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Finite Gibbs free-energy minimum

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a finite-dimensional complex Hilbert space of dimension $d \geq 1$ with inner product linear in the first argument (Hilbert space), let $H : H \to H$ be a self-adjoint linear operator with real eigenvalues $E_{1}, \dots, E_{d}$ (repeated according to multiplicity) and orthonormal eigenbasis $(v_{1}, \dots, v_{d})$ , and let $β > 0$ . Write $Z : = \sum_{j = 1}^{d} e^{- β E_{j}} \in (0, \infty), ρ_{β} : = \frac{1}{Z} \sum_{j = 1}^{d} e^{- β E_{j}} T_{v_{j}, v_{j}},$ where $T_{u, v} x = ⟨ x, v ⟩ u$ (Density operators and scalar spectral probability). Let $D$ be the set of density operators on $H$ , i.e. positive self-adjoint trace-class operators $ρ$ with $tr ρ = 1$ (Trace class operator, Trace of a trace class operator), and for $ρ \in D$ with eigenvalues $w_{1}, \dots, w_{d} \geq 0$ (counted with multiplicity and in any order) define the entropy $s (ρ) : = - \sum_{i : w_{i} > 0} w_{i} \log w_{i} \in [0, \log d]$ (The natural logarithm as the inverse of the exponential function). Then:

(a) $ρ_{β} \in D$ , $tr (ρ_{β} H) = \frac{1}{Z} \sum_{j} e^{- β E_{j}} E_{j}$ , and $tr (ρ_{β} H) - β^{- 1} s (ρ_{β}) = - β^{- 1} \log Z;$

(b) for every $ρ \in D$ , $tr (ρ H) - β^{- 1} s (ρ) \geq - β^{- 1} \log Z;$

(c) for every $ρ \in D$ , $s (ρ) \leq \log d$ , with $s (ρ) = \log d$ if and only if $ρ = d^{- 1} i d$ ; when $H = 0$ one has $Z = d$ , $ρ_{β} = d^{- 1} i d$ and the bound in (b) is attained there.

The combination $tr (ρ H) - β^{- 1} s (ρ)$ is a purely mathematical functional of $(ρ, H, β)$ ; identifying $β$ with an inverse physical temperature and $ρ_{β}$ with a thermal equilibrium state are additional physical assumptions not asserted here.

Facts & Assumptions

Given: the finite-dimensional complex Hilbert space $H$ of dimension $d \geq 1$ , the self-adjoint operator $H$ with eigenvalues $E_{j}$ and orthonormal eigenbasis $(v_{j})$ , the number $β > 0$ , the partition sum $Z$ , the Gibbs weight operator $ρ_{β}$ , and an arbitrary density operator $ρ \in D$ with eigenvalues $w_{i}$ and orthonormal eigenbasis $(u_{i})$ .

[F1]

Finite-dimensional spectral theorem: every self-adjoint (indeed normal) endomorphism of $H$ has an orthonormal basis of eigenvectors; the eigenvalues are real for $H$ , and $ρ$ has an orthonormal eigenbasis with $w_{i} \geq 0$ , $\sum_{i} w_{i} = tr ρ = 1$ (Complex spectral theorem: a normal endomorphism of a finite-dimensional complex inner product space has an orthonormal eigenbasis, and conversely, Density operators and scalar spectral probability).

[F2]

On a finite-dimensional space every operator is trace class, the trace is basis-independent, and for a diagonal operator the trace is the sum of its eigenvalues (Trace class operator, Trace of a trace class operator, Trace is absolutely convergent and basis independent).

[F3]

For $x > 0$ , $\log^{'} (x) = 1 / x$ and $\log x = \int_{1}^{x} d t / t$ ; for $x \geq 1$ , $1 - 1 / x \leq \log x \leq x - 1$ (The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t, The natural logarithm as the inverse of the exponential function, If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ ). The inequality $\log x \leq x - 1$ for every $x > 0$ , with equality only at $x = 1$ , is derived in step 1.3.

