Fermi-Dirac statistics — fermions, Fermi energy: Question bank

2.4.17 · D5 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Fermi-Dirac statistics — fermions, Fermi energy

Unke symbols ki reminder jiske baare mein hum reason karte hain (sab parent mein define hain):

$f (ε)$ — probability ki energy $ε$ ki one single-particle state occupied hai, hamesha $0 \leq f \leq 1$ .
$μ$ — chemical potential (ek particle add karne ki energy cost); $E_{F} \equiv μ (T = 0)$ .
$g (ε)$ — density of states, number of states per unit energy (dekho Density of States).
$k_{B} T$ — thermal energy scale; $T_{F} = E_{F} / k_{B}$ Fermi temperature hai.

Spot the error

Spot the error: "

f (ε)

batata hai ki kitne electrons ki energy

ε

hai."

f

ek probability per state hai, dimensionless aur

\leq 1

; electrons ki number

d N = f (ε) g (ε) d ε

hai — tumhe Density of States

g (ε)

se multiply karna hoga.

Spot the error: "Bosons aur fermions mein sirf ek harmless

\pm

sign ka farq hai, isliye unki physics basically same hai."

Sign opposite worlds encode karta hai:

- 1

(Bose-Einstein Statistics)

f \to \infty

allow karta hai aur condensation permit karta hai;

+ 1

f \leq 1

cap karta hai aur exclusion force karta hai — macroscopic consequences (superfluidity vs degeneracy pressure) bilkul alag hain.

Spot the error: "

T = 0

par Fermi surface fuzzy hai kyunki quantum uncertainty hai."

Exactly

T = 0

par surface ek sharp step hai (

f = 1

neeche,

0

upar);

\sim k_{B} T

width ki fuzziness tabhi aati hai jab

T > 0

step ko thermally smear karta hai.

Spot the error: "Humne partition function mein sirf

n = 0, 1

count kiya taaki algebra simple rahe."

Yeh simplicity ke liye nahi hai — Pauli Exclusion Principle ek state mein

n \geq 2

ko physically forbid karta hai, isliye

n \in {0, 1}

poora exact sum hai, koi truncation nahi.

Spot the error: "Fermi sphere woh real space ka region hai jahan electrons hote hain."

Yeh momentum ( $k$ )-space mein radius

k_{F}

ki ek sphere hai; yeh describe karta hai ki

T = 0

par kaun se momentum states filled hain, na ki physical space ka koi region.

Spot the error: "Kyunki

E_{F}

ek energy hai, isliye thande metals ki

E_{F}

chhoti hoti hai."

E_{F}

sirf number density

n

par depend karta hai

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

ke through,

T

par nahi; ek fixed metal ko thanda karne se

E_{F}

essentially unchanged rehta hai.

Spot the error: "Denominator mein

+ 1

spin degeneracy include karne se aata hai."

Nahi —

+ 1

ek single state ke do allowed occupations

n = 0, 1

par sum karne se aata hai; spin alag se ek degeneracy factor

g_{s} = 2

ke roop mein enter karta hai jo state count ko multiply karta hai.

Why questions

Ek fermion state mein zyaada se zyada ek particle kyun ho sakta hai, lekin ek boson state mein koi bhi number?

Fermions ka antisymmetric many-body wavefunction hota hai, jo vanish ho jaata hai agar do same state mein hon (Pauli exclusion); bosons symmetric hote hain, aisi koi limit impose nahi karte — isliye denominators mein

+ 1

- 1

hota hai.

f

derive karne ke liye hum Grand Canonical Ensemble use karte hain, canonical kyun nahi?

Kyunki yeh particle number ko fixed

μ

par fluctuate karne deta hai, isliye har single-particle state ek independent tiny subsystem ban jaata hai jo reservoir se particles exchange karta hai — wahi independence hai jo one-state sum ko trivial banati hai.

f (ε)

mein chemical potential

μ

naturally "reference energy" kyun hai?

μ

woh energy hai jahan ek particle add karne ki net cost kuch nahi; neeche ke states fill karne mein saste hain (isliye

f \to 1

), upar wale mehnge hain (isliye

f \to 0

), aur yeh exactly half-filled hai — yeh occupation ka tipping point hai.

f

\sim 0.9

\sim 0.1

tak ka transition material chahe koi bhi ho, width

\approx 4.4 k_{B} T

kyun rakhta hai?

Kyunki

f

energy par sirf dimensionless combination

(ε - μ) / k_{B} T

ke through depend karta hai; 0.9-to-0.1 crossing us variable ki ek fixed range span karta hai, isliye iska energy width purely

k_{B} T

ke saath scale karta hai.

