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Finite interacting asymptotic Wick expansion with remainder

Statement

Assume the Axiom of Choice and the hypotheses and notation of Finite Euclidean quartic interacting measure. Let $P \in R [q_{1}, \dots, q_{N}]$ , $Z (λ) = E_{γ_{C}} [e^{- λ U}]$ , $N_{P} (λ) = E_{γ_{C}} [P e^{- λ U}]$ , and $⟨ P ⟩_{λ} = N_{P} (λ) / Z (λ)$ . For every integer $M \geq 0$ ,

$Z (λ) = \sum_{j = 0}^{M} z_{j} λ^{j} + R_{M, 1} (λ), z_{j} = \frac{(- 1)^{j}}{j!} E_{γ_{C}} [U^{j}], ∣ R_{M, 1} (λ) ∣ \leq \frac{λ^{M + 1}}{(M + 1)!} E_{γ_{C}} [U^{M + 1}]$

for $λ \geq 0$ . Likewise,

$N_{P} (λ) = \sum_{j = 0}^{M} a_{j} (P) λ^{j} + R_{M, P} (λ), a_{j} (P) = \frac{(- 1)^{j}}{j!} E_{γ_{C}} [P U^{j}], ∣ R_{M, P} (λ) ∣ \leq \frac{λ^{M + 1}}{(M + 1)!} E_{γ_{C}} [∣ P ∣ U^{M + 1}] .$

Define $b_{0} (P) = a_{0} (P)$ and, for $m \geq 1$ , $b_{m} (P) = a_{m} (P) - \sum_{j = 1}^{m} z_{j} b_{m - j} (P)$ . Then

$⟨ P ⟩_{λ} = \sum_{j = 0}^{M} b_{j} (P) λ^{j} + O (λ^{M + 1}) as λ ↓ 0.$

Every coefficient is computed from finitely many Gaussian polynomial moments and hence from finitely many covariance pairings by Finite Gaussian normalization and Wick moments. These estimates hold separately for each finite $M$ ; they do not assert convergence of the infinite formal series or any continuum limit.

Facts & Assumptions

Given: Assume the Axiom of Choice, fix $N, C, U$ as in Finite Euclidean quartic interacting measure, take $P \in R [q_{1}, \dots, q_{N}]$ , and let $λ \geq 0$ .

[A1]

The Axiom of Choice is carried from the finite-dimensional normal-law construction. The Axiom of Choice

[F1]

The Borel probability $γ_{C}$ , the nonnegative quartic $U$ , $Z (λ)$ and the polynomial insertion $N_{P} (λ)$ are defined in Finite Euclidean quartic interacting measure.

[F2]

All Gaussian polynomial moments are absolutely finite and are sums over the finite set of covariance pairings. Finite Gaussian normalization and Wick moments

[F3]

The normalized measure exists, $Z (0) = 1$ , and $Z (λ) \to 1$ as $λ ↓ 0$ . Actual finite Euclidean polynomial measure and moments

[F4]

For a $C^{M + 1}$ function on an interval, the Lagrange remainder is $f^{(M + 1)} (ξ) x^{M + 1} / (M + 1)!$ for some $ξ$ between $0$ and $x$ . The Lagrange and Cauchy forms of Taylor's remainder

[F5]

The real exponential is smooth with derivative itself, and the chain rule applies to its composition with $x \mapsto - x$ . The exponential function is smooth and $(\exp)^{'} = \exp$ , The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$

[F6]

For $x \geq 0$ , the real exponential power series gives $\exp (x) \geq 1$ , and the exponential is positive with $\exp (- x) = 1 / \exp (x)$ . Thus $0 < \exp (- x) \leq 1$ . The real exponential function and the number $e$ by a power series, The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$

Proof

technique · direct finite-order remainder and formal division

1.1A1F4F5F6given

By [F5], $f (x) = \exp (- x)$ is $C^{M + 1}$ and $∣ f^{(M + 1)} (ξ) ∣ = \exp (- ξ)$ . Apply [F4] at $x = λ U (q) \geq 0$ ; [F6] gives $\exp (- ξ) \leq 1$ . Thus pointwise $∣ \exp (- λ U (q)) - \sum_{j = 0}^{M} \frac{(- λ U (q))^{j}}{j!} ∣ \leq \frac{λ^{M + 1} U (q)^{M + 1}}{(M + 1)!} .$ The case $λ U (q) = 0$ has zero remainder.

1.2givenalgebra

Let $Z_{M} (λ) = \sum_{j = 0}^{M} z_{j} λ^{j}$ and $A_{M} (λ) = \sum_{j = 0}^{M} a_{j} (P) λ^{j}$ . By the recurrence defining $b_{j} (P)$ , every coefficient through degree $M$ in $A_{M} (λ) - (\sum_{j = 0}^{M} b_{j} (P) λ^{j}) Z_{M} (λ)$ is zero, because $z_{0} = 1$ . The remaining polynomial terms have degree at least $M + 1$ , so their absolute value is bounded by a finite constant times $λ^{M + 1}$ on $0 \leq λ \leq 1$ .

2.1step 1.1F1F2algebra

Integrate the inequality in step 1.1 against $γ_{C}$ to obtain the stated partition-function expansion and bound. Multiply the same pointwise inequality by $∣ P (q) ∣$ and integrate to obtain the numerator expansion and bound. Every right-hand expectation is finite by [F2].

3.1step 2.1step 1.2F2F3algebra∎

On $0 \leq λ \leq 1$ , the polynomial $B_{M} (λ) = \sum_{j = 0}^{M} b_{j} (P) λ^{j}$ is bounded. Using the remainder bounds in step 2.1 and the polynomial identity in step 1.2 therefore gives $∣ N_{P} (λ) - B_{M} (λ) Z (λ) ∣ \leq K_{M} λ^{M + 1}$ for some finite $K_{M}$ . As $λ ↓ 0$ , $\exp (- λ U (q)) \to 1$ pointwise and $0 < \exp (- λ U (q)) \leq 1$ by [F6], so [F3] gives $Z (λ) \to Z (0) = 1$ . Choose $δ > 0$ such that $Z (λ) \geq 1 / 2$ for $0 \leq λ < δ$ . Dividing by $Z (λ)$ yields $∣ ⟨ P ⟩_{λ} - B_{M} (λ) ∣ \leq 2 K_{M} λ^{M + 1}$ there. Finally, [F2] writes every coefficient moment as a finite covariance-pairing sum; no limit in $M$ was taken.

Finite interacting asymptotic Wick expansion with remainder

Statement

Facts & Assumptions

Proof

Depends on

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