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Qm Many Body And Equilibrium Models — Examples

1 · Prerequisites

2 · Summary

This companion page works the many-body and equilibrium statements of qm-many-body-and-equilibrium-models through three exact model computations and one hypothetical scenario. The two-site bond model: exact canonical weights, mean energy, entropy and free energy of one particle on two sites solves the one-particle two-site bond model exactly: the tunnelling-split eigenbasis, the canonical weights, the mean energy −ttanh⁡(βt), the entropy and the free energy. Finite-mode non-interacting thermal ensembles: two fermions in three modes and two bosons in two modes evaluates the canonical partition sums and occupation means of two fermions in three modes and two bosons in two modes, checking the total-occupation identity and the ground energies against the page's diagonalisation theorem. Free-mode statistics at low occupation: exact dilute identities, a finite arithmetic check and two convergence examples verifies the exact Bose and Fermi dilute identities in exact rational arithmetic at q=1/10 and contrasts linearly spaced mode energies, whose product partition converges, with constant mode energies, whose product partition diverges. A site-localised preparation of the two-site bond model never becomes canonical, and its readouts need not take their thermal means is a hypothetical exactly-solved scenario: a site-localised preparation of the bond model never becomes the canonical preparation under the unitary flow, the site-occupation readout, assigned in the dimensionless particle-count unit, takes the canonical value 1/2 at some times, and the energy readout never takes its canonical value, so readout coincidence is not state convergence. What the companion examples and the thought experiment establish records what the examples do and do not establish. None of the items contains measured data, and every number is an exact model value in a declared finite-mode or countable-mode idealization.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

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The two-site bond model: exact canonical weights, mean energy, entropy and free energy of one particle on two sites

Physical scope

A finite exact computation in the declared one-particle two-site (bond) model of the finite-mode many-body realization: the hopping Hamiltonian on two sites, its two exact one-particle levels, the canonical preparation at assigned temperature, and the exact mean energy, entropy and free energy. The example is a model computation in a declared two-dimensional idealization; it asserts no physical molecule, no measured energy, no transition and no thermodynamic limit.

Example

Adopt P1, P2, P4 and P7 in the readings of Assigned sharp observables and the outcome law of the framework and Entropy, the Boltzmann constant and the free-energy functional of the declared finite-sector model, and work in the finite-mode realization of the declared fixed-N model of The declared fixed-N many-body model: species sector, one-particle data, interaction and the sector energy observable with: one-particle space K=C2, declared sites ∣1⟩,∣2⟩ an orthonormal basis of K, particle number N=1, single site-occupation sector, and the declared one-particle operator (the bond Hamiltonian) h:=−t(T∣1⟩,∣2⟩+T∣2⟩,∣1⟩),t>0 in joules. Adopt the canonical equilibrium postulate The canonical equilibrium-ensemble assumption: the Gibbs preparation of a declared model at assigned temperature at assigned temperature T>0 and inverse temperature β=1/(kBT), and read ε:=2t as the tunnelling splitting. Then:

(a) h has the orthonormal eigenbasis u−:=∣1⟩+∣2⟩2,u+:=∣1⟩−∣2⟩2,hu∓=∓t u∓, so its eigenvalues are −t and +t; equivalently h+t id is the two-level Hamiltonian of Closed formulas for the two-level canonical weights, mean energy, entropy and free energy with levels 0 and ε=2t and eigenvectors u−,u+;

(b) the canonical preparation of h+t id and hence of h is ρβ=Tu−,u−+e−2βt Tu+,u+1+e−2βt, with shifted partition sum Z0=1+e−2βt and physical partition trace Zh=tr⁡e−βh=eβtZ0;

(c) the exact means, entropy and free energy are ⟨h⟩β=2te2βt+1−t=−ttanh⁡(βt),s(ρβ)=log⁡(1+e−2βt)+2βte−2βt1+e−2βt,F=−1βlog⁡(1+e−2βt)−t, and for every β>0 one has −t<⟨h⟩β<0 and 0≤s(ρβ)<log⁡2;

(d) scope: the computation is an exact statement about the declared two-dimensional bond model at the assigned temperature; it identifies no physical double-well or molecular system, asserts no measured splitting and no transition, and the temperature assignment is the tested identification of P7.

