Exercises — Quantum statistics — distinguishable vs indistinguishable particles

Rule (from the Spin–Statistics Theorem): integer spin → boson (any number per state); half-integer spin → fermion (at most one).

• (a) electron, spin $\frac{1}{2}$ → fermion, at most one. (Pauli Exclusion Principle ko govern karta hai.)
• (b) photon, spin $1$ → boson, any number (isliye lasers kaam karte hain).
• (c) ${}^4$He: fermions ki even sankhya (6) → total spin integer → boson (isliye ye Bose–Einstein Condensation se guzarta hai).
• (d) proton, spin $\frac{1}{2}$ → fermion, at most one. Key move: composite ke andar fermions gino. Even count ⇒ boson; odd count ⇒ fermion.

$a=+1$ (Fermi–Dirac). Kyunki $e^{\beta (\epsilon -\mu )}>0$ hamesha hota hai, denominator $e^{\beta (\epsilon -\mu )}+1>1$ hota hai, isliye $⟨n⟩<1$. Yahi Pauli exclusion arithmetic mein dikhta hai: ek level mein zyada se zyada ek fermion hota hai, isliye uski average filling 0 aur 1 ke beech hoti hai. Contrast karo $a=-1$ (Bose) se: denominator $0$ ki taraf shrink ho sakta hai, isliye $⟨n⟩$ bahut bada ho sakta hai — unlimited occupation.

Neeche di gayi figure 2-in-3 case ko explicitly, box by box, draw karti hai, taaki tum dekh sako ki teen counts kyun alag hain; 2-in-4 answers phir usi formula se aate hain.

Figure s01 — kya dikhata hai: teen side-by-side panels, har ek "3 boxes" ki rows dikhata hai jisme do particles rakhe hain. Left panel (orange, distinguishable): particles ke labels A aur B hain, isliye "A in box 1, B in box 2" aur "B in box 1, A in box 2" alag rows ki tarah draw hain — total 9 rows (ek representative sample draw kiya gaya hai). Middle panel (blue, bosons): bina labels ke identical dots, aur wo rows bhi allowed hain jahan dono dots ek hi box mein hain — 6 rows. Right panel (green, fermions): identical dots lekin koi bhi box kabhi do nahi rakhta — sirf wo 3 rows bachti hain jisme ek dot do alag boxes mein hai. Left se right padho toh tum literally arrangements ko gayab hote dekhte ho jaise rules tight hote hain: $9\to 6\to 3$.

2-in-3 (figure): distinguishable $3^2=9$, bosons $\left(\genfrac{}{}{0ex}{}{3+2-1}{2}\right)=\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$, fermions $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$.

2-in-4 (posed problem):

• (a) Distinguishable: har particle independently 4 boxes mein se ek choose karta hai → $4^2=\boxed{16}$.
• (b) Bosons (unordered, repeats allowed) → $\left(\genfrac{}{}{0ex}{}{4+2-1}{2}\right)=\left(\genfrac{}{}{0ex}{}{5}{2}\right)=\boxed{10}$.
• (c) Fermions (unordered, no repeats) → $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=\boxed{6}$. Ordering notice karo $6<10<16$: fermions sabse zyada "spread out" hain, distinguishable sabse zyada numerous — wohi pattern jo figure 2-in-3 ke liye dikhati hai.

$\epsilon =\mu$ ke saath, $\beta (\epsilon -\mu )=0$, isliye $e^0=1$: $⟨n{⟩}_{FD}=\frac{1}{1+1}=\boxed{\frac{1}{2}}.$ Ye har temperature par bilkul $\frac{1}{2}$ hai — Fermi energy par level hamesha half-filled hota hai. Ye Fermi function ka pivot point hai: $\mu$ se neeche levels mostly full hain, upar mostly empty, aur bilkul $\mu$ par ye ek coin flip hai.

