Microstates, macrostates, and multiplicity

Six units, one total energy

Imagine six localized units, each with a ground level of energy 0 and an excited level of energy ε = 1.00×10⁻²¹ J. Three units belong to group A and three to group B. Exactly three units are excited, so the ideal counting energy is fixed at 3ε. A detector reports how many excitations are in group A, but not which individual units carry them. How many different microscopic arrangements can produce the same detector reading? Which readings should be common?

There are 20 allowed arrangements in this fixed-excitation model. The four possible readings qA = 0, 1, 2, 3 contain 1, 9, 9 and 1 arrangements. That counting fact will be our starting point. It becomes a prediction of frequencies only after we specify how the microscopic alternatives are weighted and whether the physical preparation or dynamics can access them.

Our ideal Hamiltonian is H₀ = εΣnᵢ, where each occupation nᵢ is 0 or 1 at a physically identified site. The counted alternatives are product-basis level assignments, with fixed total excitation number. An application may have weak, excitation-conserving interactions that redistribute excitations while leaving this counting approximation useful. The count does not specify the exact interacting spectrum, an exchange rate or a proof of equilibration. General quantum states are discussed in a later return section.

On a first reading, follow this small case, the counting rules, the probability counterexample, oscillator sharing and the typicality comparison. Return to the sections marked “Further” for asymptotics, continuous phase space, quantum state counting and dynamics. For preparation, review “Physical systems, states, boundaries, and environments”, “Work, kinetic energy, potential energy, and power”, and “Conservation of mass, momentum, and energy”. We define the needed counting and probability terms locally. Later subjects develop statistical entropy and ensembles in more depth.

A state, a record and a macrostate

Label the actual sites A1, A2, A3, B1, B2 and B3. In this model, the string 100011 means that A1, B2 and B3 are excited and the other sites are in their ground levels. This complete occupation assignment labels one microscopic basis alternative. The shorter record qA = 1 does not identify that arrangement: it also describes eight others.

A microstate specifies the microscopic variables required by the chosen physical model. A macrostate specifies a smaller set of collective quantities, within stated tolerances. Coarse-graining groups microscopic states that the chosen macroscopic record does not distinguish. Multiplicity Ω(M) is the number of allowed microscopic alternatives in a discrete macrostate M, or an appropriately normalized measure when the microscopic state space is continuous.

These are different levels of description of the same model. An observer not knowing which state was prepared does not create extra physical states. Nor does a microstate have to be experimentally known in detail for the model to assign it. An actual measurement can interact with the sample; merely deciding to keep a shorter record is a descriptive operation, not a new exchange of energy.

Count the allowed states

Without fixing the total number of excitations, each of the six sites has two alternatives, giving 2⁶ = 64 product-basis assignments. Fixing q = 3 excludes every assignment with any other total. Choose which three of the six sites are excited: Ω(6, 3) = C(6, 3) = 6!/(3!3!) = 20. The symbol n! means n×(n−1)×…×1, with 0! = 1.

For a particular reading qA = k, choose k of the three A sites and 3−k of the three B sites. These choices combine freely once both excitation totals are fixed, so Ω(k|q = 3) = C(3, k)C(3, 3−k). For k = 1, there are three choices in A and three in B, producing nine complete arrangements.

The four readings are mutually exclusive and cover the allowed sector, so their multiplicities add to 20. The readings 1 and 2 are tied for largest multiplicity; no exact reading qA = 1.5 is possible. A mean value may nevertheless be 1.5 across a statistical preparation. An average need not itself be an allowed individual outcome.

Products, sums and physical constraints

Multiply counts for choices that can be combined without an additional compatibility restriction. Add counts for disjoint alternatives. If two subsystems have fixed individual energies and no relevant interaction energy, their joint count is ΩA(EA)ΩB(EB). If only their total energy E is fixed, sum over allowed divisions: Ωtotal(E) = ΣEA ΩA(EA)ΩB(E−EA). The total constraint prevents us from multiplying unrestricted subsystem counts.

The permitted range matters. For NA and NB two-state sites with total q, max(0, q−NB) ≤ qA ≤ min(NA, q). A binomial coefficient C(N, q) is zero when q is outside 0, …, N. Extra conserved quantities or inaccessible sectors must also be imposed. Knowing the energy alone is not always enough to define the accessible state space.

For example, suppose A1 is permanently in its ground level. Only five sites can carry the three excitations, leaving C(5, 3) = 10 allowed arrangements. The qA counts become 1, 6, 3 and 0. Applying the old 20-state distribution to this constrained sample would count inaccessible states.

Changing a record is different from adding a physical constraint. If we merely merge qA = 0 and 1 into “low A” and 2 and 3 into “high A”, each coarser record contains ten of the same 20 states. We have changed the partition used to report the result, not the sample’s microscopic alternatives.

Counts need weights before they become probabilities

If all 20 allowed alternatives have equal probability, each has probability 1/20. Summing the weights within each macrostate gives P(qA) = Ω(qA)/20: 0.05, 0.45, 0.45 and 0.05. The two middle readings together have probability 0.90, but neither individual reading is overwhelmingly certain in this small system.

Equal weighting is a model of an equilibrium or preparation measure under the stated constraints. It is not a consequence of the word “microstate”, a consequence of ignorance alone, or a guarantee that one trajectory visits every state equally often. A subsystem in contact with an environment generally receives weights from that environment, even if the combined isolated system is assigned a uniform measure.

Here is a deliberate counterexample using exactly the same 20 allowed alternatives. Prepare the sole qA = 0 state with relative weight 50, and every other individual state with weight 1. The macrostate weights are then 50, 9, 9 and 1, with sum 69. The one-state macroregion is now the most probable. This is a specified nonuniform preparation, not the uniform isolated equilibrium distribution.

In general P(M) = Σi in M pᵢ. Under uniform weights this reduces to a multiplicity ratio. Under weights wᵢ, normalize by their sum first. A large number of weakly weighted states need not beat one strongly weighted state.

Another physical model: oscillators share quanta

A two-state site can carry at most one excitation. An ideal harmonic oscillator can carry any nonnegative number of energy quanta. For N physically distinguished oscillator modes with total q quanta, a microscopic occupation record is (n₁, …, nN), with Σnᵢ = q. Zero-point energy is omitted here: the number of modes is fixed, so it adds the same constant to every sharing pattern.

Count these records by putting N−1 separators among q identical quantum markers. The separator positions specify how many quanta belong to each mode. There are ΩE(N, q) = C(q+N−1, q) records for N ≥ 1 and q ≥ 0. The modes are distinguished; the quantum markers are not individually labeled particles. The formula counts occupations, not permutations of named energy packets.

Let A have three modes and B have two, sharing six quanta of the same spacing ε. At fixed qA, the product count is ΩA(qA)ΩB(6−qA). Summing all seven divisions gives ΩE(5, 6) = 210. Under a uniform preparation over those fixed-total occupation alternatives, the most probable integer division is qA = 4, with probability 45/210 = 3/14. The exact mean is 3.6 quanta in A, while the total energy stays fixed.

The larger group need not receive exactly the same total energy as the smaller group. In a suitable large-system limit, the dominant division is governed by the growth of both multiplicities, rather than by a rule that total energies must be equal. This finite example retains sizeable fluctuations and a difference between the mean and the mode.

A typical region is wider than one exact reading

Now change the preparation. Take N two-state sites and independently excite each with probability 1/2. All 2ᴺ assignments are equally weighted, and q is not fixed. Since different q have different energies, this is a preparation across energy sectors, not the single-energy sector of the opening example. The count at q is still C(N, q), but now P_N(q) = C(N, q)/2ᴺ.

For this independent preparation, the mean excitation count is N/2 and its variance is N/4. Thus the standard deviation of the fraction q/N is 1/(2√N). Absolute count fluctuations grow like √N, while fractional fluctuations shrink. This statement depends on the independent two-state model; correlations can alter the scaling.

For N = 100, the exact middle count q = 50 has probability about 0.0795892. The wider band 40 ≤ q ≤ 60 has probability about 0.9647998. These are different questions: one exact reading is much less probable than a physically specified region around equilibrium.

As N grows through 10, 100 and 1000, the probability of exactly q = N/2 decreases, yet the probability of the fixed fractional band 0.40 ≤ q/N ≤ 0.60 increases. The lesson is concentration in a region, not certainty of one integer value. Even at N = 1000 the outside probability is positive, about 1.80168×10⁻¹⁰; a plot rounded to 1 must not turn that into a mathematical impossibility.

