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Finite covariance Wick normal powers

Definition

Assume the Axiom of Choice, let $Q \sim N_{N} (0, C)$ be the centered finite-dimensional Gaussian vector of Finite Gaussian normalization and Wick moments, and write $C_{j j} = E [Q_{j}^{2}] \geq 0$ for the variance of its $j$ th coordinate. For each coordinate $j$ and integer $n \geq 0$ , its finite covariance Wick normal power is the polynomial $: Q_{j}^{n} :_{C}$ defined by the formal generating identity

$\exp ⁣ (t Q_{j} - \frac{1}{2} t^{2} C_{j j}) = \sum_{n = 0}^{\infty} : Q_{j}^{n} :_{C} \frac{t^{n}}{n!} .$

Equivalently,

$: Q_{j}^{n} :_{C} = n! \sum_{s = 0}^{⌊ n / 2 ⌋} \frac{(- C_{j j} / 2)^{s} Q_{j}^{n - 2 s}}{s! (n - 2 s)!}, : Q_{j}^{4} :_{C} = Q_{j}^{4} - 6 C_{j j} Q_{j}^{2} + 3 C_{j j}^{2} .$

This is a finite-dimensional polynomial definition; it does not define a product of unsmeared continuum fields.

Facts & Assumptions

Given: Assume the Axiom of Choice, let $Q \sim N_{N} (0, C)$ , and fix a coordinate $j$ with $1 \leq j \leq N$ .

[A1]

The Axiom of Choice is explicit in the finite Gaussian-law construction. The Axiom of Choice

[F1]

The finite Gaussian vector has covariance $C$ , so $C_{j j} = E [Q_{j}^{2}]$ is a well-defined nonnegative variance. Finite Gaussian normalization and Wick moments

Proof

technique · formal power-series coefficient extraction

1.1A1F1givenalgebra

Multiply the formal series for $e^{t Q_{j}}$ and $e^{- C_{j j} t^{2} / 2}$ . The coefficient of $t^{n}$ is the finite sum over $0 \leq s \leq ⌊ n / 2 ⌋$ of $Q_{j}^{n - 2 s} (- C_{j j} / 2)^{s} / ((n - 2 s)! s!)$ . By the defining identity, multiplying this coefficient by $n!$ gives the displayed formula for $: Q_{j}^{n} :_{C}$ . This is formal coefficient algebra and requires no convergence assertion.

2.1step 1.1algebra∎

For $n = 4$ , the terms $s = 0, 1, 2$ are respectively $Q_{j}^{4}$ , $- 6 C_{j j} Q_{j}^{2}$ , and $3 C_{j j}^{2}$ , which gives the stated quartic Wick polynomial. The formula also covers $C_{j j} = 0$ .

Finite covariance Wick normal powers

Definition

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources