition function

True or false — justify

$Z$ is dimensionless for any system. ::: For a discrete sum of $e^{- β E_{i}}$ it is dimensionless, but the classical phase-space $Z$ is made dimensionless only by dividing by $h$ per degree of freedom (and $N!$ for identical particles) — omit those factors and $Z$ carries stray units.
Adding a constant $E_{0}$ to every energy level leaves all measurable predictions unchanged. ::: False for $Z$ and $F$ (both shift: $Z \to e^{- β E_{0}} Z$ , $F \to F + E_{0}$ ), but true for observables that come from derivatives like $p$ and $S$ ; $U$ shifts by exactly $E_{0}$ , as physically expected.
$F = - k_{B} T ln Z$ is only valid at equilibrium. ::: True — $Z$ presupposes Boltzmann-weighted microstates, which is the equilibrium distribution; out of equilibrium $F = U - T S$ is defined but not equal to $- k_{B} T ln Z$ .
Because $Z$ multiplies for independent subsystems, $F$ adds for them. ::: True — that is the whole point of the logarithm (see figure 1): $ln (Z_{1} Z_{2}) = ln Z_{1} + ln Z_{2}$ , so $F$ is extensive while $Z$ is multiplicative.
$F$ is always negative. ::: False — the sign of $F$ depends on the zero of energy and on $Z$ ; only the changes in $F$ are physically meaningful, so its raw sign carries no universal meaning.
Lowering temperature always lowers $F$ . ::: False — the slope in figure 2 is $\partial F / \partial T = - S \leq 0$ , so $F$ increases as $T$ drops (since $S \geq 0$ ); $F$ decreases as $T$ rises.
The equilibrium state minimizes $F$ at fixed $T, V, N$ . ::: True — this is the defining variational property: among all conceivable arrangements at fixed $T, V, N$ , the real equilibrium is the one sitting at the lowest value of $F = U - T S$ , not the minimum of $U$ alone.
Doubling the number of microstates always doubles $F$ . ::: False — $F \propto ln Z$ , so doubling $Z$ subtracts a fixed $k_{B} T ln 2$ from $F$ ; only replicating an independent copy of the whole system doubles $F$ .

Spot the error

" $U = - \frac{\partial Z}{\partial β}$ ." ::: Missing normalization: $- \partial Z / \partial β = \sum_{i} E_{i} e^{- β E_{i}}$ is an unnormalized weighted sum; you must divide by $Z$ , giving $U = - \frac{1}{Z} \frac{\partial Z}{\partial β} = - \frac{\partial ( ln Z )}{\partial β}$ .
" $S = + \partial F / \partial T$ ." ::: Wrong sign: $d F = - S d T - p d V + μ d N$ carries a built-in minus, so $S = - \partial F / \partial T$ ; the plus sign would make entropy negative.
"For $N$ independent particles $Z = N Z_{1}$ ." ::: Wrong operation: energies add inside the exponent, which makes exponentials multiply, so $Z = Z_{1}^{N}$ (up to a $1/ N!$ for indistinguishability), not $N Z_{1}$ .
" $p = + \partial F / \partial V$ ." ::: Sign error again: from $d F = - p d V$ at fixed $T, N$ , pressure is $p = - \partial F / \partial V$ ; the vault result $p = k_{B} T \partial_{V} ln Z$ already has the correct sign baked in.
" $F = - k_{B} T Z$ (dropped the log)." ::: Without the log $F$ would multiply instead of add when systems combine, breaking extensivity; the logarithm is exactly what converts the product structure of $Z$ into an additive potential.
"Since $p_{i} = e^{- β E_{i}} / Z$ , the most probable microstate is always the ground state." ::: The most probable single microstate is the lowest-energy one, but the most probable energy level can be a high one if it has many microstates — probability of a level is $p_{i}$ times its degeneracy.

