Physics · Advanced university
Canonical ensembles and partition functions
For a system at inverse temperature β=1/(kBT), the canonical probability of a microstate of energy E is exp(−βE)/Z.
Prerequisites
Probability, exponential functions and thermodynamics.
Study notes
Canonical weights
For a system at inverse temperature β=1/(kBT), the canonical probability of a microstate of energy E is exp(−βE)/Z.
Partition function
Z sums Boltzmann weights over microstates. Degeneracy must be included when summing over distinct energy levels.
Averages
Mean energy is −∂ln Z/∂β when the energy spectrum is independent of β. Thermodynamic identities inherit the ensemble assumptions.
Physical scope
Canonical equilibrium assumes thermal contact with an appropriate reservoir. It is not a universal model for every evolving system.
Worked examples
Two nondegenerate levels have energies 0 and ε with exp(−βε)=0.25. Find Z.
Add weights 1+0.25.
1.25
For a nondegenerate two-level canonical model with energies 0 and ε and exp(−βε)=0.25, find the excited-state probability.
Divide the excited weight by total weight: 0.25/1.25.
0.2
Check your understanding
What extra factor enters a level's weight when it has degeneracy g?
- g² always
- g
- No factor in any case
- 1/g always
A ground level has degeneracy 2 and excited level degeneracy 3. With excited weight per microstate 0.5, find Z.
What quantity is −∂ln Z/∂β for a temperature-independent spectrum?
- Mean position always
- Volume always
- Mean energy
- Zero in all ensembles
Why should probabilities be normalized after assigning Boltzmann factors?
Answers and marking guidance
Answers and marking notes
- g
There are g microstates at that energy, each contributing a Boltzmann weight.
- 3.5
Z=2×1+3×0.5.
- Mean energy
Differentiating the canonical partition sum produces the weighted energy mean.
- Individual weights need not sum to one. Dividing by Z ensures a valid probability distribution across all included microstates.
Credit Z as the normalization sum.
Further reading and review references
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