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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.

Depends on

Dependency tree · 2 levels

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