Canonical ensemble — partition function Z — Hinglish notes

2.4.10 · Physics › Thermodynamics & Statistical Mechanics (Advanced)

Canonical ensemble KYA hai?

Comparison:

Microcanonical $(N, V, E)$ : energy fixed.
Canonical $(N, V, T)$ : temperature fixed, energy fluctuate karti hai. ← yahi note hai.
Grand canonical $(μ, V, T)$ : particles aur energy dono fluctuate karte hain.

Energy $E_{i}$ wale state ko $e^{- β E_{i}}$ weight KYU milta hai?

Yahi sab kuch ka core hai. Isko derive karo, kabhi memorize mat karo.

First principles se derivation:

$P_{i} \propto Ω_{R} (E_{tot} - E_{i})$

jahan $Ω_{R}$ reservoir ke microstates ki sankhya hai. Entropy $S_{R} = k_{B} ln Ω_{R}$ se kaam karo, toh $Ω_{R} = e^{S_{R} / k_{B}}$ . Kyunki $E_{i} ≪ E_{tot}$ , $S_{R}$ ko Taylor-expand karo:

$S_{R} (E_{tot} - E_{i}) \approx S_{R} (E_{tot}) - E_{i} \frac{\partial S _{R}}{\partial E}_{E_{tot}}$

Lekin thermodynamically $\frac{\partial S _{R}}{\partial E} = \frac{1}{T}$ (temperature ki definition!). Toh:

$S_{R} (E_{tot} - E_{i}) \approx S_{R} (E_{tot}) - \frac{E _{i}}{T}$

Isliye:

$P_{i} \propto exp (\frac{S _{R}}{k _{B}}) = exp (\frac{S _{R} ( E _{tot} )}{k _{B}}) e^{- E_{i} / (k_{B} T)}$

Pehla factor ek constant hai (ye $i$ par depend nahi karta). $β \equiv \frac{1}{k _{B} T}$ define karke:

$P_{i} \propto e^{- β E_{i}} — the Boltzmann factor$

Normalize kaise kare: $Z$ define karna

Probabilities ka sum 1 hona chahiye. Toh hume ek normalizer chahiye:

$P_{i} = \frac{e ^{- β E_{i}}}{j \sum e ^{- β E_{j}}}$

"Partition function" naam (German Zustandssumme = "sum over states") literal hai: yeh states par ek sum hai.

$Z$ master key KYU hai: thermodynamics kaise nikalte hain

Average energy — derive karo:

$⟨ E ⟩ = \sum_{i} E_{i} P_{i} = \frac{\sum _{i} E _{i} e ^{- β E_{i}}}{Z}$

Dhyan do ki $\sum_{i} E_{i} e^{- β E_{i}} = - \frac{\partial}{\partial β} \sum_{i} e^{- β E_{i}} = - \frac{\partial Z}{\partial β}$ . Isliye:

$⟨ E ⟩ = - \frac{1}{Z} \frac{\partial Z}{\partial β} = - \frac{\partial ln Z}{\partial β}$

Energy fluctuations (aur heat capacity tak ek raasta):

