Average energy from partition function: Question bank

2.4.11 · D5 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Average energy from partition function

Definition Is page par use hone wale symbols (ek baar define, freely use karo)

$k_{B}$ = Boltzmann's constant, temperature aur energy ke beech ka fixed conversion factor ( $k_{B} \approx 1.38 \times 1 0^{- 23} J/K$ ). Ye tab aata hai jab temperature $T$ ko energy scale $k_{B} T$ mein convert karna ho.
$β \equiv \frac{1}{k _{B} T}$ = inverse temperature. Bada $β$ matlab cold, chota $β$ matlab hot.
$Z$ = partition function. Discrete microstates $i$ ke liye jinka energy $E_{i}$ hai, $Z = \sum_{i} e^{- β E_{i}}$ ; continuous spectrum ke liye, $Z = \int d E ρ (E) e^{- β E}$ . Ye sabhi Boltzmann weights ka normalizing sum hai.
$p_{i}$ = microstate $i$ ki Boltzmann probability, yaani $p_{i} = \frac{e ^{- β E_{i}}}{Z}$ , wo fraction jitni baar system state $i$ mein rehta hai. Construction se $\sum_{i} p_{i} = 1$ .
$ρ (E)$ = density of states, yaani $ρ (E) d E$ count karta hai ki kitne microstates ki energy window $[E, E + d E]$ mein hai.
$ℏ \equiv \frac{h}{2 π}$ = reduced Planck constant ( $h$ = Planck's constant). Ye quantum energy levels ki spacing set karta hai, jaise oscillator ka $ℏ ω$ .
$F$ = Helmholtz free energy, defined by $F = - k_{B} T ln Z$ — wo thermodynamic potential jo fixed temperature par system ke liye natural hai. Dekho Helmholtz free energy F = -kT ln Z.
$C_{V}$ = heat capacity at constant volume, $C_{V} = \frac{\partial ⟨ E ⟩}{\partial T}$ — system kitna energy absorb karta hai per unit temperature rise. Dekho Heat capacity and energy fluctuations.
Is page par oscillator energy convention: hum quantum harmonic oscillator ke levels ko uske ground state se measure karte hain, $E_{n} = n ℏ ω$ jahan $n = 0, 1, 2, \dots$ , taki $E_{0} = 0$ . (Sach wala zero-point $\frac{1}{2} ℏ ω$ add karna sirf $⟨ E ⟩$ ko ek constant se shift karta hai — neeche "adding a constant" trap dekho.)

Recurring theme: $⟨ E ⟩ = - \frac{\partial ln Z}{\partial β}$ jahan $β = 1/ k_{B} T$ , aur neeche har trap ek sign, ek $1/ Z$ , ek missing Jacobian, ya ek limit mein chhupta hai.

True or false — justify

True or false:

⟨ E ⟩ = - \partial_{β} Z

average energy hai.

False — tumhe

Z

se divide karna hoga. Bare

- \partial_{β} Z = \sum_{i} E_{i} e^{- β E_{i}}

unnormalized hai; sahi object hai

- \frac{1}{Z} \partial_{β} Z = - \partial_{β} ln Z

True or false:

⟨ E ⟩

hamesha positive hota hai.

Saamanya mein False — ye

E_{i}

ka sign inherit karta hai. Agar tum sari energies neeche shift karo (jaise levels

- ε, 0

), toh

⟨ E ⟩

negative ho sakta hai; sirf tab

⟨ E ⟩ \geq 0

hota hai jab har

E_{i} \geq 0

ho.

True or false: har energy level mein ek constant

E_{0}

add karne se

⟨ E ⟩

mein

E_{0}

ka change aata hai.

True —

Z \to e^{- β E_{0}} Z

, toh

ln Z \to ln Z - β E_{0}

aur

- \partial_{β} ln Z \to ⟨ E ⟩ + E_{0}

. Energy ka koi absolute zero nahi hota; sirf differences physical hote hain.

