Exercises — Bose-Einstein statistics — bosons

Ek boson ka spin integer hota hai ($0,1,2,\dots$). Har ek ko check karo:

• Photon, spin $1$ → boson. ✓
• Electron, spin $\frac{1}{2}$ → half-integer → fermion (dekhein Fermi-Dirac statistics — fermions).
• Helium-4, net spin $0$ → boson. ✓
• Proton, spin $\frac{1}{2}$ → fermion.
• Phonon, spin $0$ → boson. ✓

Bosons: photon, He-4, phonon. Ye bilkul wahi "pile-up" particles hain jo hamare formula ko follow karte hain.

Iska signature denominator mein ==$-1$== hai. Yeh boson geometric series ${\sum }_{n=0}^{\infty }{x}^n=1/(1-x)$ ko sum karne se aata hai, jisse $\bar{n}=x/(1-x)$ milta hai. Sign ko $+1$ kar do aur tumhe Fermi-Dirac mil jaayega, jiska sum $1+x$ (sirf $n=0,1$) $\bar{n}=x/(1+x)$ deta hai.

Mnemonic: Boson → Bunch → Below by one → $-1$.

Kya karna hai: $\bar{n}=1/(e^{\beta (\epsilon -\mu )}-1)$ mein $\mu =0$ aur $\beta \epsilon =2$ plug karo. Kyun: photons ke liye number conserved nahi hota, isliye $\mu =0$ — yeh fact yaad rakhne waala hai. $\bar{n}=\frac{1}{e^2-1}=\frac{1}{7.389-1}=\frac{1}{6.389}\approx \mathrm{0.157.}$ Iska matlab: thermal energy se double energy wala mode average mein ek photon se bhi kaafi kam rakhta hai — yahi high-frequency starvation hai jo blackbody intensity ko giraati hai (dekhein Planck radiation law).

Bose: ${\bar{n}}_{BE}=\frac{1}{e^{0.05}-1}=\frac{1}{1.05127-1}=\frac{1}{0.05127}\approx \mathrm{19.50.}$ Classical: ${\bar{n}}_{MB}=e^{-0.05}=\mathrm{0.95123.}$ Ratio: ${\bar{n}}_{BE}/{\bar{n}}_{MB}\approx 19.50/0.9512\approx \mathrm{20.5.}$ Iska matlab: $\mu$ ke bahut kareeb bosons classical guess se ~20× zyada bunch karte hain — yahi "extra crowding" lasers aur Bose-Einstein condensation ka physical seed hai.

Step — kya aur kyun: ${\bar{n}}_{BE}=1/(e^y-1)$ likhon jahan $y=\beta (\epsilon -\mu )$ hai. Bade $y$ ke liye, $e^y\gg 1$, isliye $e^y-1\approx e^y$. Yeh valid hai kyunki hum ek bade number ke saamne $1$ drop kar rahe hain. ${\bar{n}}_{BE}=\frac{1}{e^y-1}=\frac{e^{-y}}{1-e^{-y}}\approx e^{-y}\left(1+e^{-y}+\dots \right)\approx e^{-y}={\bar{n}}_{MB}.$ Fractional error $y=4$ par: exact ${\bar{n}}_{BE}=1/(e^4-1)=1/53.598=0.018659$; classical $e^{-4}=0.018316$. $\frac{{\bar{n}}_{BE}-{\bar{n}}_{MB}}{{\bar{n}}_{BE}}=\frac{0.018659-0.018316}{0.018659}\approx 0.0184=1.84\%.$ Iska matlab: $\mu$ se $4k_{B}T$ upar hi quantum bunching correction $2\%$ se kam ho jaati hai — yahi "dilute limit" hai jahan teeno statistics ek jaise ho jaate hain (dekhein Maxwell-Boltzmann statistics).

