Canonical ensemble — partition function Z: Worked examples

2.4.10 · D3 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Canonical ensemble — partition function Z

Kuch bhi shuru karne se pehle, teen reminders un symbols ke jo poore raaste kaam aate hain:

Recall

$β$ kya hai aur hum $T$ ki jagah isko kyun use karte hain? $β$ ::: $β = \frac{1}{k _{B} T}$ . Hum isse isliye use karte hain kyunki $Z = \sum_{i} e^{- β E_{i}}$ mein temperature hamesha sirf energy ko multiply karte hue aata hai. $β$ ke respect mein differentiate karne par cleanly ek factor $- E_{i}$ "pull down" hota hai, jo ki exactly wahi hai jo average energy ko chahiye.

Recall

$h$ (Planck's constant) kya hai aur yeh classical $Z$ mein kyun aata hai? $h$ ::: $h$ Planck's constant hai, jiske units hain (energy $\times$ time) = (position $\times$ momentum). Ek continuous system mein hum states pe sum, position–momentum plane pe integrate karke karte hain; har genuinely distinct state area $h$ ka ek cell occupy karta hai. Har degree of freedom ke liye $h$ se divide karne par "phase space ka area" ek dimensionless count of states ban jaata hai.

Recall Teen master formulas jo hum baar baar use karenge

$Z = \sum_{i} e^{- β E_{i}}$ ::: partition function (Boltzmann weights ka sum) $⟨ E ⟩ = - \frac{\partial ln Z}{\partial β}$ ::: average energy $σ_{E}^{2} = \frac{\partial ^{2} ln Z}{\partial β ^{2}}$ ::: energy fluctuation ( $⟨ E ⟩$ ke around $E$ ka spread)

Scenario matrix

Partition function se jo bhi problem aa sakti hai, woh ek cell mein aati hai. Neeche ke examples label kiye hain ki woh kaun si cell hit karte hain; milaakar yeh poora grid cover karte hain.

Cell	Kya cheez isse alag banati hai	Example
A. Cold limit $T \to 0, β \to \infty$	system lowest state(s) mein freeze ho jaata hai	Ex 1
B. Hot limit $T \to \infty, β \to 0$	sabhi states equally likely	Ex 1
C. Degeneracy $g_{n} > 1$	levels multiplicity ke saath count hote hain	Ex 2
D. Discrete, infinite ladder	geometric series, converge karni chahiye	Ex 3
E. Continuous / classical	phase-space integral, Gaussians	Ex 4
F. Fluctuations & heat capacity	$ln Z$ ka second derivative	Ex 5
G. Many independent particles	factorization $Z_{1}^{N}$ , aur $N!$	Ex 6
H. Real-world word problem	physics se states chunna, phir compute karna	Ex 7
I. Exam twist / degenerate zero	ek level shift ya $g = 0$ -style trap	Ex 8

Ex 1 — Two-level system: dono temperature extremes (cells A + B)

Step 1 — $Z$ likho. $Z = e^{- β \cdot 0} + e^{- β ε} = 1 + e^{- β ε}$ Yeh step kyun? $Z$ by definition sabhi states pe Boltzmann weights ka sum hai — yahan exactly do hain.

Step 2 — Differentiate karo. $⟨ E ⟩ = - \frac{\partial l n Z}{\partial β} = \frac{ε e ^{- β ε}}{1 + e ^{- β ε}} = \frac{ε}{e ^{β ε} + 1}$ Yeh step kyun? Master formula "mere paas $Z$ hai" ko "mere paas average energy hai" mein ek derivative ke saath convert karta hai — haath se $E_{i} P_{i}$ sum karne ki zaroorat nahi.

Step 3 — Cold limit (cell A). $β \to \infty \Rightarrow e^{β ε} \to \infty \Rightarrow ⟨ E ⟩ \to 0$ . Yeh step kyun? Limits sabse sasta correctness test hain; unhe physics se match karna chahiye.

Step 4 — Hot limit (cell B). $β \to 0 \Rightarrow e^{β ε} \to 1 \Rightarrow ⟨ E ⟩ \to ε /2$ .

Step 5 — $k_{B} T = ε$ par $β ε = 1$ hoga, toh $⟨ E ⟩ = \frac{ε}{e + 1}$ .

Recall

$k_{B} T = ε$ par numeric value $⟨ E ⟩ / ε$ ::: $1/ (e + 1) \approx 0.2689$

Verify: Cold mein $0$ ✓, hot mein $ε /2$ ✓ (dono forecast se match). Units: $⟨ E ⟩$ ke har term mein $ε$ (ek energy) ka factor hai, baaki dimensionless hai — sahi hai.