Verification

[L1] The supplied two-level formulas. For a two-dimensional space with orthonormal basis (u0,u1) and H=εTu1,u1, ε>0, β>0: Z=1+e−βε, ρβ is the corresponding mixture, tr⁡(ρβH)=ε/(eβε+1) with values in (0,ε/2), and s(ρβ)=log⁡(1+e−βε)+βε/(eβε+1)<log⁡2, with tr⁡(ρβH)−β−1s(ρβ)=−β−1log⁡Z (Closed formulas for the two-level canonical weights, mean energy, entropy and free energy clauses (a)-(c)).

[L2] The declared model. The finite-mode realization of The declared fixed-N many-body model: species sector, one-particle data, interaction and the sector energy observable with K=C2, N=1, one-particle operator h and W=0 is a declared non-interacting model; its sector is K itself and the canonical preparation is the Gibbs operator of the adopted ensemble postulate; the free-energy functional is F(ρ)=tr⁡(ρh)−kBTs(ρ) by Entropy, the Boltzmann constant and the free-energy functional of the declared finite-sector model clause (e).

Proof technique: direct; diagonalise the two-by-two real symmetric matrix and substitute the supplied two-level formulas with ε=2t after the shift by t id.

Given: the declared bond model with coupling t>0, temperature T>0, β=1/(kBT), the vectors u∓, and [L1], [L2].

1.1L1algebra

[Diagonalisation] In the basis (∣1⟩,∣2⟩) the operator h has the matrix (0−t−t0); the vectors u−=(∣1⟩+∣2⟩)/2, u+=(∣1⟩−∣2⟩)/2 are orthogonal unit vectors and hu∓=∓t u∓ by direct evaluation, so {u−,u+} is an orthonormal eigenbasis with eigenvalues −t and t by Complex spectral theorem: a normal endomorphism of a finite-dimensional complex inner product space has an orthonormal eigenbasis, and conversely. Hence h+t id has the eigenvalues 0 on u− and 2t on u+, and it satisfies the hypothesis of [L1] with ε=2t, u0=u−, u1=u+. This proves (a).

2.1L1L2step 1.1algebra

[Canonical data] Applying [L1] with ε=2t gives Z=1+e−2βt and ρβ=Z−1(Tu−,u−+e−2βtTu+,u+) for h+t id; since h=(h+t id)−t id and both operators are functions of the same eigenbasis, the Gibbs operator of h is the same ρβ, because the normalized Gibbs weights are unchanged by the shift: e−βh=eβte−β(h+tI), so its partition trace is Zh=eβtZ and division gives the same density. The mean of h+t id is 2t/(e2βt+1)∈(0,t), so ⟨h⟩β=2t/(e2βt+1)−t=−ttanh⁡(βt), which lies in (−t,0) because e2βt>1 for βt>0; the entropy formula is that of [L1] at ε=2t and satisfies 0≤s<log⁡2 by [L1], and F=−β−1log⁡(1+e−2βt)−t is the free energy of h by [L2] and [L1]. This proves (b) and (c) apart from the scope clause.

3.1L1L2step 2.1∎

[Scope] The computation used only the declared two-dimensional model, its exact diagonalisation and the supplied two-level formulas; no physical double well, measured splitting, transition or thermodynamic limit entered, which is (d).

Remarks

  • The shift by t id. Adding a constant to the Hamiltonian multiplies Z by an exponential and leaves the canonical preparation and the entropy unchanged while shifting the free energy by the same constant; the example uses this to instantiate the supplied two-level formulas, whose levels are 0 and ε.
  • Low- and high-temperature values. The formulas give s→0 and ⟨h⟩β→−t as the exponential factor is suppressed, and s→log⁡2, ⟨h⟩β→0 as it approaches one; the example states the exact formulas and the bounds, not a limit theorem.
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Finite-mode non-interacting thermal ensembles: two fermions in three modes and two bosons in two modes

Physical scope

Two exact finite-mode thermal computations in the declared non-interacting model: the canonical partition sums, ground energies and occupation means of two spinless fermions in three modes with energies 0, epsilon, 2 epsilon, and of two spinless bosons in two modes with energies 0 and epsilon, at assigned temperature. The examples are model computations in declared finite-dimensional idealizations; they assert no physical particle species, no measured occupation and no thermodynamic limit.