Further: large counts and logarithms

Exact factorial formulas can overflow ordinary numerical arithmetic long before a probability becomes difficult to represent. Use exact integer combinations, stable recurrences or logarithms. For fixed N, the recurrence C(N, q+1)/C(N, q) = (N−q)/(q+1) builds all adjacent counts without forming separate huge factorials. Keep large exact counts as integers or decimal strings; a displayed floating-point number is not automatically an exact count.

Stirling’s approximation ln n! ≈ n ln n − n + ½ln(2πn) gives, for x = q/N with both q and N−q large, ln C(N, q) ≈ N H(x) − ½ln[2πN x(1−x)], where H(x) = −x ln x − (1−x)ln(1−x). These logarithms are natural and their arguments are dimensionless.

The leading function H(x) is largest at x = 1/2, explaining the central concentration of the independent model. Its curvature gives the familiar Gaussian shape near the maximum, but neither the central Gaussian nor the refined logarithmic formula is uniformly valid at q = 0 or N. Exact endpoint counts are one. For small samples or far tails, check an approximation against exact counts rather than extending a smooth curve beyond its domain.

Near the centre, P_N(N/2) ≈ √[2/(πN)] for even large N. It tends to zero although a fixed fractional band tends toward probability one. The finite table above illustrates both tendencies without requiring an approximation.

Further: a subsystem inherits the environment’s count

Consider a weakly interacting combined isolated system with fixed total energy E. If a particular A microstate has energy e, the number of compatible environment states is ΩB(E−e). Under a uniform measure for the whole system, that one A state therefore receives a weight proportional to ΩB(E−e). Equal weighting of whole-system alternatives does not imply equal weighting of all subsystem alternatives.

If we ask only for A’s energy, we must also count how many A states have that energy: P(EA) is proportional to ΩA(EA)ΩB(E−EA). A specified individual state and an energy macrostate are not the same object. The analogous reasoning applies to particle exchange only after total particle number and each allowed composition have been included.

For sufficiently large regular systems with a sharply dominant interior division, maximizing ln ΩA(EA)+ln ΩB(E−EA) yields equality of the two energy derivatives. Defining 1/T = kB∂lnΩ/∂E on the appropriate thermodynamic branch turns this into equality of temperature. It is equality of this growth rate, not equality of subsystem energies. Finite level spacings, multiple maxima, phase coexistence and extra constraints can invalidate a simple single-saddle approximation.

This counting argument prepares the microcanonical, canonical and grand-canonical ensemble subject. We do not assume a heat bath or derive a partition function for every example here; each preparation in this lesson keeps its own stated constraints.

Further: a classical microstate is a phase-space point

For classical particles, a microscopic state includes both positions and momenta, together with any relevant internal variables. Positions alone generally do not determine subsequent motion. With N structureless particles in d dimensions, the position–momentum phase space has 2dN coordinates. A region contains continuously many phase points, so ordinary counting of a finite list is replaced by a measure.

Take a one-particle, one-dimensional rectangle spanning 1 m in position x and 2 kg m/s in momentum p. Its phase-space area is 2 J s. A bookkeeping grid with Δx = 0.50 m and Δp = 0.50 kg m/s has eight cells. A finer grid with both widths 0.25 has 32 cells. The rectangle and its physical classical points have not changed: the representation has.

The canonical coordinate stretch x′ = 2 x, p′ = p/2 preserves dx dp. If we transform the coarse cells as well, their widths become 1 m and 0.25 kg m/s, the rectangle becomes 2 m by 1 kg m/s, and it still contains eight transformed cells. Changing coordinates consistently is not a physical increase in multiplicity.

Further: an energy shell needs a width and normalization

In a classical model, the set with exactly one energy H = E is a hypersurface and generally has zero ordinary phase-space volume. A shell E ≤ H < E+δE with δE > 0 gives a finite volume. The cumulative volume below E, the shell volume, and a density of states are different quantities. A density of states g(E) has units of inverse energy after dimensionless state normalization; g(E)δE approximates a shell count only when it varies little across the interval.

For a semiclassical Maxwell–Boltzmann description of identical particles in d dimensions, one often writes Ωshell ≈ [1/(N! h^(dN))]∫shell d^(dN)r d^(dN)p. The action scale h is Planck’s constant, so the normalized measure is dimensionless; internal degeneracies must be included if present. The factor N! removes artificial label permutations in the identical-particle phase-space representation. This formula has a classical/dilute regime and semiclassical state-counting limits; it is not an exact formula for an arbitrary finite quantum spectrum or a quantum-degenerate gas.