True or false:

⟨ E ⟩ = k_{B} T^{2} \partial_{T} ln Z

aur

⟨ E ⟩ = - \partial_{β} ln Z

ek doosre se contradict karte hain.

False — ye identical hain. Kyunki

β = 1/ k_{B} T

hai, chain rule deta hai

\partial_{β} = - k_{B} T^{2} \partial_{T}

, toh dono forms alag variables mein likhi ek hi equation hai.

True or false:

ln Z

dimensionless hota hai.

True —

Z = \sum_{i} e^{- β E_{i}}

pure dimensionless Boltzmann weights ka sum hai, toh

ln Z

ek pure number hai. (Continuous

Z

ke liye dimensionlessness ko phase-space factor jaise

h

se restore karte hain.)

True or false: ek two-level system mein

⟨ E ⟩

kabhi

ε

se zyada ho sakta hai.

False — ye

0

(cold) aur

ε /2

(hot) ke beech interpolate karta hai,

ε

tak bhi nahi pahuncha. Sirf do states hone par, mean

0

aur

ε

ka weighted average hai jo maximum

ε /2

tak pohonchta hai jab dono equally likely hote hain.

True or false:

T \to \infty

par har system ka

⟨ E ⟩

diverge karta hai.

False — ek bounded spectrum ke liye (jaise two-level system) ye levels ke arithmetic mean par saturate ho jaata hai. High

T

par divergence sirf unbounded spectra ke liye hota hai (jaise oscillator).

True or false: doosra derivative

\partial_{β}^{2} ln Z

negative ho sakta hai.

False — ye variance

⟨ E^{2} ⟩ - ⟨ E ⟩^{2} = σ_{E}^{2} \geq 0

ke barabar hota hai, toh

ln Z

β

mein convex hai. Negative value ka matlab hoga imaginary energy spread, jo impossible hai.

Spot the error

"Kyunki energy temperature ke saath badhti hai,

⟨ E ⟩ = + \partial_{β} ln Z

Sign error.

\partial_{β} e^{- β E} = - E e^{- β E}

, toh minus Boltzmann exponent mein baked in hai:

⟨ E ⟩ = - \partial_{β} ln Z

. (Aur note karo ki

T

badhane se

β

ghatta hai, toh

T

ke baare mein intuition ko

β

signs par apply nahi karna chahiye.)

⟨ E ⟩ = k_{B} T \partial_{T} ln Z

kyunki maine

\partial_{β}

\partial_{T}

mein convert kiya."

Ek factor of

T

chhoot gaya. Sahi Jacobian hai

\partial_{β} = - k_{B} T^{2} \partial_{T}

, jo deta hai

⟨ E ⟩ = k_{B} T^{2} \partial_{T} ln Z

— power

T^{2}

hai,

T

nahi.

"Oscillator ke liye main likhun ga

Z = \sum_{n} e^{- β n ℏ ω} = \frac{1}{1 + e ^{- β ℏ ω}}

Galat closed form. Geometric series

\sum_{n} r^{n} = \frac{1}{1 - r}

mein denominator mein minus hai:

Z = \frac{1}{1 - e ^{- β ℏ ω}}

1 + \dots

wala form two-level partition function hai, ek alag problem.

C_{V} = \partial ⟨ E ⟩ / \partial β

Galat variable. Heat capacity define hoti hai per unit temperature se:

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

β

mein differentiate karne par

- σ_{E}^{2}

milta hai,

C_{V}

nahi.

"Kyunki

p_{i} = e^{- β E_{i}} / Z

hai, high

T

par highest-energy state sabse zyada probable hoti hai."

Ulta emphasis. High

T

par (

β \to 0

) sare weights

e^{- β E_{i}} \to 1

, toh states equally likely ho jaati hain — koi state dominate nahi karti. Low-energy states sirf low

T

par dominate karti hain.

"Equipartition

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

deta hai per coordinate."

Imprecise. Ye

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

deta hai per quadratic term energy mein. Ek 1D oscillator mein do hain (kinetic

\frac{p ^{2}}{2 m}

aur potential

\frac{1}{2} k x^{2}

) →

k_{B} T

; ek free particle ka single coordinate bina quadratic potential ke kuch contribute nahi karta.

