Free energy from partition function: Worked examples

2.4.12 · D3 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Free energy from partition function

Shuru karne se pehle, symbols ka ek reminder taaki tum kabhi guess na karo:

Recall Symbols, simple alfazon mein

$Z$ ::: partition function, $Z = \sum_{i} e^{- β E_{i}}$ — ek number jo har microstate ka "Boltzmann weight" sum karta hai.
$β$ ::: $1/ (k_{B} T)$ ka shorthand. Bada $β$ = thanda; chhota $β$ = garam.
$k_{B}$ ::: Boltzmann's constant $\approx 1.381 \times 1 0^{- 23} J/K$ — temperature aur energy ke beech conversion factor.
$F$ ::: Helmholtz free energy, $F = - k_{B} T ln Z$ .
$U$ ::: average (internal) energy, $U = - \partial_{β} ln Z$ .
$ε$ ::: two-level system ka energy gap.

Prerequisites yahan hain Boltzmann distribution and microstates aur Entropy — Gibbs and Boltzmann definitions; master result yahan hai parent topic par.

#	Case class	Kya special hai	Example
A	Degenerate input: ek hi accessible state	$Z = 1$ , isliye $ln Z = 0$	Ex 1
B	Low- $T$ limit ( $β \to \infty$ )	System ground state mein freeze ho jaata hai	Ex 2
C	High- $T$ limit ( $β \to 0$ )	Sab states equally likely hain	Ex 2
D	Finite discrete sum, general $T$	Two-level system, sabhi $T$ par	Ex 2
E	Continuous integral sum ki jagah	Classical DOF, Gaussian integral	Ex 3
F	Extensive / many independent parts	$Z = Z_{1}^{N}$ , $F \propto N$	Ex 4
G	Real-world word problem	Numbers, units, ek actual answer	Ex 5
H	Exam twist: degeneracy factor	States with multiplicity $g$	Ex 6
I	$S, p$ ka sign / limiting sanity	$S = - \partial_{T} F$ har jagah check karna	Ex 7

Worked examples

Worked example Ex 1 — Cell A: sirf EK state wala system (degenerate)

Ek frozen crystal mein bilkul ek accessible configuration hai, energy $E_{0} = 0$ . $Z$ , $F$ , $U$ , $S$ nikalo.

Forecast: padhne se pehle guess karo — agar sirf ek hi tarika hai hone ka, toh kitna messy (entropy) ho sakta hai? Kitni free energy hai?

$Z = e^{- β \cdot 0} = 1$ . Yeh step kyun? Sum $\sum_{i} e^{- β E_{i}}$ mein sirf ek term hai; zero energy par ek state ka Boltzmann weight $e^{0} = 1$ hota hai.
$F = - k_{B} T ln Z = - k_{B} T ln 1 = 0$ . Kyun? $ln 1 = 0$ , isliye $F$ zero ho jaata hai. Ek aisi system se koi "usefulness" extract nahi hoti jisme koi choice hi nahi.
$U = - \partial_{β} ln Z = - \partial_{β} (0) = 0$ . Kyun? $ln Z$ $β$ par depend nahi karta, isliye uska slope zero hai. Woh ek state energy $0$ par baith gayi hai.
$S = (U - F) / T = (0 - 0) / T = 0$ . Kyun? Entropy choices count karta hai; ek state matlab zero disorder — yeh miniature mein thermodynamics ka third law hai.

Verify: $S = 0$ , Boltzmann form $S = k_{B} ln Ω = k_{B} ln 1 = 0$ se agree karta hai. Har number zero hai, jo ek aisi system ke liye bilkul sahi hai jo "sirf ek hi tarah ho sakti hai." ✓

Worked example Ex 2 — Cells B, C, D: two-level system SABHI temperatures par

Ek single spin ki energies hain $0$ aur $ε > 0$ . $Z (T)$ nikalo, phir cold ( $β \to \infty$ ), hot ( $β \to 0$ ), aur ek numeric mid-point examine karo.

Forecast: bahut low $T$ par spin ground state mein chhup jaayega ( $U \to 0$ ); bahut high $T$ par dono states equally likely honge isliye $U \to ε /2$ . Har end par entropy guess karo.

