Worked examples — Quantum statistics — distinguishable vs indistinguishable particles

2.4.15 · D3 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Quantum statistics — distinguishable vs indistinguishable pa

Shuru karne se pehle, woh ek master formula yaad karo jo parent ne derive kiya tha:

Ek convenient shorthand yahan define kar leta hoon taaki baad mein chori chori na dalaoon: $x\equiv \beta (\epsilon -\mu )=\frac{\epsilon -\mu }{k_{B}T}.$ Yeh ek pure number hai (energy over energy). Positive $x$ matlab "yeh level going rate $\mu$ se zyada costly hai"; negative $x$ matlab "yeh level sasta hai, particle ki price se neeche hai."

Neeche ki picture ek saath teeno master curves dikhati hai — jab bhi hum koi cell work karein, isko dekhte rehna.

Figure padhna: $x$ ke against plot karke, blue Fermi–Dirac curve $0$ aur $1$ ke beech trapped rehti hai (exclusion), pink Bose–Einstein curve $x\to 0^{+}$ par rocket ki tarah upar jaati hai (bunching), aur pale-yellow Maxwell–Boltzmann curve $e^{-x}$ unke beech se nikalti hai. Bade $x$ par (right side) teeno merge ho jaati hain — yahi classical corner hai.

---

The scenario matrix

Is topic ke har question ka ek cell mein jaana tay hai. Har example ko uska cell tag kiya gaya hai.

Cell	Regime jis par test ho raha hai	Kaun sa example
A	$x$ ka sign: level $\mu$ se upar ($x>0$)	Ex 1
B	$x$ ka sign: level exactly $\mu$ par ($x=0$)	Ex 2
C	$x$ ka sign: level $\mu$ se neeche ($x<0$)	Ex 3
D	Limit $T\to 0$ (cold): Fermi step aur Bose freeze-out	Ex 4
E	Limit $T\to \infty$ / dilute: teeno classical par collapse	Ex 5
F	Degenerate edge: boson at $\epsilon \to \mu$ (divergence / BEC trigger)	Ex 6
G	Scratch se pure counting (microstate census)	Ex 7
H	Real-world word problem (photon gas, $\mu =0$)	Ex 8
I	Exam twist: Gibbs $1/N!$ aur entropy of mixing	Ex 9

Prerequisite links agar koi cell shaky lage: Grand Canonical Ensemble, Fermi Gas & Fermi Energy, Bose–Einstein Condensation, Pauli Exclusion Principle, Blackbody Radiation (Photon Gas), Gibbs Paradox & Entropy of Mixing, Maxwell–Boltzmann Distribution — aur hamesha parent topic note.

---

Example 1 — Cell A: chemical potential se upar ka level

Forecast: level $\mu$ se $0.2\text{eV}$ upar hai, jo thermal energy se do guna hai. Guess: half-full se kaafi kam — ek chota number.

1. $x$ compute karo. $x=\frac{\epsilon -\mu }{k_{B}T}=\frac{5.2-5.0}{0.10}=2$. Yeh step kyun? Har distribution sirf dimensionless ratio $x$ par depend karti hai. $x$ nikaalna ek physics problem ko arithmetic mein convert kar deta hai.
2. Fermi–Dirac mein daalo ($a=+1$): $⟨n⟩=\frac{1}{e^2+1}$. Yeh step kyun? Electrons fermions hote hain (half-integer spin, Pauli Exclusion Principle), isliye hum $a=+1$ use karte hain.
3. Evaluate karo. $e^2\approx 7.389$, toh $⟨n⟩=\frac{1}{8.389}\approx 0.119$. Yeh step kyun? Yahi actual number hai jo us level ki average filling ka measurement dega.

Verify: $x>0$ fermions ke liye $⟨n⟩<\frac{1}{2}$ dena chahiye ($\mu$ se upar ke levels empty zyada hote hain full se kam). $0.119<0.5$ ✓. Saath hi $0<⟨n⟩<1$, exclusion ka respect ✓

---

Example 2 — Cell B: exactly chemical potential par

Forecast: "$\mu$" ko yahan often Fermi level kaha jaata hai. Guess: exactly half-full, $T$ se independent.

1. $x$ compute karo. $x=\frac{\mu -\mu }{k_{B}T}=0$. Yeh step kyun? Notice karo $T$ cancel ho jaata hai — isliye answer temperature-independent hoga.
2. Daalo: $⟨n⟩=\frac{1}{e^0+1}=\frac{1}{1+1}=\frac{1}{2}$. Yeh step kyun? $e^0=1$ exactly hai, toh Fermi function $\frac{1}{2}$ se guzarti hai $\epsilon =\mu$ par sab $T>0$ ke liye.

