Exercises — Canonical ensemble — partition function Z

Hum kya karte hain: har microstate ke liye ek Boltzmann factor ka sum. Yahan do hain. $Z=e^{-\beta (+\epsilon )}+e^{-\beta (-\epsilon )}=e^{-\beta \epsilon }+e^{+\beta \epsilon }=2\cosh (\beta \epsilon )$ $\cosh$ kyun? Kyunki $\cosh x=\frac{1}{2}(e^x+e^{-x})$ literally do exponentials ka average times 2 hai — yeh natural shape tab aati hai jab do states zero energy ke symmetrically dono taraf baithte hain. "Up" ki probability: $P_{\uparrow }=\frac{e^{+\beta \epsilon }}{2\cosh (\beta \epsilon )}$ Sanity check: jab $T\to 0$ ($\beta \to \infty$) toh $e^{+\beta \epsilon }$ term doosre ko crush kar deta hai, isliye $P_{\uparrow }\to 1$ — spin apni lowest energy $-\epsilon$ mein freeze ho jaata hai. ✓

Microstates ka sum (teen terms, ek har microstate ke liye): $Z=e^{-\beta \cdot 0}+e^{-\beta \epsilon }+e^{-\beta \epsilon }=1+2e^{-\beta \epsilon }$ Degeneracy $g_n$ use karke (ek energy $E_n$ share karne wale microstates ki count): yahan energy $0$ par $g_0=1$, aur energy $\epsilon$ par $g_1=2$. $Z={\sum }_n{g}_n{e}^{-\beta E_n}=1\cdot e^0+2\cdot e^{-\beta \epsilon }=1+2e^{-\beta \epsilon }$ Dono agree karte hain. Degeneracy sirf bookkeeping hai: yeh tumhe har energy ke liye ek term likhne deti hai instead of har microstate ke liye.

Hum kya karte hain: $⟨E⟩=-{\partial }_{\beta }\ln Z$ apply karo. $\ln Z=\ln 2+\ln \cosh (\beta \epsilon )$ $\ln \cosh$ differentiate kyun karein? Kyunki $\frac{d}{du}\ln \cosh u=\tanh u$ — $\cosh$ ka derivative $\sinh$ hai, $\cosh$ se divide karne par $\tanh$ milta hai. Toh: $⟨E⟩=-\frac{\partial }{\partial \beta }[\ln 2+\ln \cosh (\beta \epsilon )]=-\epsilon \tanh (\beta \epsilon )$ Limits:

• $T\to 0$ ($\beta \to \infty$): $\tanh \to 1$, isliye $⟨E⟩\to -\epsilon$ ✓ (ground state $-\epsilon$ mein lock ho gaya).
• $T\to \infty$ ($\beta \to 0$): $\tanh \to 0$, isliye $⟨E⟩\to 0$ ✓ (dono states equally likely, $+\epsilon$ aur $-\epsilon$ ka average zero hai).

$⟨E⟩$ vs $T$ ki shape dekho:

Parent se, $⟨E⟩=\frac{\epsilon }{e^{\beta \epsilon }+1}$. $T$ ke w.r.t. differentiate karo. $\beta =1/(k_{B}T)$ use karna zyada clean hai, isliye $\frac{d\beta }{dT}=-\frac{1}{k_B{T}^2}$. $\frac{\partial ⟨E⟩}{\partial \beta }=\epsilon \cdot \left(-\frac{\epsilon e^{\beta \epsilon }}{(e^{\beta \epsilon }+1{)}^2}\right)=-\frac{{\epsilon }^2{e}^{\beta \epsilon }}{(e^{\beta \epsilon }+1{)}^2}$ $T$ ke liye chain rule: $C_V=\frac{\partial ⟨E⟩}{\partial \beta }\cdot \frac{d\beta }{dT}=\frac{{\epsilon }^2{e}^{\beta \epsilon }}{(e^{\beta \epsilon }+1{)}^2}\cdot \frac{1}{k_B{T}^2}=k_B{\left(\frac{\epsilon }{k_{B}T}\right)}^2\frac{e^{\beta \epsilon }}{(e^{\beta \epsilon }+1{)}^2}$ Shape (Schottky anomaly): $C_V\to 0$ dono $T\to 0$ par (excite karne ke liye kuch nahi) aur $T\to \infty$ par (dono states already full), beech mein $k_{B}T\sim \epsilon$ ke paas ek bump ke saath. Ek finite level gap hamesha hump banata hai, plateau nahi.