Neeche ki figure exactly yahi $⟨ E ⟩ (T)$ plot karti hai. Blue curve ko follow karo jo $0$ (cold) se uthkar red dashed ceiling $ε /2$ (hot) ki taraf jaati hai; yellow dot hamare special point $k_{B} T = ε$ ko mark karta hai jahan $⟨ E ⟩ \approx 0.269 ε$ hai. Dhyaan do ki curve kabhi $ε /2$ se zyada nahi jaati — energy ke liye 50/50 mixture se garam koi jagah nahi hai.

Figure — Canonical ensemble — partition function Z

Ex 2 — Ek degenerate three-level system (cell C)

Step 1 — $Z$ multiplicity ke saath banao. $Z = ground 1 \cdot e^{0} + 3 excited states 3 \cdot e^{- β ε} = 1 + 3 e^{- β ε}$ Yeh step kyun? Har microstate apna Boltzmann factor contribute karta hai. Energy $ε$ par teen microstates $3 e^{- β ε}$ contribute karte hain. Yeh $Z = \sum_{n} g_{n} e^{- β E_{n}}$ rule hai action mein.

Step 2 — Average energy. $⟨ E ⟩ = - \frac{\partial l n Z}{\partial β} = \frac{3 ε e ^{- β ε}}{1 + 3 e ^{- β ε}} = \frac{3 ε}{e ^{β ε} + 3}$ Yeh step kyun? Same master formula; degeneracy bas $Z$ ke andar saath chalti hai.

Step 3 — Hot limit. $β \to 0 \Rightarrow ⟨ E ⟩ \to \frac{3 ε}{1 + 3} = \frac{3}{4} ε$ . Yeh step kyun? "Sabhi chaar states equal" wali prediction confirm karta hai.

Verify: Cold limit $β \to \infty$ pe $0$ deta hai ✓. Hot limit $\frac{3}{4} ε$ ✓ forecast se match. Agar hum $g = 1$ set karein toh Ex 1 ka $ε / (e^{β ε} + 1)$ wapas aata hai ✓ — formula gracefully degrade karta hai.

Ex 3 — Quantum harmonic oscillator: ek infinite ladder (cell D)

Step 1 — Sum likho. $Z = \sum_{n = 0}^{\infty} e^{- β ℏ ω (n + 1/2)} = e^{- β ℏ ω /2} \sum_{n = 0}^{\infty} (e^{- β ℏ ω})^{n}$ Yeh step kyun? Constant $\frac{1}{2} ℏ ω$ ko bahar nikaalne par common ratio $r = e^{- β ℏ ω}$ wali ek plain geometric series expose hoti hai.

Step 2 — Check karo ki ratio 1 se neeche hai, phir sum karo. Temperature hamesha positive hoti hai, toh $β = 1/ k_{B} T > 0$ , aur $ℏ ω > 0$ ; isliye exponent $- β ℏ ω$ negative hai, jo $r = e^{- β ℏ ω}$ ko strictly $0$ aur $1$ ke beech ka number banata hai. $0 < r < 1$ wali geometric series $\sum_{n \geq 0} r^{n} = \frac{1}{1 - r}$ mein converge karti hai: $Z = \frac{e ^{- β ℏ ω /2}}{1 - e ^{- β ℏ ω}}$ Yeh step kyun? Series converge isliye karti hai kyunki $r < 1$ — yeh fact ki energies bina bound ke badhti hain, exactly wahi hai jo $r$ ko $1$ se neeche force karti hai aur ek finite answer guarantee karti hai.

Step 3 — $sinh$ ki tarah rewrite karo. Upar aur neeche $e^{+ β ℏ ω /2}$ se multiply karo. Numerator $e^{- β ℏ ω /2} \cdot e^{+ β ℏ ω /2} = 1$ hoga; denominator $e^{+ β ℏ ω /2} - e^{- β ℏ ω /2}$ ban jaata hai, jo definition $sinh x = \frac{1}{2} (e^{x} - e^{- x})$ se exactly $2 sinh (β ℏ ω /2)$ hai: $Z = \frac{1}{e ^{β ℏ ω /2} - e ^{- β ℏ ω /2}} = \frac{1}{2 s i n h ( β ℏ ω /2 )}$ Yeh step kyun? $sinh$ form compact hai aur uska logarithm agli step mein cleanly differentiate hota hai.