\sigma_E^2 = \langle E^2\rangle - \langle E\rangle^2 = \frac{\partial^2 \ln Z}{\partial\beta^2} = -\frac{\partial \langle E\rangle}{\partial \beta}$$ $C_V = \dfrac{\partial\langle E\rangle}{\partial T}$ aur $\dfrac{\partial}{\partial\beta} = -k_BT^2\dfrac{\partial}{\partial T}$ use karke: $$\boxed{\sigma_E^2 = k_B T^2 C_V}$$ **Helmholtz free energy** — classical thermodynamics se sabse saaf connection: $$\boxed{F = -k_B T \ln Z}$$ $F$ se sab kuch milta hai: $S = -\left(\dfrac{\partial F}{\partial T}\right)_{V}$, $\;P = -\left(\dfrac{\partial F}{\partial V}\right)_{T}$, $\;\langle E\rangle = F + TS$. ![[2.4.10-Canonical-ensemble-—-partition-function-Z.png]] --- ## Worked Example 1 — Two-level system ("qubit") Ek single particle ke do states hain: energy $0$ aur energy $\varepsilon$. **Step 1 — $Z$ likho.** *Kyun?* $Z$ sirf do states par Boltzmann factors ka sum hai. $$Z = e^{-\beta\cdot 0} + e^{-\beta\varepsilon} = 1 + e^{-\beta\varepsilon}$$ **Step 2 — Average energy.** *$-\partial_\beta \ln Z$ kyun use karein?* Yeh derived shortcut hai. $$\ln Z = \ln(1+e^{-\beta\varepsilon}), \quad \langle E\rangle = -\frac{\partial \ln Z}{\partial\beta} = \frac{\varepsilon e^{-\beta\varepsilon}}{1+e^{-\beta\varepsilon}} = \frac{\varepsilon}{e^{\beta\varepsilon}+1}$$ **Step 3 — Limits check karo.** *Kyun?* Limits algebra errors turant pakad lete hain. - $T\to 0\;(\beta\to\infty)$: $\langle E\rangle\to 0$ ✓ (ground state mein freeze). - $T\to\infty\;(\beta\to 0)$: $\langle E\rangle\to \varepsilon/2$ ✓ (dono states equally likely). --- ## Worked Example 2 — Ek classical harmonic oscillator (1D) $H = \dfrac{p^2}{2m} + \dfrac{1}{2}m\omega^2 x^2$. **Step 1 — Classical $Z$.** *Integrate kyun karte hain?* Phase space continuous hai. $$Z = \frac{1}{h}\int_{-\infty}^{\infty}\!\!\int_{-\infty}^{\infty} e^{-\beta\left(\frac{p^2}{2m}+\frac{1}{2}m\omega^2 x^2\right)}\,dx\,dp$$ **Step 2 — Gaussian integrals.** *Yeh kyun kaam karta hai?* Yeh do $\int e^{-ax^2}dx=\sqrt{\pi/a}$ mein factorize ho jaata hai. $$\int e^{-\beta p^2/2m}dp = \sqrt{\frac{2\pi m}{\beta}}, \qquad \int e^{-\beta m\omega^2 x^2/2}dx = \sqrt{\frac{2\pi}{\beta m\omega^2}}$$ $$Z = \frac{1}{h}\sqrt{\frac{2\pi m}{\beta}}\sqrt{\frac{2\pi}{\beta m\omega^2}} = \frac{2\pi}{h\beta\omega} = \frac{k_B T}{\hbar\omega}$$ **Step 3 — Energy.** *Differentiate kyun karte hain?* Wahi master formula. $$\langle E\rangle = -\frac{\partial \ln Z}{\partial\beta} = -\frac{\partial}{\partial\beta}\big(-\ln\beta + \text{const}\big) = \frac{1}{\beta} = k_B T$$ Yeh **equipartition** hai: 2 quadratic DOF $\times\, \tfrac12 k_BT = k_BT$. ✓ --- ## Worked Example 3 — $N$ independent particles & factorization > [!intuition] Products KYU? > Agar particles independent hain, toh total energy ek **sum** hai $E = \sum_k \epsilon_k$. Sum ka exponential **product** hota hai exponentials ka, aur saari configurations par sum factorize ho jaata hai. $N$ **distinguishable** independent identical subsystems ke liye, jahan har ek ka single-particle partition function $Z_1$ hai: $$Z_N = Z_1^N$$ $N$ **indistinguishable** particles ke liye (classical ideal gas), overcounting $N!