Localized physical site labels in our first example are different from fictitious labels on otherwise identical gas particles. Swapping occupations between A1 and B1 can change an observable state. Dividing the six-site count by 6! would remove genuine alternatives. Conversely, counting identical-particle relabelings as different gas states can introduce a fictitious mixing contribution.

The choice of a small energy interval and macroscopic tolerances is part of the statistical description. For a large regular system, reasonable nearby choices often differ only by subleading corrections to the extensive logarithm, but finite systems and singular regimes need their actual definitions retained. A shell count must not silently become a cumulative count in the middle of a derivation.

Further: quantum state counting counts a subspace dimension

A quantum state may be a superposition, not one definite occupation string. Our 20-state opening count means that the fixed-q sector has 20 orthogonal product-basis alternatives. It does not mean that only 20 normalized quantum vectors exist in that sector. There are continuously many superpositions, and nonorthogonal vectors cannot all be counted as mutually exclusive perfectly distinguishable alternatives.

More generally, let Π be the projector onto a finite allowed Hilbert subspace. Its dimension is Ω = Tr Π, independent of which orthonormal basis is used inside it. Under the maximally mixed preparation ρ = Π/Ω, a compatible macro-projector ΠM within the allowed sector has probability Tr(ρΠM) = dim(HM)/Ω. This is the quantum version of a multiplicity ratio with its weighting assumption displayed.

For a general prepared density operator ρ, the probability is instead Tr(ρΠM); for a pure vector ψ, it is ⟨ψ|ΠM|ψ⟩. A pure state entirely in a small macro-subspace can assign that region probability one. Counting alone cannot replace the prepared state. Compatible coarse observables can define a joint partition; noncommuting observables cannot in general be given arbitrary simultaneous sharp values.

A basis, a state vector, a mixed statistical description and a macroscopic record therefore have distinct jobs. The later ensemble and quantum-statistics subjects develop these objects further. This section only prevents a count of orthogonal alternatives from being misread as a count of all possible wavefunctions.

Further: what makes two arrangements physically different?

Consider a separate comparison with two particles and three orthogonal one-particle modes. No total-energy constraint is imposed, and spin or other internal degeneracy is not left implicit. Two physically distinguished particles have nine ordered product-mode assignments. Two identical spinless bosons have six occupation-basis alternatives. Two identical fully spin-polarized fermions have three, because each single-particle mode may be occupied at most once.

For N such particles in g modes, the corresponding dimensions are gᴺ for physically distinguished particles, C(N+g−1, N) for bosons, and C(g, N) for these fermions. They count different physical state spaces. The boson expression resembles the oscillator stars-and-bars formula, but its particles and modes have a different interpretation; mathematical similarity is not physical identity.

For example, occupation (2, 0, 0) corresponds to one ordered assignment for distinguished particles, while (1, 1, 0) corresponds to two. Summing over the six occupancy triples gives 1+1+1+2+2+2 = 9 distinguished assignments. Each bosonic occupation triple labels one basis alternative; only the final three triples obey the fermionic restriction.

Dividing 3² by 2! gives 4.5, which is not the six-state bosonic dimension. The repeated-mode and separated-mode assignments have different permutation-orbit sizes. The continuum Gibbs factor must not be turned into a universal instruction to divide every finite discrete count by N!.

Coarse-graining and the entropy bridge

A macrostate must specify which collective variables and tolerances define it. Resolving all six occupations gives singleton cells in the opening basis model. Keeping only qA gives four cells of sizes 1, 9, 9 and 1. Keeping only “low A” or “high A” gives two cells of size ten. The underlying allowed states remain the same, while the reported multiplicities depend on the chosen partition.

Boltzmann’s entropy for a specified discrete macrostate is S_B(M) = kB ln Ω(M), with kB = 1.380649×10⁻²³ J/K. The logarithm is dimensionless. A qA = 1 macrostate has S_B/kB = ln 9 ≈ 2.197225; recording only the total excitation constraint gives ln 20 ≈ 2.995732 for the whole allowed sector. These refer to different specified records and must not be silently interchanged.

For thermodynamic interpretation, use physically appropriate collective observables and keep the coarse-graining convention fixed across comparisons. Simply forgetting a detail, choosing a different grid or changing coordinate units does not demonstrate heat transfer or irreversible microscopic dynamics. An actual relaxation changes the physical state within a defined description; changing the description is a different operation.

