Canonical ensemble — partition function Z: Question bank

2.4.10 · D5 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Canonical ensemble — partition function Z

Questions se pehle, ek shared vocabulary reminder taaki koi cheez define hone se pehle use na ho:

Definition Symbols jo kaam mein hain

$Z = \sum_{i} e^{- β E_{i}}$ — partition function, har microstate $i$ ke weights ka plain sum.
$β = 1/ (k_{B} T)$ — "coldness"; bada $β$ matlab low temperature $T$ . Yahan $k_{B}$ Boltzmann's constant hai (temperature aur energy ke beech conversion).
$e^{- β E_{i}}$ — Boltzmann factor, state $i$ ka unnormalized weight.
$P_{i} = e^{- β E_{i}} / Z$ — actual probability (yeh 1 tak sum hota hai, $Z$ nahi).
$⟨ E ⟩ = - \partial_{β} ln Z$ — average energy jo $Z$ se differentiation ke zariye nikali jaati hai.
$F = - k_{B} T ln Z$ — Helmholtz free energy.
$C_{V} \equiv (\frac{\partial ⟨ E ⟩}{\partial T})_{V}$ — heat capacity at fixed volume: average energy kitni badhti hai per degree warming.
$h$ — Planck's constant, ek fixed quantity jiska physical dimension "energy $\times$ time" hai (jise action kehte hain). Nature phase space — positions $x$ aur momenta $p$ ki space — ko $h$ size ke cells mein bundle karta hai per degree of freedom, toh ek state $x$ – $p$ plane ka ek chunk $h$ occupy karta hai, na ki ek mathematical point. Phase-space integral ko $h$ se divide karna literally in cells ko count karta hai.
$μ$ — chemical potential: system mein ek aur particle add karne ki energy cost. Yeh sirf tab aata hai jab particle number change hone diya jaata hai (grand canonical), ordinary canonical $Z$ mein kabhi nahi.

Neeche, jawab dete waqt dimag mein rakho ek picture — teen temperatures par same energy ladder:

Figure — Canonical ensemble — partition function Z

Spot the error

⟨ E ⟩ = + \frac{\partial ln Z}{\partial β}

Sign error.

e^{- β E_{i}}

differentiate karne par

- E_{i}

neeche aata hai, toh

⟨ E ⟩ = - \partial_{β} ln Z

. Minus physical hai:

β

badhana (cooling) average energy ghatata hai.

"Kyunki

F = - k_{B} T ln Z

aur

⟨ E ⟩ = - \partial_{β} ln Z

, hum keh sakte hain

F = ⟨ E ⟩

Galat.

F = ⟨ E ⟩ - T S

; free energy average energy ke barabar sirf

T = 0

par hoti hai jahan entropy term

T S

vanish ho jaata hai. Yeh genuinely alag quantities hain jo $F$ ke zariye

S = - \partial F / \partial T

se linked hain.

⟨ E^{2} ⟩

nikalne ke liye bas

⟨ E ⟩

ko square karo."

Galat.

⟨ E^{2} ⟩ \neq = ⟨ E ⟩^{2}

generally; unka difference hi variance

σ_{E}^{2}

hai.

⟨ E^{2} ⟩

nikalne ke liye

Z

ka second

β

-derivative chahiye, mean ko square karna nahi.

"Variance

σ_{E}^{2} = k_{B} T^{2} C_{V}

negative ho sakta hai jab

T

small ho."

Galat. Variance squared deviations ka sum hai aur kabhi negative nahi hota; correspondingly

C_{V} \geq 0

hamesha, jo yeh relation actually guarantee karta hai. Jo chain unhe jodhti hai woh hai

σ_{E}^{2} = \partial^{2} ln Z / \partial β^{2} = - \partial ⟨ E ⟩ / \partial β = k_{B} T^{2} \partial ⟨ E ⟩ / \partial T = k_{B} T^{2} C_{V}

— toh

C_{V}

literally

ln Z

ka second derivative hai (dekho Heat capacity and fluctuations, jahan yeh identity poori kahaani hai).

"Continuous oscillator ke liye hum

x

aur

p

par integrate karte hain, toh

h

ka koi factor zaruri nahi."

Error.

1/ h

(per degree of freedom) zaruri hai

Z

ko dimensionless banane ke liye aur quantum state count se match karne ke liye — yaad rakho

h

ek phase-space cell ka size hai, toh isse divide karna "

x

–

p

plane mein area" ko "states ki sankhya" mein convert karta hai. Isse drop karne par

F

aur

S

additive constants se change ho jaate hain aur comparisons toot jaati hain.

Z = \sum_{n} e^{- β E_{n}}

tab bhi kaam karta hai jab kai states ek energy share karti hain."

Incomplete. Sum microstates par hai, toh degenerate levels ko ek ek baar count karna hoga; energy levels par sum ki tarah likha jaaye toh degeneracy

g_{n}

insert karni hogi:

Z = \sum_{n} g_{n} e^{- β E_{n}}

Why questions

Higher-energy state ko exponentially smaller weight kyun milta hai, sirf linearly smaller kyun nahi?

Kyunki weight reservoir ka microstate count

Ω_{R} = e^{S_{R} / k_{B}}

track karta hai, aur reservoir entropy mein linear drop (

S_{R} \to S_{R} - E_{i} / T

) exponentiate hone ke baad exponential drop ban jaata hai — exponential entropy se inherit hota hai, assume nahi kiya jaata.

Boltzmann factor derive karte waqt

\partial S_{R} / \partial E = 1/ T

kyun aata hai?

