Exercises — Bose-Einstein condensation — concept

(b). Bosons ke liye $\bar{n}(\epsilon )=\frac{1}{e^{(\epsilon -\mu )/k_{B}T}-1}$ non-negative hona chahiye, jo $\mu \le 0$ force karta hai. Capacity finite kyun hai? Integral dekho. Excited states mein sabse zyada atoms jo aa sakte hain (jab $\mu =0$) woh is curve ke neeche ka area hai: $N_{\text{exc}}^{\max }={\int }_0^{\infty }\underset{\propto \sqrt{\epsilon }}{\underbrace{g(\epsilon )}}\underset{\text{occupation}}{\underbrace{\frac{1}{e^{\epsilon /k_{B}T}-1}}}d\epsilon .$ $x=\epsilon /k_{B}T$ substitute karne par yeh ek pure number times temperature factor ban jaata hai: ${\int }_0^{\infty }\frac{\sqrt{x}}{e^x-1}dx=\Gamma (3/2)\zeta (3/2)$, jo ek finite value hai. Neeche ki figure dikhati hai kyun yeh finite hai: $x=0$ ke paas occupation $\frac{1}{e^x-1}\approx \frac{1}{x}$ blast kar jaata hai, lekin density-of-states factor $\sqrt{x}$ product ko $x^{-1/2}$ tak le aata hai, jiska zero ke paas area finite hai (shaded region mein koi infinite spike nahi). Finite ceiling ke saath, extra atoms ke paas $\epsilon =0$ ground state ke alaawa koi jagah nahi. Koi attraction nahi, koi gravity nahi — pure statistics.

False. Agar $\mu >0$ hota toh ground state (${\epsilon }_0=0$) ke liye hame $e^{-\mu /k_{B}T}<1$ milta, jo denominator $e^{-\mu /k_{B}T}-1<0$ banata aur isse $\bar{n}({\epsilon }_0)<0$ — negative number of atoms, jo impossible hai. Isliye $\mu$ apne aap ko $0^{-}$ par pin kar leta hai, aur $N_0=\frac{1}{e^{-\mu /k_{B}T}-1}$ ka $\mu \to 0^{-}$ par diverge karna woh "escape valve" hai jo surplus ko absorb kar leta hai.

$n{\lambda }_T^3=\zeta (3/2)\approx \mathrm{2.612.}$ $n{\lambda }_T^3$ ek "quantum volume" ${\lambda }_T^3$ ke andar baithe atoms ki number hai. Jab yeh $\sim 2.6$ tak pahunch jaata hai, neighbouring atoms ke matter-wave clouds overlap kar lete hain — atoms ko alag wavepackets ke roop mein pehchana nahi ja sakta — aur condensation shuru hoti hai.

$k_B{T}_c=\frac{2\pi {\hslash }^2}{m}{\left(\frac{n}{g\zeta (3/2)}\right)}^{2/3}$ use karo. Inner ratio: $\frac{n}{\zeta (3/2)}=\frac{1.0\times 10^{19}}{2.612}=3.83\times 10^{18}{\text{m}}^{-3}$. Iska $2/3$ power: $(3.83\times 10^{18}{)}^{2/3}=2.44\times 10^{12}{\text{m}}^{-2}$. Prefactor: $\frac{2\pi {\hslash }^2}{m}=\frac{2\pi (1.0546\times 10^{-34}{)}^2}{1.44\times 10^{-25}}=4.85\times 10^{-43}{\text{J⋅m}}^2$. Toh $k_B{T}_c=4.85\times 10^{-43}\times 2.44\times 10^{12}=1.18\times 10^{-30}\text{J}$. $k_B$ se divide karo: $T_c=\frac{1.18\times 10^{-30}}{1.381\times 10^{-23}}\approx 8.6\times 10^{-8}\text{K}\approx 86\text{nK}$. Order of magnitude ~$10^{-7}$ K — sub-microkelvin, bilkul jaisa parent note ne kaha tha.

$\frac{N_0}{N}=1-{\left(\frac{T}{T_c}\right)}^{3/2}=1-(0.5{)}^{3/2}=1-0.3536=\mathrm{0.6464.}$ ≈ 64.6% atoms already condensed hain critical temperature ke aadhe par. Steep early rise fraction curve ki infinite initial slope ko reflect karti hai $T_c$ ke theek neeche.

