2.4.11 · D3 · HinglishThermodynamics & Statistical Mechanics (Advanced)
Worked examples — Average energy from partition function
2.4.11 · D3· Physics › Thermodynamics & Statistical Mechanics (Advanced) › Average energy from partition function
Ye page parent formula ki drill floor hai Yahan Boltzmann's constant hai (sirf ek number jo temperature ko energy mein convert karta hai), aur "coldness" hai — bada matlab thanda, chota matlab garam. Definition dhyan se dekho: iska matlab hai , ek aisi baat jis par hum tab takate hain jab koi limit deti hai. Neeche use kiya gaya har symbol parent note mein kamaya gaya hai; agar derivation chahiye, pehle wahan wapas jao. Yahan hum sirf tool use karte hain, har us situation mein jo ye face kar sakta hai.
Scenario matrix
Kuch bhi solve karne se pehle, aao list karte hain ki ye ek formula kitne tarah ki problems se mil sakta hai. Formula kabhi nahi badalta — lekin (Boltzmann weights ka sum) situation ke hisaab se alag dikhta hai, aur limits jo hum sanity-check karte hain wo bhi situation par depend karti hain.
| Cell | Case class | Kya special hai | Example |
|---|---|---|---|
| A | Finite discrete sum | = thodi si terms | Ex. 1 (two-level), Ex. 2 (three-level) |
| B | Degenerate levels | ek energy par kai states hit hoti hain → count se multiply karo | Ex. 3 |
| C | Infinite discrete sum | geometric/known series chahiye | Ex. 4 (oscillator) |
| D | Continuous integral | = ek integral, Gaussian use karo | Ex. 5 (equipartition) |
| E | Cold limit () | system ground state mein freeze ho jaata hai | checks in Ex. 1, 4 |
| F | Hot limit () | saari states equally likely | checks in Ex. 1, 3 |
| G | Shifted zero of energy | har mein ek constant add karo | Ex. 6 |
| H | Real-world word problem | physics → energies → mein translate karo | Ex. 7 |
| I | Exam twist (given , find ) | second derivative / fluctuations | Ex. 8 |
Game ka rule: chhe se aath worked examples jo milke har cell ko touch karte hain. Har ek batata hai ki wo kis cell(s) mein aata hai.
Example 1 — Two-level system · cells A, E, F
Step 1 — banao. Do microstates hain, toh sum mein do terms hain: Ye step kyun? states par sum hai, aur yahan "saari states" ka matlab exactly do hai.
Step 2 — Log lo aur differentiate karo.
= -\frac{-\varepsilon e^{-\beta\varepsilon}}{1+e^{-\beta\varepsilon}} = \frac{\varepsilon\,e^{-\beta\varepsilon}}{1+e^{-\beta\varepsilon}}.$$ *Ye step kyun?* $\partial_\beta \ln Z = \frac1Z\partial_\beta Z$, aur $\partial_\beta e^{-\beta\varepsilon}=-\varepsilon e^{-\beta\varepsilon}$ — exponent mein minus formula ke $-$ ko cancel karta hai, ek positive answer deta hai (energies $\ge0$ hain, toh ye *zaroor* positive hona chahiye). **Step 3 — Saaf karo.** Top aur bottom ko $e^{\beta\varepsilon}$ se multiply karo: $$\boxed{\;\langle E\rangle = \frac{\varepsilon}{e^{\beta\varepsilon}+1}\;}$$ *Ye step kyun?