Direct variance. $P_0=\frac{1}{Z}$, $P_1=\frac{e^{-\beta \epsilon }}{Z}$, $Z=1+e^{-\beta \epsilon }$ ke saath: $⟨E⟩=\frac{\epsilon e^{-\beta \epsilon }}{Z},⟨E^2⟩=\frac{{\epsilon }^2{e}^{-\beta \epsilon }}{Z}$ ${\sigma }_E^2=⟨E^2⟩-⟨E{⟩}^2=\frac{{\epsilon }^2{e}^{-\beta \epsilon }}{Z}-\frac{{\epsilon }^2{e}^{-2\beta \epsilon }}{Z^2}=\frac{{\epsilon }^2{e}^{-\beta \epsilon }}{Z^2}(Z-e^{-\beta \epsilon })=\frac{{\epsilon }^2{e}^{-\beta \epsilon }}{Z^2}$ kyunki $Z-e^{-\beta \epsilon }=1$. $k_B{T}^2{C}_V$ se compare karo. L2.2 se, $C_V=\frac{{\epsilon }^2{e}^{\beta \epsilon }}{k_B{T}^2(e^{\beta \epsilon }+1{)}^2}$. Toh: $k_B{T}^2{C}_V=\frac{{\epsilon }^2{e}^{\beta \epsilon }}{(e^{\beta \epsilon }+1{)}^2}=\frac{{\epsilon }^2{e}^{-\beta \epsilon }}{(1+e^{-\beta \epsilon }{)}^2}=\frac{{\epsilon }^2{e}^{-\beta \epsilon }}{Z^2}$ (upar aur neeche $e^{-2\beta \epsilon }$ se multiply karo). Dono match karte hain. ✓ Energy fluctuations asal mein heat capacity disguise mein hain: jo system heat easily absorb karta hai, uski energy zyada jitter karti hai.

Free energy: $F=-k_{B}T\ln Z=-k_{B}T\ln \left(\frac{k_{B}T}{\hslash \omega }\right)$ Entropy $S=-\partial F/\partial T$. $F=-k_{B}T(\ln T+c)$ likho jahan $c=\ln (k_B/\hslash \omega )$: $S=-\frac{\partial F}{\partial T}=k_B(\ln T+c)+k_{B}T\cdot \frac{1}{T}=k_B\ln \left(\frac{k_{B}T}{\hslash \omega }\right)+k_B$ $F$ se energy: $⟨E⟩=F+TS=-k_{B}T\ln \left(\frac{k_{B}T}{\hslash \omega }\right)+T\left[k_B\ln \left(\frac{k_{B}T}{\hslash \omega }\right)+k_B\right]=k_{B}T$ Log terms cancel ho jaate hain, $⟨E⟩=k_{B}T$ bachta hai ✓ — equipartition result se match karta hai (2 quadratic degrees of freedom $\times \frac{1}{2}{k}_{B}T$).

Factorization. Independent subsystems ⇒ total energy ek sum hai ⇒ sum ka Boltzmann factor ek product hai ⇒ saari configurations ka sum factorize ho jaata hai: $Z_N=Z_1^N=(2\cosh \beta \epsilon {)}^N$ Total energy — note karo $\ln Z_N=N\ln Z_1$, isliye: $⟨E{⟩}_N=-\frac{\partial \ln Z_N}{\partial \beta }=N\cdot (-\epsilon \tanh \beta \epsilon )=-N\epsilon \tanh \beta \epsilon$ Exactly $N$ copies single-spin answer ke, jaisa hona chahiye — independent parts add hote hain. Low-$T$: $\beta \to \infty$, $\tanh \to 1$, $⟨E{⟩}_N\to -N\epsilon$ (saare spins ground state mein). ✓

$1/N!$ kyun? Do identical atoms ko swap karna ek nayi microstate nahi hai; plain product $Z_1^N$ $N!$ se overcount karta hai. Divide karne se yeh theek ho jaata hai aur (jaise parent note batata hai) Gibbs paradox resolve ho jaata hai. $\ln Z_N=N\ln Z_1-\ln N!\approx N\ln \left(\frac{V}{{\lambda }^3}\right)-N\ln N+N$ Free energy: $F=-k_{B}T\ln Z_N$. Pressure $P=-{\left(\frac{\partial F}{\partial V}\right)}_T=k_{B}T{\left(\frac{\partial \ln Z_N}{\partial V}\right)}_T$. $\ln Z_N$ ka sirf $N\ln V$ wala piece $V$ par depend karta hai (yaad karo $Z_1=V/{\lambda }^3$ aur $\lambda$ mein koi $V$ nahi): $\frac{\partial \ln Z_N}{\partial V}=\frac{\partial }{\partial V}(N\ln V+\text{(V-independent)})=\frac{N}{V}$ $\boxed{P=\frac{N{k}_{B}T}{V}}⟺PV=N{k}_{B}T$ Ideal gas law seedha nikal aata hai. $1/N!$ ne $V$ ko kabhi touch nahi kiya, isliye yeh $P$ change nahi karta — lekin correct, extensive entropy ke liye yeh zaroori hai.