$Z = 1 + e^{- β ε}$ . Kyun? Dono Boltzmann weights sum karo.
$U = - \partial_{β} ln Z = \frac{ε e ^{- β ε}}{1 + e ^{- β ε}} = \frac{ε}{e ^{β ε} + 1}$ . Kyun? $ln (1 + e^{- β ε})$ ko $β$ ke respect mein differentiate karo; chain rule $- ε e^{- β ε}$ neeche laata hai, aur overall minus ise positive kar deta hai.
Cell B (thanda, $β \to \infty$ ): $e^{β ε} \to \infty$ , isliye $U \to 0$ aur $Z \to 1$ , $F \to 0$ , $S \to 0$ . Kyun? $T = 0$ par spin apne ground state mein freeze ho jaata hai — ek state, koi entropy nahi.
Cell C (garam, $β \to 0$ ): $e^{β ε} \to 1$ , isliye $U \to ε /2$ ; aur $Z \to 2$ , $S \to k_{B} ln 2$ . Kyun? Dono states equally likely → average energy midpoint hai, aur entropy hai " $k_{B}$ times do choices ka log."
Cell D (mid, numeric): $β ε = 1$ lo. Tab $U / ε = 1/ (e^{1} + 1) = 1/ (e + 1) \approx 0.2689$ , aur $Z = 1 + e^{- 1} \approx 1.3679$ , isliye $F / ε = - \frac{1}{1} ln Z \cdot (k_{B} T / ε)$ ... explicitly $F = - k_{B} T ln Z$ ; $k_{B} T = ε$ ke saath (kyunki $β ε = 1$ ), $F = - ε ln (1.3679) \approx - 0.3133 ε$ .

Verify (figure dekho): $U$ smoothly $0$ se $ε /2$ tak badhta hai jab $T$ increase hota hai — kabhi $ε /2$ se upar nahi, kyunki equilibrium mein excited state ko ground state se zyada likely banana possible nahi. Units: $U / ε$ dimensionless hai, achha hai. ✓

Figure — Free energy from partition function

Figure cell D ka poora curve dikhata hai aur uspar dono limiting cells B aur C ko mark karta hai.

Worked example Ex 3 — Cell E: continuous integral (1D particle in a box, translational

$Z$ ) Mass $m$ ka ek single classical particle 1D mein length $L$ par freely move karta hai. Uski energy purely kinetic hai, $E = p^{2} /2 m$ . Phase-space integral ko measure $d x d p / h$ ke saath use karke $Z$ nikalo, phir $U$ .

Forecast: ek quadratic degree of freedom hai (momentum), isliye equipartition predict karta hai $U = \frac{1}{2} k_{B} T$ . Guess karo $Z$ mein $T$ ki kaun si power aayegi.

Integral set up karo. Sum integral kyun banta hai? Classical position aur momentum continuous hain, isliye " $\sum_{i}$ " $\frac{1}{h} \int d x d p$ ban jaata hai — $1/ h$ har phase-space cell ko ek state banata hai (Planck's constant cell size set karta hai). $Z = \frac{1}{h} \int_{0}^{L} d x \int_{- \infty}^{\infty} e^{- β p^{2} /2 m} d p .$
Position integral karo: $\int_{0}^{L} d x = L$ . Kyun? Energy ka $x$ par koi dependence nahi, isliye position trivially integrate ho jaata hai.
Gaussian momentum integral karo: $\int_{- \infty}^{\infty} e^{- a p^{2}} d p = π / a$ jahan $a = β /2 m$ . Yeh tool kyun? Bell-shaped $e^{- a p^{2}}$ ka koi elementary antiderivative nahi, lekin poori line par uska area standard Gaussian result hai — infinitely many continuous states sum karne ka yahi ek clean tarika hai. $\int_{- \infty}^{\infty} e^{- β p^{2} /2 m} d p = \frac{2 π m}{β} .$
Assemble karo: $Z = \frac{L}{h} \frac{2 π m}{β} \propto β^{- 1/2}$ .
Energy: $ln Z = - \frac{1}{2} ln β + const$ , isliye $U = - \partial_{β} ln Z = \frac{1}{2 β} = \frac{1}{2} k_{B} T$ . Kyun? Sirf $ln β$ term ka $β$ -slope hai; uska derivative $- 1/ (2 β)$ hai, jo positive ho jaata hai.

Verify: $U = \frac{1}{2} k_{B} T$ bilkul equipartition hai ek quadratic DOF ke liye. $Z$ ki $β^{- 1/2}$ scaling ek single continuous quadratic mode ki pehchaan hai. Contrast karo: parent note mein oscillator ke do quadratic DOF the ( $x$ aur $p$ ), jisse $Z \propto β^{- 1}$ aur $U = k_{B} T$ mila. ✓

Worked example Ex 4 — Cell F: extensive many-particle system

$N$ independent, distinguishable two-level spins lo (har ek ka gap $ε$ hai). Dikhao ki $F$ $N$ ke saath scale karta hai, aur $U$ aur $S$ per spin nikalo.