Woh derivative temperature ki thermodynamic definition hi hai; yeh exactly woh coefficient hai jo reservoir ki entropy ke Taylor expansion mein

E_{i}

ke saath aata hai, aur isliye

T

(aur hence

β

) weight mein enter karta hai.

Averages ke liye

β

ke saath differentiate karna (rather than

T

) natural move kyun hai?

Kyunki

β

har

E_{i}

ko exponent ke andar linearly multiply karta hai, toh

\partial_{β}

cleanly har term mein se ek factor

- E_{i}

neeche laata hai —

T

se differentiate karne par extra

1/ T^{2}

chain-rule ka jhanjhat hota.

Probabilities ko

Z

se divide karna zaroori kyun hai?

Raw Boltzmann factors sirf relative weights hain; unke total

Z

se divide karne par

\sum_{i} P_{i} = 1

force hota hai, jo kisi bhi probability distribution ki defining requirement hai — yahi

Z

ka poora kaam hai.

Classical harmonic oscillator

⟨ E ⟩ = k_{B} T

kyun deta hai bina

ω

ke kisi reference ke?

Kyunki

ln Z = - ln β + const

, aur

- ln β

term (jo saari

β

-dependence carry karta hai)

ω

se independent hai; do quadratic degrees of freedom mein se har ek

\frac{1}{2} k_{B} T

contribute karta hai Equipartition theorem ke according.

Z

ko "master key" kyun kehte hain?

Kyunki har thermodynamic quantity — energy, entropy, pressure, heat capacity, free energy —

ln Z

β

T

, ya

V

ke respect mein differentiate karke milti hai;

Z

jaanna matlab system ki poori thermodynamics jaanna hai.

Haan — agar infinitely many states pile up hon aur unki energies itni tezi se na badhen (jaise classical hydrogen atom ka bound spectrum), toh sum converge nahi karta, jo signal deta hai ki naive model unphysical hai aur cutoff ya quantum correction chahiye.

Two-level

⟨ E ⟩

T \to \infty

par kya hoga?

Yeh

ε /2

approach karta hai,

ε

nahi: infinite temperature par dono states equally likely hain, toh average exactly

0

aur

ε

ke beech mein hota hai.

Two-level

⟨ E ⟩

T \to 0

par kya hoga?

Yeh

0

approach karta hai, ground-state energy: excited state ka Boltzmann factor

e^{- β ε} \to 0

, particle ko lower level mein freeze kar deta hai.

Kya temperature — aur hence

β

— negative ho sakta hai?

Haan, lekin sirf aisi system ke liye jisme bounded, capped energy spectrum ho (jaise two-level system, jiska maximum energy

ε

hai). Agar zyada particles upper level mein hon lower se zyada, toh

\partial S / \partial E < 0

, toh

T < 0

aur

β < 0

; Boltzmann factor

e^{- β E_{i}}

tab high energy ko prefer karta hai. Negative-temperature state kisi bhi positive-temperature state se hotter hoti hai, aur yeh harmonic oscillator ya free particle ke liye exist nahi kar sakti jinka energy

+ \infty

tak jaati hai (wahan

β < 0

ke liye

Z

diverge ho jaata).

Two-level system mein

β < 0

par

⟨ E ⟩

kahan hota hai?

ε /2

se upar: jaise

β \to - \infty

upper level dominate karta hai aur

⟨ E ⟩ \to ε

, low-temperature limit ka exact mirror image — yeh population inversion hai, lasers ki physics.

Kya canonical ensemble kabhi microcanonical ensemble ke equivalent hota hai?

Thermodynamic limit mein (

N \to \infty

) energy fluctuations

σ_{E} / ⟨ E ⟩ \sim 1/ N \to 0

, toh canonical energy effectively sharp ho jaati hai aur dono ensembles identical thermodynamics predict karte hain.

Canonical aur grand canonical ensemble mein

Z

ke level par kya fark hai?

Canonical

N

fix karta hai aur sirf energy states par Boltzmann factors sum karta hai; grand canonical

N

ko bhi fluctuate hone deta hai, ek factor

e^{β μ N}

insert karta hai — jahan

μ

(upar define kiya) ek extra particle ki energy cost hai — aur particle number par bhi sum karta hai.

Agar ek level

g

-fold degenerate hai, toh low-temperature entropy par kya asar hota hai?

Jaise

T \to 0

system

g

equally likely ground states par spread ho jaata hai, ek residual entropy

S \to k_{B} ln g

deta hai instead of zero — Boltzmann distribution se directly traceable ek genuine physical consequence.

Recall Fast self-test

Ek word mein reason do ki $Z_{N} = Z_{1}^{N} / N!$ gas ke liye kyun. ::: Indistinguishability. True/false: $Z$ 1 se kam ho sakta hai. ::: True — jaise ek state $E_{0} > 0$ ke saath $Z = e^{- β E_{0}} < 1$ deta hai; $Z$ ek weight sum hai jiska 1 ka koi upper ya lower bound nahi. $⟨ E ⟩ = - \partial_{β} ln Z$ mein minus sign kahan se aata hai? ::: Woh $- E_{i}$ se jo $\partial_{β}$ ke $e^{- β E_{i}}$ par hit karne par girata hai. $T < 0$ physically kab possible hai? ::: Sirf un systems ke liye jinmein maximum (capped) energy ho, jab upper levels lower se zyada populated hon.

Question bank — Canonical ensemble — partition function Z

True or false — justify karo

Spot the error

Why questions

Edge cases