$ se divide karo: $$\boxed{Z_N = \frac{Z_1^N}{N!}}$$ *$N!$ kyun?* Do identical particles swap karne se same physical state milti hai; integral ne ise $N!$ baar count kiya tha. (Yeh **Gibbs paradox** fix karta hai.) --- ## Forecast-then-Verify > [!example] Compute karne se pehle predict karo > **Q:** Two-level system ke liye, kis $T$ par $\langle E\rangle = \varepsilon/4$ hoga? > **Forecast:** Chahiye $\dfrac{\varepsilon}{e^{\beta\varepsilon}+1}=\dfrac{\varepsilon}{4}\Rightarrow e^{\beta\varepsilon}=3\Rightarrow \beta\varepsilon=\ln 3\Rightarrow T = \dfrac{\varepsilon}{k_B\ln 3}$. > **Verify:** Plug back karo: $\langle E\rangle = \varepsilon/(3+1)=\varepsilon/4$ ✓. --- ## Common mistakes (Steel-manned) > [!mistake] "$Z$ ek probability hai, toh yeh ≤ 1 honi chahiye." > **Kyun sahi lagta hai:** Boltzmann factors probabilities jaisi dikhti hain. **Reality:** $Z$ ek *normalizer* hai, weights ka sum — yeh bahut bada ho sakta hai (roughly thermally accessible states count karta hai). Probability hai $e^{-\beta E_i}/Z$, *woh* ≤ 1 hoti hai. > [!mistake] "$\langle E\rangle = +\partial_\beta \ln Z$ use karo." > **Kyun sahi lagta hai:** Differentiation energy deta hai, sign chhootna easy hai. **Fix:** Har term $e^{-\beta E_i}$ hai; $\partial_\beta$ se $-E_i$ neeche aata hai. Minus sign physical hai — ek baar derive karo aur kabhi nahi bhoologe: $\langle E\rangle = -\partial_\beta \ln Z$. > [!mistake] "Ideal gas ke liye $1/N!$ bhool jaana." > **Kyun sahi lagta hai:** $Z_1^N$ complete lagta hai. **Fix:** Classical identical particles indistinguishable hote hain; $1/N!$ ke bina entropy extensive nahi hoti (Gibbs paradox). Hamesha poochho: *kya main particles alag pehchaan sakta hoon?* > [!mistake] "$Z$ ko energy **levels** par sum karo lekin degeneracy bhool jao." > **Kyun sahi lagta hai:** Levels "states" jaisi lagti hain. **Fix:** $Z=\sum_i e^{-\beta E_i}$ **microstates** par hai. Agar levels par sum karo, har ek ko $g_n$ se weight karo: $Z=\sum_n g_n e^{-\beta E_n}$. --- ## #flashcards/physics Canonical ensemble define karo (controlled variables). ::: Ek system ki copies jo heat bath ke thermal contact mein hain; $(N,V,T)$ fixed, energy $E$ fluctuate karti hai. Microstate $i$ ko weight $e^{-\beta E_i}$ kyun milta hai? ::: Reservoir entropy $S_R(E_{tot}-E_i)$ ko Taylor-expand karne par $\partial S_R/\partial E = 1/T$ se milta hai $P_i \propto e^{-E_i/k_BT}$. Partition function $Z$ define karo. ::: $Z=\sum_i e^{-\beta E_i}$, saare microstates par sum; Boltzmann probabilities ka normalizer. $\beta$ kya hai? ::: $\beta = 1/(k_B T)$. Microstate $i$ ki probability? ::: $P_i = e^{-\beta E_i}/Z$. $Z$ se average energy? ::: $\langle E\rangle = -\partial \ln Z/\partial\beta$. $Z$ se Helmholtz free energy? ::: $F = -k_B T \ln Z$. Energy fluctuation ka heat capacity se relation? ::: $\sigma_E^2 = \langle E^2\rangle-\langle E\rangle^2 = k_B T^2 C_V$. $N$ indistinguishable independent classical particles ka $Z$? ::: $Z_N = Z_1^N / N!$. Two-level system $(0,\varepsilon)$ ka $Z$? ::: $Z = 1+e^{-\beta\varepsilon}$. 