Quantum statistics — distinguishable vs indistinguishable particles

2.4.15 · Physics › Thermodynamics & Statistical Mechanics (Advanced)

HUM POOCH KYA RAHE HAIN?

KYUN indistinguishability aati hai? Kyunki probability density $|\psi {|}^2$ tab bhi unchanged rehni chahiye jab hum do identical particles ko swap karein. Ye force karta hai ki wavefunction ya to symmetric ho ($+$, bosons) ya antisymmetric ($-$, fermions) exchange ke under.

WAVEFUNCTION ISKO KAISE ENCODE KARTI HAI (scratch se derive karo)

Do particles lo jo single-particle states $a$ aur $b$ mein hain. Maano $\hat{P}$ particle 1 aur 2 ko swap karta hai.

Step 1 — physical requirement. Identical particles ko swap karna koi bhi measurable cheez nahi badal sakta: $|\psi (1,2){|}^2=|\psi (2,1){|}^2.$ Ye step kyun? Probability density position ke baare mein ek maatra observable hai; identical particles ko identical predictions deni chahiye.

Step 2 — solve karo. Iska matlab hai $\psi (2,1)=e^{i\varphi }\psi (1,2)$. Do baar swap karo → wapas shuruaat pe, isliye $e^{2i\varphi }=1\Rightarrow e^{i\varphi }=\pm 1$. Ye step kyun? Do swaps identity operation hai, ${\hat{P}}^2=1$.

Step 3 — states banao. ${\psi }_{\pm }=\frac{1}{\sqrt{2}}[{\psi }_a(1){\psi }_b(2)\pm {\psi }_a(2){\psi }_b(1)].$ $+$ → symmetric (bosons), $-$ → antisymmetric (fermions). Ye step kyun? Ye product states ke sirf wahi combinations hain jo $\psi (2,1)=\pm \psi (1,2)$ satisfy karte hain.

Step 4 — Pauli exclusion free mein nikal aata hai. Antisymmetric case mein $a=b$ set karo: ${\psi }_{-}=\frac{1}{\sqrt{2}}[{\psi }_a(1){\psi }_a(2)-{\psi }_a(2){\psi }_a(1)]=0.$ Ye step kyun? Jo state har jagah zero ho uski probability zero hai — do fermions ek hi single-particle state mein nahi reh sakte.

Wo counting jo statistics define karti hai

Socho 2 particles ko 3 single-particle states (boxes) mein daalna hai. Allowed microstates count karo.

Occupation-number distributions (key results derive karo)

Grand canonical ensemble use karo: energy $\epsilon$ ki har single-particle level temperature $T$, chemical potential $\mu$ wale reservoir ke saath particles exchange karti hai. Maano $\beta =1/k_{B}T$. Ek level ka grand partition function uski occupation $n$ ke upar sum hai: $\mathcal{Z}={\sum }_n{e}^{-\beta (\epsilon -\mu )n}.$ Ye step kyun? Levels independent hain, isliye hum ek level treat karte hain aur baad mein sum/multiply karte hain. Boltzmann factor har occupation ko uski energy aur particle number se weight karta hai.

Mean occupation hai $⟨n⟩=\frac{{\sum }_{n}n{e}^{-\beta (\epsilon -\mu )n}}{{\sum }_n{e}^{-\beta (\epsilon -\mu )n}}=-\frac{1}{\beta }\frac{\partial }{\partial \mu }\ln \mathcal{Z}\text{(equivalently)}.$

Fermions: $n\in \{0,1\}$. $\mathcal{Z}=1+e^{-\beta (\epsilon -\mu )},⟨n⟩=\frac{e^{-\beta (\epsilon -\mu )}}{1+e^{-\beta (\epsilon -\mu )}}.$

Bosons: $n\in \{0,1,2,\dots \}$, ek geometric series (zaroori hai $\epsilon >\mu$): $\mathcal{Z}={\sum }_{n=0}^{\infty }{e}^{-\beta (\epsilon -\mu )n}=\frac{1}{1-e^{-\beta (\epsilon -\mu )}}.$ $⟨n⟩=\frac{e^{-\beta (\epsilon -\mu )}}{1-e^{-\beta (\epsilon -\mu )}}.$

Ek-line unifier: $⟨n⟩=\frac{1}{e^{\beta (\epsilon -\mu )}+a}$, jahan $a=+1$ (Fermi), $a=-1$ (Bose), $a=0$ (classical).

Gibbs ka $1/N!$ patch (kyun classical counting galat thi)

Common mistakes (Steel-man + fix)

Flashcards

Connections

• Bose–Einstein Condensation
• Fermi Gas & Fermi Energy
• Grand Canonical Ensemble
• Gibbs Paradox & Entropy of Mixing
• Pauli Exclusion Principle
• Blackbody Radiation (Photon Gas)
• Spin–Statistics Theorem
• Maxwell–Boltzmann Distribution

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