Justification of Metropolis Sampling

Why does the Metropolis algorithm generate configurations with weight exp(-U/kT)?

(not rigorous)

For simplicity we will assume that only a finite number of configurations are possible.
consider a very large ensemble of systems

let k(a) be the number of systems in configuration C(a)
- what we want to show is that k(a) is proportional to exp(-U(a)/kT)
Do one step for each system in the ensemble and let:
- P(a->b) = probability that the trial configuration for C(a) is C(b)
- P(b->a) = probability that the trial configuration for C(b) is C(a)
according to step(1), P(a->b) = P(b->a), whenever C(b) and C(a) are separated by the movement of one atom in a box of side s. If this is not the case, the probabilites are zero.
assume that U(a) > U(b) then the number of systems actually moving from C(a) -> C(b) is:

k(a)*P(a->b) (as all trials are allowed, step(3))
the number of systems actually moving from C(b) -> C(a) is:

k(b)*P(b->a)*exp(-delta_U/kT) ; where delta_U = U(a) - U(b)

as the fraction of moves accepted in step(5) is exp(-delta_U/kT).
At equilibrium, we assume that the rates of:

C(a) -> C(b)
C(b) -> C(a)

are equal (microscopic reversibility), hence:

k(a)*P(a->b) = k(b)*P(b->a)*exp(-delta_U/kT)

so that:

k(a)/k(b) = exp(-delta_U/kT)= exp(-[U(a)-U(b)]/kT)

as P(a->b) = P(b->a), and:

k(a)/k(b) = exp(-U(a)/kT) / exp(-U(b)/kT)

i.e. k(a) is proportional to exp(-U(a)/kT)

so that the configurations are weighted by exp(-U/kT) at "equilibrium" as required.

In the above we have implicitly assumed that all equilibrium configurations can be reached from the starting configuration. (This is the ergodic assumption.)