1D classical harmonic oscillator ka $\langle E\rangle$? ::: $k_BT$ (equipartition, 2 quadratic DOF). Identical particles ke liye $N!$ se kyun divide karte hain? ::: Indistinguishability permutations ko $N!$ baar overcount karta hai; Gibbs paradox fix karta hai / $S$ ko extensive banata hai. $Z$ se entropy kaise nikalein? ::: $S=-(\partial F/\partial T)_V$ jahan $F=-k_BT\ln Z$. --- > [!recall]- Feynman: 12-saal ke bachche ko samjhao > Ek chhota sa toy imagine karo jo alag-alag "settings" mein ho sakta hai, aur har setting mein kuch energy lagti hai. Toy ek garam kamre mein baitha hai. Kamra energy ka ek bada piggy bank hai. Agar kisi setting mein bahut zyada energy lage, toh piggy bank ke paas kam bacha aur khud ko arrange karne ke kam tarike — toh toy kabhi-kabhi hi woh mehngi setting choose karta hai. Garam kamra → toy bahut saari settings try karta hai; thanda kamra → toy sabse sasti setting mein rehta hai. Partition function $Z$ bas **har setting ke "likeliness scores" ko add karna** hai. Ek baar add kar lo, toh average energy, kitna jitter hai, sab kuch pata chal jaata hai — bina har setting dobara check kiye. Yeh ek number hai jo sab jaanta hai. > [!mnemonic] Toolbox yaad rakho > **"Z Builds Free Energy, Fluctuations Follow."** > $Z$ → **B**oltzmann weights → **F**ree energy $F=-k_BT\ln Z$ → **F**luctuations $\sigma_E^2=k_BT^2 C_V$. > Aur master derivative: *"minus log-Z by beta gives E"* → $\langle E\rangle=-\partial_\beta\ln Z$. ## Connections - [[Boltzmann distribution]] — woh $e^{-\beta E_i}$ weights jo $Z$ normalize karta hai. - [[Microcanonical ensemble]] — fixed-energy cousin; canonical us par reduce ho jaata hai jab bath badhta hai. - [[Grand canonical ensemble — grand partition function]] — $\mu$ ke zariye particle exchange add karo. - [[Helmholtz free energy F]] — thermodynamic potential $-k_BT\ln Z$. - [[Equipartition theorem]] — har quadratic DOF $\tfrac12 k_BT$ contribute karta hai, $Z$ se derivable. - [[Gibbs paradox]] — $1/N!$ kyun matter karta hai. - [[Heat capacity and fluctuations]] — $\sigma_E^2 = k_BT^2 C_V$. ## 🖼️ Concept Map ```mermaid flowchart TD POST[Fundamental postulate: equal microstates] -->|applied to| ISO[System plus reservoir isolated] ISO -->|Pi proportional to| OMEGA[Reservoir microstates Omega_R] OMEGA -->|via S equals kB ln Omega| ENT[Reservoir entropy S_R] ENT -->|Taylor expand small Ei| EXP[S_R minus Ei over T] TEMP[Temperature: dS/dE equals 1/T] -->|defines slope| EXP EXP -->|exponentiate| BOLTZ[Boltzmann factor e^-beta Ei] BOLTZ -->|normalize sum to 1| Z[Partition function Z] Z -->|weights states in| CANON[Canonical ensemble N,V,T] CANON -->|contrast fixed E| MICRO[Microcanonical N,V,E] CANON -->|contrast variable N| GRAND[Grand canonical mu,V,T] Z -->|differentiate| THERMO[All thermodynamic quantities] ```

Canonical ensemble KYA hai?

Energy Ei​ wale state ko e−βEi​ weight KYU milta hai?

Normalize kaise kare: Z define karna

Z master key KYU hai: thermodynamics kaise nikalte hain

Energy $E_{i}$ wale state ko $e^{- β E_{i}}$ weight KYU milta hai?

Normalize kaise kare: $Z$ define karna

$Z$ master key KYU hai: thermodynamics kaise nikalte hain