Average energy from partition function: Exercises — Hinglish deep dive

2.4.11 · D4 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Average energy from partition function

Prerequisites jo tumhare paas open rehne chahiye: Partition function (canonical ensemble), Boltzmann distribution, Helmholtz free energy F = -kT ln Z, Equipartition theorem, Heat capacity and energy fluctuations, Quantum harmonic oscillator — thermal, Two-level system / Schottky anomaly.

Level 1 — Recognition

Ye check karte hain ki kya tum $Z$ dekh ke sahi operation kar sakte ho — koi lambi algebra nahi.

Recall Solution L1.1

Kya karna hai: $Z$ ka logarithm lo, $β$ ke saath differentiate karo, sign palto: $⟨ E ⟩ = - \frac{\partial l n Z}{\partial β} .$ Har piece kyun:

$ln$ "divide by $Z$ " ko derivative ke andar fold kar deta hai, kyunki $\frac{1}{Z} \partial_{β} Z = \partial_{β} ln Z$ .
$\partial_{β}$ jab $e^{- β E_{i}}$ par act karta hai toh $- E_{i}$ ka factor banta hai — bilkul wahi energy jo hum average karna chahte hain.
minus sign us $- E_{i}$ ko cancel karta hai, jab sab $E_{i} \geq 0$ hon toh positive energy milti hai. Mnemonic: "Log it, slope it, flip it."

Recall Solution L1.2

Kya karna hai: $\frac{\partial ln Z}{\partial β} = - \frac{1}{2 β}$ (constant $C$ derivative mein zero ho jaata hai). Isliye: $⟨ E ⟩ = - \frac{\partial l n Z}{\partial β} = + \frac{1}{2 β} = \frac{1}{2} k_{B} T .$ Iska matlab: koi bhi $Z \propto β^{- 1/2}$ ek quadratic degree of freedom hai — equipartition ka result $\frac{1}{2} k_{B} T$ . Tumhe system ki details jaanne ki zaroorat hi nahi thi.

Level 2 — Application

Ab ek real $Z$ ko machine se guzaaro.

Recall Solution L2.1

Setup: $Z = 1 + e^{- β ε}$ , aur standard result hai $⟨ E ⟩ = \frac{ε}{e ^{β ε} + 1} .$ $β ε = 2$ plug karo: $\frac{⟨ E ⟩}{ε} = \frac{1}{e ^{2} + 1} = \frac{1}{7.389 + 1} = \frac{1}{8.389} = 0.1192.$ Toh $⟨ E ⟩ \approx 0.119 ε$ . System mostly ground state (energy $0$ ) mein hi hai kyunki $ε$ thermal scale $k_{B} T$ se thoda bada hai.

Recall Solution L2.2

Setup: $Z = \frac{1}{1 - e ^{- β ℏ ω}}$ deta hai (dekho Quantum harmonic oscillator — thermal) $⟨ E ⟩ = \frac{ℏ ω}{e ^{β ℏ ω} - 1} .$ $β ℏ ω = 1$ plug karo: $\frac{⟨ E ⟩}{ℏ ω} = \frac{1}{e ^{1} - 1} = \frac{1}{1.71828} = 0.5820.$ Toh $⟨ E ⟩ \approx 0.582 ℏ ω$ . (Yahan convention $E_{0} = 0$ use ki gayi hai; sahi ground state $\frac{1}{2} ℏ ω$ ke saath $0.5$ aur add karna hoga.)

Recall Solution L2.3

Kya karna hai: teen microstates hain, isliye $Z = 1 + e^{- β ε} + e^{- 2 β ε} .$ Differentiate karo: $\partial_{β} Z = - ε e^{- β ε} - 2 ε e^{- 2 β ε}$ , toh

=\frac{\varepsilon e^{-\beta\varepsilon}+2\varepsilon e^{-2\beta\varepsilon}} {1+e^{-\beta\varepsilon}+e^{-2\beta\varepsilon}}.$$ **$\beta\varepsilon=1$ par:** $e^{-1}=0.36788$, $e^{-2}=0.13534$ ke saath: $$\frac{\langle E\rangle}{\varepsilon}=\frac{0.36788+2(0.13534)}{1+0.36788+0.13534} =\frac{0.63856}{1.50322}=0.4248.$$ Toh $\langle E\rangle\approx 0.425\,\varepsilon$.