Verify: Yeh fermions ke liye chemical potential ki defining property hai — woh level jo exactly half-occupied hai. Upar ke master figure mein, blue curve exactly $x=0$ par $\frac{1}{2}$ cross karti hai. ✓

---

Example 3 — Cell C: chemical potential se neeche ka level

Forecast: $\mu$ se neeche matlab ek "sasta" level; fermions isko almost completely fill kar denge. Guess: 1 ke karib.

1. $x$ compute karo. $x=\frac{4.8-5.0}{0.10}=-2$. Yeh step kyun? Negative $x$ Ex 1 ka mirror image hai — symmetry dekhna achha lagta hai.
2. Daalo: $⟨n⟩=\frac{1}{e^{-2}+1}=\frac{1}{0.1353+1}=\frac{1}{1.1353}\approx 0.881$. Yeh step kyun? "Below $\mu$" cell fill karta hai taaki tumne dono signs dekhe hon.
3. Symmetry notice karo. $⟨n(x)⟩+⟨n(-x)⟩=1$: yahan $0.119+0.881=1.000$. Yeh step kyun? Fermi function $(\mu ,\frac{1}{2})$ ke baare mein antisymmetric hai; yeh ek fast exam sanity check hai.

Verify: $0.881>\frac{1}{2}$ ($\mu$ se neeche ka level half se zyada full hai) ✓ aur Ex 1 ke saath pair karke exactly 1 sum deta hai ✓.

---

Example 4 — Cell D: cold limit $T\to 0$ — Fermi step aur Bose freeze-out

Forecast: fermions lowest levels mein pack ho jaate hain — $\mu$ ke neeche full, upar empty, $\mu$ par ek cliff. Bosons, jo pile up kar sakte hain, excited levels chhod kar lowest mein crash kar jaate hain — har excited level empty ho jaata hai.

Part (a) — Fermi step.

1. $\mu$ ke neeche: $\epsilon <\mu \Rightarrow x<0\Rightarrow x\to -\infty$ jab $T\to 0$. Tab $e^x\to 0$, toh $⟨n⟩=\frac{1}{0+1}=1$. Yeh step kyun? Cold matlab $\beta$ bahut bada; yeh $(\epsilon -\mu )$ ke sign ko $\pm \infty$ mein magnify kar deta hai.
2. $\mu$ ke upar: $\epsilon >\mu \Rightarrow x\to +\infty$, toh $e^x\to \infty$ aur $⟨n⟩=\frac{1}{\infty +1}=0$. Yeh step kyun? Wahi logic, opposite sign — yahi step produce karta hai.
3. The step. Fermi function ek unit step ban jaati hai: $\mu$ tak 1, phir 0. $\mu$ par value $\frac{1}{2}$ rehti hai (Ex 2). $T=0$ par, $\mu$ ko Fermi energy ${\epsilon }_F$ kaha jaata hai. Yeh step kyun? Yahi step white-dwarf aur metal-electron physics ki poori neenv hai.

Part (b) — Bose freeze-out.

4. Koi bhi excited level jahan $\epsilon >\mu$: $x>0$ hai aur $x\to +\infty$ jab $T\to 0$, toh $e^x\to \infty$ aur $⟨n{⟩}_{BE}=\frac{1}{e^x-1}\to \frac{1}{\infty -1}=0$. Yeh step kyun? Har excited boson level $T\to 0$ par empty ho jaata hai — yahi "freeze-out" hai. Particles gayab nahi hote; woh sab lowest level (ground state) mein gir jaate hain.
5. Lowest level: sab $N$ particles rakhne ke liye uska $⟨n⟩$ macroscopic hona chahiye, jo (Ex 6 se) $x\to 0^{+}$ chahiye, yaani $\mu \to {\epsilon }_{\min }$ neeche se. Yeh macroscopic ground-state pile-up hi Bose–Einstein Condensation hai. Yeh step kyun? Yeh cell close karta hai: fermions ek step banate hain (spread out), bosons condense karte hain (ek state mein collapse) — do cold limits bilkul ulte hain.

Verify: Neeche ke figure mein, blue cold Fermi curve almost ek vertical cliff hai $\epsilon =\mu$ par $\frac{1}{2}$ se guzarti hui; pink cold Bose curve excited levels ke liye $0$ ke paas pinned hai except jahan $\epsilon \to \mu$ hota hai, wahan spike — exactly wahi do behaviours jo abhi derive kiye. ✓

---

Example 5 — Cell E: classical corner (teeno agree karte hain)

Forecast: bada $x$ matlab $\pm 1$ sirf $e^5\approx 148$ ke paas rounding error hai. Guess: teeno answers almost $1\%$ ke andar.