Step 4 — Average energy. $⟨ E ⟩ = - \frac{\partial l n Z}{\partial β} = \frac{ℏ ω}{2} + \frac{ℏ ω}{e ^{β ℏ ω} - 1}$ Yeh step kyun? Master derivative phir se; pehla term zero-point energy hai, doosra thermal Planck occupation hai.

Step 5 — High- $T$ limit. $β ℏ ω ≪ 1$ ke liye, $e^{β ℏ ω} - 1 \approx β ℏ ω$ , toh $⟨ E ⟩ \approx \frac{ℏ ω}{β ℏ ω} = k_{B} T$ . Yeh step kyun? Classical oscillator result se match karna chahiye — quantum aur classical agree karte hain jab thermal energy level spacing se zyada ho.

Verify: Low- $T$ limit $β \to \infty$ pe $⟨ E ⟩ \to ℏ ω /2$ (pure zero-point) ✓. High- $T$ pe $k_{B} T$ ✓, parent Example 2 se match.

Figure dono curves ko contrast karti hai. Blue quantum curve thandi hone par green dotted floor $ℏ ω /2$ par flatten ho jaati hai (energy zero-point value par freeze ho jaati hai), jabki yellow dashed classical line $k_{B} T$ origin se seedha nikalta hai. Dekho blue aur yellow right side pe kaise merge hoti hain: jab $k_{B} T ≫ ℏ ω$ ho, quantization invisible hai.

Ex 4 — Box mein free particle: continuous phase space (cell E)

Step 1 — Phase-space integral set up karo. $Z_{1} = \frac{1}{h} \int_{0}^{L} d x \int_{- \infty}^{\infty} d p e^{- β p^{2} /2 m}$ Yeh step kyun? Continuous system ke liye hum phase space $(x, p)$ pe integrate karke states pe sum karte hain, Planck's constant $h$ se divide karte hain (upar recall box dekho) taaki area $h$ wala har cell ek state count ho.

Step 2 — Integrals karo. $x$ -integral bas $L$ hai; $p$ -integral ek Gaussian hai $\int e^{- a p^{2}} d p = π / a$ jahan $a = β /2 m$ hai: $Z_{1} = \frac{L}{h} \frac{2 π m}{β}$ Yeh step kyun? Gaussian tool isliye hai kyunki $e^{- a p^{2}}$ exactly Gaussian shape hai; uska area ek jaana-maana standard result hai, toh koi numerical kaam nahi chahiye.

Step 3 — Average energy. $ln Z_{1} = - \frac{1}{2} ln β + const$ likho: $⟨ E ⟩ = - \frac{\partial l n Z _{1}}{\partial β} = \frac{1}{2 β} = \frac{1}{2} k_{B} T$ Yeh step kyun? Derivative ke liye sirf $β$ -dependence matter karta hai; box length $L$ aur $h$ dono constant mein baith jaate hain aur drop out ho jaate hain.

Verify: Ek quadratic DOF $\Rightarrow \frac{1}{2} k_{B} T$ ✓ (forecast aur Equipartition theorem se match). Units: $⟨ E ⟩$ ke units $k_{B} T$ = energy hain ✓.

Ex 5 — Two-level fluctuations aur heat capacity (cell F)

Is example se pehle humein ek naya quantity chahiye.

Step 1 — $ln Z$ se fluctuation. $ln Z = ln (1 + e^{- β ε})$ ke saath, $σ_{E}^{2} = \frac{\partial ^{2} l n Z}{\partial β ^{2}} = \frac{ε ^{2} e ^{- β ε}}{( 1 + e ^{- β ε} ) ^{2}}$ Yeh step kyun? $ln Z$ ka second derivative energy ka variance hai — woh identity (parent note se) "spread compute karo" ko "do baar differentiate karo" mein convert karti hai.

Step 2 — Heat capacity via $σ_{E}^{2} = k_{B} T^{2} C_{V}$ . $C_{V} = \frac{σ _{E}^{2}}{k _{B} T ^{2}} = k_{B} (β ε)^{2} \frac{e ^{- β ε}}{( 1 + e ^{- β ε} ) ^{2}}$ Yeh step kyun? Fluctuation–response link se hum seedha $T$ ke respect mein $⟨ E ⟩$ differentiate kiye bina $C_{V}$ nikal sakte hain.

Step 3 — Limits. $β \to \infty$ par, factor $(β ε)^{2} e^{- β ε} \to 0$ . $β \to 0$ par, prefactor $(β ε)^{2} \to 0$ . Toh dono ends par $C_{V} \to 0$ . Yeh step kyun? Anomaly ke dono shoulders confirm karta hai.

Step 4 — Peak numerically locate karo. $x = β ε$ ke upar $C_{V}$ maximize karne par peak $x \approx 2.399$ par milta hai, yaani $k_{B} T \approx ε /2.399 \approx 0.417 ε$ .

Recall Schottky heat capacity ki peak

$k_{B} T_{peak} / ε$ ::: $\approx 0.417$ (jahan $x = β ε \approx 2.399$ )

Verify: Dono limits mein $C_{V} \to 0$ ✓. Peak $k_{B} T \sim ε$ ke paas ✓ jaise forecast mein.

Figure mein red curve $C_{V} / k_{B}$ hai: yeh thandi left side par zero se shuru hoti hai, ek single hump tak uthti hai, phir garam right side par wapas zero par decay hoti hai. Yellow dot peak ko $k_{B} T \approx 0.417 ε$ par mark karta hai. Yeh akela bump — koi step nahi, koi plateau nahi — real heat-capacity data mein ek two-level system ka fingerprint hai.

Ex 6 — $N$ independent atoms: factorization aur $N!$ trap (cell G)

Step 1 — Factorize karo. Total energy $E = \sum_{k = 1}^{N} ϵ_{k}$ , toh $Z_{N} = \sum_{all configs} e^{- β \sum_{k} ϵ_{k}} = \prod_{k = 1}^{N} (\sum_{states of k} e^{- β ϵ_{k}}) = Z_{1}^{N}$ Yeh step kyun? $e^{sum} = \prod e^{terms}$ , aur independent labels pe sum independent sums ke product mein split ho jaata hai. Isliye $Z$ large systems ke liye itna powerful hai.

Step 2 — Total energy. $ln Z_{N} = N ln Z_{1}$ , toh $⟨ E ⟩_{N} = - \frac{\partial l n Z _{N}}{\partial β} = N ⟨ E ⟩_{1} = \frac{N ε}{e ^{β ε} + 1}$ Yeh step kyun? Logarithm product ko sum mein badal deta hai, aur energy extensive hai — yeh $N$ ke saath linearly scale karti hai.

Step 3 — Indistinguishable case. Agar yeh free identical gas particles hote, toh hum divide karte: $Z_{N} = Z_{1}^{N} / N!$ . Yeh step kyun? Do identical particles ko swap karna ek naya microstate nahi hai; product $N!$ baar overcount karta hai. Yeh factor drop karna entropy ki extensivity tod deta hai — Gibbs paradox.

Verify: Distinguishable atoms ke liye energies simply add hoti hain, toh $⟨ E ⟩_{N} = N ⟨ E ⟩_{1}$ ✓. $1/ N!$ , $Z$ ko ek constant se multiply karta hai ( $β$ se independent), toh yeh free energy aur entropy badalta hai (Ex 8 aur Helmholtz free energy F mein introduce) lekin $⟨ E ⟩ = - \partial_{β} ln Z$ nahi — ek useful sanity check.

Ex 7 — Real-world word problem: magnetic field mein paramagnet (cell H)

Step 1 — Physics se states aur energies chuno. Do states: aligned energy $E_{+} = - μ B$ ke saath aur anti-aligned energy $E_{-} = + μ B$ ke saath. Toh $Z = e^{- β (- μ B)} + e^{- β (+ μ B)} = e^{+ β μ B} + e^{- β μ B} = 2 cosh (β μ B)$ Yeh step kyun? Word problem humein koi formula nahi deta — koi bhi machinery apply karne se pehle humein "aligned/anti-aligned" ko concrete energies mein translate karna hoga. Do exponentials $2 cosh$ mein combine ho jaate hain definition $cosh x = \frac{1}{2} (e^{x} + e^{- x})$ se.

Step 2 — Observable $m$ ka average nikalo. Magnetization $⟨ m ⟩ = \sum_{i} m_{i} P_{i}$ hai, jahan $P_{i} = e^{- β E_{i}} / Z$ : $⟨ m ⟩ = \frac{( + μ ) e ^{+ β μ B} + ( - μ ) e ^{- β μ B}}{2 c o s h ( β μ B )} = μ \frac{e ^{β μ B} - e ^{- β μ B}}{e ^{β μ B} + e ^{- β μ B}} = μ tanh (β μ B)$ Yeh step kyun? Boltzmann distribution kisi bhi observable ko weight karta hai, sirf energy ko nahi. Aligned state (zyada weight, kyunki lower energy hai) $+ μ$ carry karta hai; $sinh$ -like top aur $cosh$ -like bottom ka ratio exactly $tanh$ hai.

