A thermal bath favors lower-energy states, while the number of states at each energy also matters.
By the end: normalize Boltzmann weights and distinguish the probability of one state from the population of an energy level.
Include the thermal surroundings in the count
The equal left/right weights earlier came from equal-sized compartments and symmetric assumptions. Different energies need not have equal probability for a small system exchanging energy with a large thermal bath. Giving an extra energy Δ to the system removes that energy from the bath. At fixed bath temperature T, the bath’s entropy falls by approximately Δ/T. Since multiplicity depends exponentially on entropy divided by kB, the higher-energy system state receives the relative weight exp of −Δ divided by kB T.
The exponential exp is the inverse of the natural logarithm. For example exp(−1) is about 0.368. An ensemble is a collection of possible states with assigned probabilities. A microcanonical description fixes energy, volume, and particle number for an isolated system. A canonical description instead fixes bath temperature, volume, and particle number while allowing the system’s energy to fluctuate. These Boltzmann weights belong to the canonical description. The weights are not yet probabilities: add the weights of all accessible states, then divide each weight by that sum. This sum is the partition function. It is a normalization count weighted by energy, not a probability and not a partition wall.
Follow the energy across the bath boundary
Choose the small system and its bath together as an isolated total. The small system can gain energy, but that energy must come from the bath. Suppose its ground energy is zero and its excited energy is Δ. Comparing these two possibilities requires counting bath alternatives as well as system alternatives. When the small system is excited, the bath has Δ less energy available. At positive temperature, that usually leaves fewer bath states compatible with the fixed energy of the combined total.
The temperature tells us how strongly the bath count changes with energy. For a small removal Δ, its entropy change is approximately −Δ/T. The bath loses energy without appreciably changing temperature. Dividing by kB gives the change in the logarithm of its multiplicity. Undoing the logarithm gives the ratio exp of −Δ divided by kB T. That ratio compares two individual system states through the number of compatible bath alternatives.
An excitation removes energy from the bath
Schematic closed small system and large bath inside an isolated outer boundary. The 300 K label is Δ/kB, not a second bath temperature. Arrow width does not encode energy.
In this spectrum, Δ/kB equals 300 K, so the exponent is simply −300 K divided by T. At 150 K the exponent is −2 and the weight ratio is about 0.1353. At 300 K it is −1 and the ratio is about 0.3679. Warming the bath makes surrendering the gap energy less costly in bath entropy. The gap itself has not become smaller. It remains approximately 4.142 × 10⁻²¹ J for one system, using the stated value of kB.
A weight ratio does not yet say what fraction of systems are excited. Imagine a probability strip whose two lengths are initially in the ratio 1 to 0.3679. Its total is 1.3679. To turn it into a strip of total probability one, divide both lengths by 1.3679. The ground part becomes about 0.7311 and the excited part about 0.2689. Their sum is one because normalization rescales both parts, rather than treating the ground weight as a probability of certainty.
Rescale both weights by the same total
Widths encode the ratio 1 : exp(−1) at 300 K with one excited state. The same division changes weights into probabilities without changing their ratio.
Changing the zero of energy cannot change those predictions. Add the same energy E0 to every system state. Both weights acquire the common factor exp of −E0 divided by kB T. Their sum acquires that factor too, so it cancels when each weight is divided by the sum. Only energy differences affect these probabilities. By contrast, adding another accessible state changes the sum without multiplying every existing term equally. That changes probabilities even if the original energies stay fixed.
The boundary also determines the appropriate ensemble. Here the small system has fixed particle number and exchanges energy with the bath, which is the canonical setting. It is not isolated merely because the combined total is isolated. No piston motion or mechanical work appears in this two-level energy account. An excitation receives energy from the bath and a relaxation returns it. The equilibrium formula describes the distribution after equilibration, not a sequence in which every small system must alternately take and return exactly one gap.
Compare two levels with different degeneracies
Use a ground energy of zero with one state and an excited energy Δ. Set the gap divided by kB to 300 K. At 300 K, each excited state has weight exp(−1), while the ground state has weight one. With one state at each level, the excited probability is about 26.89%. At 600 K it rises to about 37.75%; it does not become 100%.