Products of compatible subsystem multiplicities become sums of logarithms. This explains why a logarithm is useful for an extensive entropy description. The next subject treats probability and statistical entropy, including the distinctions needed for nonuniform preparations; this lesson does not identify every uncertainty measure with the thermodynamic entropy of an individual sample.

Further: typicality is conditional on a measure and a question

A state is typical when it belongs to a set taking up most of a specified measure on the allowed state space. Thus a typicality claim names the conserved constraints, the measure, the macrovariables, their tolerances and the system-size regime. It does not say that all microstates have similar detailed configurations or that any arbitrary partition has one dominant cell.

For many large systems with suitable short-range interactions and regular equilibrium conditions, physically coarse equilibrium regions occupy nearly all of the allowed equilibrium measure. The binomial example demonstrates one exact concentration result under independent preparation. Strong correlations, phase coexistence, criticality, long-range interactions or extra conserved quantities can change the shape, split the dominant regions or invalidate a simple √N fluctuation law.

A uniformly chosen state can be typical while every exact microscopic alternative has a tiny probability. Conversely, a deliberately prepared atypical state can be physically real. An improbable outcome is not a forbidden state. The measured question—an exact integer, a finite-width band or a detailed pattern—must stay fixed when probabilities are compared.

Further: counting is not a clock

Multiplicity says how much state space belongs to a record. It supplies no time scale. Six noninteracting units that neither flip nor exchange excitations can remain in their initially prepared configuration forever, even though the model lists many other arrangements with the same total excitation number. Extra dynamics is needed to connect those alternatives.

In classical Hamiltonian mechanics, Liouville’s theorem preserves the fine phase-space volume of an evolving set of initial conditions. A physical point can nevertheless move from a small predefined macroregion into a larger one. The larger volume of the destination macroregion does not mean that the microscopic evolution has created phase-space volume.

Time-reversal-invariant equations also permit specially prepared trajectories that retrace a relaxation. For measure-preserving dynamics on a finite-measure bounded region, recurrence statements allow almost every state eventually to return close to its earlier state; they supply neither a short universal recurrence time nor a failure of a probabilistic typicality statement. An arrow-of-time account must include the preparation and low-entropy boundary conditions, not counts alone.

Ergodicity concerns long-time averages relative to an invariant measure. It does not by itself give a practical relaxation rate, and useful macroscopic relaxation need not require the strongest form of full phase-space mixing. The transport and nonequilibrium/statistical subjects develop such dynamical questions. Do not infer a transition rate from a ratio of multiplicities.

Use the right state space for the application

Localized two-state units are useful models for dilute excitations and simple spin occupations. Oscillator occupations model quantized vibrational energy sharing. Classical phase-space volumes are useful in dilute-gas calculations, while bosonic and fermionic occupation restrictions matter when identical-particle quantum statistics is essential. Each model keeps different microscopic variables and different allowed records. A vacancy model can count which physical lattice sites are empty, but a formation-energy law and the atom-reservoir constraints are needed before predicting a vacancy fraction.

Before counting, specify the system, its physical degrees of freedom, conserved quantities, allowed energy window and any constraints imposed by walls, symmetry or preparation. Then choose the macrorecord. Decide whether alternative choices can be multiplied, must be summed, or are incompatible. Finally state the weighting measure and the assumptions connecting it to a physical prediction.

Applications to magnetism, molecular vibrations, gas mixing and state-resolved computation often fail through overcounting or counting inaccessible states rather than through difficult arithmetic. Repeating an identical-particle label, omitting a degeneracy, confusing a marginal weight with a subsystem count, or treating a rounded probability as exactly one can each change the conclusion.

What the counts establish

The six-site model shows how one conserved energy can contain many microscopic alternatives, how a macrostate groups them, and why multiplicity ratios require equal microscopic weights. The biased preparation and frozen-site constraint show two distinct ways that the same naive probability prediction can fail.

The large-N comparison shows concentration in a region while the most probable exact reading becomes less likely. The oscillator and particle-mode examples show why the physical meaning of an occupation matters. The phase-space comparison shows why a finite bookkeeping grid is not a new finite set of fundamental classical states.

“Probability and statistical entropy” develops the weighting and information questions. “Microcanonical, canonical, and grand canonical ensembles” develops isolated and reservoir-controlled preparations. “Bose–Einstein and Fermi–Dirac statistics” develops identical-particle populations, and the later transport and nonequilibrium subjects connect state counting to actual motion.