Kya aur kyun: lowest level $\epsilon ={\epsilon }_{\min }$ par, jab $\mu \to {\epsilon }_{\min }^{-}$ hota hai toh ${\epsilon }_{\min }-\mu \to 0^{+}$ hota hai, isliye $x=e^{-\beta ({\epsilon }_{\min }-\mu )}\to 1^{-}$ ho jaata hai. Tab $\bar{n}=x/(1-x)$ ka denominator $\to 0$ ho jaata hai, toh $\bar{n}\to +\infty$. Numeric check $\beta ({\epsilon }_{\min }-\mu )=0.001$ par: ${\bar{n}}_0=\frac{1}{e^{0.001}-1}=\frac{1}{1.0010005-1}=\frac{1}{0.0010005}\approx \mathrm{999.5.}$ Iska matlab: ek almost-infinite bheed single lowest state mein ghus jaati hai — ${\mathcal{Z}}_0=1/(1-x)$ ki mathematical divergence hi Bose-Einstein condensation ki shuruat hai. Isliye $\mu$ ${\epsilon }_{\min }$ ke paas aa sakta hai par usse touch nahi kar sakta. Neeche ki figure ka left edge dekho.

Step 1 — kya aur kyun. Probabilities hain $P(n)=x^n/{\mathcal{Z}}_i$ jahan ${\mathcal{Z}}_i=1/(1-x)$ hai. Hum pehle se jaante hain $\bar{n}=x{\partial }_x\ln {\mathcal{Z}}_i=x/(1-x)$ (parent note se generating-function trick). Ab second moment tak pahunchne ke liye $x{\partial }_x$ dobara apply karte hain. Step 2 — machinery. Ek clean identity: $\mathrm{Var}(n)=x\frac{\partial \bar{n}}{\partial x}$. Yeh kyun kaam karta hai: $\bar{n}=x{\partial }_x\ln \mathcal{Z}$ hai, aur log-partition function ko doosri baar differentiate karne par fluctuation milta hai. Apply karo: $\bar{n}=\frac{x}{1-x}⟹\frac{\partial \bar{n}}{\partial x}=\frac{(1-x)+x}{(1-x{)}^2}=\frac{1}{(1-x{)}^2}.$ $\mathrm{Var}(n)=x\cdot \frac{1}{(1-x{)}^2}=\frac{x}{(1-x{)}^2}.$ Step 3 — isse $\bar{n}$ form mein factor karo. $\frac{x}{(1-x{)}^2}=\frac{x}{1-x}\cdot \frac{1}{1-x}=\bar{n}\cdot (1+\bar{n}),$ $\frac{1}{1-x}=1+\frac{x}{1-x}=1+\bar{n}$ use karke. $\boxed{\mathrm{Var}(n)=\bar{n}(1+\bar{n}).}$ Interpretation: ek classical (Poisson) count mein $\mathrm{Var}=\bar{n}$ hota hai. Bosons ek extra ${\bar{n}}^2$ carry karte hain: inke fluctuations zyada hote hain. Woh excess photon bunching hai — bosons clumps mein aate hain, jo Hanbury Brown–Twiss experiment mein measure kiya gaya statistical fingerprint hai.

Kya aur kyun: har level ke liye $y_j=\beta ({\epsilon }_j-\mu )$ calculate karo, phir ${\bar{n}}_j=1/(e^{y_j}-1)$.

• Level 1: $y_1=\beta ({\epsilon }_1-\mu )=0.2$. ${\bar{n}}_1=1/(e^{0.2}-1)=1/0.22140=\mathrm{4.5167.}$
• Level 2: $y_2=\beta ({\epsilon }_2-\mu )=1.2$. ${\bar{n}}_2=1/(e^{1.2}-1)=1/2.32012=\mathrm{0.43101.}$ Ratio: $\frac{{\bar{n}}_1}{{\bar{n}}_2}=\frac{4.5167}{0.43101}\approx \mathrm{10.48.}$ Iska matlab: level 1 (sirf $0.2k_{B}T$ $\mu$ se upar) crowded near-$\mu$ regime mein hai; level 2 ($1.2k_{B}T$ upar) dilute classical tail ki taraf ja raha hai. Energy mein sirf $k_{B}T$ ka fark occupancy ko ek order of magnitude gira deta hai — bosons disproportionately saste states mein pile up karte hain.

