PHY.8:5.1 - Predict an excited population from a reservoir argument

A device contains N=400 distinguishable, weakly interacting units. Each has a nondegenerate ground state of energy 0 and a nondegenerate excited state of energy Delta. The units have equilibrated with a large reservoir at temperature T. Treat interactions between units and their contribution to the reservoir’s temperature change as negligible. An ideal readout counts excited units before appreciable relaxation changes that count.

The physical question is the mean count and its variation between independently repeated equilibrium preparations. The reservoir argument weights a unit’s state by the number of compatible reservoir states. With reservoir entropy S_R and Boltzmann constant k_B,

Omega_R(E-Delta)/Omega_R(E) approximately exp(-Delta/(k_B T)),

using the first-order entropy change and dS_R/dE=1/T. This approximation needs the reservoir’s temperature to remain effectively constant over the exchanged energies. The equal weighting used for the combined equilibrium energy shell is a premise of this construction.

Choose Delta=k_B T ln(3). The excited-to-ground weight ratio is 1/3, so the excited probability is p=1/4, not 1/2. Two possible energy values do not receive equal weights under this preparation.

Under the stated independence approximation, the count K is binomial:

P(K=k) = choose(400,k) (1/4)^k (3/4)^(400-k).

It has mean 100 and variance 75, giving a standard deviation about 8.66 units. If the receiving requirement is that the count differs from 100 by less than 50 in at least 96% of preparations, Chebyshev gives P(|K-100|>=50)<=75/2500=0.03. That bound already satisfies the requirement; calculating every binomial probability is unnecessary for this decision.

Now suppose the excited energy has three distinguishable states at the same Delta, with the same equilibrium and independence premises. Their total weight is three times larger. The excited probability becomes 1/2, the mean count 200 and the variance 100. The energy gap alone no longer supplies the previous count.

If the device is instead prepared with precisely 100 excited units and isolated during readout, the count variance is zero. Individual units can still each have excited probability 1/4 across a permutation-symmetric preparation. The fixed-total restriction prevents the binomial independence assumption. Use the preparation that the work actually supplies.