Bose-Einstein condensation — concept

2.4.18 · Physics › Thermodynamics & Statistical Mechanics (Advanced)

BEC hota kyun hai?

Bosons special kyun hain? Woh Bose–Einstein statistics follow karte hain: koi bhi number of bosons ek hi single-particle state mein ho sakta hai. Energy $\epsilon$ wali state ki mean occupation hai

$\bar{n}(\epsilon )=\frac{1}{e^{(\epsilon -\mu )/k_{B}T}-1}.$

$\mu$ ko hum constrain kaise karte hain? Occupations non-negative honi chahiye, isliye $e^{(\epsilon -\mu )/k_{B}T}>1$ sabhi states ke liye hona chahiye. Lowest energy ${\epsilon }_0=0$ hai, isliye humein chahiye ki

$\mu \le 0\text{(bosons ke liye, ground state }0\text{ par)}.$

Ground-state occupation hai $N_0=\frac{1}{e^{-\mu /k_{B}T}-1}.$ Jaise hi $\mu \to 0^{-}$, $N_0\to \infty$. Yahi escape valve hai: jab excited states saturate ho jaati hain, $\mu$ khud ko $0$ se thoda neeche pin kar leta hai aur ground state baaki sab ko chupchap soak kar leta hai.

Saturation number derive karna (scratch se)

Step 1 — States count karo. Excited states par sum karo → density of states ke saath integral. Yeh step kyun? Ek bade box mein levels dense hote hain; sum ko integral se replace karna continuum approximation hai. Pakad (yaad rakhna!): integral $\epsilon =0$ state ko miss kar deta hai kyunki density of states $g(\epsilon )\propto \sqrt{\epsilon }$ wahan zero ho jaati hai — isliye hum $N_0$ ko alag track karna zaroori hai.

Density of states ($k$-shells count karne se derive kiya, $N(k)=\frac{V}{(2\pi {)}^3}\cdot \frac{4}{3}\pi k^3\cdot g$): $g(\epsilon )d\epsilon =\frac{gV}{4{\pi }^2}{\left(\frac{2m}{{\hslash }^2}\right)}^{3/2}\sqrt{\epsilon }d\epsilon .$

Step 2 — Maximum excited population $\mu =0$ par hoti hai: $N_{\text{exc}}^{\max }(T)={\int }_0^{\infty }\frac{g(\epsilon )d\epsilon }{e^{\epsilon /k_{B}T}-1}.$

Step 3 — Substitute karo $x=\epsilon /k_{B}T$. Kyun? Taaki saari $T$-dependence aage aa jaye aur ek pure number expose ho. $N_{\text{exc}}^{\max }=\frac{gV}{4{\pi }^2}{\left(\frac{2m{k}_{B}T}{{\hslash }^2}\right)}^{3/2}\underset{\Gamma (3/2)\zeta (3/2)}{\underbrace{{\int }_0^{\infty }\frac{\sqrt{x}dx}{e^x-1}}}.$

$\Gamma (3/2)=\frac{\sqrt{\pi }}{2}$ aur thermal de Broglie wavelength ${\lambda }_T=\frac{h}{\sqrt{2\pi m{k}_{B}T}}$ use karte hue, yeh khoobsurti se collapse ho jaata hai

Matlab kya hai: excited states zyada se zyada $\sim g\zeta (3/2)$ particles per thermal volume ${\lambda }_T^3$ hold kar sakti hain. Real density isse zyada kar do (ya $T$ cool karo taaki ${\lambda }_T$ bade) aur baaki condense ho jaate hain.

Transition temperature

$n_{\text{exc}}^{\max }(T_c)=n$ set karo (actual density) — woh point jahan excited states bilkul full hain: $n=g\frac{\zeta (3/2)}{{\lambda }_{T_c}^3}\Rightarrow {\lambda }_{T_c}^3=\frac{g\zeta (3/2)}{n}.$

$T$ ke liye solve karte hue (${\lambda }_T\propto T^{-1/2}$ use karke):

$T_c$ ke neeche condensate fraction

$T<T_c$ ke liye, $\mu \approx 0$, isliye $N_{\text{exc}}(T)\propto T^{3/2}$. $T_c$ par uski value se compare karo (jahan $N_{\text{exc}}=N$): $\frac{N_{\text{exc}}(T)}{N}={\left(\frac{T}{T_c}\right)}^{3/2}.$ Baaki sab ground state mein hai:

Worked examples

Common mistakes (Steel-manned)

Active recall

Connections

• Bose-Einstein statistics — occupation law jo condensation ko possible banati hai.
• Thermal de Broglie wavelength — ${\lambda }_T=h/\sqrt{2\pi m{k}_{B}T}$ overlap criterion set karta hai.
• Density of states — iski $\sqrt{\epsilon }$ form ground-state subtlety ka kaaran hai.
• Chemical potential — $0^{-}$ par pinning condensation ka mechanism hai.
• Riemann zeta function — $\zeta (3/2)$ saturation integral se aata hai.
• Phase transitions — BEC ek continuous transition hai jisme interactions se koi latent heat nahi.
• Superfluidity — BEC superfluid ${}^4$He aur atomic condensates ke neeche hai.
• Fermi-Dirac statistics — contrast: Pauli exclusion fermions ke liye condensation forbid karta hai.

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