$e^{\ln 2}=2$, isliye $⟨n{⟩}_{BE}=\frac{1}{2-1}=\boxed{1}.$ Average mein ek particle. Usi argument par Fermi se compare karo: $1/(2+1)=\frac{1}{3}$. Same energy, lekin bosons zyada pile up karte hain — $+1$ ki jagah $-1$ count badhata hai.

Upar di gayi definition callout se classical form yaad karo: $⟨n{⟩}_{MB}=e^{-\beta (\epsilon -\mu )}$ (master formula with $a=0$). Yahan $\beta (\epsilon -\mu )=3$ hai, isliye $e^3\approx 20.0855$.

• Classical: $⟨n{⟩}_{MB}=e^{-3}\approx \boxed{0.0498}$.

• Fermi: $\frac{1}{20.0855+1}\approx \boxed{0.0474}$.

• Bose: $\frac{1}{20.0855-1}\approx \boxed{0.0524}$. Teeno $\sim 5\%$ ke andar agree karte hain. Kyun? Jab $e^{\beta (\epsilon -\mu )}\gg 1$ ho, toh $\pm 1$ ek choti si correction hai — ye bilkul classical regime hai: dilute, kam double occupations, isliye distinguishable-vs-not matter karna band ho jaata hai. Fermi classical se neeche baitha hai (exclusion suppress karta hai), Bose upar (bunching enhance karta hai).

Figure s02 — kya dikhata hai: teeno occupation curves single variable $x=\beta (\epsilon -\mu )$ ke against plot ki gayi hain horizontal axis par (right jaana = zyada thanda ya higher-energy level), aur $⟨n⟩$ vertical axis par hai. Blue hai Bose $1/(e^x-1)$, jo $x\to 0^{+}$ par infinity ki taraf shoot karta hai (condensation divergence, red dashed line se $x=0$ par mark kiya gaya hai). Green hai Fermi $1/(e^x+1)$, jo door left par dotted ceiling $⟨n⟩=1$ par flat ho jaata hai aur marked point $⟨n⟩=\frac{1}{2}$ se $x=0$ par guzarta hai. Orange dashed classical $e^{-x}$ hai, unke beech mein. $x=3$ par ek gray dashed line is problem mein use kiye gaye point ko mark karti hai: wahaan teeno curves visually merge ho gayi hain — classical limit. $x=0$ ke paas ye fan apart hoti hain, yahi jagah hai jahan quantum differences dominate karte hain.

Requirement: occupation negative nahi ho sakti, $⟨n{⟩}_{BE}=\frac{1}{e^{\beta (\epsilon -\mu )}-1}\ge 0$. Iske liye denominator positive chahiye: $e^{\beta (\epsilon -\mu )}-1>0\Rightarrow e^{\beta (\epsilon -\mu )}>1\Rightarrow \beta (\epsilon -\mu )>0\Rightarrow \mu <\epsilon .$ Kyunki ye har level ke liye hold karna chahiye including lowest, $\mu <{\epsilon }_{\min }$. Jab $\mu \to {\epsilon }_{\min }$: ground level ke liye $\beta ({\epsilon }_{\min }-\mu )\to 0^{+}$, isliye $e^{(\cdots )}\to 1^{+}$ aur $⟨n⟩\to \infty$ — particles ki macroscopic sankhya single lowest state mein crash kar jaati hai. Yahi hai Bose–Einstein Condensation. Fermions ke liye aisa koi bound nahi: unka denominator $+1$ hai, hamesha positive, isliye $\mu$ (the Fermi level) koi bhi sign ho sakta hai. Edge case — inequality kitni strict hai? Thermodynamic limit (infinite system) mein strict $\mu <{\epsilon }_{\min }$ hold karta hai aur $\mu$ sirf ${\epsilon }_{\min }$ se neeche se approach kar sakta hai, kabhi touch nahi kar sakta. Finite real system mein lowest level mein finite maximum number hota hai, isliye $\mu$ microscopically close aa sakta hai, ${\epsilon }_{\min }$ ke $O(1/N)$ ke andar, aur divergence ek large-but-finite ground-state occupation se replace ho jaata hai. Toh "$\mu <{\epsilon }_{\min }$ strictly" sirf $N\to \infty$ par exact hai; finite $N$ ke liye ye "utna close jitna ground level ki capacity allow kare" hai. Isliye real BEC ek sharp-but-smoothed crossover hai, mathematically perfect singularity nahi.