Step 3 — Cold limit (cell A flavour). $T \to 0 \Rightarrow β \to \infty \Rightarrow tanh (β μ B) \to 1$ , toh $⟨ m ⟩ \to μ$ : har spin field par lock ho jaata hai. Yeh step kyun? Zero temperature par energy borrow karne ke liye kuch nahi, toh system apni lowest state mein baith jaata hai — fully aligned.

Step 4 — Hot limit (cell B flavour). $T \to \infty \Rightarrow β \to 0 \Rightarrow tanh (β μ B) \to 0$ , toh $⟨ m ⟩ \to 0$ : dono orientations equally likely ho jaati hain aur cancel kar deti hain. Yeh step kyun? Infinite temperature energy bias mitaata hai; up aur down equiprobable ho jaate hain.

Verify: $⟨ m ⟩ = μ tanh (β μ B) > 0$ for $B > 0$ ✓ (field ke saath aligned, forecast se match). Field reverse karne par $B \to - B$ se $⟨ m ⟩$ ka sign flip hota hai (function $B$ mein odd hai) ✓ — physically magnetization field direction follow karta hai. Units: $⟨ m ⟩$ moment $μ$ carry karta hai aur $tanh$ dimensionless hai ✓.

Ex 8 — Exam twist: energy zero shift karna (cell I)

Pehle, do thermodynamic quantities jo yeh example check karta hai:

Step 1 — Naya partition function. $Z^{'} = \sum_{i} e^{- β (E_{i} + E_{0})} = e^{- β E_{0}} \sum_{i} e^{- β E_{i}} = e^{- β E_{0}} Z$ Yeh step kyun? Common factor $e^{- β E_{0}}$ sum se seedha bahar nikal jaata hai, toh $Z^{'}$ bas $Z$ times ek constant weight hai.

Step 2 — Probabilities invariant hain. $P_{i}^{'} = \frac{e ^{- β (E_{i} + E_{0})}}{Z ^{'}} = \frac{e ^{- β E_{0}} e ^{- β E_{i}}}{e ^{- β E_{0}} Z} = \frac{e ^{- β E_{i}}}{Z} = P_{i}$ . Yeh step kyun? Shift upar aur neeche cancel ho jaata hai — isliye sirf energy differences kabhi bhi physical hote hain.

Step 3 — Energy aur free energy $E_{0}$ se shift hote hain. $ln Z^{'} = ln Z - β E_{0}$ use karte hue: $⟨ E ⟩^{'} = - \frac{\partial l n Z ^{'}}{\partial β} = - \frac{\partial}{\partial β} (ln Z - β E_{0}) = ⟨ E ⟩ + E_{0}$ $F^{'} = - k_{B} T ln Z^{'} = - k_{B} T (ln Z - β E_{0}) = F + E_{0}$ Yeh step kyun? $β E_{0}$ term energy ke liye differentiate hoke $E_{0}$ deta hai, aur $F$ ke liye $- k_{B} T \cdot (- β E_{0}) = E_{0}$ deta hai (kyunki $k_{B} T β = 1$ ). Dono sea-level change se shift hote hain, jaise hona chahiye.

Step 4 — Entropy unchanged rehti hai. $S^{'} = \frac{⟨ E ⟩ ^{'} - F ^{'}}{T} = \frac{(⟨ E ⟩ + E _{0} ) - ( F + E _{0} )}{T} = \frac{⟨ E ⟩ - F}{T} = S$ . Yeh step kyun? $E_{0}$ , difference $⟨ E ⟩ - F$ mein cancel ho jaata hai, toh entropy — accessible states ka count — energy zero kahan hai iske par depend nahi kar sakta.

Step 5 — $g = 0$ trap. Degeneracy $g = 0$ wala level $0 \cdot e^{- β E} = 0$ contribute karta hai: woh simply exist nahi karta aur $Z$ mein kuch nahi add karta. Koi divergence nahi, koi special case nahi — ek "missing" level sum ke liye invisible hai. Yeh step kyun? Exam questions ek phantom level insert karna pasand karte hain; sum ise automatically handle karta hai.

Verify: Probabilities unchanged ✓ (ek constant energy shift unphysical hai). $⟨ E ⟩$ aur $F$ dono $E_{0}$ se shift ✓; entropy $S$ unchanged ✓ — sab forecast se match.