Level 3 — Analysis

Yahan tum limits, signs aur shapes ke baare mein sochte ho, sirf numbers nahi.

Recall Solution L3.1

Thanda, $T \to 0 \Rightarrow β \to \infty$ : $e^{β ε} \to \infty$ , toh $⟨ E ⟩ \to ε /\infty = 0$ . System ground state mein freeze ho jaata hai — $ε$ tak chadhne ke liye koi thermal energy available nahi. Garam, $T \to \infty \Rightarrow β \to 0$ : $e^{β ε} \to e^{0} = 1$ , toh $⟨ E ⟩ \to \frac{ε}{1 + 1} = \frac{ε}{2} .$ Dono states ab equally likely hain, aur ${0, ε}$ ka mean $ε /2$ hai. Poora curve $0$ se $ε /2$ tak monotonically badhta hai — figure dekho.

Figure — Average energy from partition function

Recall Solution L3.2

$C_{V}$ measure karta hai ki system ko garam karne par $⟨ E ⟩$ kitni tezi se change hoti hai.

Thanda ( $T \to 0$ ): sab kuch ground state mein hai; thoda sa warm karne se kuch bhi gap $ε$ par nahi chad sakta, toh $⟨ E ⟩$ almost nahi hilta ⟹ $C_{V} \to 0$ .
Garam ( $T \to \infty$ ): $⟨ E ⟩$ pehle se $ε /2$ par saturate ho chuka hai; aur warm karne se kuch nahi hota ⟹ $C_{V} \to 0$ .
Beech mein ( $k_{B} T \sim ε$ ): thermal energy bilkul itni hai ki spins gap ke paar promote hone lage, toh $⟨ E ⟩$ sabse tezi se change hoti hai — $C_{V}$ mein maximum aata hai. Jo function dono ends par $0$ ho aur beech mein positive ho, usme peak zaroori hai. Yahi bump hai Schottky anomaly (dekho Two-level system / Schottky anomaly).

Recall Solution L3.3

Kaunsa tool aur kyun: jab $ℏ ω ≪ k_{B} T$ , toh quantity $x \equiv β ℏ ω$ chhoti hoti hai, toh hum exponential ka Taylor series $e^{x} = 1 + x + \frac{x ^{2}}{2} + \dots$ se expand karte hain — sahi tool hai kyunki yeh ek mushkil function ko ek leading power mein badal deta hai jo hum read kar sakte hain. Karo: $e^{x} - 1 = x + \frac{x ^{2}}{2} + \dots \approx x$ chhote $x$ ke liye. Isliye $⟨ E ⟩ = \frac{ℏ ω}{e ^{β ℏ ω} - 1} \approx \frac{ℏ ω}{β ℏ ω} = \frac{1}{β} = k_{B} T .$ Matlab: discrete quantum ladder blur hokar continuum ban jaata hai aur classical oscillator ka $k_{B} T$ recover hota hai — woh do quadratic modes hain (kinetic + potential), har ek $\frac{1}{2} k_{B} T$ deta hai, bilkul equipartition.

Level 4 — Synthesis

Average-energy machine ko fluctuations, free energy aur multiple modes ke saath combine karo.

Recall Solution L4.1

Direct route (cross-check): probabilities $p_{0} = \frac{1}{1 + e ^{- β ε}}$ aur $p_{1} = \frac{e ^{- β ε}}{1 + e ^{- β ε}}$ ke saath, aur energies $0, ε$ : $⟨ E ⟩ = ε p_{1}, ⟨ E^{2} ⟩ = ε^{2} p_{1},$ $σ_{E}^{2} = ε^{2} p_{1} - ε^{2} p_{1}^{2} = ε^{2} p_{1} (1 - p_{1}) = ε^{2} p_{0} p_{1} .$ Derivative route: $ln Z = ln (1 + e^{- β ε})$ same $σ_{E}^{2} = ε^{2} \frac{e ^{- β ε}}{( 1 + e ^{- β ε} ) ^{2}} = ε^{2} p_{0} p_{1}$ deta hai. Dono agree karte hain — jaise fluctuation identity guarantee karti hai. $β ε = 1$ par: $p_{1} = \frac{e ^{- 1}}{1 + e ^{- 1}} = \frac{0.36788}{1.36788} = 0.26894$ , $p_{0} = 0.73106$ , toh $\frac{σ _{E}^{2}}{ε ^{2}} = p_{0} p_{1} = 0.73106 \times 0.26894 = 0.19661.$