Forecast: independent parts → unke $Z$ 's multiply hote hain. Phir $F$ ke liye $ln$ kya karta hai?

$Z_{tot} = Z_{1}^{N} = (1 + e^{- β ε})^{N}$ . Kyun? Independent systems ke liye $\sum e^{- β (E_{a} + E_{b} + \dots)}$ product mein factorize ho jaata hai; parent note ka mistake #4 dekho. (Distinguishable hain, isliye koi $1/ N!$ nahi.)
$F = - k_{B} T ln Z_{1}^{N} = - N k_{B} T ln (1 + e^{- β ε})$ . Kyun? $ln (a^{N}) = N ln a$ — logarithm product ko sum mein badal deta hai, jisse $F$ extensive ban jaata hai ( $\propto N$ ).
$U_{tot} = N \frac{ε}{e ^{β ε} + 1}$ . Kyun? $ln Z_{tot} = N ln Z_{1}$ , aur $\partial_{β}$ linear hai, isliye total energy Ex 2 ki single-spin energy ka $N$ guna hai.
Numeric check, $β ε = 1$ , $N = 1000$ : $U_{tot} / ε = 1000/ (e + 1) \approx 268.9$ .

Verify: $N$ double karne se $F$ aur $U$ exactly double ho jaate hain — yeh extensive quantity ki pehchaan hai. Agar humne (galat) likha hota $F = - k_{B} T Z$ , toh $F$ $(1 + e^{- β ε})^{N}$ jaisa scale karta, $N$ mein exponential — clearly extensive nahi. $ln$ hi hame bachata hai. ✓ (Dekho Helmholtz vs Gibbs free energy ki extensivity kyun matter karti hai.)

Worked example Ex 5 — Cell G: real-world word problem (units aur ek number)

Ek crystal mein colour centre ka ground state aur ek excited state hai jo $ε = 0.10 eV$ upar hai. Room temperature $T = 300 K$ par, in centres ka kitna fraction excited hai, aur eV mein free energy per centre kya hai?

Forecast: $0.10$ eV versus $k_{B} T \approx 0.0259$ eV at $300$ K, isliye $β ε \approx 3.9$ — gap kuch $k_{B} T$ ka hai, matlab excited state kaafi rare hai. Guess karo "kuch percent."

$β ε$ calculate karo. Kyun? Yeh dimensionless ratio sab kuch control karta hai. $k_{B} T = 8.617 \times 1 0^{- 5} eV/K \times 300 K = 0.02585 eV$ , isliye $β ε = 0.10/0.02585 = 3.868$ .
Excited fraction $p_{exc} = \frac{e ^{- β ε}}{1 + e ^{- β ε}} = \frac{1}{e ^{β ε} + 1}$ . Kyun? Upper state ki Boltzmann probability, $Z$ se normalize karke. $p_{exc} = \frac{1}{e ^{3.868} + 1} = \frac{1}{47.83 + 1} = 0.02048 \approx 2.0%.$
Free energy per centre: $F = - k_{B} T ln (1 + e^{- β ε}) = - 0.02585 eV \times ln (1 + 0.02090)$ . $F = - 0.02585 \times ln (1.02090) = - 0.02585 \times 0.020684 = - 5.35 \times 1 0^{- 4} eV .$

Verify: $\approx 2%$ excited hamara forecast match karta hai ("kuch percent"). Units: eV mein $k_{B} T$ times dimensionless log eV deta hai, theek hai. $F$ chhota aur negative hai, jaisa expect tha jab excited state barely contribute karta ho. ✓

Worked example Ex 6 — Cell H: exam twist degeneracy factor ke saath

Ek two-level system jahan excited level 3-fold degenerate hai: ground energy $0$ (1 state), excited energy $ε$ (3 states). $Z$ nikalo, aur excited-population fraction; note karo ki degeneracy high- $T$ limit ko kaise shift karta hai.

Forecast: teen excited states versus ek ground state ke saath, bahut high $T$ par system zyada time excited rehega — population $3/4$ ki taraf tilt honi chahiye, $1/2$ nahi.

