PHY.8:5 - Archetypal Grounding

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.

PHY.8:5.2 - Recover collective fluctuations from a prepared joint state

A readout measures the z component of N=100 spin-1/2 systems along one common axis. Write each normalized outcome as s_i=+1 or -1; the physical angular momentum is (hbar/2) s_i. The collective normalized signal is M=sum_i s_i.

Three ideal preparations give every individual spin equal probabilities of +1 and -1:

For a singlet pair, the state is (|up down>-|down up>)/sqrt(2). Its ideal common-axis measurements give opposite results, so each pair contributes zero to M. For the last preparation, M itself is +100 or -100 with equal probabilities. The variance entries follow from these joint properties and from addition of independent variances in the first preparation.

A readout designed only from the individual mean would predict the same zero signal in all three cases. Their root-mean-square collective signals are 10, 0 and 100. If saturation occurs when |M| exceeds 50, the last preparation always saturates. The first has probability at most 100/50²=0.04 by the variance bound, and the ideal paired preparation never saturates.

Now retain the paired preparation but read only one spin from each pair. The sum of those fifty outcomes has variance 50, because different pairs were prepared independently. The zero-variance conclusion applied to a complete-pair sum, not to an arbitrary selected subset.

This calculation uses joint states and a stated measurement. Common-axis anticorrelation alone would also be compatible with other preparations; it does not identify the singlet uniquely. Recovering entanglement would require a different question and suitable measurements. Detector errors or correlations between pairs would change the recording law or the preparation premise.

PHY.8:5.3 - Derive transport from persistent microscopic motion

Particles move on an unbounded line with speed v>0. Each reverses direction at independent Poisson events of rate alpha>0. Initially each particle is at the origin, with either direction equally likely. This is a physical stochastic model for the motion; its validity must come from the selected mechanism and regime.

Let p_+(x,t) and p_-(x,t) describe position probabilities with positive and negative velocity. The point preparation leaves probability atoms at the unreversed fronts x=+vt and x=-vt; interpret the following density equations in the distributional sense. The transition and transport laws give

partial_t p_+ = -v partial_x p_+ - alpha p_+ + alpha p_-,

partial_t p_- = v partial_x p_- + alpha p_+ - alpha p_-.

Define total density n=p_++p_- and probability current j=v(p_+-p_-). Adding and subtracting give

partial_t n = -partial_x j,

partial_t j = -v² partial_x n - 2 alpha j.

The current retains the directional persistence. Eliminating it gives

partial_tt n + 2 alpha partial_t n = v² partial_xx n.

For times long compared with 1/(2 alpha) and spatial variation slow enough for the current to relax, use j approximately -D partial_x n, with D=v²/(2 alpha). The resulting diffusion equation is an approximation with physical grounds, not a consequence of fitting a bell-shaped histogram.

The mean position stays zero. For the stated initial preparation, the mean-square displacement is

E[x(t)²] = (v²/alpha) [t - (1-exp(-2 alpha t))/(2 alpha)].

With v=2 cm/s and alpha=1/s, it is about 0.03746 cm² at t=0.1 s. A diffusion calculation would give 0.4 cm², more than ten times as much. At t=20 s, the values are about 78 and 80 cm²: diffusion overestimates this observable by about 2.56% relative to the microscopic model. A 3% tolerance admits the later approximation but not the earlier one.

Changing the requested time therefore changes the needed description. Retain the density-current equations for the early response; use the diffusion reduction when its error is adequate for the requested observable. Boundary arrival probabilities would need their own comparison, since this mean-square agreement does not validate every feature of the distribution.

A simulation that flips velocity with the specified rate implements physical time. A different Markov chain designed to sample a position distribution has no such interpretation merely because its histogram agrees.