1. Fermi–Dirac: $\frac{1}{e^5+1}=\frac{1}{148.413+1}=\frac{1}{149.413}\approx 0.006693$.
2. Bose–Einstein: $\frac{1}{e^5-1}=\frac{1}{147.413}\approx 0.006784$.
3. Maxwell–Boltzmann: $e^{-5}=\frac{1}{148.413}\approx 0.006738$. Yeh steps kyun? Wahi $x$, sirf $a$ change hota hai ($+1,-1,0$). Poora point answers ko converge hote dekhna hai.
4. Relative spread. $\frac{0.006784-0.006693}{0.006738}\approx 0.0135$, lagbhag $1.35\%$. Yeh step kyun? "Woh agree karte hain" ko quantify karta hai. Jaise $x$ badhta hai spread $e^{-x}$ ki tarah sirta hai.

Verify: Ordering honi chahiye $⟨n{⟩}_{BE}>⟨n{⟩}_{MB}>⟨n{⟩}_{FD}$ (bosons bunch karte hain, fermions avoid karte hain), aur indeed $0.006784>0.006738>0.006693$ ✓. Yahi classical regime hai jo parent ne flag kiya tha, yaani master figure ka far-right.

---

Example 6 — Cell F: boson edge $\epsilon \to \mu$ (divergence)

Forecast: parent ne warn kiya tha ki $⟨n{⟩}_{BE}$ diverge karta hai jab $\epsilon \to \mu$. Guess: numbers $1/x$ ki tarah blow up honge.

1. $x=1.0$: $⟨n⟩=\frac{1}{e^1-1}=\frac{1}{1.71828}\approx 0.5820$.
2. $x=0.1$: $⟨n⟩=\frac{1}{e^{0.1}-1}=\frac{1}{0.10517}\approx 9.508$.
3. $x=0.01$: $⟨n⟩=\frac{1}{e^{0.01}-1}=\frac{1}{0.010050}\approx 99.50$. Yeh steps kyun? Dekho har bar $x$ dashguna chhota hone par $⟨n⟩$ roughly das guna badh jaata hai — yahi $1/x$ divergence hai.
4. Small-$x$ law. Chote $x$ ke liye, $e^x-1\approx x$, toh $⟨n⟩\approx 1/x\to \infty$. Yeh step kyun? Runaway explain karta hai: lowest level mein unbounded pile-up hi Bose–Einstein Condensation hai.
5. $x<0$ kyun forbidden hai. Agar $x<0$ toh $e^x-1<0$ aur $⟨n⟩<0$ — ek impossible negative occupation. Isliye bosons ke liye hamesha $\mu <{\epsilon }_{\min }$ hota hai. Yeh step kyun? Parent ke mistake-callout ko fix karta hai: boson $\mu$ ground-state energy se upar bounded hai.

Verify: $⟨n⟩$ positive hai aur $x\to 0^{+}$ par increasing hai: $0.582<9.508<99.50$ ✓, aur chote $x$ ke liye har ek $\approx 1/x$ ke barabar hai ($1/0.1=10\approx 9.5$, $1/0.01=100\approx 99.5$) ✓.

---

Example 7 — Cell G: scratch se microstate census

Forecast: distinguishable sabse bada hona chahiye ($g^N=4^3$), phir bosons, phir fermions. Guess $64,20,4$.

1. Distinguishable. $N=3$ labeled particles mein se har ek independently $g=4$ boxes mein se 1 pick karta hai: $g^N=4^3=64$. Yeh step kyun? "Order matters, repetition allowed" — parent ka classical count.
2. Bosons. Unordered, repetition allowed (stars-and-bars): $\left(\genfrac{}{}{0ex}{}{N+g-1}{N}\right)=\left(\genfrac{}{}{0ex}{}{3+4-1}{3}\right)=\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$. Yeh step kyun? Indistinguishable particles → unordered fillings count karo; bosons ek box share kar sakte hain.
3. Fermions. Unordered, no repetition: $\left(\genfrac{}{}{0ex}{}{g}{N}\right)=\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$. Yeh step kyun? Pauli Exclusion Principle: ek box mein zyada se zyada ek → bas choose karo ki $g=4$ boxes mein se kaun se $N=3$ occupied hain.

Verify: Monotone ordering $64>20>4$ ✓ (har restriction microstates remove karti hai). Fermion count ka doosre tarike se sanity check: choose karo kaun sa single box empty hai, woh bhi $\left(\genfrac{}{}{0ex}{}{4}{1}\right)=4$ deta hai ✓.

---

Example 8 — Cell H: real-world word problem (photon gas, $\mu =0$)

Forecast: photons bunching pasand karte hain (bosons), lekin is mode ki cost thermal energy se do guni hai, toh zyada nahi. Guess: 1 se thoda kam.

