Questions: Monte Carlo Methods in Statistical Mechanics

Question 1

Why can the Metropolis algorithm compute thermal averages without ever calculating the partition function Z?

• A. Z is always equal to 1 for normalized probability distributions, so it doesn't need to be computed
• B. The acceptance probability min(1, exp(−ΔE/kT)) involves only energy differences, so Z cancels in the ratio of Boltzmann factors for any two states
• C. The Metropolis algorithm samples configuration space uniformly, making Z irrelevant
• D. Z is computed implicitly in the thermalization step and stored for later use

Answer: (B)

The acceptance probability for a Metropolis move is min(1, P(s_new)/P(s_old)) = min(1, exp(−E_new/kT) / exp(−E_old/kT)) = min(1, exp(−ΔE/kT)). Since Z appears in both numerator and denominator of the ratio P(s_new)/P(s_old), it cancels exactly. This is the crucial insight: you never need to enumerate all states or sum the full partition function. You only need to compare the relative probabilities of the current and proposed states, and those ratios depend only on the energy difference ΔE. The algorithm samples from the correct Boltzmann distribution without normalizing it.

Question 2

In a Metropolis simulation, a proposed move increases the system's energy by ΔE = 2 kT. What happens?

• A. The move is always rejected because it increases energy, violating the second law of thermodynamics
• B. The move is always accepted because the system must be able to explore higher-energy states at finite temperature
• C. The move is accepted with probability exp(−2) ≈ 0.135 and rejected otherwise
• D. The move is accepted with probability exp(+2) ≈ 7.4, but capped at 1

Answer: (C)

The Metropolis rule: if ΔE ≤ 0, accept; if ΔE > 0, accept with probability exp(−ΔE/kT). For ΔE = 2kT, the acceptance probability is exp(−2) ≈ 0.135 — about 1 in 7 attempts succeed. This is essential: a simulation that always rejected energy-increasing moves would be trapped in local minima and never reach the true equilibrium distribution. At finite temperature T, the Boltzmann distribution assigns nonzero probability to higher-energy states. The Metropolis rule ensures that higher-energy states are visited proportionally to their Boltzmann weight, allowing the algorithm to sample the full canonical distribution.

Question 3

The Metropolis algorithm is expected to enumerate most possible microstates to correctly sample the Boltzmann distribution.

• T. True
• F. False

Answer: False

Enumerating all microstates is precisely what Monte Carlo methods are designed to avoid. For a system of N ~ 10²³ particles, the number of microstates is astronomically large — direct enumeration is inconceivable. The Metropolis algorithm instead generates a Markov chain through configuration space, spending more time in low-energy (high-probability) states by accepting energy-decreasing moves always and energy-increasing moves with probability exp(−ΔE/kT). After thermalization, the fraction of time spent in any state converges to its Boltzmann weight without any global enumeration.

Question 4

After a Metropolis simulation has thermalized, configurations generated by the algorithm are samples drawn from the canonical (Boltzmann) distribution, regardless of the initial configuration chosen.

• T. True
• F. False

Answer: True

This is the mathematical guarantee of the Metropolis algorithm: the Boltzmann distribution is the unique stationary distribution of the Markov chain defined by the acceptance rule. The proof relies on detailed balance — for any two states s and s', the probability flux from s to s' equals the flux from s' to s when weighted by their Boltzmann probabilities. Any Markov chain satisfying detailed balance converges to its stationary distribution. 'Thermalization' is the time needed for the chain to 'forget' its starting state; after that, subsequent samples are effectively drawn from the target distribution.

Question 5

What does 'detailed balance' mean in the context of the Metropolis algorithm, and why is it important for correctness?

Model answer: Detailed balance means that for any two states s and s', the rate of transitions from s to s' (weighted by P(s)) equals the rate of transitions from s' to s (weighted by P(s')). Formally: P(s) × T(s → s') = P(s') × T(s' → s), where T is the transition probability. The Metropolis acceptance rule enforces detailed balance with respect to the Boltzmann distribution: it is easy to verify that P(s) × min(1, P(s')/P(s)) = min(P(s), P(s')) = P(s') × min(1, P(s)/P(s')). Detailed balance is sufficient to guarantee that the Boltzmann distribution is the stationary (equilibrium) distribution of the Markov chain — meaning that a simulation run long enough will sample states in the exact proportions required by the canonical ensemble.

Detailed balance is a stronger condition than necessary (global balance suffices), but it is easier to verify and is satisfied by Metropolis. It is what distinguishes a correct Monte Carlo algorithm from one that merely looks like it's sampling the right distribution. Without it, there is no guarantee that the simulation converges to the physically correct equilibrium, regardless of how long it runs.