Edge cases

T = 0

par (

β \to \infty

), kisi bhi system ke liye

⟨ E ⟩

kya hoga jiska unique ground state ho?

Ye ground-state energy

E_{0}

ke barabar hoga. Saare excited weights

e^{- β E_{i}}

ground term se exponentially tezi se die karte hain, toh system apni lowest state mein freeze ho jaata hai.

Quantum oscillator ke liye

Z

ka kya hoga jab

T \to 0

e^{- β ℏ ω} \to 0

, toh

Z \to \frac{1}{1 - 0} = 1

— sirf

n = 0

ground state bachti hai. Correspondingly

⟨ E ⟩ \to 0

, jo ek frozen oscillator se match karta hai (hamare convention se

E_{0} = 0

Two-level system ke liye high

T

par

⟨ E ⟩

kya hai, aur ye

ε

kyun nahi hai?

ε /2

hai,

0

aur

ε

ka plain average, kyunki dono states equally probable ho jaati hain. Do equally likely values ka mean unka midpoint hota hai, kabhi bada wala nahi.

Unbounded spectrum ke liye, finite

T

par

Z

diverge kyun nahi karta?

Exponential suppression

e^{- β E_{i}}

, states

ρ (E)

ki growth ko beat kar deta hai, toh sum (ya integral

\int d E ρ (E) e^{- β E}

) kisi bhi

β > 0

ke liye converge karta hai. Convergence sirf tab fail ho sakti hai jab

β \leq 0

ho.

Oscillator formula

⟨ E ⟩ = \frac{ℏ ω}{e ^{β ℏ ω} - 1}

mein

β \to 0

exactly try karne par kya toot jaata hai?

Denominator

\to 0

ho jaata hai, toh expression diverge karta hai — lekin limit carefully lete hue (expand

e^{x} \approx 1 + x

) milta hai

⟨ E ⟩ \to k_{B} T

, jo classical equipartition value hai. Apparent singularity removable hai.

Kya

β \leq 0

(negative temperature) hamesha sirf mathematical pathology hai?

Nahi — ek bounded spectrum ke liye,

β < 0

ek genuine physical state hai jisme inverted populations hoti hain (upper level mein zyada atoms), jo nuclear-spin aur laser systems mein realize hoti hai; ye "hotter than infinity" hota hai. Sirf unbounded ladder ke liye

β \leq 0

Z

ko diverge karta hai aur unphysical ho jaata hai.

Ek system ka degenerate ground level ho (multiplicity

g

). Kya

T = 0

par

⟨ E ⟩

badal jaata hai?

Nahi —

⟨ E ⟩ \to E_{0}

phir bhi, kyunki saare

g

degenerate states ek hi energy share karte hain. Degeneracy

Z

ke prefactor aur entropy ko affect karta hai, lekin

T = 0

par mean energy ko nahi.

Kya

⟨ E ⟩ = - \partial_{β} ln Z

continuous spectrum ke liye valid hai jahan

Z

ek integral hai?

Haan — derivation ne sirf derivative ki linearity aur

e^{- β E}

ki structure use ki, jo

\sum_{i}

\int d E ρ (E)

se replace karne par bhi survive karta hai. Equipartition example exactly yahi case hai.

Recall Pure trap set ka one-line summary

Lagbhag har trap teen mein se ek sin hai: ek giraya hua minus sign, ek bhula hua $1/ Z$ , ek missing $k_{B} T^{2}$ Jacobian, ya ek illegal limit bounded-vs-unbounded spectrum par. Unhe guard karo aur tum safe ho.

Related: Boltzmann distribution · Helmholtz free energy F = -kT ln Z · Heat capacity and energy fluctuations · Equipartition theorem · Two-level system / Schottky anomaly · Quantum harmonic oscillator — thermal

Question bank — Average energy from partition function

True or false — justify

Spot the error

Why questions

Edge cases