Worked examples

Try it yourself

1. Read the occupation string 100011 in the stated site order. Give qA, qB and the ideal energy.

   Show the answer and reasoning

   qA = 1, qB = 2 and E = 3ε = 3.00×10⁻²¹ J.

   The first three entries refer to A sites and sum to 1; the final three sum to 2. The energy depends on their total, while the macrorecord qA omits the individual site identities.

2. How many six-site occupation alternatives exist before fixing q? Explain why the opening denominator 20 no longer describes that whole set.

   Show the answer and reasoning

   2⁶ = 64 alternatives.

   Each physical site has two independent basis alternatives. Fixing q = 3 retains only 20 of those 64; omitting the constraint changes the allowed set before any probability is assigned.

3. Under the uniform opening preparation, what is P(qA = 0 or 3), and how does it compare with P(qA = 1 or 2)?

   Show the answer and reasoning

   Extreme readings: 2/20 = 0.10. Middle readings: 18/20 = 0.90.

   Each event is a disjoint union of macrostates. Add their multiplicities before dividing by the common uniform total; do not treat the four macrorecords as four equally weighted states.

4. A learner says that qA = 1 names one microstate. Give a concrete correction.

   Show the answer and reasoning

   It names a macrostate containing nine occupation alternatives, including 100011 and 010011.

   Both strings have one excitation in A and two in B, but different physical A sites are excited. A collective count does not specify the full microscopic assignment.

5. The uniform opening preparation has mean qA = 1.5. What integer outcomes can actually be observed, and which are modes?

   Show the answer and reasoning

   Outcomes 0, 1, 2, 3; modes 1 and 2.

   An expectation is a weighted average and need not be a possible individual value. The two modal multiplicities are both 9, so this small system has a tie rather than one exact balanced integer.

6. Merge 0 and 1 into low A and 2 and 3 into high A. Find their uniform probabilities and say what physically changed.

   Show the answer and reasoning

   Both probabilities are 10/20 = 1/2; only the reporting partition changed.

   The underlying 20 assignments are the same. Adding the disjoint cell counts changes the record’s multiplicity without imposing a new dynamical constraint or producing new microscopic alternatives.

7. With A1 permanently unexcited, compare the uniform probabilities of qA = 1 and qA = 2.

   Show the answer and reasoning

   P(qA = 1) = 6/10 = 0.60 and P(qA = 2) = 3/10 = 0.30; their ratio is 2.

   The constrained sector has 10 alternatives. Its counts 6 and 3 retain the same common denominator, so the probability ratio equals the count ratio. The inaccessible qA = 3 record contributes nothing.

8. A proposed biased normalization is 50+1+1+1 = 53. Diagnose and correct the error.

   Show the answer and reasoning

   The correct normalization is 50×1+1×9+1×9+1×1 = 69.

   The inputs specify weights per microscopic alternative. The two middle macrostates each contain nine alternatives, so summing just four per-state weights forgets multiplicity.

9. In the biased preparation, find P(qA = 1 or 2).

   Show the answer and reasoning

   18/69 = 6/23 ≈ 0.260870.

   Each middle macrostate has total weight 9. Add their weights and divide by the full 69. The state-space count has not changed; its nonuniform preparation has.

10. For two spinless bosons in the stated three modes, list all occupations and check the count.

    Show the answer and reasoning

    (2, 0, 0), (0, 2, 0), (0, 0, 2), (1, 1, 0), (1, 0, 1), (0, 1, 1): six.

    The particles are identical and the orthogonal modes are distinguished. The occupation-basis count is C(2+3−1, 2) = C(4, 2) = 6, agreeing with the explicit list without adding particle-name permutations.

11. For the five-mode oscillator case, calculate P(qA = 5) under the uniform total measure.

    Show the answer and reasoning

    ΩA = 21, ΩB = 2, so P = 42/210 = 1/5.

    At qA = 5 the three A modes have C(7, 5) = 21 alternatives; the two B modes share the remaining one quantum in two ways. Multiplication applies at fixed compatible subgroup energies.

12. Why is equal total subsystem energy not a universal equilibrium rule in the 3-mode/2-mode example?

    Show the answer and reasoning

    The larger A group has more occupation alternatives; its mode is qA = 4 rather than an equal 3–3 division.

    The product of the two multiplicities controls the weighted division. In an appropriate large-system limit it is equality of entropy growth rates, or temperature, that matters, while small systems retain fluctuations.