(a) $Z$ sum karo. Constant $\frac{1}{2}$ offset bahar nikalo: $Z={\sum }_{n=0}^{\infty }{e}^{-\beta \hslash \omega (n+1/2)}=e^{-\beta \hslash \omega /2}{\sum }_{n=0}^{\infty }{\left(e^{-\beta \hslash \omega }\right)}^n$ Geometric series kyun? Har term pichle ka ek fixed ratio $r=e^{-\beta \hslash \omega }$ times hai — yeh exactly wahi hai jo geometric series hoti hai, aur ${\sum }_{n\ge 0}{r}^n=\frac{1}{1-r}$ jab bhi $0<r<1$ (yahan sach hai kyunki $\beta \hslash \omega >0$). Toh: $Z=\frac{e^{-\beta \hslash \omega /2}}{1-e^{-\beta \hslash \omega }}=\frac{1}{2\sinh (\beta \hslash \omega /2)}$ (doosra form use karta hai $e^{-x/2}/(1-e^{-x})=1/(e^{x/2}-e^{-x/2})=1/(2\sinh \frac{x}{2})$). (b) Energy. $\ln Z=-\ln \left(2\sinh \frac{\beta \hslash \omega }{2}\right)$. Kyunki $\frac{d}{du}\ln \sinh u=\coth u$: $⟨E⟩=-\frac{\partial \ln Z}{\partial \beta }=\frac{\hslash \omega }{2}\coth \left(\frac{\beta \hslash \omega }{2}\right)$ (c) Limits:

• $T\to 0$ ($\beta \to \infty$): $\coth \to 1$, isliye $⟨E⟩\to \frac{1}{2}\hslash \omega$ — zero-point energy, ek purely quantum floor jo classical oscillator (Example 2) ke paas kabhi nahi tha. ✓
• $T\to \infty$ ($\beta \to 0$): small $x$ ke liye $\coth x\approx 1/x$ use karo, isliye $\coth \frac{\beta \hslash \omega }{2}\approx \frac{2}{\beta \hslash \omega }$, jo deta hai $⟨E⟩\approx \frac{\hslash \omega }{2}\cdot \frac{2}{\beta \hslash \omega }=\frac{1}{\beta }=k_{B}T$ — classical equipartition result. ✓

Poori curve in dono ke beech smoothly interpolate karti hai, $k_{B}T\sim \hslash \omega$ se neeche freeze out hoti hai:

Number: $x\equiv \beta \hslash \omega =2$ par, $\frac{⟨E⟩}{\hslash \omega }=\frac{1}{2}\coth (1)=\frac{1}{2}\cdot \frac{e^1+e^{-1}}{e^1-e^{-1}}\approx \frac{1}{2}(1.31304)\approx 0.65652$ Toh $⟨E⟩\approx 0.657\hslash \omega$: zero-point $0.5\hslash \omega$ se upar, kyunki kuch higher levels populated hain, lekin classical $k_{B}T=\hslash \omega /2=0.5\hslash \omega$ se kaafi neeche... ruko — yahan $k_{B}T=\hslash \omega /x=\hslash \omega /2=0.5\hslash \omega$, aur quantum answer $0.657\hslash \omega$ use exceed karta hai: zero-point energy quantum oscillator ko is cold temperature par naive classical value se upar push karti hai. Woh gap quantum correction hai. Myth: yahan $Z=1/(2\sinh 1)\approx 1/2.3504\approx 0.4255$, jo 1 se kam hai — phir bhi system "ek se kam states mein" nahi hai. $Z$ Boltzmann factors ka weighted sum hai, probability nahi; $\frac{1}{2}\hslash \omega$ offset bake in hone ke saath, iska numerical value 1 se bhi neeche ja sakta hai. Probabilities $e^{-\beta E_n}/Z$ hain, har ek safely $[0,1]$ mein.