Canonical ensemble — partition function Z: Question bank

Before the questions, one shared vocabulary reminder so nothing is used before it is defined:

Definition Symbols in play

$Z = \sum_{i} e^{- β E_{i}}$ — the partition function, a plain sum of weights over every microstate $i$ .
$β = 1/ (k_{B} T)$ — "coldness"; big $β$ means low temperature $T$ . Here $k_{B}$ is Boltzmann's constant (the conversion between temperature and energy).
$e^{- β E_{i}}$ — the Boltzmann factor, the unnormalized weight of state $i$ .
$P_{i} = e^{- β E_{i}} / Z$ — the actual probability (this is what sums to 1, not $Z$ ).
$⟨ E ⟩ = - \partial_{β} ln Z$ — average energy pulled out of $Z$ by differentiation.
$F = - k_{B} T ln Z$ — Helmholtz free energy.
$C_{V} \equiv (\frac{\partial ⟨ E ⟩}{\partial T})_{V}$ — heat capacity at fixed volume: how much the average energy rises per degree of warming.
$h$ — Planck's constant, a fixed quantity of physical dimension "energy $\times$ time" (called action). Nature bundles phase space — the space of positions $x$ and momenta $p$ — into cells of size $h$ per degree of freedom, so a state occupies a chunk $h$ of the $x$ – $p$ plane rather than a mathematical point. Dividing a phase-space integral by $h$ literally counts these cells.
$μ$ — chemical potential: the energy cost of adding one more particle to the system. It appears only when particle number is allowed to change (grand canonical), never in the ordinary canonical $Z$ .

Below, a picture to keep in your head while you answer — the same energy ladder at three temperatures:

Figure — Canonical ensemble — partition function Z

Spot the error

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

Sign error. Differentiating

e^{- β E_{i}}

brings down

- E_{i}

, so

⟨ E ⟩ = - \partial_{β} ln Z

. The minus is physical: raising

β

(cooling) lowers the average energy.

"Since

F = - k_{B} T ln Z

and

⟨ E ⟩ = - \partial_{β} ln Z

, we have

F = ⟨ E ⟩

Wrong.

F = ⟨ E ⟩ - T S

; free energy equals average energy only at

T = 0

where the entropy term

T S

vanishes. They are genuinely different quantities linked through $F$ via

S = - \partial F / \partial T

"To get

⟨ E^{2} ⟩

just square

⟨ E ⟩

Wrong.

⟨ E^{2} ⟩ \neq = ⟨ E ⟩^{2}

in general; their difference is the variance

σ_{E}^{2}

. Getting

⟨ E^{2} ⟩

requires the second

β

-derivative of

Z

, not squaring the mean.

"The variance

σ_{E}^{2} = k_{B} T^{2} C_{V}

can be negative when

T

is small."

Wrong. A variance is a sum of squared deviations and is never negative; correspondingly

C_{V} \geq 0

always, which this relation actually guarantees. The chain that ties them together is

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}

— so

C_{V}

literally emerges as a second derivative of

ln Z

(see Heat capacity and fluctuations, where this identity is the whole story).

"For the continuous oscillator we integrate over

x

and

p

, so no factor of

h

is needed."

Error. The

1/ h

(per degree of freedom) is required to make

Z

dimensionless and to match the quantum count of states — recall

h

is the size of one phase-space cell, so dividing by it converts "area in the

x

–

p

plane" into "number of states." Dropping it changes

F

and

S

by additive constants and breaks comparisons.

Z = \sum_{n} e^{- β E_{n}}

works even when several states share an energy."

Incomplete. The sum is over microstates, so degenerate levels must be counted once each; written as a sum over energy levels you must insert the degeneracy

g_{n}

Z = \sum_{n} g_{n} e^{- β E_{n}}

Why questions

Why does a higher-energy state get an exponentially smaller weight, not just a linearly smaller one?

Because the weight tracks the reservoir's microstate count

Ω_{R} = e^{S_{R} / k_{B}}

, and a linear drop in reservoir entropy (

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

) becomes an exponential drop after exponentiating — the exponential is inherited from entropy, not assumed.

Why does

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

appear in deriving the Boltzmann factor?

That derivative is the thermodynamic definition of temperature; it is exactly the coefficient of

E_{i}

in the Taylor expansion of the reservoir's entropy, which is why

T

(and hence

β

) enters the weight at all.