Example

Adopt P1, P2 and P6 in the readings of The declared finite-mode model and its occupation readouts and Assigned sharp observables and the outcome law of the framework, work in the finite-mode realization of the declared fixed-N model of The declared fixed-N many-body model: species sector, one-particle data, interaction and the sector energy observable, and adopt the canonical equilibrium postulate The canonical equilibrium-ensemble assumption: the Gibbs preparation of a declared model at assigned temperature at assigned temperature T>0 with β=1/(kBT). Fix ε>0 in joules.

Fermionic model. One-particle space K=C3 with modes f1,f2,f3, fermionic statistics σ=−, N=2, zero interaction W=0, and the declared one-particle operator h:=ε (0⋅Tf1,f1+1⋅Tf2,f2+2⋅Tf3,f3).

Bosonic model. One-particle space K=C2 with modes g1,g2, bosonic statistics σ=+, N=2, zero interaction W=0, and h′:=ε(0⋅Tg1,g1+1⋅Tg2,g2).

Then:

(a) in the fermionic model the antisymmetric sector has orthonormal eigenbasis e{1,2}∧,e{1,3}∧,e{2,3}∧ with energies ε,2ε,3ε; the partition sum and the exact occupation means are ZF=e−βε+e−2βε+e−3βε, ⟨N1⟩β=e−βε+e−2βεZF,⟨N2⟩β=e−βε+e−3βεZF,⟨N3⟩β=e−2βε+e−3βεZF, with ⟨N1+N2+N3⟩β=2 and ground energy ε (configuration {1,2});

(b) in the bosonic model the symmetric sector has orthogonal eigenbasis m(2,0),m(1,1),m(0,2) with energies 0,ε,2ε; the partition sum and the exact occupation means are ZB=1+e−βε+e−2βε,⟨N1⟩β=2+e−βεZB,⟨N2⟩β=e−βε+2e−2βεZB, with ⟨N1+N2⟩β=2 and ground energy 0=Nε1;

(c) scope: the computations are exact statements about the declared finite-mode models at the assigned temperature; they assert no physical species, no measured occupation, no thermalisation and no thermodynamic limit.

Verification

[L1] Diagonalisation theorem. In the non-interacting finite-mode model the labelled operator ∑ih(i) is diagonal on the product basis with eigenvalue ∑kεik, the Slater vectors eJ∧ and the symmetric vectors mλ are eigenvectors of the sector operator with eigenvalues ∑j∈Jεj and ∑jλjεj, and the partition sum is the corresponding finite sum (Diagonalisation of the non-interacting finite-mode sector: occupation eigenvalues, partition sum and ground energy clauses (a)-(c)).

[L2] Occupation operators. The occupation operators Nj act on the sector basis by NjeJ∧=1J(j)eJ∧ and Njmλ=λjmλ, and the sector dimensions are three in both the bosonic and fermionic cases; the occupations are assigned sharp observables with finite means in every preparation (Slater and symmetric occupation-number bases of the finite-dimensional permutation sectors clauses (c), (d), Finite-mode occupation configurations: fermionic exclusion counts, bosonic counts and truncated dimensions clauses (a), (b), Mean, second moment and variance of an assigned observable in a preparation).

Proof technique: direct; list the occupation configurations, read their energies from the diagonalisation theorem, and compute the weighted means in the canonical weights.

Given: the declared fermionic and bosonic finite-mode models, the energies 0,ε,2ε, the assigned temperature data β>0, and [L1], [L2].

1.1L1L2algebra

[Fermionic configuration sum] The two-element subsets of {1,2,3} are {1,2},{1,3},{2,3} with energies 0+ε=ε, 0+2ε=2ε, ε+2ε=3ε by [L1]; the canonical weights are qJ=e−βλJ/ZF, and the occupation means are ∑JqJ1J(j) by [L2], giving the three displayed formulas. The total occupation is ∑JqJ⋅2=2 because every configuration has two occupied modes, and the smallest energy is that of {1,2}, namely ε, by [L1] clause (c). This proves (a).

1.2L1L2algebra

[Bosonic configuration sum] The occupation vectors of total two in two modes are (2,0),(1,1),(0,2) with energies 0,ε,2ε by [L1]; the canonical weights are qλ=e−βλλ/ZB and the occupation means are ∑λqλλj by [L2], giving ⟨N1⟩=(2⋅1+1⋅e−βε+0)/ZB and ⟨N2⟩=(0+1⋅e−βε+2⋅e−2βε)/ZB, and the total occupation is 2 because every configuration has ∑jλj=2; the ground energy is 0=Nε1 by [L1] clause (c). This proves (b).