[F4]

For positive $x, y$ , $\log (x y) = \log x + \log y$ , $\log (x / y) = \log x - \log y$ and $\log (1 / x) = - \log x$ (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

Proof

1.1F1F2F4algebra

[The Gibbs state and its entropy] The weights $q_{j} : = e^{- β E_{j}} / Z$ are positive with $\sum_{j} q_{j} = 1$ , so by [F1] and [F2] $ρ_{β} = \sum_{j} q_{j} T_{v_{j}, v_{j}}$ is a positive self-adjoint operator with orthonormal eigenbasis $(v_{j})$ , eigenvalues $q_{j}$ and trace $1$ ; hence $ρ_{β} \in D$ . Its entropy is $s (ρ_{β}) = - \sum_{j} q_{j} \log q_{j} = \sum_{j} q_{j} (β E_{j} + \log Z) = β tr (ρ_{β} H) + \log Z$ by [F2], where $tr (ρ_{β} H) = \sum_{j} q_{j} E_{j}$ . Rearranging gives $tr (ρ_{β} H) - β^{- 1} s (ρ_{β}) = - β^{- 1} \log Z$ , which is (a).

1.2F1algebra

[The doubly stochastic matrix] Let $p_{i j} : = ∣ ⟨ u_{i}, v_{j} ⟩ ∣^{2}$ . Since $(u_{i})$ and $(v_{j})$ are orthonormal bases, [F1] gives $p_{i j} \geq 0$ , $\sum_{j} p_{i j} = ∥ u_{i} ∥^{2} = 1$ and $\sum_{i} p_{i j} = ∥ v_{j} ∥^{2} = 1$ : the matrix $(p_{i j})$ is doubly stochastic. For $r_{i} : = \sum_{j} p_{i j} q_{j}$ one has $r_{i} > 0$ because all $q_{j} > 0$ and $\sum_{j} p_{i j} = 1$ , and $\sum_{i} r_{i} = \sum_{j} q_{j} \sum_{i} p_{i j} = 1$ .

1.3F1F3algebra

[The entropy bound and equality case] For $x > 1$ , the integral formula in [F3] gives $\log x = \int_{1}^{x} d t / t < x - 1$ , because $1 / t < 1$ for $t > 1$ ; for $0 < x < 1$ , put $y = 1 / x > 1$ . On $[1, y)$ one has $1 / t > 1 / y = x$ , so $\log y = \int_{1}^{y} d t / t > x (y - 1) = 1 - x$ , and the reciprocal law gives $\log x = - \log y < x - 1$ . Thus $\log x \leq x - 1$ for every $x > 0$ , with equality exactly at $x = 1$ . Let $S : = {i : w_{i} > 0}$ and $k : = ∣ S ∣$ , so $1 \leq k \leq d$ . Then Since $0 < w_{i} \leq 1$ , every term $- w_{i} \log w_{i} = w_{i} \log (1 / w_{i})$ is nonnegative. Also $s (ρ) - \log d = \sum_{i \in S} w_{i} \log \frac{1}{d w_{i}} \leq \sum_{i \in S} w_{i} (\frac{1}{d w_{i}} - 1) = \frac{k}{d} - 1 \leq 0.$ If $s (ρ) = \log d$ , then $k = d$ and equality holds in every positive-weight logarithm inequality, since each individual difference $w_{i} [(1 / (d w_{i}) - 1) - \log (1 / (d w_{i}))]$ is nonnegative and their finite sum is zero. The equality characterization above forces $1 / (d w_{i}) = 1$ for all $i$ , so $w_{i} = 1 / d$ and $ρ = \sum_{i} (1 / d) T_{u_{i}, u_{i}} = d^{- 1} i d$ . Conversely $ρ = d^{- 1} i d$ has entropy $\log d$ . When $H = 0$ all $E_{j} = 0$ , so $Z = d$ and $ρ_{β} = d^{- 1} i d$ , and the bound of (b) is attained there.