Electron degeneracy pressure

T = 0

par bhi kyun exist karta hai?

Exclusion electrons ko

k_{F}

tak high-momentum states mein force karta hai, unhe badi kinetic energy aur isliye pressure deta hai — ek purely quantum effect jo temperature se independent hai (dekho Electron Degeneracy Pressure & White Dwarfs).

E_{F}

density ke saath linearly nahi,

n^{2/3}

ke saath kyun scale karta hai?

Filled states ek Fermi sphere banate hain jiska volume

\propto k_{F}^{3}

hai aur

N

states hold karta hai, isliye

k_{F} \propto n^{1/3}

; kyunki

ε \propto k^{2}

, top energy

k_{F}^{2} \propto n^{2/3}

ke saath scale karti hai.

T

badhne par kisi metal ke liye

μ (T)

thoda kyun decrease karta hai?

Density of states typically energy ke saath badhti hai, isliye thermal smearing

μ

ke neeche wale states ko upar wale states fill hone se zyada empty karta hai;

N

fixed rakhne ke liye

μ

ko neeche shift karna padta hai (leading term

- \frac{π ^{2}}{12} (k_{B} T / E_{F})^{2} E_{F}

Sommerfeld Expansion se). Neeche ki figure yeh geometric imbalance dikhati hai.

Edge cases

Edge case:

T \to 0

limit mein exactly

ε = μ

par

f

kya hai?

Finite-

T

curves sab

ε = μ

par

\frac{1}{2}

se guzarti hain, lekin exactly

T = 0

par function ek discontinuous step hai aur wahaan iska value undefined hai;

\frac{1}{2}

assign karna ek convention hai (jump ka midpoint / common limit), koi derived fact nahi — upar definition box dekho.

Edge case: Classical (non-degenerate) limit

n \to

small ya

T \to

large mein

f (ε)

kya ban jaata hai?

Yeh Boltzmann factor

f \approx e^{- (ε - μ) / k_{B} T} ≪ 1

mein reduce ho jaata hai, kyunki states itni sparsely occupied hoti hain ki exclusion kabhi bite nahi karta — Maxwell-Boltzmann Statistics recover ho jaata hai.

Edge case: Finite

T

ke liye

μ

se kaafi neeche,

ε ≪ μ

par

f

kya hota hai?

Exponent large aur negative hota hai, isliye

f \to 1

: yeh deep states essentially fully occupied hain aur thermally frozen hain — yeh conduction ya heat capacity mein kuch contribute nahi karte.

Edge case: Kya

μ

negative ho sakta hai?

Haan — dilute ya classical gas ke liye

μ < 0

hota hai (Boltzmann limit mein bahut negative ho jaata hai); degenerate metal ke liye

μ \approx E_{F} > 0

μ

ka sign batata hai ki gas classical hai ya degenerate.

Edge case:

T = 0

par

E_{F}

ke paas thermally active shell mein kitne fraction electrons hote hain?

Exactly zero width ki active shell — shell ki width

\sim k_{B} T

hai, jo

T = 0

par vanish ho jaati hai, isliye koi bhi electron "thermally active" nahi hota; active fraction

\sim k_{B} T / E_{F}

sirf

T > 0

ke liye appear hota hai.

Edge case: Agar do alag metals ki density

n

same hai, toh unki Fermi energies ki comparison kaise hogi?

Woh equal hongi (free-electron model ke liye), kyunki

E_{F}

sirf

n

aur electron mass

m

par depend karta hai; band-structure effects (effective mass) real materials mein ise shift kar sakte hain, lekin free-gas value identical hai.

Recall Quick self-test

$μ$ vs $E_{F}$ : kaun sa temperature-dependent hai? ::: $μ (T)$ , $T$ par depend karta hai; $E_{F} = μ (0)$ ek fixed reference value hai. Woh single fact jo poora " $+ 1$ " produce karta hai: ::: Ek state sirf $n = 0$ ya $1$ hold karta hai, isliye $Z = 1 + e^{- (ε - μ) / k_{B} T}$ , jiska average occupation $+ 1$ carry karta hai. Exactly $T = 0$ par $f (μ)$ kya hai? ::: Undefined (step wahaan discontinuous hai); $\frac{1}{2}$ ek convention hai.

Question bank — Fermi-Dirac statistics — fermions, Fermi energy

True or false — justify

Spot the error

Why questions

Edge cases