Fermi-Dirac statistics — fermions, Fermi energy

2.4.17 · Physics › Thermodynamics & Statistical Mechanics (Advanced)

Fermi-Dirac statistics exist KYUN karta hai?

HUM kya problem solve kar rahe hain? Classical Maxwell-Boltzmann statistics humein batati hai ki distinguishable particles energy levels par kaise spread hote hain. Lekin quantum particles indistinguishable hote hain, aur fermions ke paas ek extra rule hai: Pauli exclusion principle. Humein ek aisi counting rule chahiye jo ise respect kare.

YEH rule matter KYUN karta hai? Exclusion ke bina, ek metal ke saare electrons $T = 0$ par ground state mein collapse ho jaate aur metals bilkul alag behave karte (na electron degeneracy pressure hoti, na white dwarfs hote, na semiconductor band gaps utne mayne rakhte jitne rakhte hain).

HOW hum Fermi-Dirac distribution derive karte hain (scratch se)

Hum grand canonical ensemble use karte hain: system temperature $T$ aur chemical potential $μ$ par ek reservoir ke saath particles aur energy exchange karta hai.

Step 1 — Energy $ε$ ke ek single quantum state par focus karo. Kyun? Kyunki fermion states occupation mein independent hoti hain — har single-particle state apni ek chhoti subsystem hai jo $n = 0$ ya $n = 1$ particle rakh sakti hai (exclusion!).

Step 2 — Us ek state ke liye grand partition function likho. $n$ particles aur energy $n ε$ wale microstate ka grand canonical weight $e^{- (n ε - μ n) / k_{B} T}$ hai. $Z = \sum_{n = 0}^{1} e^{- n (ε - μ) / k_{B} T} = 1 + e^{- (ε - μ) / k_{B} T}$ Sirf $n = 0, 1$ kyun? Pauli exclusion: ek single state mein do identical fermions nahi ho sakte.

Step 3 — Average occupation number. $\overset{n}{ˉ} = \frac{\sum _{n} n e ^{- n (ε - μ) / k_{B} T}}{Z} = \frac{0 \cdot 1 + 1 \cdot e ^{- (ε - μ) / k_{B} T}}{1 + e ^{- (ε - μ) / k_{B} T}}$

Upar aur neeche dono ko $e^{- (ε - μ) / k_{B} T}$ se divide karo:

Fermi energy: $T = 0$ limit

$E_{F}$ KYA hai? Fermi energy $E_{F} \equiv μ (T = 0)$ define karte hain.

$f (ε)$ mein $T \to 0$ lo:

$ε < E_{F}$ ke liye: exponent $\to e^{- \infty} = 0 \Rightarrow f = 1$ .
$ε > E_{F}$ ke liye: exponent $\to e^{+ \infty} = \infty \Rightarrow f = 0$ .

Figure — Fermi-Dirac statistics — fermions, Fermi energy

Free electron gas (3D) ke liye $E_{F}$ derive karna

Step 1 — States count karo (density of states). Volume $V$ ke ek box mein, allowed momenta quantized hain; $k$ -space mein states uniformly density $V / (2 π)^{3}$ ke saath fill hoti hain. Spin factor $g_{s} = 2$ include karke: $# states with ∣ k ∣ < k_{F} = 2 \cdot \frac{V}{( 2 π ) ^{3}} \cdot \frac{4}{3} π k_{F}^{3}$ $\frac{4}{3} π k_{F}^{3}$ kyun? $k_{F}$ radius wale sphere ka volume — $T = 0$ par filled states ek sphere banati hain (the Fermi sphere).

Step 2 — Ise electrons ki number $N$ ke barabar set karo. $N = \frac{V}{3 π ^{2}} k_{F}^{3} \Rightarrow k_{F} = (3 π^{2} \frac{N}{V})^{1/3}$

Step 3 — Energy mein convert karo $ε = ℏ^{2} k^{2} /2 m$ use karke:

Hum Fermi temperature $T_{F} = E_{F} / k_{B}$ aur Fermi velocity $v_{F} = ℏ k_{F} / m$ bhi define karte hain.