1. $\mu =0$ set karo. Tab $x=\beta \epsilon =\epsilon /k_{B}T=2$. Yeh step kyun? Photons ka $\mu =0$ hota hai (woh freely create/destroy ho sakte hain), isliye hum cavity mein photon BEC nahi dekhte.
2. Bose–Einstein occupation ($a=-1$): $⟨n⟩=\frac{1}{e^2-1}$. Yeh step kyun? Yahi Planck occupation number hai — Planck ke radiation law ka seed.
3. Evaluate karo. $e^2-1=6.389$, toh $⟨n⟩=\frac{1}{6.389}\approx 0.1565$. Yeh step kyun? Yahi us specific mode mein us temperature par average photon count hai.

Verify: Ex 1 ke wahi $x=2$ par fermion answer ($0.119$) se compare karo: boson value $0.1565$ zyada hai, jaisa bunching demand karta hai $⟨n{⟩}_{BE}>⟨n{⟩}_{FD}$ ✓. Positive aur finite ($x>0$ hai, koi divergence nahi) ✓.

---

Example 9 — Cell I: exam twist — Gibbs $1/N!$ aur mixing entropy

Forecast: agar gases sach mein identical hain, toh unhe mix karna physically kuch nahi badalta, isliye $\Delta S=0$. Guess: naive un-corrected count ek spurious positive $\Delta S$ deta hai; $1/N!$ ise cancel kar deta hai.

1. Naive (labeled) count. Particles ko distinguishable treat karte hue, partition hatane par har particle ko double volume milta hai, toh naive $\Delta S_{\text{naive}}=2N{k}_B\ln 2$. Yeh step kyun? Yahi galat, non-extensive answer hai jo Gibbs paradox create karta hai.
2. Gibbs correction apply karo. $Z_N$ ko $N!$ se divide karna relabeling over-count subtract karta hai. $2N$ identical particles ko do labeled halvon mein split karne ke tarike $\left(\genfrac{}{}{0ex}{}{2N}{N}\right)$ hain, toh correction $k_B\ln \left(\genfrac{}{}{0ex}{}{2N}{N}\right)$ remove karti hai. Yeh step kyun? Relabelings ka over-count exactly yahi binomial factor hai.
3. Stirling se cancel karo. Bade $N$ ke liye, $\ln \left(\genfrac{}{}{0ex}{}{2N}{N}\right)\approx 2N\ln 2$, toh $\Delta S=2N{k}_B\ln 2-k_B\ln \left(\genfrac{}{}{0ex}{}{2N}{N}\right)\approx 2N{k}_B\ln 2-2N{k}_B\ln 2=0$. Yeh step kyun? Identical gases ke liye spurious mixing entropy gayab ho jaati hai — extensivity restore hoti hai.
4. Numbers, $N=100$. $\ln \left(\genfrac{}{}{0ex}{}{200}{100}\right)=135.80$; aur $2N\ln 2=200\times 0.6931=138.63$. Ratio $135.80/138.63=0.980$. Yeh step kyun? Hum actual finite-$N$ binomial compute karte hain aur Stirling estimate $2N\ln 2$ se compare karte hain taaki reader dekhe ki cancellation finite $N$ ke liye exact nahi hai — yeh sirf tab exact hoti hai jab $N\to \infty$, jahan residual ($\frac{1}{2}\ln (\pi N)$ wala Stirling term) $N$ ke saath badhne wale $2N\ln 2$ ke muqable mein negligible ho jaata hai.

Verify: Distinguishable gases ke liye (ya do alag gases ke liye) correction apply nahi hoti aur $\Delta S=2N{k}_B\ln 2>0$ — ek real, positive mixing entropy — jabki identical gases $0$ deti hain. Numeric check $\ln \left(\genfrac{}{}{0ex}{}{200}{100}\right)\approx 135.80$ vs $138.63$ ratio $0.980$ confirm karta hai, jo $N$ badhne par $1$ ki taraf jaata hai. ✓

---

Recall Rapid self-test

Level above $\mu$, fermion, $x=2$: occupation ::: $1/(e^2+1)\approx 0.119$ Level exactly at $\mu$, fermion: occupation ::: exactly $1/2$, har $T$ ke liye Boson with $x\to 0^{+}$: occupation ::: $1/x$ ki tarah diverge karta hai → BEC Bose excited levels as $T\to 0$: occupation ::: sab $0$ ho jaate hain (freeze-out) — particles ground state mein condense ho jaate hain Photon mode ($\mu =0$) at $\epsilon =2k_{B}T$: occupation ::: $1/(e^2-1)\approx 0.157$ 3 particles, 4 states, fermions: microstates ::: $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$ 3 particles, 4 states, bosons: microstates ::: $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$ Identical gases ke liye mixing entropy kyun zero hoti hai ::: $1/N!$ Gibbs factor relabeling over-count remove kar deta hai