Question bank — Bose-Einstein statistics — bosons

True. $\bar{n}$ ensemble par ek level ki average occupancy hai, toh $0.58$ ya $9.5$ jaisi value bilkul theek hai, chahe kisi bhi single measurement mein integer count mile.

False. Agar $\mu >{\epsilon }_{\min }$ toh ground level ke liye $x=e^{(\mu -\epsilon )/k_{B}T}>1$ aur geometric sum $\sum x^n$ diverge kar jaata hai — distribution ka wahan koi matlab nahi. Chemical potential dekho.

True. Dono ka integer spin hai aur symmetric total wavefunctions hain, toh dono unlimited occupancy allow karte hain; sirf unka $\mu$ alag hota hai (photons $\mu =0$, He-4 $\mu <{\epsilon }_{\min }$).

True. Jab $\epsilon -\mu \gg k_{B}T$ tab $-1$ us bade $e^{(\epsilon -\mu )/k_{B}T}$ ke against negligible hai, toh $\bar{n}\approx e^{-(\epsilon -\mu )/k_{B}T}$, classical MB tail.

False. $-1$ exact hai aur poora boson signature yahi hai; precisely $\mu$ ke paas (denominator $\to 0$) hi ye dominate karta hai aur $\bar{n}\to \infty$ pile-up produce karta hai.

False. Low energy ka kisi cap se koi lena-dena nahi — bosons ki kabhi koi occupancy limit nahi hoti. Low energy sirf average occupancy ko bada banata hai.

False. BEC ek finite critical temperature $T_c=\frac{2\pi {\hslash }^2}{m}(\frac{N/V}{\zeta (3/2)}{)}^{2/3}/k_B$ par shuru hota hai; usse neeche particles ka ek macroscopic fraction ground state occupy karta hai. Bose-Einstein condensation dekho.

Ulta hai. Pauli exclusion fermions (antisymmetric wavefunction) par apply hota hai. Integer-spin bosons ka koi exclusion nahi hota — isliye hi wo share kar sakte hain.

Ye fermion sum hai (sirf $n=0,1$). Bosons $n=0\to \infty$ sum karte hain: ${\mathcal{Z}}_i=1/(1-x)$, jisse $\bar{n}=x/(1-x)$ milta hai aur $-1$ aata hai.

Fermions ke liye $\mu =\epsilon$ harmless hai. Bosons ke liye $\mu ={\epsilon }_{\min }$ se ${\mathcal{Z}}_0=1/(1-x)$ diverge ho jaata hai — mathematical breakdown hi BEC ka onset hai, koi non-event nahi.

Masslessness reason nahi hai. $\mu =0$ isliye hai kyunki photon number conserved nahi hai (photons freely create/destroy hote hain), toh ek add karne ka koi cost nahi lagta. Grand canonical ensemble dekho.

Occupancy badhta hai lekin har particle phir bhi ${\epsilon }_i$ carry karta hai; total energy $\bar{n}{\epsilon }_i$ finite rehti hai jab tak $\mu <{\epsilon }_{\min }$ strictly ho. Divergence us strict inequality se avoid hota hai.

Clean trick $\bar{n}=x{\partial }_x\ln {\mathcal{Z}}_i$ (ya $-\frac{1}{\beta }{\partial }_{\mu }\ln {\mathcal{Z}}_i$) hai, jo generating function se mean count bahar kheench laata hai; sahi prefactors ke bina raw ${\partial }_{\epsilon }$ constants miss kar deta hai.

Nahi — ye Grand canonical ensemble mein derive hota hai precisely isliye kyunki $N$ fixed nahi hota; particles chemical potential $\mu$ par reservoir se flow karte hain.

Ye sirf $x=e^{\beta (\mu -\epsilon )}<1$, yaani ${\epsilon }_i-\mu >0$ ke liye converge hota hai; physically ye $\mu <{\epsilon }_i$ har level ke liye force karta hai, toh $\mu$ sabse low level ke neeche rehna chahiye.

Symmetry ka matlab hai $\psi (1,2)=+\psi (2,1)$, toh $\psi$ vanish nahi karta jab do particles ek state share karte hain, antisymmetric fermion case ke unlike jahan ye $\psi =0$ force karta hai.

$-1$ denominator ko classical $e^{-\beta (\epsilon -\mu )}$ se faster shrink karta hai (yaad raho $\beta =1/k_{B}T$), toh chhote $\epsilon -\mu$ par ${\bar{n}}_{BE}\gg {\bar{n}}_{MB}$ — lasing aur condensation ka seed.

$\mu =0$ ke saath, $\bar{n}=1/(e^{\beta \epsilon }-1)$; photon energy aur density of states se multiply karne par blackbody spectral energy density milti hai.

Jab $\mu \to {\epsilon }_{\min }^{-}$ tab ground occupancy unbounded badhti hai jabki excited states par sum ($g(\epsilon )\propto \sqrt{\epsilon }$ se weighted, jo $\epsilon =0$ par vanish karta hai) ek finite maximum par saturate ho jaata hai, toh $N$ finite rakhne ke liye ground term ko haath se nikalna padta hai.

Density of states $g(\epsilon )\propto \sqrt{\epsilon }\to 0$ jab $\epsilon =0$, toh integral ground state ko zero weight deta hai; discrete sum wahan phir bhi ek macroscopic ${\bar{n}}_0$ hold karta hai, toh integral undercount karta hai aur tumhe ground state alag se add karna padta hai.

Dono geometric-style sum se aate hain, lekin fermion sum $n=1$ par ruk jaata hai (giving $+1$) jabki boson sum infinity tak jaata hai (giving $-1$) — ek single sign jo exclusion ko bunching mein flip kar deta hai.

Tab $x=e^{\beta (\mu -\epsilon )}=1$ aur denominator $e^0-1=0$ hai, toh $\bar{n}\to \infty$ — yahi reason hai ki $\mu =\epsilon$ forbidden hai aur BEC threshold mark karta hai (Figure 3 dekho).

$\beta =1/k_{B}T\to \infty$ toh $\beta (\epsilon -\mu )\to \infty$, jisse $\bar{n}\to 0$: excited levels khaali ho jaate hain aur particles ground state ki taraf collapse karte hain.

Nahi. $\mu <\epsilon$ ke saath $e^{\beta (\epsilon -\mu )}>1$ hai, toh denominator positive hai aur $\bar{n}>0$ hamesha.

$\bar{n}=1/(e^{\beta \epsilon }-1)\approx 1/(\beta \epsilon )=k_{B}T/\epsilon \to \infty$: low-frequency modes heavily populated hote hain, classical equipartition (Rayleigh-Jeans) limit.

Nahi. Dono corrections bade exponential $e^{\beta (\epsilon -\mu )}$ se dab jaate hain, toh BE, FD aur MB teeno $\bar{n}\approx e^{-\beta (\epsilon -\mu )}$ mein collapse ho jaate hain.

Intrinsically nahi — har level ki occupancy unbounded hai; total sirf tab fixed hoti hai jab tum particle-number constraint impose karte ho jo $\mu$ set karta hai.

Kyunki $T_c\propto (N/V{)}^{2/3}$, $N/V$ double karne se $T_c$ factor $2^{2/3}\approx 1.59$ se badh jaata hai — denser gases zyaada high temperature par condense karte hain.