Degeneracy means that distinct states have the same energy. If the excited level contains three states, each still has the same individual Boltzmann weight, but the level’s total weight is tripled. At 300 K this raises the excited-level population to about 52.46%. Each individual excited state remains less likely than the single ground state. The diagrams show separate state lines, not three different excited energies.
Count equal-energy states before combining them
An energy level is a collection of states with the same energy. Degeneracy is the number of distinct states in that collection. The model's excited degeneracy of three therefore means three separate alternatives at Δ, not three successive steps of energy Δ. Write their weights separately as w, w, and w. Together with the ground weight one, the partition function is 1 + 3w. Combining the excited weights after listing them prevents a common error in which one level is mistakenly counted as one state.
At 300 K, put w = 0.367879 into that sum. The excited weights total 1.103638 and the full sum is 2.103638. Dividing the ground weight by the full sum gives about 0.4754. Dividing one excited weight by the same sum gives about 0.1749. Adding the three excited probabilities gives about 0.5246. Thus the excited level wins the comparison of levels, although every one of its states loses the comparison with the ground state.
Three alternatives share one excited energy
Vertical height encodes energy in units of Δ. Horizontal placement separates states only. At 300 K, the three equal-energy probabilities sum to 52.46%.
For many independent copies of this small system at equilibrium, probability gives an expected fraction, not an exact head count. Among 1,000 copies at 300 K with three excited states, the expected excited count is about 524.6. An actual count must be an integer and can fluctuate around that expectation. It is not necessary for exactly the same copies to remain excited. A state-population diagram groups possible alternatives; its neighboring boxes do not represent neighboring particles or a spatial arrangement inside a container.
The crossover between ground-level and excited-level majority follows from equal total weights. Set 3w equal to one, which gives exp of −300/T equal to one third. Taking natural logarithms gives 300/T = ln 3, so T is about 273.1 K. This number comes from the gap and degeneracy. Its proximity to water's melting temperature is accidental. This spectrum contains no phase transition. With one excited state there is no finite positive-temperature majority crossover.
A level majority appears when 3w exceeds one
Three excited states and one ground state. The crossover is 300/ln 3 ≈ 273.1 K. This graph concerns populations, not a water phase transition.
Temperature and degeneracy change the population by different mechanisms. Raising T reduces the penalty for each individual excited state relative to the ground state. Raising degeneracy adds alternatives without reducing that penalty. At a given temperature the ratio of one excited state's probability to the ground probability remains w in both degeneracy settings. Both individual probabilities can decrease when new states enter the normalization sum. The excited-level probability can nevertheless increase because its total now includes more terms, each of which contributes positive probability.
Compare these weighted alternatives with the earlier left/right assignment count. Six balanced assignments out of sixteen had probability six sixteenths because every assignment had equal weight. Here three excited states and one ground state do not immediately give an excited probability of three quarters. Equal counting would discard the energy penalty. Three quarters is the limiting result when the gap becomes small relative to kB T. At finite temperature the bath distinguishes energies, even though it does not distinguish between the three equal-energy excited alternatives.
Work through the mechanism
Weight of one excited state relative to the ground
Δ is an energy gap in J, not a temperature. The model sets Δ/kB = 300 K. exp(x) means e raised to x.
Normalize a level containing g states
The ground level has one state of weight one; the excited level has g states of weight w each. Z = 1 + gw is the partition function.
Count states at each energy before normalizing
Partition function Z = 2.104. Excited level 52.46%; each excited state 17.49%; ground 47.54%.
State lines at the same height have the same energy; their horizontal placement is schematic. The gap divided by kB is 300 K in both panels. Probability bars share a 0–100% scale. Temperature controls change only T; state-count buttons change only excited degeneracy. The fixed reference is 300 K with one state per level. These are equilibrium probabilities, not transition trajectories.
Temperature
Excited states
Excited-level probability
150 K
1
11.92%
300 K
1
26.89%
600 K
1
37.75%
300 K
3
52.46%
Ground degeneracy is one; the energy gap divided by kB is fixed at 300 K.
Use the population controls to separate two causes
Begin with the 300 K temperature preset and select one excited state. Before reading the result, calculate a weight ratio of exp of −1 and predict an excited minority. The selected panel should now match the fixed reference, with about 26.89% excited. Next select three excited states without changing temperature. Predict that the energy-line heights will remain unchanged while the total excited bar grows. Check the displayed per-state probability as well as the larger level total.