$Z = 1 + 3 e^{- β ε}$ . Kyun? Teeno excited microstates mein se har ek ka weight $e^{- β ε}$ hai; degeneracy multiply karta hai, koi naya exponent add nahi hota. Yeh general rule hai $Z = \sum_{levels} g_{ℓ} e^{- β E_{ℓ}}$ degeneracy $g_{ℓ}$ ke saath.
Excited fraction $p_{exc} = \frac{3 e ^{- β ε}}{1 + 3 e ^{- β ε}}$ . Kyun? Total excited weight over $Z$ .
High- $T$ limit ( $β \to 0$ ): $p_{exc} \to \frac{3}{1 + 3} = \frac{3}{4} = 0.75$ . Kyun? Jab Boltzmann weights sab $\approx 1$ hon, states purely count hoti hain ki kitni hain: 3 excited vs 1 ground.
Numeric, $β ε = 1$ : $p_{exc} = \frac{3 e ^{- 1}}{1 + 3 e ^{- 1}} = \frac{1.1036}{2.1036} = 0.5246$ .

Verify: non-degenerate Ex 2 se compare karo $β ε = 1$ par, jahan $p_{exc} = 1/ (e + 1) = 0.2689$ tha. Excited degeneracy teen guna karne se population roughly double ho gayi — sensible hai. High- $T$ limit $3/4$ ≠ $1/2$ , exactly wahi twist hai. ✓

Worked example Ex 7 — Cell I:

$S$ ka sign & limiting sanity Non-degenerate two-level system (Ex 2) use karke verify karo ki $S = - \partial F / \partial T$ ek non-negative entropy deta hai jo $k_{B} ln 2$ par peak karta hai, notorious sign check karte hue.

Forecast: entropy har jagah $\geq 0$ honi chahiye, $T = 0$ par $0$ ho, aur $T \to \infty$ par $k_{B} ln 2$ (do equally likely states) par saturate kare.

$F (T) = - k_{B} T ln (1 + e^{- ε / k_{B} T})$ likho aur correct sign $S = - \partial F / \partial T$ se differentiate karo. Minus kyun? Kyunki $d F = - S d T - p d V + μ d N$ ; minus differential mein built-in hai (parent note, mistake #3).
Result: $S = k_{B} ln (1 + e^{- β ε}) + \frac{ε / T}{e ^{β ε} + 1}$ . Do terms kyun? Pehla term hai $k_{B} ln Z$ ; doosra hai $U / T$ , exactly $S = U / T + k_{B} ln Z$ .
Numeric, $β ε = 1$ (isliye $k_{B} T = ε$ ): pehla term $k_{B} ln (1.3679) = 0.3133 k_{B}$ ; doosra term $= \frac{1}{e + 1} k_{B} = 0.2689 k_{B}$ . Sum $S = 0.5822 k_{B}$ .
Limits: $T \to 0 \Rightarrow S \to 0$ ; $T \to \infty \Rightarrow S \to k_{B} ln 2 = 0.6931 k_{B}$ . Kyun? Dono Ex 2 mein upar compute kiye gaye hain, aur $0.5822 < 0.6931$ , isliye mid-value ceiling se neeche sahi jagah par hai.

Verify: $S > 0$ poore time, monotone rise karta hua $k_{B} ln 2$ tak, aur agar humne galat sign $S = + \partial F / \partial T$ use kiya hota toh negative entropy milta — ek instant contradiction. ✓

Recall Har cell ka one-line map

Kaun se examples cover karte hain: degenerate, cold, hot, general discrete, continuous, extensive, word-problem, degeneracy, sign? ::: Ex1 (A), Ex2 (B,C,D), Ex3 (E), Ex4 (F), Ex5 (G), Ex6 (H), Ex7 (I) — saatein nove cells covered.

Active recall

Two-level system, excited fraction at $β ε = 1$ ? ::: $1/ (e + 1) \approx 0.269$ .
1D free particle $U$ Gaussian $Z$ se? ::: $\frac{1}{2} k_{B} T$ (ek quadratic DOF, equipartition).
Independent parts ke liye $F$ $N$ jaisa kyun scale karta hai? ::: $Z = Z_{1}^{N}$ aur $ln (Z_{1}^{N}) = N ln Z_{1}$ — log $F$ ko extensive banata hai.
Excited level degeneracy 3 ke saath — high- $T$ population? ::: $3/4$ , kyunki jab weights $\to 1$ ho jaayein tab states multiplicity se count hoti hain.
$F$ se entropy ka correct sign? ::: $S = - \partial F / \partial T$ (minus $d F$ mein built-in hai).

Worked examples — Free energy from partition function

Scenario matrix

Worked examples

Active recall