13. For N = 1000 independent 50:50 sites, find the standard deviations of q and q/N.

    Show the answer and reasoning

    σq = √1000/2 ≈ 15.811388; σ(q/N) ≈ 0.0158114.

    Independence gives variance N/4 for the count. Dividing the count by N divides its standard deviation by N, so absolute count fluctuations grow while fractional fluctuations shrink.

14. Use the table to compare the exact balanced-point probabilities for N = 10 and 1000. What mistaken conclusion should be avoided?

    Show the answer and reasoning

    They decrease from 0.24609375 to about 0.0252250; equilibrium-like concentration has not disappeared.

    The fixed fractional band becomes more probable even as any one exact integer reading becomes less probable. A band contains increasingly many allowed readings near the mean.

15. For N = 100, what is the exact-model numerical probability outside 40 ≤ q ≤ 60?

    Show the answer and reasoning

    Approximately 0.035200200218.

    The band and its complement exhaust all integer excitation counts, so subtract 0.964799799782 from 1. This is not the probability of just one extreme string.

16. Why should the refined Stirling formula not be evaluated literally at q = 0? What is the exact count?

    Show the answer and reasoning

    Its logarithmic prefactor is singular there; the exact count is C(N, 0) = 1.

    The approximation requires both q and N−q to be large. Its endpoint failure does not mean that a real count diverges; the endpoint has one all-ground assignment.

17. A classical record gives every particle position but no momenta. What information needed for a microstate is missing?

    Show the answer and reasoning

    The momenta, and any other relevant internal variables.

    Within a classical Hamiltonian model, identical positions with different momenta can evolve differently. A positional record alone is a coarser description, even if its spatial resolution is very fine.

18. By what factor does the cell count change from the stated coarse grid to the stated fine grid? What physical quantity stays fixed?

    Show the answer and reasoning

    It increases by 4, from 8 to 32; the phase-space area stays 2 J s.

    Both cell widths halve, so the cell area shrinks from 0.25 to 0.0625 J s. The rectangle and its continuous physical points remain the same; only the bookkeeping resolution changes.

19. A density of states has units J⁻¹. What extra quantity is needed to estimate a dimensionless shell multiplicity?

    Show the answer and reasoning

    An energy interval δE, using Ωshell ≈ g(E)δE when g varies little across it.

    A density per unit energy and a count within an interval are distinct objects. An exact classical energy hypersurface is not a shell of finite ordinary phase-space volume.

20. The two-fermion count 3 assumes fully spin-polarized particles. Why must it be revised if extra spin modes become available?

    Show the answer and reasoning

    The list of one-particle modes changes, including the internal spin alternatives.

    Pauli exclusion applies to complete single-particle states, not merely a spatial label. Omitting allowed internal modes changes g and can exclude genuine many-particle alternatives.

21. A learner counts every normalized superposition in the fixed-q sector as a separate perfectly distinguishable alternative. What count should replace that claim?

    Show the answer and reasoning

    The orthogonal subspace dimension is 20, while normalized pure vectors form a continuous family.

    Nonorthogonal quantum states are not all mutually exclusive perfectly distinguishable outcomes. A basis count, a prepared vector and a probability distribution play different roles.

22. An isolated system has a large equilibrium macroregion. What additional information is needed before predicting its relaxation time from an unusual initial state?

    Show the answer and reasoning

    Its dynamical laws, accessible sector, preparation and relevant transport or transition rates.

    A state-space volume ratio contains no unit of time. Reversible dynamics, invariant subsets and atypical preparations can change the trajectory even when the count of possible alternatives is large.

What to carry forward

• Multiplicity counts or measures alternatives compatible with a specified physical model, macrorecord and constraints; changing the description does not create physical states.
• Disjoint alternatives add, compatible independent choices multiply, and shared conservation laws require constrained products and sums.
• A multiplicity ratio is a probability only under the stated uniform microscopic measure; nonuniform preparation and inaccessible sectors can change the prediction.
• Large systems can concentrate in a narrow fractional region even while the probability of any one exact integer reading decreases.
• Classical continua require a measure and normalization; quantum counts are dimensions of allowed orthogonal subspaces, with the physical particle statistics included.
• Counting provides neither a relaxation time nor a proof of ergodicity or a universal entropy increase; dynamics, preparation and macroscopic coarse-graining remain necessary.