2.1L1L2step 1.1step 1.2algebra∎

[Checks and scope] In both models Z is a finite sum of positive terms, hence 0<Z<∞; the displayed occupation sums equal the particle number N=2 identically in β by steps 1.1-1.2, and the ground energies agree with the general formulas of [L1] clause (c) for the sorted energy lists (0,ε,2ε) and (0,ε). Every quantity is a spectral datum of the declared finite-dimensional model; no physical species, measured occupation, thermalisation or thermodynamic limit entered, which is (c).

Remarks

  • Dimension counts. The fermionic sector is three-dimensional by exclusion and the bosonic sector is three-dimensional by the configuration count; the two examples are therefore directly comparable finite checks of the diagonalisation theorem at different statistics.
  • Exactness. The formulas are exact rational functions of e−βε; no expansion in ε/(kBT) is used, and the dilute-limit approximations are treated separately on this page.
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Free-mode statistics at low occupation: exact dilute identities, a finite arithmetic check and two convergence examples

Physical scope

A finite check of the declared free-mode thermal occupation formulas at low occupation q = e^{-beta epsilon}: the exact Bose and Fermi means, their dilute remainders, an exact rational evaluation at q = 1/10, and the product convergence for linearly spaced and for constant free-mode energies. The example is a model computation at assigned temperature and chemical potential; it asserts no measured occupation, no dilute gas, no thermodynamic limit and no condensation.

Example

Adopt the mode models and thermal readings of Thermal occupation means of the declared free-mode models: Bose and Fermi factors, product convergence and dilute remainders at assigned temperature T>0 with β=1/(kBT). For a canonical single mode of energy ε>0, put x:=βε>0 and q:=e−x∈(0,1). In the grand-canonical reading put x:=β(ε−μ) and q:=e−x>0; the bosonic formulas below apply only when μ<ε (so q<1), while the fermionic formulas hold for every μ∈R. Then:

(a) the exact single-mode means and their dilute identities are ⟨N⟩B=q1−q=q+q21−q,⟨N⟩F=q1+q=q−q21+q, so the bosonic mean exceeds q by q2/(1−q) and the fermionic mean falls below q by q2/(1+q) exactly;

(b) at q=1/10 the exact means are ⟨N⟩B=1/9 and ⟨N⟩F=1/11, and the exact identities 1/9=1/10+1/90 and 1/11=1/10−1/110 verify the two remainder formulas;

(c) for the free-mode energies εj=jε, j≥1, the sums ∑je−βεj=e−βε/(1−e−βε) converge, so the bosonic and fermionic product partitions are finite (bosonic case with μ<ε); for the constant energies εj≡ε the sum ∑je−βεj=+∞ diverges, so both product partitions diverge;

(d) scope: the identities and the check are exact statements about the declared mode models at the assigned temperature and, where stated, chemical potential; no measured occupation, dilute gas, thermodynamic limit or condensation is asserted.

Verification

[L1] Supplied occupation formulas. For a single bosonic mode of energy ε>0 the canonical mean occupancy is 1/(eβε−1), for a single fermionic orbital with parameters ε1,μ it is 1/(eβ(ε1−μ)+1), and the dilute identities and remainder bounds of clause (d) of the cited theorem hold with q=e−β(ε−μ) (Thermal occupation means of the declared free-mode models: Bose and Fermi factors, product convergence and dilute remainders clauses (a), (b), (d), Fermion and dilute grand-canonical formulas clause (c)).

[L2] Supplied product criterion. Countably many free modes with energies εj>0 have finite bosonic product partition exactly when ∑je−βεj<∞ and μ<inf⁡jεj, and finite fermionic product partition exactly when ∑je−βεj<∞ (Countable-mode product criterion clauses (d), (e), Thermal occupation means of the declared free-mode models: Bose and Fermi factors, product convergence and dilute remainders clause (c)); the geometric series gives ∑j≥1rj=r/(1−r) for 0<r<1 (For ∣r∣<1, ∑k≥0rk=1/(1−r), and for ∣r∣≥1 the series diverges).

Proof technique: direct; substitute q=e−x in the supplied means, evaluate the rational identities at q=1/10 in exact fractions, and apply the product criterion to the two energy sequences.

