Physlib.QFT.PerturbationTheory.WickContraction.IsFull

A Wick contraction $c$ is defined to be full if the set of its uncontracted indices, $c.\text{uncontracted}$, is the empty set $\emptyset$. This means that every index in the contraction is part of a contracted pair and no field remains uncontracted.

For any Wick contraction $c$, the property of being a full contraction is decidable. A contraction $c$ is considered full if its set of uncontracted indices is empty, denoted as $c.\text{uncontracted} = \emptyset$. This decidability is established by checking the equality of the finite set $c.\text{uncontracted}$ with the empty set $\emptyset$.

A Wick contraction $c$ on $n$ indices is full if and only if its associated involution $f: \{0, 1, \dots, n-1\} \to \{0, 1, \dots, n-1\}$ has no fixed points, which is to say that $f(i) \neq i$ for every index $i$. In this context, a contraction is full when the set of its uncontracted indices is empty.

Given a natural number $n$, there is a bijective correspondence (equivalence) between the set of full Wick contractions on $n$ indices and the set of fixed-point free involutions on the set $\{0, 1, \dots, n-1\}$. A Wick contraction $c$ is considered **full** if every index is part of a contracted pair, meaning the set of uncontracted indices $c.\text{uncontracted}$ is empty. A **fixed-point free involution** is a function $f: \{0, 1, \dots, n-1\} \to \{0, 1, \dots, n-1\}$ such that $f(f(i)) = i$ and $f(i) \neq i$ for all indices $i$.

If $n$ is an even natural number, then the number of full Wick contractions on $n$ indices is $(n - 1)!!$, where $!!$ denotes the double factorial. A Wick contraction is considered full if every index is part of a contracted pair, meaning the set of uncontracted indices is empty.

If $n$ is an odd natural number, then the number of full Wick contractions on $n$ indices is $0$. A Wick contraction is considered full if every index is part of a contracted pair, such that the set of uncontracted indices is empty.