2.1F2F3F4step 1.1step 1.2step 1.3algebra

[Relative entropy is nonnegative] With $w_{i}$ the eigenvalues of $ρ$ in the orthonormal eigenbasis $(u_{i})$ of [F1], $\log ρ = \sum_{i : w_{i} > 0} (\log w_{i}) T_{u_{i}, u_{i}}$ and $\log ρ_{β} = \sum_{j} (\log q_{j}) T_{v_{j}, v_{j}}$ (finite-dimensional spectral calculus). For the product with $ρ$ , the symbol $\log ρ$ here denotes the logarithm on the positive spectral support extended by zero on $\ker ρ$ ; it does not assign a value to $\log 0$ . Thus by [F2] $tr (ρ \log ρ) = \sum_{i : w_{i} > 0} w_{i} \log w_{i}$ and $tr (ρ \log ρ_{β}) = \sum_{i : w_{i} > 0} w_{i} \sum_{j} p_{i j} \log q_{j}$ . For each $i$ with $w_{i} > 0$ , $r_{i} = \sum_{j} p_{i j} q_{j} > 0$ and $\sum_{j} p_{i j} = 1$ by step 1.2. Applying $\log x \leq x - 1$ from step 1.3 to $x = q_{j} / r_{i}$ and summing with weights $p_{i j}$ gives $\sum_{j} p_{i j} \log (q_{j} / r_{i}) \leq \sum_{j} p_{i j} (q_{j} / r_{i} - 1) = 0$ , hence $\sum_{j} p_{i j} \log q_{j} \leq \log r_{i}$ . Therefore $tr (ρ \log ρ) - tr (ρ \log ρ_{β}) \geq \sum_{i : w_{i} > 0} w_{i} (\log w_{i} - \log r_{i}) = \sum_{i : w_{i} > 0} w_{i} \log (w_{i} / r_{i})$ . Writing $w_{i} \log (w_{i} / r_{i}) = - w_{i} \log (r_{i} / w_{i})$ and again applying step 1.3 to $x = r_{i} / w_{i} > 0$ gives $w_{i} \log (w_{i} / r_{i}) \geq w_{i} - r_{i}$ ; summing, $\sum_{i : w_{i} > 0} (w_{i} - r_{i}) = 1 - \sum_{i : w_{i} > 0} r_{i} = \sum_{i : w_{i} = 0} r_{i} \geq 0$ , since $\sum_{i} r_{i} = 1$ by step 1.2. Therefore $tr (ρ \log ρ) \geq tr (ρ \log ρ_{β})$ .

3.1F2F4step 1.1step 2.1algebra

[The lower bound] By the spectral description of $ρ_{β}$ in step 1.1, $tr (ρ \log ρ_{β}) = \sum_{j} (\log q_{j}) tr (ρ T_{v_{j}, v_{j}}) = \sum_{j} (- β E_{j} - \log Z) ⟨ ρ v_{j}, v_{j} ⟩ = - β tr (ρ H) - \log Z$ , since $\sum_{j} ⟨ ρ v_{j}, v_{j} ⟩ = tr ρ = 1$ and $tr (ρ H) = \sum_{j} E_{j} ⟨ ρ v_{j}, v_{j} ⟩$ (because $H = \sum_{j} E_{j} T_{v_{j}, v_{j}}$ ). Combining with step 2.1 gives $tr (ρ \log ρ) \geq - β tr (ρ H) - \log Z$ ; multiplying by $β^{- 1}$ and rearranging, using $s (ρ) = - tr (ρ \log ρ)$ , gives the bound (b).

4.1step 1.1step 1.2step 2.1step 3.1step 1.3algebra∎

[Conclusion and scope] Steps 1.1-3.1 prove (a), (b) and (c), including the asserted equality in (a) and the attainment in (c). The final paragraph of the statement records that the thermodynamic reading of the functional and of $ρ_{β}$ requires additional physical assumptions, which are not part of this mathematical statement.

Finite Gibbs free-energy minimum

Statement

Facts & Assumptions

Proof

Depends on

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