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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 $β = 1 / (k_{B} T)$ . For a canonical single mode of energy $ε > 0$ , put $x : = β ε > 0$ and $q : = e^{- x} \in (0, 1)$ . 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 $μ \in R$ . Then:

(a) the exact single-mode means and their dilute identities are $⟨ N ⟩_{B} = \frac{q}{1 - q} = q + \frac{q^{2}}{1 - q}, ⟨ N ⟩_{F} = \frac{q}{1 + q} = q - \frac{q^{2}}{1 + q},$ so the bosonic mean exceeds $q$ by $q^{2} / (1 - q)$ and the fermionic mean falls below $q$ by $q^{2} / (1 + q)$ 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 $ε_{j} = j ε$ , $j \geq 1$ , the sums $\sum_{j} e^{- β ε_{j}} = e^{- β ε} / (1 - e^{- β ε})$ converge, so the bosonic and fermionic product partitions are finite (bosonic case with $μ < ε$ ); for the constant energies $ε_{j} \equiv ε$ the sum $\sum_{j} e^{- β ε_{j}} = + \infty$ 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 $1 / (e^{β ε} - 1)$ , for a single fermionic orbital with parameters $ε_{1}, μ$ it is $1 / (e^{β (ε_{1} - μ)} + 1)$ , 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 $ε_{j} > 0$ have finite bosonic product partition exactly when $\sum_{j} e^{- β ε_{j}} < \infty$ and $μ < \inf_{j} ε_{j}$ , and finite fermionic product partition exactly when $\sum_{j} e^{- β ε_{j}} < \infty$ (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 $\sum_{j \geq 1} r^{j} = r / (1 - r)$ for $0 < r < 1$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ 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 $β > 0$ , $q = e^{- β ε}$ , the energy sequences $ε_{j} = j ε$ and $ε_{j} \equiv ε$ , and [L1], [L2].

1.1L1algebra

[Dilute identities] In the canonical reading, $q = e^{- β ε} \in (0, 1)$ ; 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 $q / (1 - q) = q + q^{2} / (1 - q)$ for $0 < q < 1$ and $q / (1 + q) = q - q^{2} / (1 + q)$ for $q > 0$ are exact; the bosonic correction is positive and the fermionic correction is negative, with absolute values $q^{2} / (1 - q)$ and $q^{2} / (1 + q)$ . 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 $q^{2} / (1 - q) = (1 / 100) / (9 / 10) = 1 / 90$ and $q^{2} / (1 + q) = (1 / 100) / (11 / 10) = 1 / 110$ , 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 $ε_{j} = j ε$ the sum is $\sum_{j \geq 1} (e^{- β ε})^{j} = e^{- β ε} / (1 - e^{- β ε}) < \infty$ by [L2], and for the bosonic product one needs in addition $μ < \inf_{j} ε_{j} = ε$ ; for $ε_{j} \equiv ε$ every term of the sum equals $e^{- β ε} > 0$ , 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.

Free-mode statistics at low occupation: exact dilute identities, a finite arithmetic check and two convergence examples

Example

Verification

Remarks

Depends on

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