Step 1 — pieces assemble karo. Energy density = (photon per energy) $\times$ (us energy par modes ki sankhya) $\times$ (mode per photons): $u(\epsilon )\propto \underset{\text{energy}}{\underbrace{\epsilon }}\cdot \underset{g(\epsilon )}{\underbrace{{\epsilon }^2}}\cdot \underset{\bar{n},\mu =0}{\underbrace{\frac{1}{e^{\epsilon /k_{B}T}-1}}}=\frac{{\epsilon }^3}{e^{\epsilon /k_{B}T}-1}.$ Yahi Planck's law hai (dekhein Planck radiation law). Step 2 — kyun maximise karte hain. $u$ ka peak hume kisi hot object ka colour batata hai. $z=\epsilon /k_{B}T$ (dimensionless) lene par $u\propto z^3/(e^z-1)$ ho jaata hai. $\frac{d}{dz}\left[\frac{z^3}{e^z-1}\right]=0$ set karo. Step 3 — transcendental condition. Differentiate karo (quotient rule): $\frac{3z^2(e^z-1)-z^3{e}^z}{(e^z-1{)}^2}=0⟹3(e^z-1)=z{e}^z⟹3-3e^{-z}=z.$ Rearrange karo: $\boxed{(3-z)e^z=3}$, yaani $z=3(1-e^{-z})$. Step 4 — numerically solve karo (kyun: koi closed form nahi). $z_{k+1}=3(1-e^{-z_k})$ iterate karo $z_0=3$ se shuru karke: $3(1-e^{-3})=2.851;3(1-e^{-2.851})=2.827;\to \mathrm{2.821.}$ Converge hota hai: $z^{\star }={\epsilon }^{\star }/k_{B}T\approx \mathrm{2.821.}$ Iska matlab: blackbody energy density ${\epsilon }^{\star }\approx 2.821k_{B}T$ par peak karti hai — zyada hot body ⇒ zyada high peak energy ⇒ neela glow. Yeh single number Wien's displacement law hai, jo purely boson $-1$ aur ${\epsilon }^2$ density of states se nikalta hai.

Convergence condition: geometric sum $\mathcal{Z}=\sum x^n=1/(1-x)$ ke liye $x=e^{-\beta (\epsilon -\mu )}<1$ chahiye, yaani har level ke liye $\mu <\epsilon$. BEC tab trigger hota hai jab $\mu$ ko utha ke ground-state energy ${\epsilon }_{\min }$ tak pahunchaya jaata hai, jo macroscopic ground-state occupancy force karti hai.

Massive bosons ke liye: particle number $N$ conserved hai. $T$ girne par $N$ fixed rakhne ke liye, $\mu$ ko ${\epsilon }_{\min }$ ki taraf upar jaana padta hai — aur ek finite critical temperature par woh usse touch kar leta hai. Condensation hota hai.

Photons ke liye: number conserved nahi hai, isliye equilibrium $\mu$ ko $0$ par pin kar deta hai (yeh kabhi free knob nahi hota jise utha sako). Yeh ${\epsilon }_{\min }$ ki taraf nahi chadh sakta; balki, $T$ girne par photons simply walls mein samai jaate hain. $\mu$ $0$ par frozen rehne se, ground-state occupancy $1/(e^{\beta {\epsilon }_{\min }}-1)$ finite rehti hai (jab ${\epsilon }_{\min }>0$ ho). Koi divergence nahi ⇒ koi condensate nahi.

Ek line mein: BEC ke liye conserved $N$ chahiye taaki $\mu$ ko ${\epsilon }_{\min }$ tak drive kiya ja sake; photons mein woh conservation nahi hai, isliye woh condense hone ki jagah thin out ho jaate hain.

1.1 photon, He-4, phonon · 1.2 woh $-1$ · 2.1 $0.157$ · 2.2 $19.5$ (ratio $20.5$) · 3.1 $1.84\%$ error · 3.2 $\approx 999.5$, diverges · 4.1 $\mathrm{Var}=\bar{n}(1+\bar{n})$ · 4.2 ratio $\approx 10.5$ · 5.1 ${\epsilon }^{\star }\approx 2.821k_{B}T$ · 5.2 photons: $\mu =0$ fixed ⇒ no BEC.