Systems in a Heat Bath: The Boltzmann Distribution
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You should be comfortable with the concept of a microstate and the definition of temperature from the previous lecture. Familiarity with basic probability theory, specifically normalisation conditions, is assumed.
Why this matters
The Boltzmann distribution is the fundamental link between microscopic energy levels and macroscopic thermal behaviour.
In the previous lecture, we derived temperature and pressure by counting microstates for isolated systems. However, most real systems are not isolated; they exchange energy with their surroundings. To describe a system in contact with a heat bath, we cannot simply count states of the system alone. We must consider the combined system-plus-bath. The result of this statistical argument is the Boltzmann distribution, which tells us the probability of finding a system in a specific energy state. This distribution is the cornerstone of statistical mechanics, allowing us to calculate observable quantities like heat capacity and magnetisation from first principles.
The Physical Picture
Imagine a small system, such as a single molecule, immersed in a large reservoir of heat. The reservoir is so large that its temperature remains constant even if it exchanges a tiny amount of energy with the system. The system can occupy various discrete energy levels, \( E_1, E_2, E_3, \dots \). Because the system is in thermal equilibrium with the bath, it constantly exchanges energy with it. At any given instant, the system is in one specific state. Over time, it jumps between states. The key question is: how likely is it to be in a high-energy state versus a low-energy state? Intuitively, low-energy states are more probable because the bath has more ways to arrange its own energy if the system takes less energy. The Boltzmann distribution quantifies this intuition, showing that the probability of a state decreases exponentially with its energy relative to the thermal energy scale \( k_B T \).
Deriving the Probability
To find the probability \( p_i \) of the system being in state \( i \) with energy \( E_i \), we look at the total number of microstates of the combined system (system + bath). Let \( \Omega_S(E_i) \) be the number of microstates of the system in state \( i \) (which is 1 for a specific state) and \( \Omega_B(E_{\text{total}} - E_i) \) be the number of microstates of the bath when the system has energy \( E_i \). The total number of microstates is the product of these two. Since the bath is much larger than the system, \( E_i \) is a tiny perturbation to the bath's total energy. We can expand the entropy of the bath, \( S_B = k_B \ln \Omega_B \), around the total energy. The probability is proportional to the number of bath microstates, which leads to an exponential dependence on \( E_i \).
To turn this proportionality into a probability, we must normalise it so that the sum of probabilities over all states equals 1. This introduces the partition function \( Z \), which acts as a normalisation constant.
Here, \( k_B \) is the Boltzmann constant and \( T \) is the absolute temperature. The partition function \( Z \) sums the Boltzmann factors over all possible states. It is a crucial quantity because it encodes all the thermodynamic information about the system.
Interpreting the Boltzmann Factor
The term \( e^{-E_i / k_B T} \) is known as the Boltzmann factor. It compares the energy of the state \( E_i \) to the typical thermal energy available, \( k_B T \). If \( E_i \) is much smaller than \( k_B T \), the exponent is close to zero, and the factor is close to 1. This means low-energy states are highly probable. If \( E_i \) is much larger than \( k_B T \), the exponent is a large negative number, and the factor becomes very small. This means high-energy states are unlikely. The ratio of probabilities for two states \( i \) and \( j \) is given by:
This ratio depends only on the energy difference and the temperature. It shows that temperature controls the "spread" of the distribution. At low temperatures, the distribution is sharply peaked at the ground state. At high temperatures, the distribution becomes flatter, and higher-energy states become accessible.
Connection to Thermodynamics
Although derived from counting microstates, the Boltzmann distribution reproduces the laws of thermodynamics. For example, the average energy of the system, \( \langle E \rangle \), can be calculated as the sum of energies weighted by their probabilities:
This average energy is the internal energy \( U \) of the system in the thermodynamic limit. By differentiating the partition function with respect to temperature, we can derive other thermodynamic potentials. The partition function is thus the bridge between the microscopic world of quantum states and the macroscopic world of heat and work. It allows us to calculate observables without needing to know the detailed dynamics of the system, only its energy spectrum.
The derivation
The Boltzmann Distribution
A small system exchanging energy with a large reservoir occupies each microstate with probability proportional to exp(-E/kT).
A common misconception is that the Boltzmann distribution needs the system to be small or microscopic. It does not: it applies to any system, however large, in contact with a much larger bath, with \(E_i\) the energy of a microstate of the whole system. What must be small is the system compared with the bath, so that the bath's temperature does not change. The combined system plus bath is treated microcanonically; the Boltzmann factor is what that gives for the system's states. C
Worked example
Problem. A two-level system has levels \(E_0 = 0\) and \(E_1 = \Delta E = 0.050\ \mathrm{eV}\). Find the probability of the excited state at 300 K and at 1000 K. Use \(k_B = 8.617 \times 10^{-5}\ \mathrm{eV\,K^{-1}}\).
- Thermal energy at 300 K: \(k_BT = 8.617 \times 10^{-5} \times 300 = 0.02585\ \mathrm{eV}\).
- Ratio: \(\Delta E/k_BT = 0.050/0.02585 = 1.934\), so the Boltzmann factor of the upper level is \(e^{-1.934} = 0.1445\).
- Partition function: \(Z = 1 + 0.1445 = 1.1445\); probability \(p_1 = 0.1445/1.1445 = 0.126\).
- At 1000 K: \(k_BT = 0.08617\ \mathrm{eV}\), \(\Delta E/k_BT = 0.580\), \(e^{-0.580} = 0.560\), so \(p_1 = 0.560/1.560 = 0.359\).
- Even at very high temperature \(p_1\) only approaches \(1/2\): a two-level system in equilibrium can never have more systems up than down at positive temperature.
Units check. \(\Delta E\) and \(k_BT\) are both in eV, so the exponent is dimensionless and \(p_1\) is a pure number.
Three quick tests
- What happens to the probability of the excited state as \( T \to 0 \)?
Answer
It approaches zero, as the system falls into the ground state. - How does the partition function \( Z \) change as temperature increases?
Answer
It increases, as more high-energy states become thermally accessible. - If two states have the same energy, what is the ratio of their probabilities?
Answer
The ratio is 1, meaning they are equally probable.
Problem set for PU-203 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes checked line by line by the editor; every number in the worked example was recomputed.