Step A — entropy formula kahan se aata hai. Upar define ki gayi free energy use karke, $S=-\partial F/\partial T$ with $F=-k_{B}T\ln Z_N$ deta hai $S=k_B(\ln Z_N+\beta ⟨E⟩)$. Ek ideal gas ke liye single-particle partition function factorise hota hai $Z_1=V/{\lambda }^3$ ki tarah, jahan $\lambda =h/\sqrt{2\pi m{k}_{B}T}$ thermal wavelength hai, isliye $Z_1$ $V$ ke proportional hai. Step B — $N!$ patch apply karo. $Z_N=Z_1^N/N!$ ke saath, $\ln Z_N=N\ln Z_1-\ln N!\approx N\ln \frac{V}{{\lambda }^3}-(N\ln N-N)=N\ln \frac{V}{N{\lambda }^3}+N,$ Stirling $\ln N!\approx N\ln N-N$ use karke. Energy term $\frac{3}{2}N$ add karta hai (equipartition, $⟨E⟩=\frac{3}{2}N{k}_{B}T$). Constants collect karne par Sackur–Tetrode entropy milti hai: $S=N{k}_B\left[\ln \frac{V}{N}+\frac{3}{2}\ln \frac{2\pi m{k}_{B}T}{h^2}+\frac{5}{2}\right].$ Is derivation ka crucial feature ye hai jo ye earn karta hai: $-\ln N!$ ne $\ln V$ ko $\ln (V/N)$ mein badal diya — entropy density $V/N$ par depend karti hai, raw volume par nahi. Step C — identical gas mix karo.

• Pehle: do boxes, har ek $(N,V)$: $S_i=2\cdot N{k}_B\left[\ln \frac{V}{N}+\cdots \right]$.
• Baad mein: ek box, $(2N,2V)$: $S_f=2N{k}_B\left[\ln \frac{2V}{2N}+\cdots \right]=2N{k}_B\left[\ln \frac{V}{N}+\cdots \right]$.
• $\Delta S=S_f-S_i=\boxed{0}$. Sahi hai — jab tum ek gas ko khud ke saath mix karte ho toh kuch bhi physically nahi badla. $N!$ ke bina entropy $\ln (V/N)$ ki jagah $\ln V$ rakhti, fake mixing term $\Delta S_{\text{wrong}}=2N{k}_B\ln 2>0$ deti — Gibbs paradox. $N!$ classical bookkeeping mein leaking indistinguishability hai.

Parent derivation se, ${\psi }_{-}=\frac{1}{\sqrt{2}}[{\psi }_a(1){\psi }_b(2)-{\psi }_a(2){\psi }_b(1)].$ $b=a$ set karo: ${\psi }_{-}=\frac{1}{\sqrt{2}}[{\psi }_a(1){\psi }_a(2)-{\psi }_a(2){\psi }_a(1)]=\frac{1}{\sqrt{2}}\cdot 0=0.$ Ek wavefunction jo identically zero hai uski probability $|\psi {|}^2=0$ har jagah hoti hai — configuration exist hi nahi karta. Yahi hai Pauli Exclusion Principle, aur note karo ye koi force nahi hai: koi potential term nahi aata, sirf counting ki antisymmetry hai. Isliye Q2.1 mein fermion count ne $\left(\genfrac{}{}{0ex}{}{N}{k}\right)$ (no repeats) use kiya tha.