Worked Examples

Common Mistakes (Steel-manned)

Flashcards

Fermion ko kaunsi property define karti hai?

Half-integer spin aur antisymmetric exchange wavefunction (Pauli exclusion follow karta hai).

Fermi-Dirac distribution likho.

f (ε) = \frac{1}{e ^{(ε - μ) / k_{B} T} + 1}

Denominator mein

+ 1

kyun hai (bosons ke

- 1

ke comparison mein)?

Yeh occupation ko

f \leq 1

par cap karta hai, Pauli exclusion enforce karta hai (ek fermion per state).

ε = μ

par exactly

f

kya hota hai?

1/2

, kisi bhi

T > 0

ke liye.

Fermi energy define karo.

E_{F} = μ (T = 0)

; woh energy jiske neeche saari states filled hain aur jiske upar saari empty hain

T = 0

par.

T = 0

par

f (ε)

kaisa dikhta hai?

Ek step function:

ε < E_{F}

ke liye

1

ε > E_{F}

ke liye

0

3D free electron gas ki Fermi energy?

E_{F} = \frac{ℏ ^{2}}{2 m} (3 π^{2} n)^{2/3}

jahan

n = N / V

k_{F}

ka electron density se kya relation hai?

k_{F} = (3 π^{2} n)^{1/3}

Fermi temperature kya hai aur yeh kyun matter karta hai?

T_{F} = E_{F} / k_{B}

; agar

T ≪ T_{F}

toh gas degenerate hai (metals:

T_{F} \sim 1 0^{4}

–

1 0^{5}

K).

Finite T par smeared transition ki width kitni hai?

Approximately

\sim k_{B} T

(≈

4.4 k_{B} T

f = 0.9

0.1

tak).

f

se particle number paane ke liye tum kya multiply karte ho?

Density of states

g (ε)

d N = f (ε) g (ε) d ε

White dwarf star ko kya rok ke rakhta hai?

Pauli exclusion ki wajah se electron degeneracy pressure (electrons saare ground state mein nahi baith sakte).

Recall Feynman: ek 12-saal ke bachche ko explain karo

Ek movie theater imagine karo jahan har seat par exactly ek hi insaan baith sakta hai — lap par baithna allowed nahi (yahi Pauli rule hai). Log hamesha pehle saste front-row seats chahte hain. Agar tum ek bada crowd laao chahe kitni bhi thandi ho (zero temperature), woh phir bhi front se seat-by-seat fill karte hain, kyunki front seats already bhari hain. Jo last filled row hai woh Fermi level hai. Jab thoda warm hota hai, sirf us last row ke aas-paas wale log restless hote hain aur thodi si upar wali row mein jump karte hain — andar front mein sab waise hi rehte hain. Isliye metals ko garam karne par bhi zyada change nahi hota: sirf Fermi level ke paas wale electrons move kar sakte hain.

Connections

Pauli Exclusion Principle — $+ 1$ aur step function ki microscopic origin.
Maxwell-Boltzmann Statistics — tab recover hoti hai jab $f ≪ 1$ (dilute / high- $T$ limit).
Bose-Einstein Statistics — $- 1$ wala sibling; condensation allow karta hai.
Density of States — $f$ ko actual particle counts mein convert karne ke liye zaroori.
Grand Canonical Ensemble — woh framework jisse $f$ derive kiya jaata hai.
Electron Degeneracy Pressure & White Dwarfs — Fermi sea ka astrophysical consequence.
Chemical Potential — $μ$ , half-filling energy; $T = 0$ par $E_{F}$ ke barabar hai.
Sommerfeld Expansion — $f$ se $μ (T)$ aur heat capacity kaise follow karte hain.