Let particles exchange with a reservoir
Derive grand-canonical weights and occupation laws while distinguishing an imposed chemical potential from a fixed particle number.
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The question for today
What you’ll understand
- Identify the fixed variables of the grand canonical ensemble.
- Differentiate the grand partition function to obtain mean number and fluctuations.
- Derive distinct one-state Bose and Fermi occupations with their allowed chemical potentials.
A small observation volume inside a gas can exchange particles as well as energy with its surroundings. A canonical description with exactly fixed particle number is then the wrong constraint, even if its average density is useful. The grand canonical ensemble fixes reservoir temperature and chemical potential instead. Its state weight contains energy minus chemical potential times particle number. That extra term leads directly to occupancy laws and number fluctuations, provided each particle-number sector is counted consistently.
The reservoir pays both an energy and a particle cost
The entropy differential for a reservoir at fixed volume contains energy divided by temperature minus chemical potential times particle number divided by temperature. If the subsystem gains energy and particles, the reservoir loses both. Expanding its entropy therefore yields a weight exponential in minus beta times subsystem energy minus chemical potential times subsystem number. The sign of the chemical-potential term follows from this exchange argument; it is not chosen by analogy with the energy term.
Sum that weight over every allowed particle-number sector and every physical microstate within it. If fixed-number canonical partition functions already include the Gibbs factorial, do not divide by another factorial in the outer number sum. Fugacity is the exponential of beta times chemical potential and is dimensionless. Changing the one-particle energy zero requires shifting chemical potential by the same amount to preserve occupations. Only differences between a level energy and chemical potential enter the one-state weights.
The normalizer generates number response
At fixed temperature and volume, differentiating the logarithm of the grand partition function with respect to chemical potential gives mean number times beta. A second derivative gives the number variance times beta squared. Equivalently, number variance is thermal energy times the response of mean number to chemical potential. It is nonnegative in a stable grand-canonical state. These derivatives must not be taken along a path where chemical potential is simultaneously adjusted to hold number fixed.
For a classical ideal gas with one internal state, summing its canonical functions produces an exponential grand partition function. Its number distribution is Poisson, with variance equal to mean. This differs from a binomial distribution of occupants in finitely many exclusion-limited states. Both can become sharply concentrated in relative terms for large mean populations, yet they describe different state spaces. The ensemble determines what fluctuates, while the particle statistics determine the allowed occupation count for each state.
Allowed occupations set the quantum distribution
A single fermionic state permits occupation zero or one, so its grand sum has two terms. A bosonic state permits every nonnegative occupation, so its sum is geometric. Differentiation gives denominators differing by a sign. A fermionic chemical potential may lie above or below the lowest level. At positive temperature, the bosonic grand sum instead requires chemical potential strictly below the lowest level for a finite system; approaching that boundary can generate a macroscopic ground occupation.
Independent equal-energy fermionic states give binomial total occupancy and connect directly to the probability lab. Bosonic occupation has a geometric distribution with variance larger than its mean, and cannot be represented by that same binomial instrument. Calling both outputs a generic quantum probability would hide the decisive difference. Derive the occupation probability first, then use the lab only for the fermionic set whose zero-or-one occupation assumptions actually match its exact binomial model.
Work it through
Occupy four independent fermionic states
Four distinct fermionic states have the same energy ε and exchange particles with a reservoir such that exp[β(ε−μ)]=3. Find each occupation probability, the total grand partition function, mean particle number, and number variance.
Each state has occupied-to-empty weight exp[−β(ε−μ)]=1/3. Its occupation probability is (1/3)/(1+1/3)=1/4.
The one-state grand function is 4/3. Four independent states give Ξ=(4/3)⁴=256/81, multiplying state functions rather than dividing by an extra particle factorial.
Total occupancy is binomial with four states and probability 1/4. Its mean is 1 and variance is 4×(1/4)×(3/4)=3/4. Fixing the mean to one has not fixed the actual particle number.
Each state is occupied with probability 1/4; Ξ=256/81, mean number is 1, and number variance is 3/4. The reservoir fixes μ, not the exact number.
The living notebook
Explore the model
Interactive lab · probability
Inspect exclusion-limited number fluctuations
For fresh comparisons Reset and set BOTH p=0.25, trials n=4; then Reset and set p=0.25, n=8. Predict mean counts 1 and 2 and variances 0.75 and 1.5 for independent equal-energy fermionic states. The model does not compute chemical potential or bosonic geometric occupation.
- 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 forGrand-canonical fermion occupations can fluctuate even though each individual state allows at most one particle.
Pause & reason
Check your understanding
Your turn
Try it yourself
For a single state with exp[β(ε−μ)]=2, compute its mean occupation and variance for bosons and for fermions.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Use the opposite denominator signs, then the corresponding occupation-variance formulas.
Compare with the solution
Bosons have mean 1/(2−1)=1 and variance 1×2=2. Fermions have mean 1/(2+1)=1/3 and variance (1/3)(2/3)=2/9. The bosonic mean may exceed one for other parameters because there is no exclusion restriction; the fermionic state always remains between zero and one.
Keep the idea
Take it with you
- Grand-canonical weights use E−μN and include all allowed number sectors.
- Number variance measures response to μ at fixed T and V.
- Bose and Fermi statistics differ through their allowed single-state occupations.
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