* $e^{\beta\varepsilon}/e^{\beta\varepsilon}=1$ se multiply karna (value mein koi change nahi) numerator se chota exponential saaf kar deta hai, aur tidy "Fermi-like" form $\varepsilon/(e^{\beta\varepsilon}+1)$ deta hai jiske limits aankhon se padhe aasaan hain. > [!verify] Verify — do forecasts > - **Cold, $\beta\to\infty$:** $e^{\beta\varepsilon}\to\infty$, toh $\langle E\rangle\to 0$. ✓ (ground state mein frozen) > - **Hot, $\beta\to 0$:** $e^{\beta\varepsilon}\to 1$, toh $\langle E\rangle\to \varepsilon/2$. ✓ (dono states ka average) > [!intuition] Neeche wala figure padhna > Figure $\langle E\rangle/\varepsilon$ (vertical axis, $0$ se $\varepsilon/2$ tak) ko temperature $k_BT/\varepsilon$ (horizontal axis, left par thanda, right par garam) ke against plot karta hai. **Mint curve** > humara result $1/(e^{\varepsilon/k_BT}+1)$ hai. Iske do ends dekho: bahut left par (cold) lavender > arrow use $0$ ke paas pinned dikhata hai — spin ground state mein frozen hai; bahut right par (hot) coral dashed line ceiling $\varepsilon/2$ mark karti hai jis par curve flat ho jaata hai — dono states equally > likely. Pura graph do forecasts ko visible banata hai: $0$ se $\varepsilon/2$ tak ek smooth climb. ![[deepdives/dd-physics-2.4.11-d3-s01.png]] --- ## Example 2 — Three-level ladder · cell A > [!example] Statement > Energies $0,\ \varepsilon,\ 2\varepsilon$ (har ek state). $\langle E\rangle$ us special temperature par dhundho > jahan $\beta\varepsilon = \ln 2$ (yaani $e^{-\beta\varepsilon}=\tfrac12$) ho. > [!intuition] Forecast > Weights $1,\ \tfrac12,\ \tfrac14$ ke saath top level sabse kam likely hai, toh average middle energy $\varepsilon$ se **neeche** hona chahiye. Kuch $0.5\varepsilon$–$0.7\varepsilon$ jaisa expect karo. **Step 1 — $Z$ banao.** Teen terms, aur $x=e^{-\beta\varepsilon}=\tfrac12$ set karo: $$Z = 1 + x + x^2 = 1 + \tfrac12 + \tfrac14 = \tfrac74.$$ *Ye step kyun?* Energy $n\varepsilon$ ki har level weight $e^{-\beta n\varepsilon}=x^n$ contribute karti hai. **Step 2 — Directly average karo (derivative ka cross-check).** Kyunki states $0,\varepsilon,2\varepsilon$ hain weights $1,\tfrac12,\tfrac14$ ke saath: $$\langle E\rangle = \frac{0\cdot 1 + \varepsilon\cdot\tfrac12 + 2\varepsilon\cdot\tfrac14}{7/4} = \frac{\varepsilon}{7/4} = \frac{4\varepsilon}{7}.$$ *Ye step kyun?* $\langle E\rangle=\sum_i p_i E_i$ hamesha available hai; yahan ye quick hai aur $Z$-method ko confirm karta hai. > [!verify] Verify > $4/7 \approx 0.571$, jo $\varepsilon$ se neeche hai — forecast se match karta hai. Aur $\varepsilon$ se neeche hona sense banta hai: sasta ground state mean ko neeche kheenchta hai. --- ## Example 3 — Degenerate excited level + hot limit · cells B, F > [!example] Statement > Ground energy $0$ (ek state) aur ek excited energy $\varepsilon$ jo **three-fold degenerate** hai > (teen alag states saari energy $\varepsilon$ share karti hain). $\langle E\rangle$ aur uski hot-limit value dhundho. > [!intuition] Forecast > Excited level tak teen doors jaati hain lekin ground level tak sirf ek. Toh hot limit bhi > $50$–$50$ split **nahi** karegi; excited level $3:1$ se favoured hai, hot average ko $\tfrac34\varepsilon$ ki taraf kheenchta hai. **Step 1 — Multiplicity ke saath $Z$ banao.** $$Z = 1 + 3e^{-\beta\varepsilon}.$$ *Ye step kyun?* Degeneracy $g$ ka matlab hai $g$ identical terms, toh hum us Boltzmann weight ko sirf $3$ se multiply karte hain. **Step 2 — Differentiate karo.