Recall Solution L4.2

Kya karna hai: $C_{V} = k_{B} β^{2} σ_{E}^{2}$ . $σ_{E}^{2} = ε^{2} p_{0} p_{1}$ ke saath aur $β ε = 1$ (toh $β^{2} ε^{2} = 1$ ): $\frac{C _{V}}{k _{B}} = β^{2} ε^{2} p_{0} p_{1} = 1 \times 0.19661 = 0.19661.$ Toh $C_{V} \approx 0.197 k_{B}$ per spin is temperature par — L3.2 ke Schottky bump ki rising flank par.

Recall Solution L4.3

Key fact: product ka $ln$ sum hota hai: $ln Z = ln Z_{spin} + ln Z_{mode}$ . Machine apply karo: derivative linear hai, toh

=\langle E\rangle_{\text{spin}}+\langle E\rangle_{\text{mode}}.$$ **Result:** $\langle E\rangle=\dfrac{\varepsilon}{e^{\beta\varepsilon}+1}+\dfrac{1}{2}k_BT$. Independent subsystems ⟹ $Z$ multiply karo ⟹ energies add karo. Isliye log form itna powerful hai: factorization addition ban jaati hai.

Level 5 — Mastery

Open-ended reasoning aur ek degenerate/limiting case.

Recall Solution L5.1

Kya karna hai: $g$ excited microstates mein se har ek ka weight $e^{- β ε}$ hai, toh $Z = 1 + g e^{- β ε} .$ Differentiate karo: $\partial_{β} Z = - g ε e^{- β ε}$ , isliye

=\frac{\varepsilon}{\tfrac1g e^{\beta\varepsilon}+1}.$$ **High-$T$ limit ($\beta\to0$):** $e^{\beta\varepsilon}\to1$, toh $$\langle E\rangle\to\frac{g\varepsilon}{1+g}.$$ **Interpretation:** infinite $T$ par saare $1+g$ microstates equally likely hain; excited band mein fraction $g/(1+g)$ hai, aur unme se har ek energy $\varepsilon$ carry karta hai. $g=1$ set karne par ordinary two-level answer $\varepsilon/2$ wapas aata hai. Degeneracy high-$T$ average ko $\varepsilon$ ki taraf tilt kar deti hai.

Recall Solution L5.2

Reasoning: $⟨ E ⟩ = \sum_{i} p_{i} E_{i}$ ek weighted average hai jisme weights $p_{i} \geq 0$ hain aur $1$ tak sum karte hain. Numbers ka weighted average kabhi bhi unme se sabse chhote se chhota ya sabse bade se bada nahi ho sakta: $E_{m i n} = \sum_{i} p_{i} E_{m i n} \leq \sum_{i} p_{i} E_{i} \leq \sum_{i} p_{i} E_{m a x} = E_{m a x} .$ Consequences: jab $T \to 0$ toh sara weight $E_{m i n}$ par jaata hai isliye $⟨ E ⟩ \to E_{m i n}$ ; jab $T \to \infty$ toh saare $p_{i}$ equal ho jaate hain isliye $⟨ E ⟩ \to$ levels ka plain arithmetic mean (degeneracy se weighted). Derivative machine hamesha is window ke andar land karegi — kisi bhi answer ke liye useful sanity check.

Recall Master checklist

$ln Z$ lo ::: $Z$ se divide karna ek clean derivative mein aa jaata hai aur factorization addition ban jaati hai. $β$ mein slope ::: $\partial_{β}$ har weight $e^{- β E_{i}}$ se factor $- E_{i}$ nikaalti hai. Sign palto ::: $- E_{i}$ cancel hota hai, $E_{i} \geq 0$ ke liye positive energy milti hai. Doosra slope ::: $\partial_{β}^{2} ln Z = σ_{E}^{2}$ , aur $C_{V} = k_{B} β^{2} σ_{E}^{2}$ . Hamesha sanity-check karo ::: finite spectra ke liye $E_{m i n} \leq ⟨ E ⟩ \leq E_{m a x}$ .