Why is differentiating with respect to

β

(rather than

T

) the natural move for averages?

Because

β

multiplies each

E_{i}

linearly inside the exponent, so

\partial_{β}

cleanly pulls down one factor of

- E_{i}

per term — differentiating by

T

would introduce an extra

1/ T^{2}

chain-rule mess.

Why must probabilities be divided by

Z

at all?

The raw Boltzmann factors are only relative weights; dividing by their total

Z

forces

\sum_{i} P_{i} = 1

, which is the defining requirement of any probability distribution — this is the whole job of

Z

Why does the classical harmonic oscillator give

⟨ E ⟩ = k_{B} T

with no reference to

ω

Because

ln Z = - ln β + const

, and the

- ln β

term (which carries all the

β

-dependence) is independent of

ω

; the two quadratic degrees of freedom each contribute

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

by the Equipartition theorem.

Why is

Z

called the "master key"?

Because every thermodynamic quantity — energy, entropy, pressure, heat capacity, free energy — is obtained from

ln Z

by differentiation with respect to

β

T

, or

V

; knowing

Z

is knowing the entire thermodynamics of the system.

Yes — if infinitely many states pile up without their energies growing fast enough (e.g. a classical hydrogen atom's bound spectrum), the sum fails to converge, signalling that the naive model is unphysical and needs a cutoff or quantum correction.

What happens to the two-level

⟨ E ⟩

T \to \infty

It approaches

ε /2

, not

ε

: at infinite temperature both states are equally likely, so the average sits exactly midway between

0

and

ε

What happens to the two-level

⟨ E ⟩

T \to 0

It approaches

0

, the ground-state energy: the Boltzmann factor of the excited state

e^{- β ε} \to 0

, freezing the particle into the lower level.

Can temperature — and hence

β

— be negative?

Yes, but only for a system with a bounded, capped energy spectrum (like the two-level system, which has a maximum energy

ε

). If more particles occupy the upper level than the lower one,

\partial S / \partial E < 0

, so

T < 0

and

β < 0

; the Boltzmann factor

e^{- β E_{i}}

then favours high energy. A negative-temperature state is hotter than any positive-temperature one, and it cannot exist for the harmonic oscillator or free particle whose energies run to

+ \infty

(there

Z

would diverge for

β < 0

For a two-level system at

β < 0

, where does

⟨ E ⟩

sit?

Above

ε /2

: as

β \to - \infty

the upper level dominates and

⟨ E ⟩ \to ε

, the exact mirror image of the low-temperature limit — this is population inversion, the physics behind lasers.

Is the canonical ensemble ever equivalent to the microcanonical one?

In the thermodynamic limit (

N \to \infty

) the energy fluctuations

σ_{E} / ⟨ E ⟩ \sim 1/ N \to 0

, so the canonical energy is effectively sharp and the two ensembles predict identical thermodynamics.

What distinguishes the canonical from the grand canonical ensemble at the level of

Z

Canonical fixes

N

and sums Boltzmann factors over energy states only; grand canonical also lets

N

fluctuate, inserting a factor

e^{β μ N}

— where

μ

(defined above) is the energy price of one extra particle — and summing over particle number too.

If a level is

g

-fold degenerate, how does that change the low-temperature entropy?

T \to 0

the system spreads over

g

equally likely ground states, giving a residual entropy

S \to k_{B} ln g

rather than zero — a genuine physical consequence traceable directly to the Boltzmann distribution over those

g

states.

Recall Fast self-test

Give the one-word reason $Z_{N} = Z_{1}^{N} / N!$ for a gas. ::: Indistinguishability. True/false: $Z$ can be less than 1. ::: True — e.g. a single state with $E_{0} > 0$ gives $Z = e^{- β E_{0}} < 1$ ; $Z$ is a weight sum with no upper or lower bound of 1. The minus sign in $⟨ E ⟩ = - \partial_{β} ln Z$ comes from where? ::: The $- E_{i}$ that falls out when $\partial_{β}$ hits $e^{- β E_{i}}$ . When is $T < 0$ physically possible? ::: Only for systems with a maximum (capped) energy, when the upper levels are more populated than the lower ones.

Question bank — Canonical ensemble — partition function Z

True or false — justify

Spot the error

Why questions

Edge cases