** $$\langle E\rangle = -\frac{\partial}{\partial\beta}\ln\!\big(1+3e^{-\beta\varepsilon}\big) = \frac{3\varepsilon\,e^{-\beta\varepsilon}}{1+3e^{-\beta\varepsilon}} = \frac{3\varepsilon}{e^{\beta\varepsilon}+3}.$$ *Ye step kyun?* Same log-slope-flip; $3$ bina derivative se chhua untouched saath chala jaata hai. > [!verify] Verify — hot limit > $\beta\to 0$: $e^{\beta\varepsilon}\to1$, toh $\langle E\rangle\to \dfrac{3\varepsilon}{1+3}=\dfrac34\varepsilon$. ✓ $3:1$ forecast se match karta hai (teen excited doors vs ek ground door). > Cold limit $\beta\to\infty$: $\langle E\rangle\to 0$ ✓ (ground state tab bhi jeet jaata hai jab wo akela sasta option ho). Dekho [[Two-level system / Schottky anomaly]] ki degeneracy heat-capacity bump ko kaise reshape karti hai. --- ## Example 4 — Quantum oscillator: infinite sum + cold freeze · cells C, E > [!example] Statement > Ek 1D quantum oscillator ke levels $E_n = n\hbar\omega$, $n=0,1,2,\dots$ (infinitely many) hain. > $\langle E\rangle$ aur $T\to 0$ par uska behaviour dhundho. > [!intuition] Forecast > Infinitely many levels hain, lekin unche wale exponentially unlikely hain. Jab $T\to0$ sab kuch > ground level $n=0$ par collapse ho jaana chahiye, $\langle E\rangle\to 0$ deta hua. **Step 1 — Geometric series sum karo.** $r=e^{-\beta\hbar\omega}$ aur $|r|<1$ ke saath: $$Z = \sum_{n=0}^{\infty} r^n = \frac{1}{1-r} = \frac{1}{1-e^{-\beta\hbar\omega}}.$$ *Ye step kyun?* Ye akela jagah hai jahan hume ek **jaani hui infinite sum** chahiye: $\sum r^n = 1/(1-r)$. Ye converge karti hai kyunki $r<1$ hai (positive energies, positive $\beta$). **Step 2 — Log lo aur differentiate karo.** $\ln Z = -\ln(1-e^{-\beta\hbar\omega})$, toh $$\langle E\rangle = -\frac{\partial}{\partial\beta}\Big[-\ln(1-e^{-\beta\hbar\omega})\Big] = \frac{\hbar\omega\, e^{-\beta\hbar\omega}}{1-e^{-\beta\hbar\omega}} = \frac{\hbar\omega}{e^{\beta\hbar\omega}-1}.$$ *Ye step kyun?* Log par chain rule numerator $\hbar\omega e^{-\beta\hbar\omega}$ deta hai. > [!verify] Verify > - **Cold, $\beta\to\infty$:** denominator $e^{\beta\hbar\omega}-1\to\infty$, toh $\langle E\rangle\to 0$. ✓ > - **Hot, $\beta\to 0$:** $e^{\beta\hbar\omega}-1\approx\beta\hbar\omega$, toh $\langle E\rangle\to \hbar\omega/(\beta\hbar\omega)=1/\beta=k_BT$. ✓ (ek oscillator ke liye classical [[Equipartition theorem]] value; [[Quantum harmonic oscillator — thermal]] mein gehri treatment). --- ## Example 5 — Continuous quadratic mode (equipartition) · cell D > [!example] Statement > Ek degree of freedom energy $E=\tfrac12 a x^2$, $x\in(-\infty,\infty)$ ke saath (ek "spring" coordinate). > $\langle E\rangle$ dhundho. > [!intuition] Forecast > Ye classic case hai jise exactly $\tfrac12 k_BT$ per quadratic term dena chahiye — equipartition result. **Step 1 — $Z$ ab ek integral hai.** $$Z = \int_{-\infty}^{\infty} e^{-\beta a x^2/2}\,dx.$$ *Ye step kyun?* Continuous states → microstates par sum coordinate $x$ par integral ban jaata hai. **Step 2 — Substitution se Gaussian integral karo.