$\frac{N_0}{N}=\frac{1}{2}$ set karo: $\frac{1}{2}=1-{\left(\frac{T}{T_c}\right)}^{3/2}\Rightarrow {\left(\frac{T}{T_c}\right)}^{3/2}=\frac{1}{2}.$ Dono sides ko $2/3$ power tak raise karo: $\frac{T}{T_c}={\left(\frac{1}{2}\right)}^{2/3}=\mathrm{0.630.}$ Toh $T\approx 0.63T_c$ par gas exactly half condensed hai.

$T_c\propto n^{2/3}$ (baaki sab fixed). Toh $\frac{T_{c,B}}{T_{c,A}}={\left(\frac{n_B}{n_A}\right)}^{2/3}=8^{2/3}=(2^3{)}^{2/3}=2^2=4.$ $T_c$ 4 guna bada hai. Dense gas → chota interparticle spacing → matter waves zyada temperature par overlap karte hain (utna bada ${\lambda }_T$ zaruri nahi), isliye condensation pehle hoti hai (zyada garam par).

2D mein, ek $k$-disc mein states count karne par $N(k)\propto k^2\propto \epsilon$ milta hai, toh $g(\epsilon )=\frac{dN}{d\epsilon }=$ constant (koi $\sqrt{\epsilon }$ factor nahi). $\mu =0$ par maximum excited population toh yeh hai $N_{\text{exc}}^{\max }\propto {\int }_0^{\infty }\frac{g(\epsilon )d\epsilon }{e^{\epsilon /k_{B}T}-1}\propto {\int }_0^{\infty }\frac{dx}{e^x-1}.$ $x=0$ ke paas, $e^x-1\approx x$, isliye integrand $1/x$ jaisa behave karta hai, aur ${\int }_0\frac{dx}{x}=\ln x{|}_0$ diverge karta hai. Divergent maximum ka matlab hai excited states unbounded atoms absorb kar sakti hain — koi saturation nahi, hence koi overflow nahi, koi condensate nahi. 3D se contrast karo, jahan $g(\epsilon )\propto \sqrt{\epsilon }$ se extra $\sqrt{x}$ small-$x$ behaviour ko tame karta hai ($\sqrt{x}/x=x^{-1/2}$ integrable hai), ek finite number $\Gamma (3/2)\zeta (3/2)$ deta hai. Yahan $\Gamma$ Gamma function hai, simply factorial ka smooth continuation non-integer arguments tak — yeh usi integral se defined hota hai jo hum kar rahe hain, $\Gamma (s)={\int }_0^{\infty }{x}^{s-1}{e}^{-x}dx$, aur yeh $\Gamma (n)=(n-1)!$ satisfy karta hai whole numbers ke liye; specific value jo humein chahiye woh hai $\Gamma (3/2)=\frac{\sqrt{\pi }}{2}\approx 0.886$. Toh "finite number" $\Gamma (3/2)\zeta (3/2)$ bas (integral ka ek fixed shape factor) × (zeta value $2.612$) hai. Dimensionality sab decide karti hai.

Pehle intuition (hum kar kya rahe hain?). Do boxed formulas alag lagti hain — ek overlapping wavelengths ke baare mein hai, doosri temperature ke baare mein — lekin dono ek hi physics do baar likha hua hona chahiye. Humara kaam hai "matter waves overlap" se shuru karna aur crank ghoomana tab tak jab tak "temperature" bahar na aa jaye. Neeche ke har algebra step mein sirf $T_c$ ko isolate kiya ja raha hai; dekho kaise ${\lambda }_T$ (jisme $T$ ek square root ke andar chhupi hai) layer by layer unwrap hoti hai, jaise temperature ko uske packaging se nikalna. Step 1 — overlap condition state karo. $n{\lambda }_{T_c}^3=\zeta (3/2)$, yaani ${\lambda }_{T_c}^3=\frac{\zeta (3/2)}{n}$. Kyun: yeh physical starting point hai — woh moment jab clouds touch karte hain. Step 2 — hidden $T$ expose karo. ${\lambda }_{T_c}=\frac{h}{\sqrt{2\pi m{k}_B{T}_c}}$ insert karo, kyunki $T$ andar ${\lambda }_T$ mein rehta hai aur hum use baahar open mein chahte hain: ${\lambda }_{T_c}^3=\frac{h^3}{(2\pi m{k}_B{T}_c{)}^{3/2}}=\frac{\zeta (3/2)}{n}.$ Step 3 — $3/2$ power undo karo. Dono sides ko $2/3$ power tak raise karo taaki temperature bracket se exponent hat jaye (isliye final formula mein $2/3$ aata hai): $2\pi m{k}_B{T}_c={\left(\frac{n{h}^3}{\zeta (3/2)}\right)}^{2/3}={\left(\frac{n}{\zeta (3/2)}\right)}^{2/3}{h}^2.$ Step 4 — $k_B{T}_c$ isolate karo aur $h\to \hslash$ convert karo. $2\pi m$ se divide karo; phir $h=2\pi \hslash$ use karo isliye $h^2=4{\pi }^2{\hslash }^2$ aur $\frac{h^2}{2\pi m}=\frac{2\pi {\hslash }^2}{m}$: $k_B{T}_c=\frac{h^2}{2\pi m}{\left(\frac{n}{\zeta (3/2)}\right)}^{2/3}=\frac{2\pi {\hslash }^2}{m}{\left(\frac{n}{\zeta (3/2)}\right)}^{2/3}✓$ Do boxed formulas ek hi statement hain — ek overlapping waves ki geometric language mein, doosra critical temperature ki thermal language mein.

(a) $\frac{N_0}{N}=1-(0.30{)}^{3/2}=1-0.1643=0.8357$, toh ≈ 83.6%. (b) $N_0=0.8357\times 2.0\times 10^4=1.67\times 10^4$ atoms, yaani ≈ 16 700 atoms single lowest state share karte hain. (c) $N_{\text{exc}}=N-N_0=2.0\times 10^4-1.67\times 10^4=3.29\times 10^3\approx 3300$ atoms (equivalently $N(T/T_c{)}^{3/2}=2.0\times 10^4\times 0.1643$).

Fixed $n$ aur $g$ par, $T_c\propto \frac{1}{m}$ ($k_B{T}_c=\frac{2\pi {\hslash }^2}{m}(\cdots )$ se). $\frac{T_{c,\text{Na}}}{T_{c,\text{Rb}}}=\frac{m_{\text{Rb}}}{m_{\text{Na}}}=\frac{1.44\times 10^{-25}}{3.82\times 10^{-26}}=\mathrm{3.77.}$ Sodium ka $T_c$ zyada hai, rubidium se approximately 3.8 guna. Halke atoms ka ${\lambda }_T$ same temperature par bada hota hai, isliye unke matter waves pehle (zyada garam par) overlap karte hain. Isliye halki species condense karna asaan hota hai.

Onset par $n{\lambda }_{T_c}^3=\zeta (3/2)$, toh ${\lambda }_{T_c}={\left(\frac{\zeta (3/2)}{n}\right)}^{1/3}={\left(\frac{2.612}{1.0\times 10^{19}}\right)}^{1/3}=(2.612\times 10^{-19}{)}^{1/3}\text{m}.$ $(2.612\times 10^{-19}{)}^{1/3}=6.39\times 10^{-7}\text{m}\approx 0.64\mu \text{m}$. Interparticle spacing: $d=n^{-1/3}=(1.0\times 10^{19}{)}^{-1/3}=4.64\times 10^{-7}\text{m}\approx 0.46\mu \text{m}$. Ratio $\frac{{\lambda }_{T_c}}{d}=\zeta (3/2{)}^{1/3}=2.612^{1/3}=1.38$. Wavelength spacing se ~1.4 guna hai — yaani matter waves abhi ek dusre ko overlap karne bhar bade ho gaye hain. Yahi "phase-space density of order one" ka geometric matlab hai.

$x=\epsilon /k_{B}T$ substitute karne par, shape integral hai ${\int }_0^{\infty }\frac{x^{s-1}}{e^x-1}dx=\Gamma (s)\zeta (s)$, jahan $\Gamma (s)={\int }_0^{\infty }{x}^{s-1}{e}^{-x}dx$ woh Gamma function hai jo L3.2 mein introduce ki gayi thi (factorial extended to real $s$).

