Derive Boltzmann weights from a reservoir
Expand reservoir entropy to obtain canonical probabilities, finite-reservoir corrections, and energy-fluctuation identities.
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The question for today
What you’ll understand
- Derive the canonical weight from a large reservoir state count.
- Include level degeneracy exactly once.
- Relate canonical energy variance to heat capacity and identify its fixed parameters.
A subsystem in thermal contact does not have fixed energy even when the entire isolated universe does. Its probability of occupying one microstate is proportional to how many states the reservoir can occupy with the remaining energy. Expanding that reservoir entropy yields the Boltzmann factor and exposes the approximation behind a constant-temperature bath. The partition function then does more than normalize probabilities: its derivatives generate energy averages and fluctuations at fixed volume and particle number.
The reservoir supplies the exponential weight
Assume the coupling is weak enough to neglect interaction energy when assigning subsystem and reservoir energies. If a subsystem microstate has energy Ei, the reservoir has total energy minus Ei. All compatible composite microstates are equally likely, so the subsystem weight is the reservoir multiplicity at that remaining energy. Expand the logarithm of this multiplicity around a reference reservoir energy. Its first derivative is inverse temperature divided by Boltzmann's constant, producing the exponential suppression of higher subsystem energies.
The second derivative contains the reservoir heat capacity. For an ordinary stable bath it is negative, making a quadratic correction to the exponent. The canonical approximation neglects that term over the subsystem's relevant energy range. A bath need not have infinitely much total energy, but exchanges must change its temperature negligibly. Strong coupling also needs a different treatment because interaction energy and subsystem equilibrium cannot generally be represented by the bare subsystem Hamiltonian alone.
States and energy levels are different sums
The index in the partition function labels microstates, not necessarily distinct energy values. If an energy level has degeneracy g, its total probability includes g equal Boltzmann weights. One can either sum each microstate or multiply each distinct level by its degeneracy. Doing both counts the same states twice. In a classical description the corresponding integral must include its measure and Gibbs factor; those are part of the definition of the state sum.
Shifting every subsystem energy by the same constant multiplies the partition function by a common exponential factor. Normalized probabilities remain unchanged, while free energy and mean energy shift consistently. This check separates physical energy differences from an arbitrary energy zero. For independent subsystems, partition functions multiply because joint energies add and state choices factorize. Their logarithms therefore add, a feature that makes extensive thermodynamic quantities accessible from many simple constituents.
Differentiate the normalizer
Differentiate the logarithm of the partition function with respect to beta while holding the Hamiltonian, volume, and particle number fixed. The first derivative gives minus mean energy; the second gives its variance. Converting the temperature derivative of mean energy gives the heat capacity relation. A nonnegative variance implies nonnegative canonical heat capacity for a temperature-independent Hamiltonian. This statement does not rule out every finite isolated microcanonical model with a different energy constraint.
For many independent two-level sites, the total energy is the excitation gap times a binomial count. The lab can therefore show exact canonical energy-count fluctuations once the excitation probability has been calculated from the Boltzmann weight. The probability control is not a temperature dial and the animation is not a thermalization history. Changing the probability recomputes the distribution; independent comparisons must reset the instrument and specify both its probability and number of sites.
Work it through
Include a degenerate excited level
A system has one ground microstate at energy zero and two excited microstates at energy ε. At the chosen positive temperature exp(−βε)=1/4. Find its partition function, excited-level probability, mean energy, and energy variance.
The ground weight is one and the two excited microstates each have weight 1/4. Thus Z=1+2/4=3/2.
The total excited-level probability is (2/4)/(3/2)=1/3. Each individual excited microstate has probability 1/6; the ground state has probability 2/3.
The energy takes values zero and ε, so its mean is ε/3 and second moment ε²/3. Subtracting the mean squared gives variance 2ε²/9, which has energy-squared units.
Z=3/2, the excited level has probability 1/3, mean energy is ε/3, and energy variance is 2ε²/9. Degeneracy was included once.
The living notebook
Explore the model
Interactive lab · probability
Inspect canonical site fluctuations
For fresh runs Reset and set BOTH p=0.25, trials n=8; then Reset and set p=0.5, n=8. Predict excitation means 2 and 4 and variances 1.5 and 2. These binomial counts model independent sites after deriving p from a reservoir; no temperature or bath dynamics is present.
- Mean · np
- 5
- Variance · np(1−p)
- 2.5
- Total probability
- 1
These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forCanonical energy fluctuates even though reservoir temperature is held fixed.
Pause & reason
Check your understanding
Your turn
Try it yourself
Two independent nondegenerate two-level sites have gap ε and exp(−βε)=1/3. Find the one-site excitation probability and the probability that exactly one site is excited.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Normalize the one-site weights before using the binomial count.
Compare with the solution
One-site probability is (1/3)/(1+1/3)=1/4. Exactly one excitation has probability 2×(1/4)×(3/4)=3/8. The factor two counts which site is excited; it is not an additional excited-level degeneracy for each site. The two-site partition function is (4/3)²=16/9, consistent with the same result.
Keep the idea
Take it with you
- Canonical weights arise from the linear term in reservoir entropy.
- A level probability includes its degeneracy exactly once.
- At fixed V and N, canonical energy variance is kBT²CV.
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