** Master fact $\int_{-\infty}^{\infty} e^{-u^2}\,du=\sqrt{\pi}$ se shuru karo. Hamara exponent $-\tfrac12\beta a\,x^2$ hai, toh $u = \sqrt{\tfrac{\beta a}{2}}\,x$ lo, jo $du = \sqrt{\tfrac{\beta a}{2}}\,dx$ deta hai, yaani $dx = \sqrt{\tfrac{2}{\beta a}}\,du$. Tab $$Z = \int_{-\infty}^{\infty} e^{-u^2}\,\sqrt{\tfrac{2}{\beta a}}\,du = \sqrt{\tfrac{2}{\beta a}}\int_{-\infty}^{\infty} e^{-u^2}\,du = \sqrt{\tfrac{2}{\beta a}}\,\sqrt{\pi} = \sqrt{\frac{2\pi}{\beta a}}\ \propto\ \beta^{-1/2}.$$ *Ye step kyun?* Substitution hamara integral standard wale mein badal deta hai; prefactor $\sqrt{2/(\beta a)}$ exactly $dx/du$ bookkeeping hai, isliye answer $\sqrt{2\pi/(\beta a)}$ hai aur sirf $\sqrt{\pi}$ nahi. **Step 3 — Sirf $\beta$-dependence matter karti hai.** $\ln Z = -\tfrac12\ln\beta + \text{const}$, toh $$\langle E\rangle = -\frac{\partial \ln Z}{\partial\beta} = -\left(-\frac{1}{2\beta}\right)=\frac{1}{2\beta}=\tfrac12 k_BT.$$ *Ye step kyun?* Koi bhi $\beta$-independent constant ($\ln(2\pi/a)/2$ wala part) $\partial_\beta$ ke under gayab ho jaata hai — hume sirf $\beta$ ki power chahiye thi. Aakhri equality definition $\beta\equiv 1/(k_BT)$ use karti hai, toh $1/\beta = k_BT$ exactly hai (is page ke top se restate kiya gaya), $1/(2\beta)$ ko $\tfrac12 k_BT$ mein badalta hua. > [!verify] Verify > Kisi energy unit mein $k_BT=4$ wale $T$ par, $\langle E\rangle = 2$. Units: $[k_BT]=$ energy ✓. Har quadratic term = $\tfrac12 k_BT$ — do lines mein [[Equipartition theorem]]. --- ## Example 6 — Energy ka zero shift karo · cell G > [!example] Statement > Example 1 ka two-level system lo lekin energies ek shifted origin se measure karo: states ki energies > $E_0$ aur $E_0+\varepsilon$ hain (dono ek constant $E_0$ se upar uthaye gaye). Dikhao ki $\langle E\rangle$ = (purana answer) $+\,E_0$. > [!intuition] Forecast > Har energy ko $E_0$ se upar slide karne se average bhi exactly $E_0$ se upar slide hona chahiye — physically kuch nahi badla, sirf ruler ka zero badla. **Step 1 — Naya partition function.** $$Z' = e^{-\beta E_0} + e^{-\beta(E_0+\varepsilon)} = e^{-\beta E_0}\big(1 + e^{-\beta\varepsilon}\big) = e^{-\beta E_0} Z,$$ jahan $Z=1+e^{-\beta\varepsilon}$ purana wala hai. *Ye step kyun?* Common factor $e^{-\beta E_0}$ har term se bahar aa jaata hai. **Step 2 — Logs products ko sums mein badal dete hain.** $$\ln Z' = -\beta E_0 + \ln Z.$$ *Ye step kyun?* Exactly isliye $\ln$ form itna powerful hai — constant shift ek additive $-\beta E_0$ term ban jaata hai. **Step 3 — Differentiate karo.** $$\langle E\rangle' = -\frac{\partial}{\partial\beta}\big(-\beta E_0 + \ln Z\big) = E_0 - \frac{\partial \ln Z}{\partial\beta} = E_0 + \frac{\varepsilon}{e^{\beta\varepsilon}+1}.$$ *Ye step kyun?* $\partial_\beta(-\beta E_0)=-E_0$ hai, aur formula ka leading minus use flip karke $+E_0$ bana deta hai. > [!verify] Verify > $E_0=5$ ke saath, $\varepsilon$ aisa ki $e^{\beta\varepsilon}=1$ (hot), purana answer $=\varepsilon/2$; naya $=5+\varepsilon/2$. Shift exactly $E_0$ hai. ✓ Yahan bhi note karo ki $C_V$ constant shift se **unchanged** hai (second derivative $-\beta E_0$ ko maar deta hai). --- ## Example 7 — Word problem: ek molecule ke do conformations · cell H > [!example] Statement > Ek protein hinge ka ek **folded** state hai (energy $0$) aur ek **open** state hai (energy $\varepsilon = 2\,k_B T_{\text{room}}$, > yaani room temperature par $\beta\varepsilon=2$). Room $T$ par $\langle E\rangle/\varepsilon$ kitna stored hai? > [!intuition] Forecast > Open state $2k_BT$ cost karta hai — thermal energy se kaafi zyada — toh ye bahut rare hona chahiye. Expect karo ki $\langle E\rangle/\varepsilon$ chota hoga, shayad $0.1$ ke aas paas. **Step 1 — Energies mein translate karo.