Given: the declared single-mode model with energy ε>0, assigned β>0, q=e−βε, the energy sequences εj=jε and εj≡ε, and [L1], [L2].

1.1L1algebra

[Dilute identities] In the canonical reading, q=e−βε∈(0,1); in the grand-canonical reading, q=e−β(ε−μ)>0, with the bosonic formula restricted to μ<ε and hence q<1. By [L1], the bosonic and fermionic means on these respective domains are q/(1−q) and q/(1+q). The algebraic identities q/(1−q)=q+q2/(1−q) for 0<q<1 and q/(1+q)=q−q2/(1+q) for q>0 are exact; the bosonic correction is positive and the fermionic correction is negative, with absolute values q2/(1−q) and q2/(1+q). This proves (a).

1.2L1algebra

[The finite check at q=1/10] At q=1/10 the bosonic mean is (1/10)/(9/10)=1/9 and the fermionic mean is (1/10)/(11/10)=1/11; the remainders are q2/(1−q)=(1/100)/(9/10)=1/90 and q2/(1+q)=(1/100)/(11/10)=1/110, so the exact identities 1/9=1/10+1/90=10/90+1/90 and 1/11=1/10−1/110=11/110−1/110 hold. This proves (b).

1.3L2algebra

[The two convergence examples] For εj=jε the sum is ∑j≥1(e−βε)j=e−βε/(1−e−βε)<∞ by [L2], and for the bosonic product one needs in addition μ<inf⁡jεj=ε; for εj≡ε every term of the sum equals e−βε>0, so the sum diverges and [L2] makes both product partitions infinite. This proves (c).

2.1L1L2step 1.1step 1.2step 1.3algebra∎

[Checks and scope] Steps 1.1-1.3 give the exact identities, the exact rational check, and the two convergence outcomes; every quantity is computed in the declared mode models at the assigned temperature, and no measured occupation, dilute gas, thermodynamic limit or condensation entered, which is (d).

Remarks

  • What the check tests. The check confirms the two remainder identities in exact arithmetic at a fixed value of q; it is a finite verification of algebraic identities, not an empirical test and not a statement about any gas.
  • Two ways to fail convergence. Constant free-mode energies have divergent product partition because infinitely many modes share the same positive weight; linearly increasing energies have geometrically decaying weights and finite product partition. The bosonic criterion additionally requires the chemical potential below the ground-mode energy.
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A site-localised preparation of the two-site bond model never becomes canonical, and its readouts need not take their thermal means

Physical scope

A hypothetical exactly solved scenario in the declared two-site bond model: the site-localised pure preparation at a site, its unitary evolution, the stationary canonical preparation of the adopted ensemble, and two readouts - the site occupation and the energy - whose means either do or do not coincide with their canonical values at specified times. The scenario is a model deduction from the declared preparation and P4; it asserts no observed equilibration, no measured occupation or energy, and no statement about any laboratory system.

Statement

Adopt P1, P2, P4 and P7 in the readings of Assigned energy observable, unitary flow and the two pictures of autonomous evolution, Assigned sharp observables and the outcome law of the framework and Entropy, the Boltzmann constant and the free-energy functional of the declared finite-sector model, work in the two-site bond model of The two-site bond model: exact canonical weights, mean energy, entropy and free energy of one particle on two sites with coupling t>0, sites ∣1⟩,∣2⟩, bond Hamiltonian h=−t(T∣1⟩,∣2⟩+T∣2⟩,∣1⟩) and eigenbasis u−=(∣1⟩+∣2⟩)/2, u+=(∣1⟩−∣2⟩)/2 with hu∓=∓tu∓, and adopt the canonical equilibrium postulate The canonical equilibrium-ensemble assumption: the Gibbs preparation of a declared model at assigned temperature at assigned temperature T>0, β=1/(kBT). Consider the hypothetical declared preparation ρ(0):=T∣1⟩,∣1⟩ (the particle localised at site 1) evolving by ρ(τ)=U(τ)ρ(0)U(τ)∗ with U(τ)=exp⁡(−iτh/ℏ). Then:

(a) ρ(0)≠ρβ, and for every τ∈R one has ρ(τ)≠ρβ: the unitary evolution of the model never produces the canonical preparation from this preparation;