Why questions

Why a logarithm in $F$ rather than $Z$ itself? ::: Independent systems multiply their $Z$ ; taking $ln$ turns that product into a sum (figure 1), so $F$ becomes extensive (scales with system size) — the essential requirement of a thermodynamic potential.
Why does $U$ cancel exactly in $F = U - T S$ ? ::: Because $S = U / T + k_{B} ln Z$ , the $T \cdot (U / T) = U$ term precisely subtracts the $U$ in $F = U - T S$ , leaving pure $- k_{B} T ln Z$ — a structural fact, not a coincidence.
Why is entropy $- \partial F / \partial T$ and not something involving $Z$ directly by counting? ::: The two are the same thing: start from Gibbs entropy $S = - k_{B} \sum_{i} p_{i} ln p_{i}$ with $p_{i} = e^{- β E_{i}} / Z$ , so $ln p_{i} = - β E_{i} - ln Z$ ; substituting gives $S = k_{B} β U + k_{B} ln Z = U / T + k_{B} ln Z$ , and differentiating $F = - k_{B} T ln Z$ reproduces exactly this — the thermodynamic derivative and the microscopic count agree because $F$ was built from those same $p_{i}$ .
Why does the classical harmonic oscillator give $U = k_{B} T$ regardless of $ω$ ? ::: Because $Z \propto 1/ β$ makes $ln Z = - ln β + const$ , and $ω$ sits only in the constant; the equipartition energy $k_{B} T$ (two quadratic degrees of freedom) is independent of stiffness — see Equipartition theorem.
Why can we build all thermodynamics from $Z$ by differentiation instead of returning to microstates? ::: Every observable is an average or fluctuation of energy, and those are encoded as $β$ - and $V$ -derivatives of $ln Z$ ; the single function $ln Z$ therefore stores all the microstate information we need.
Why is $F$ (not $U$ ) minimized at fixed $T$ ? ::: At fixed $T$ the system can exchange energy with the bath, so it trades internal energy against entropy; minimizing $F = U - T S$ correctly balances "low energy" against "high disorder," which minimizing $U$ alone ignores — see Helmholtz vs Gibbs free energy.

Edge cases

What is $F$ as $T \to 0$ ? ::: Only the ground state survives in $Z$ , so $Z \to g_{0} e^{- β E_{0}}$ and $F \to E_{0} - k_{B} T ln g_{0} \to E_{0}$ ; the entropy term vanishes and $F$ approaches the ground-state energy.
What happens to $U$ for the two-level system as $T \to \infty$ ? ::: $β \to 0$ makes both states equally likely, so $U \to ε /2$ (the average of $0$ and $ε$ ), not $ε$ — high temperature means maximal ignorance, not maximal energy.
Can $Z$ diverge, and what does that mean? ::: Yes — e.g. a free particle in infinite volume or a hydrogen atom summed over all bound states; a divergent $Z$ signals that a normalizable equilibrium doesn't exist without an additional constraint (volume cutoff, regularization).
For a two-level system, how does the value of entropy behave at $T \to 0$ versus $T \to \infty$ ? ::: At $T \to 0$ only the ground state is occupied so $S \to 0$ (one accessible state, zero disorder); at $T \to \infty$ both states are equally likely so $S \to k_{B} ln 2$ (a maximum, not zero) — the two limits are physically opposite, so $S$ does not vanish at high $T$ .
Is $F = - k_{B} T ln Z$ valid in the grand canonical ensemble? ::: No — there the controlled potential is the grand potential $Φ = - k_{B} T ln Z$ with $Z$ summing over particle number too; see Grand canonical ensemble and grand potential.
What if all microstates have the same energy $E$ ? ::: Then $Z = Ω e^{- β E}$ and $F = E - k_{B} T ln Ω$ , so $S = k_{B} ln Ω$ recovers the Boltzmann entropy exactly — the canonical result collapses to the microcanonical one.
Does the formula assume distinguishable particles? ::: The single-system $F = - k_{B} T ln Z$ makes no such assumption, but building $Z$ for many identical particles requires a $1/ N!$ (Gibbs correction) to avoid the entropy-of-mixing paradox — see Ideal gas from the partition function.

Recall One-line summary of the traps

Almost every error is one of three things: dropping the $ln$ (breaks extensivity), dropping the $1/ Z$ normalization (unnormalized averages), or dropping a minus sign (from $d F = - S d T - p d V + μ d N$ ).