** Do states, $0$ aur $\varepsilon$ — Example 1 jaisi hi shape. *Ye step kyun?* Word problems "states aur unki energies list karo" mein reduce ho jaate hain. **Step 2 — Example 1 ka result use karo** $\beta\varepsilon = 2$ ke saath: $$\frac{\langle E\rangle}{\varepsilon} = \frac{1}{e^{\beta\varepsilon}+1} = \frac{1}{e^{2}+1}.$$ *Ye step kyun?* Hum pehle hi $\langle E\rangle=\varepsilon/(e^{\beta\varepsilon}+1)$ derive kar chuke hain; sirf number plug in karo. **Step 3 — Evaluate karo.** $e^2\approx 7.389$, toh $\dfrac{1}{7.389+1}=\dfrac{1}{8.389}\approx 0.119$. *Ye step kyun?* Hum symbolic fraction ko number mein badal dete hain taaki use apne forecast se aur ek measurable probability se compare kar sakein — ek formula tabhi *prediction* banta hai jab wo ek number ho. Hume $e^2$ chahiye kyunki $\beta\varepsilon=2$ exponent mein baitha hai, aur $e^2$ sirf fixed constant $\approx 7.389$ hai. > [!verify] Verify > $\langle E\rangle/\varepsilon\approx 0.119$ — lagbhag $12\%$, forecast ki tarah chota. Open-state Boltzmann probability ke against cross-check: $p_{\text{open}}=\dfrac{e^{-\beta\varepsilon}}{1+e^{-\beta\varepsilon}}=\dfrac{e^{-2}}{1+e^{-2}}\approx 0.119$. Ye match karte hain kyunki **sirf open state energy carry karta hai** ($E_{\text{folded}}=0$), toh $\langle E\rangle=\varepsilon\,p_{\text{open}}$, isliye $\langle E\rangle/\varepsilon=p_{\text{open}}$. ✓ Prediction–verification cycle saaf close hoti hai: energy fraction *open-state occupation probability hi hai*. --- ## Example 8 — Exam twist: given $Z$, heat capacity dhundho · cell I > [!example] Statement > Two-level system ($0,\varepsilon$) ke liye, heat capacity $C_V = \partial\langle E\rangle/\partial T$ > compute karo aur wo temperature dhundho (terms of $\varepsilon/k_B$ mein) jahan ye peak karta hai — **Schottky** peak. > [!intuition] Forecast > Heat capacity measure karta hai ki average energy $T$ mein ek nudge ke jawaab mein kitna respond karti hai. Ye dono extremes par vanish honi chahiye (cold par frozen, hot par saturated) aur isliye **beech mein kahin peak** honi chahiye, $\varepsilon/k_B$ ki order ke temperature par. > [!definition] $\sigma_E^2$ kya hai? > $\sigma_E^2$ **energy variance** hai: energy apne average ke aas paas kitna jitter karti hai. Symbols mein > $\sigma_E^2 \equiv \langle E^2\rangle - \langle E\rangle^2$, jahan $\langle E^2\rangle=\sum_i p_i E_i^2$ > *squared* energy ka average hai. Ye hamesha $\ge 0$ hota hai (spread negative nahi ho sakta). > Parent note ne dikhaya ki $\ln Z$ ka **second** $\beta$-derivative exactly ye spread manufacture karta hai: do baar differentiate karne se ek $E_i^2$ aur ek $(\sum E_i)^2$ piece nikalti hai, deta hua > $$\partial^2_\beta \ln Z = \langle E^2\rangle - \langle E\rangle^2 = \sigma_E^2.