(b) the site-1 occupation readout (P1,1) with P1:=T∣1⟩,∣1⟩, assigned in the dimensionless particle-count unit, has the exact mean ⟨P1⟩(τ)=cos⁡2 ⁣(tτℏ), and tr⁡(ρβP1)=1/2; in particular ⟨P1⟩(0)=1≠1/2, so at the initial time the readout does not take its canonical mean;

(c) the energy readout h has the constant mean tr⁡(ρ(τ)h)=0 for every τ, while its canonical mean is tr⁡(ρβh)=−ttanh⁡(βt)<0; hence the energy readout never takes its canonical thermal value;

(d) scope and interpretation: the scenario is a hypothetical, exactly-solved model deduction in one declared preparation; it shows that agreement or disagreement of individual readout means with their canonical values is not convergence of the preparation to the canonical one, and that the model's P4 dynamics does not produce the canonical ensemble. It asserts no observed equilibration or non-equilibration of a sample, no measured occupation or energy, and no claim about environments or open systems.

Facts & Assumptions

Given: the declared bond model with t>0, the canonical preparation ρβ, the site-localised preparation ρ(0)=T∣1⟩,∣1⟩, the flow U(τ), and the times τ∈R.

[F1]

Spectral flow of the bond Hamiltonian. For the assigned energy observable h with distinct eigenvalues ∓t and orthonormal eigenvectors u∓, the spectral projection-valued measure has the atoms E({−t})=Tu−,u− and E({t})=Tu+,u+, so the flow of Assigned energy observable, unitary flow and the two pictures of autonomous evolution clause (b) is U(τ)=eitτ/ℏTu−,u−+e−itτ/ℏTu+,u+ and maps ψ0=c−u−+c+u+ to c−eitτ/ℏu−+c+e−itτ/ℏu+; the mean of a bounded observable in a pure preparation is ⟨Aψ,ψ⟩ (Spectral theorem for unbounded self-adjoint operators (PVM form), Projection valued measure, Assigned energy observable, unitary flow and the two pictures of autonomous evolution, Mean, second moment and variance of an assigned observable in a preparation clause (a)).

[F2]

Euler formula and parity. For real θ, eiθ=cos⁡θ+isin⁡θ, cos⁡(−θ)=cos⁡θ, sin⁡(−θ)=−sin⁡θ, and cos⁡2θ+sin⁡2θ=1, so (eiθ+e−iθ)/2=cos⁡θ and (eiθ−e−iθ)/2=isin⁡θ (exp⁡(x+iy)=ex(cos⁡y+isin⁡y), ∣exp⁡(x+iy)∣=ex, and eiπ+1=0, Parity and the Pythagorean identity for sine and cosine).

[F3]

Canonical data. The canonical preparation of the bond model is ρβ=Z−1(Tu−,u−+e−2βtTu+,u+) with Z=1+e−2βt, its mean energy is −ttanh⁡(βt)∈(−t,0) for every β>0, and tr⁡(ρβTu−,u−)≠tr⁡(ρβTu+,u+) (The two-site bond model: exact canonical weights, mean energy, entropy and free energy of one particle on two sites clauses (b), (c), Closed formulas for the two-level canonical weights, mean energy, entropy and free energy).

[F4]

Invariance statement. In the finite-sector class, if ρ(0)≠ρβ then ρ(τ)≠ρβ for every τ (Unitary evolution preserves entropy and cannot produce the canonical preparation from a different one clause (c)).

Proof

technique · direct; express the initial site vector in the energy eigenbasis, insert the supplied two-level flow, and compare the readout means with the canonical data

Given: the data of the statement and [F1]-[F3].

1.1F1F2algebra

[The evolved preparation and the site readout] With ψ0=∣1⟩=(u−+u+)/2, [F1] gives ψ(τ)=U(τ)ψ0=(eiθu−+e−iθu+)/2, where θ:=tτ/ℏ; by [F2] and u−=(∣1⟩+∣2⟩)/2, u+=(∣1⟩−∣2⟩)/2 this equals cos⁡θ ∣1⟩+isin⁡θ ∣2⟩, a unit vector. Since T∣1⟩,∣1⟩=12(I+σz) with σz=T∣1⟩,∣1⟩−T∣2⟩,∣2⟩, and σzψ(τ)=cos⁡θ ∣1⟩−isin⁡θ ∣2⟩, the site-occupation mean is ⟨P1⟩(τ)=∣cos⁡θ∣2=cos⁡2θ=cos⁡2(tτ/ℏ), in particular ⟨P1⟩(0)=1.