Keep three excited states and choose the 150 K preset. The weight ratio becomes exp of −2, about 0.1353. The partition function is about 1.4060, giving ground probability about 71.12% and excited-level probability about 28.88%. Each excited state contributes about 9.63%. Although the extra states still help the excited level, they no longer outweigh its energy penalty. Compare this with the 300 K majority case. Temperature caused the reversal, not a changed spectrum.
Now choose 600 K while retaining three excited states. The weight ratio is exp of −0.5, about 0.6065, and the excited-level probability is about 64.53%. Predict what selecting one excited state will do before pressing that button. It reduces the excited total to about 37.75%, even though the weight ratio is unchanged. Then compare 600 K with one excited state against 150 K with three. The warmer system has the larger excited fraction here, so state count alone cannot order populations across different temperatures.
Predict the effect of each replacement
T
Excited states
Z
Excited total
Each excited state
300 K
1
1.3679
26.89%
26.89%
300 K
3
2.1036
52.46%
17.49%
150 K
3
1.4060
28.88%
9.63%
600 K
3
2.8196
64.53%
21.51%
600 K
1
1.6065
37.75%
37.75%
Values use Δ/kB = 300 K and one ground state. Each row is a separate equilibrium calculation, not a timed stage.
Use the temperature slider to bracket the majority crossover with three excited states. At 250 K the excited level is about 47.47% populated; at 300 K it is about 52.46%. The slider's 50 K increments do not select the exact 273.1 K crossover. The calculation between those settings explains it. Return to one excited state and check that every available temperature still leaves the ground level more populated. Approaching equal populations is different from passing through equality at a finite positive temperature.
Population also determines mean energy. Every ground outcome contributes zero and every excited outcome contributes Δ, regardless of which excited state occurred. Therefore the mean energy is Δ times the total excited probability. In units divided by kB, the mean is 300 K times that probability. At 300 K this is about 80.68 K for one excited state and 157.39 K for three. These are energy-per-kB values, not new bath temperatures. The bath temperature remains 300 K in both calculations.
Probability areas give mean energy
Each rectangle has height Δ and width equal to its excited-level probability at 300 K. Its area represents mean energy in units of Δ. Horizontal strips use the same probability scale; only the excited part is shaded.
This mean-energy calculation explains how excitation can store energy without implying that all systems share the mean. A single system occupies one of the allowed states, not a fractional energy halfway between levels. An ensemble average can lie between them because it averages distinct outcomes. Changing T in the controls selects a different equilibrium distribution immediately. It does not supply a heating duration, coupling strength, or transition mechanism. Bar changes describe expected populations, not molecular trajectories.
Keep equilibrium probabilities separate from kinetics
For one state at each level and a very high positive temperature, the two probabilities approach one half. With three excited states and one ground state, the excited-level population instead approaches three quarters. At very low positive temperature, a nonzero gap suppresses excitation. This explains why a degree of freedom can contribute little heat capacity when thermal energy cannot appreciably populate its excited states.
These are equilibrium probabilities for weak coupling to a thermal bath and a deliberately bounded spectrum. They do not predict transition times, a particular particle’s next jump, or a real material’s full heat capacity. A driven population inversion need not obey this positive-temperature equilibrium model. Thermodynamic favorability and kinetic accessibility remain separate questions.
Explain energy storage without predicting transition speed
At low positive temperature, almost every copy occupies the ground state. A small temperature increase then adds little mean energy through this gap because the excited probability is still tiny. At intermediate temperature the population changes more, so this mode can absorb appreciable energy as its excited fraction increases. At high temperature its fractions approach their limits and change little. Its excitation contribution to heat capacity therefore becomes small again, rather than remaining a constant classical value.
A finite comparison makes the energy change explicit without calculus. With one excited state, the excited probability rises from about 0.1192 at 150 K to 0.2689 at 300 K. The mean energy per kB rises from about 35.76 K to 80.68 K. Subtracting gives 44.92 K over a 150 K bath-temperature change. Their ratio is an average excitation capacity of about 0.2995 kB per copy over that interval. It is an interval average, not the capacity at every temperature within the interval.