$$ > Ise $C_V=\partial\langle E\rangle/\partial T$ mein feed karke aur $\beta=1/k_BT$ use karke (toh $\partial_T = -k_B\beta^2\partial_\beta$) fluctuation identity milti hai > $$C_V = k_B\beta^2\,\sigma_E^2.$$ **Step 1 — $C_V$ lo.** Example 1 se, $\langle E\rangle=\varepsilon/(e^{\beta\varepsilon}+1)$. $T$ mein differentiate karne par (ya equivalently box se $C_V=k_B\beta^2\sigma_E^2$ use karke) milta hai $$C_V = k_B\,(\beta\varepsilon)^2\,\frac{e^{\beta\varepsilon}}{(e^{\beta\varepsilon}+1)^2}.$$ *Ye step kyun?* $\langle E\rangle$ $T$ par sirf combination $\beta\varepsilon$ ke through depend karta hai; chain rule $(\beta\varepsilon)^2$ prefactor neeche laati hai (ye [[Heat capacity and energy fluctuations]] identity concrete banti hui hai). **Step 2 — Peak dhundho.** $u=\beta\varepsilon$ lo, toh $C_V(u)=k_B\,u^2 e^{u}/(e^{u}+1)^2$. Maximum locate karne ke liye $dC_V/du=0$ set karo. Wo derivative (product + quotient rule) carry out karke aur common positive factor $u\,e^{u}/(e^{u}+1)^2$ divide karke tidy condition milti hai $$u = 2\,\frac{e^{u}-1}{e^{u}+1} = 2\tanh(u/2).$$ *Ye step kyun?* Peak wahan hai jahan slope zero ho, toh hum $dC_V/du=0$ solve karte hain. Messy derivative factorise hoti hai: $\tfrac{e^u-1}{e^u+1}$ combination exactly $\tanh(u/2)$ hai ($\tanh$ ki definition), isliye condition $u=2\tanh(u/2)$ mein collapse ho jaati hai. Numerically solve karne par: $u^\ast \approx 2.399$, yaani $k_BT^\ast \approx \varepsilon/2.399 \approx 0.417\,\varepsilon$. > [!verify] Verify > Peak $k_BT^\ast\approx 0.417\,\varepsilon$ par — $\varepsilon/k_B$ ki order ka aur do zero limits ke beech firmly, jaise forecast tha. $u^\ast=2.399$ wapas plug karne par $C_V^\ast\approx 0.439\,k_B$ milta hai. Dono cold ($u\to\infty$) aur hot ($u\to0$) limits $C_V\to0$ deti hain ✓ — hallmark [[Two-level system / Schottky anomaly]] bump. ✓ --- > [!recall]- Har example ne kaun sa cell fill kiya? > Ex.1 ::: A, E, F (finite discrete + dono limits) > Ex.2 ::: A (three-level finite sum) > Ex.3 ::: B, F (degeneracy + hot limit) > Ex.4 ::: C, E (infinite geometric sum + cold freeze) > Ex.5 ::: D (continuous Gaussian integral, equipartition) > Ex.6 ::: G (shifted energy zero) > Ex.7 ::: H (real-world word problem) > Ex.8 ::: I (given $Z$ → heat capacity, Schottky peak) > [!mnemonic] Universal recipe > **"List → weight → sum → log → slope → flip."** States list karo, $e^{-\beta E}$ se weight karo, $Z$ mein sum karo, > $\ln$ lo, $\beta$-slope lo, sign flip karo: upar har cell ke liye same chhe moves. ## Active recall > [!recall] Self-test > quick-1 ::: Three-fold degeneracy hot-limit energy ko $\varepsilon/2$ se upar kyun push karti hai? Kyunki excited level ke teen doors hain vs ground ke ek, toh hot average $\tfrac34\varepsilon$ ki taraf trend karta hai. > quick-2 ::: Saari energies ko $E_0$ shift karne se $C_V$ unchanged kyun rehta hai? Shift $\ln Z$ mein $-\beta E_0$ add karta hai, jiska second $\beta$-derivative zero hai. > quick-3 ::: Oscillator mein, hot limit $k_BT$ kyun deta hai $\tfrac12 k_BT$ nahi? Ek vibrational mode ke *do* quadratic terms hote hain (kinetic + potential), har ek $\tfrac12 k_BT$.