1.2F3algebra

[Canonical means and the initial comparison] By [F3], tr⁡(ρβh)=−ttanh⁡(βt)<0; moreover ρβ is diagonal in the basis (u−,u+) with strictly positive diagonal weights, so tr⁡(ρβσz)=0 because σz exchanges the two basis vectors, and tr⁡(ρβP1)=12; and tr⁡(ρ(0)P1)=1≠12, so ρ(0)≠ρβ.

2.1F4step 1.2algebra

[The preparation is never canonical] Step 1.2 shows ρ(0)≠ρβ; applying [F4] to the declared finite-sector model gives ρ(τ)≠ρβ for every τ∈R, which is (a).

2.2F3step 1.1step 1.2algebra

[The energy readout never takes its thermal value] Since h commutes with U(τ), tr⁡(ρ(τ)h)=tr⁡(ρ(0)h)=⟨hΨ,Ψ⟩ for Ψ:=∣1⟩, and this vanishes because hΨ=−t∣2⟩ is orthogonal to Ψ; by step 1.2 the canonical mean is −ttanh⁡(βt)<0, so the two values never agree. This proves (c), and step 1.1 proves (b) together with the comparison of step 1.2.

3.1step 2.1step 2.2∎

[Conclusion and scope] Steps 2.1-2.2 prove (a)-(c) for the declared hypothetical preparation. The scenario is an exact model deduction from the declared preparation, the bond Hamiltonian and the canonical data; no observed equilibration, measured readout or laboratory system entered, which is (d).

Remarks

  • What the scenario shows. The preparation never equals the canonical one, yet the site-occupation readout has the same mean 1/2 as the canonical preparation at the times when cos⁡2(tτ/ℏ)=1/2; the energy readout meanwhile never takes its canonical mean. Readout coincidence is therefore not state convergence, and different readouts behave differently under the same unitary flow.
  • Hypothetical status. No apparatus, sampling or detection model is specified; the scenario is a conditional statement about the declared model, and any comparison with an experiment would require the declared preparation, isolation, detection and sampling models of P7.
RemarkRemark: AI-adaptedProof: Not applicableOpen item page →

What the companion examples and the thought experiment establish

Physical scope

The recorded scope of the companion examples and thought experiment of the many-body and equilibrium pair: exact finite-mode or countable-mode model computations of canonical weights, occupations and readouts; the limits of their interpretation with respect to physical systems, empirical data and thermodynamic limits.

Recorded scope

The companion items contain three conditional examples and one hypothetical scenario in declared finite-mode or countable-mode models:

What they do not establish

None of these items contains measured data, a measured uncertainty, a calibration, a sampling model or a detector model; all numbers are exact model values in the declared finite-mode or countable-mode idealizations, and the assignment of a physical system to any of these models is a tested identification under P7 (Preparation and apparatus postulate P7: declared preparations, isolation and sampling assumptions). In particular:

  • the bond model is not any particular molecule, and its splitting 2t is an assigned parameter, not a measured spectroscopic value;
  • the finite-mode examples are not ideal or interacting gases; they make no statement about a thermodynamic limit, pressure, condensation or a continuum of modes;
  • the comparison of readout means with canonical values in the thought experiment is a statement about the declared preparation and the exact unitary flow; it is not evidence for or against thermalisation in nature, and the canonical ensemble remains the adopted assumption of The canonical equilibrium-ensemble assumption: the Gibbs preparation of a declared model at assigned temperature;
  • the thought experiment uses the no-unitary-production conclusion (c) of Unitary evolution preserves entropy and cannot produce the canonical preparation from a different one. The separate entropy-conservation conclusion (a) of that theorem concerns the same finite closed-system class but is not a premise of the readout calculation; neither conclusion forbids or asserts open-system or environment-induced behaviour.

Source limits

The cited treatments supply the model formulas used here at the recorded locators; they do not supply measured values for these particular finite-mode or countable-mode models, and no source is used as an empirical premise on this page. Where a comparison with data would be attempted, the preparation, isolation, detection and sampling models plus the uncertainty of every assigned parameter would have to be declared and carried through, as P7 requires.

5 · Examples, experiments and counterexamples

None yet.

Sources