A finite temperature interval changes the mean energy
One excited state. The curve is ⟨E⟩/kB = 300/[1 + exp(300/T)]. The dashed secant measures the interval-average capacity, not a constant capacity at all temperatures.
Do not replace a monatomic gas's whole-sample heat capacity by that single-copy result. That gas has classical translational energy 3nRT/2 and whole-sample constant-volume capacity 3nR/2. It describes different accessible energies and includes the amount of gas. A molecular rotational or vibrational contribution can become suppressed when quantum gaps exceed the available thermal scale, but its spectrum is not generally this two-level spectrum. The two-level comparison explains the mechanism of population suppression without supplying a real molecule's full heat-capacity curve.
The equilibrium ratio constrains transition balance but not absolute speed. For a simple two-state kinetic description, let the upward transition rate per ground-state copy be a and the downward rate per excited copy be b. At equilibrium, ground probability times a equals excited probability times b. Consequently a/b equals the equilibrium excited-to-ground ratio. Multiplying both rates by ten preserves that ratio while changing how quickly a disturbed population returns. Equilibrium probabilities alone cannot determine which of these rate pairs applies to a physical system.
At 300 K with one excited state, take hypothetical rates a = exp of −1 per second, approximately 0.3679, and b = 1 per second. Ground probability 0.7311 times a and excited probability 0.2689 times b both give about 0.2689 transitions per copy per second. Rates ten times larger give about 2.689 in each direction with the same populations. Opposing transitions cancel in the population account; unchanged fractions do not require microscopic inactivity.
Equal fluxes do not fix the rate scale
Rate pair
Upward rate a
Downward rate b
Upward flux
Downward flux
Slower
0.3679 s⁻¹
1 s⁻¹
0.2689 s⁻¹
0.2689 s⁻¹
Tenfold faster
3.679 s⁻¹
10 s⁻¹
2.689 s⁻¹
2.689 s⁻¹
Hypothetical two-state rates at 300 K, rounded from a = exp(−1) s⁻¹ and b = 1 s⁻¹, then both multiplied by ten. Each flux is probability times rate and has units of transitions per copy per second. These rates are not controls or outputs of the population model.
Finally, distinguish probability of a level from probability of an exact population count. A 52.46% excited probability does not give a 52.46% chance that exactly half of 1,000 copies are excited. That latter event needs a separate counting calculation with independence assumptions, just as exact left/right balance differed from a near-balanced band. In real coupled systems, correlations can invalidate independent-copy counting. The bath model specifies the single-system equilibrium weights; additional questions about fluctuations of many systems require specifying what is independent and what total is constrained.
Exercises
Normalize the weights
For one ground state of weight one and one excited state of weight exp(−1), what is the excited probability? Why is 36.8% not the answer?
Hint
The weights must sum to one after normalization.
Worked answer
Quantity
Value
Excited / ground weight
0.3679
Total weight Z
1.3679
Excited probability
0.3679 / 1.3679 ≈ 26.89%
Ground probability
1 / 1.3679 ≈ 73.11%
A relative weight is not yet a probability.
The weight ratio is about 0.3679, but the total weight is 1.3679. Dividing gives an excited probability of about 0.2689. The ground probability is about 0.7311. Using the excited weight directly would omit the normalization supplied by the partition function.
Resolve an apparent contradiction
At 300 K, the three-state excited level is more populated than the one-state ground level. Does that mean each excited state is more probable than the ground state?
Hint
Divide the level population among its three equally weighted states.
Worked answer
Object counted
Probability
One ground state
47.54%
Each of three excited states
17.49%
Excited level in total
52.46%
Count states and levels separately.
No. The excited level has total probability about 52.46%, or about 17.49% per excited state. The single ground state has about 47.54%. More states can outweigh a smaller probability per state. This is consistent with a positive-temperature Boltzmann distribution.
Sources and further reading
These optional references support the stated laws and model assumptions. The numerical examples and diagrams are authored teaching models, not experimental measurements or engineering specifications. Selected relevant passages were reviewed, not entire textbooks or a comprehensive research literature.
Secondary reference: selected canonical weights and partition-function normalization. The two-level spectrum and degeneracies are constructed examples, not a material fit.
Selected free-expansion entropy calculation and third-law discussion. This section uses qualitative disorder language; our counting model is instead supported by the separate Chemistry entropy section and explicit enumeration.