• Small-$x$ convergence: $x=0$ ke paas, $e^x-1\approx x$, toh integrand $\approx x^{s-1}/x=x^{s-2}$. Integral ${\int }_0{x}^{s-2}dx$ converge karta hai iff $s-2>-1$, yaani $\boxed{s>1}$.
• Large-$x$: $e^x$ tail ko hamesha kill kar deta hai. Toh only condition $s>1$ hai (aur $\zeta (s)$ finite hai sirf $s>1$ ke liye — same threshold). (i) uniform 3D: $g\propto \sqrt{\epsilon }={\epsilon }^{1/2}$ toh $s-1=1/2\Rightarrow s=3/2>1$ → BEC exists ✓. (ii) uniform 2D: $g=$ const $={\epsilon }^0$ toh $s-1=0\Rightarrow s=1$, nahi $>1$ → borderline divergent, no BEC ✓ (L3.2 se match karta hai). (iii) 3D harmonic trap: $g\propto {\epsilon }^2$ toh $s-1=2\Rightarrow s=3>1$ → BEC exists (aur sach mein harmonic-trap BEC hi woh hai jo experiments actually dekhte hain). Trap $s$ ko raise karke help karta hai.

General $s$ ke liye, $N_{\text{exc}}(T)\propto {\int }_0^{\infty }\frac{{\epsilon }^{s-1}}{e^{\epsilon /k_{B}T}-1}d\epsilon \propto T^s$ (substitution $x=\epsilon /k_{B}T$ mein $(k_{B}T{)}^s$ bahar nikalo; bacha hua integral pure number $\Gamma (s)\zeta (s)$ hai). Toh $\frac{N_{\text{exc}}}{N}={\left(\frac{T}{T_c}\right)}^s,\frac{N_0}{N}=1-{\left(\frac{T}{T_c}\right)}^s.$ Isliye $\alpha =s$. Uniform 3D: $s=3/2$ (parent-note exponent recover karta hai ✓). Harmonic trap: $\alpha =s=3$. $T=0.5T_c$ par $\alpha =3$ ke saath evaluate karo: $\frac{N_0}{N}=1-(0.5{)}^3=1-0.125=\mathrm{0.875.}$ ≈ 87.5% — uniform gas ke 64.6% se compare mein, trap same reduced temperature par substantially zyada fraction condense karta hai. Reason: iska steeper $g(\epsilon )\propto {\epsilon }^2$ kam low-energy excited states chhod'ta hai jo atoms ko hide karne dein, isliye atoms ground state mein pehle force ho jaate hain.

(a) Order parameter: condensate fraction $N_0/N$. Yeh $T\ge T_c$ ke liye exactly $0$ hai aur $T<T_c$ ke liye continuously (zero se) grow karta hai — yeh continuous (second-order-like) transition ki hallmark hai (dekho Phase transitions). (b) Non-analyticity: function $N_0/N=1-(T/T_c{)}^{3/2}$ ka $T_c^{-}$ par infinite slope hai: $\frac{d}{dT}\left[1-{\left(\frac{T}{T_c}\right)}^{3/2}\right]=-\frac{3}{2T_c}{\left(\frac{T}{T_c}\right)}^{1/2}\to -\frac{3}{2T_c}\text{as}T\to T_c^{-},$ jabki $T>T_c$ ke liye yeh identically $0$ hai. Derivative jump karta hai — $T_c$ par ek kink jo koi analytic function nahi rakh sakta. (Zyada precisely, heat capacity ka $T_c$ par ek cusp hota hai.) Yahi non-analyticity phase transition ki definition hai. (c) Classical invisibility: Boltzmann statistics deta hai $\bar{n}=e^{-(\epsilon -\mu )/k_{B}T}$, jo kisi bhi $\mu$ ke liye finite aur positive hai — $\mu$ par koi cap nahi, hence koi saturation nahi, hence koi overflow nahi. Special ingredient hai $\frac{1}{e^{(\epsilon -\mu )/k_{B}T}-1}$ mein $-1$, jo $\mu \le 0$ force karta hai aur finite ceiling create karta hai. Zero interactions ki zarurat; transition poori tarah quantum statistics mein hai. (Contrast karo Fermi-Dirac statistics se, jahan $+1$ pile up ko forbid karta hai aur koi condensate nahi deta — balki ek filled Fermi sea milta hai. Condensate ki coherence Superfluidity ko underlie karti hai.)

Q: Ek breath mein, $T_c$ kya set karta hai, aur neeche condensate kaise grow karta hai (general $s$)? A: $T_c$ $n{\lambda }_{T_c}^3=g\zeta (3/2)$ se fix hota hai (matter waves overlap karte hain); neeche $N_0/N=1-(T/T_c{)}^s$ jahan $s=3/2$ (uniform 3D) ya $s=3$ (harmonic trap), aur $s>1$ nahi hone par BEC nahi hoti.