(a) $\mu =0$ kyun — free energy ke zariye. Grand Canonical Ensemble mein ek level ki occupation $e^{-\beta (\epsilon -\mu )n}$ se weighted hoti hai; yahan $\mu$ woh Lagrange multiplier hai jo chosen average particle number $⟨N⟩$ ko enforce karta hai. Lekin photons ke liye number conserved constraint nahi hai — cavity walls photons freely create aur destroy karti hain, isliye koi fixed $⟨N⟩$ enforce karne ko nahi hai. Equilibrium is jagah set hota hai jahan system apna $N$ khud choose karta hai Helmholtz free energy $F$ (upar define ki gayi) minimize karne ke liye: minimum condition hai ${\left(\partial F/\partial N\right)}_{T,V}=0$, aur kyunki $\mu =\partial F/\partial N$, ye padhta hai $\mu =0$ photons ke liye identically. (b) Occupation. Bose–Einstein mein $\mu =0$ daalo: $⟨n⟩=\frac{1}{e^{\beta \epsilon }-1}=\frac{1}{e^{\epsilon /k_{B}T}-1}.$ (c) $\epsilon =\hslash \omega$ ke saath aur photon energy aur mode density se multiply karke, ye exactly Planck blackbody spectrum hai. Koi $\mu$ nahi aata, aur kyunki $\epsilon >0=\mu$ hamesha, photons ka koi condensation nahi hota.

Set-up. $u=e^x$ lo. Kyunki $x>0$ hai isliye $u>1$ hai. Teeno occupations likho (page ke top se $a=0$ master formula use karke $⟨n{⟩}_{MB}=e^{-x}=1/u$): $⟨n{⟩}_{BE}=\frac{1}{u-1},⟨n{⟩}_{MB}=\frac{1}{u},⟨n{⟩}_{FD}=\frac{1}{u+1}.$ Kyun reciprocals order reverse karte hain. Teeno denominators compare karo: $u-1<u<u+1$ (har ek pichle ka plus one hai). Ab function $f(y)=1/y$ $y>0$ ke liye strictly decreasing hai: iska derivative $f^{\prime }(y)=-1/y^2<0$ hai, isliye bada input chota output deta hai. Teeno denominators positive hain (kyunki $u>1$), isliye $f$ apply karne se inequalities ki chain flip ho jaati hai: $u-1<u<u+1\stackrel{1/y\text{ decreasing}}{\Rightarrow }\frac{1}{u-1}>\frac{1}{u}>\frac{1}{u+1},$ jo exactly $⟨n{⟩}_{BE}>⟨n{⟩}_{MB}>⟨n{⟩}_{FD}$ hai. Interpretation: bosons hamesha over-occupy karte hain, fermions hamesha under-occupy, classical middle ke relative — $\pm 1$ ka pura content yahi hai. $x=\ln 3$ par evaluate karo. Tab $u=e^{\ln 3}=3$:

• Bose: $\frac{1}{3-1}=\frac{1}{2}=\boxed{0.5}$.
• Classical: $\frac{1}{3}\approx \boxed{0.3333}$.
• Fermi: $\frac{1}{3+1}=\frac{1}{4}=\boxed{0.25}$. Ordering $\frac{1}{2}>\frac{1}{3}>\frac{1}{4}$ confirmed, general proof ke saath consistent.

Do alag ideas, alag rakhe gaye:

• Microstates count karna (kitne labelled/unlabelled arrangements): distinguishable $N^k$, bosons $\left(\genfrac{}{}{0ex}{}{N+k-1}{k}\right)$, fermions $\left(\genfrac{}{}{0ex}{}{N}{k}\right)$.
• Ek level ki mean occupation (ek level average mein kitna full hota hai): ek single formula $⟨n⟩=1/(e^{\beta (\epsilon -\mu )}+a)$ with $a=+1,-1,0$, bosons ke liye $\mu <{\epsilon }_{\min }$ aur photons ke liye $\mu =0$ se bounded. Pehla jawab deta hai "kitne tarike?"; doosra jawab deta hai "kitna full?" — alag questions, same underlying indistinguishability.