Physlib.QFT.PerturbationTheory.WickContraction.ExtractEquiv

Let $\mathcal{W}_n$ be the set of Wick contractions of $n$ elements. For any contraction $c \in \mathcal{W}_n$, let $U(c)$ denote the set of its uncontracted indices. Let $c_1$ and $c_2$ be elements of the dependent sum $\sum_{c \in \mathcal{W}_n} \text{Option}(U(c))$, which represents pairs $(c, o)$ where $c$ is a contraction and $o$ is an optional uncontracted index (either an element of $U(c)$ or a "none" value). If the Wick contractions are equal ($c_1.1 = c_2.1$) and the optional indices are equal ($c_1.2 = c_2.2$, identifying the sets $U(c_1.1)$ and $U(c_2.1)$), then the pairs are equal ($c_1 = c_2$).

For a fixed index $i \in \{0, \dots, n\}$, let $\mathcal{W}_n$ be the set of Wick contractions on $n$ indices, and for any $c \in \mathcal{W}_n$, let $U(c)$ denote the set of its uncontracted indices. This definition establishes an equivalence (a bijection) between the set of Wick contractions on $n+1$ indices and the dependent sum of optional uncontracted indices over $\mathcal{W}_n$: \[ \mathcal{W}_{n+1} \cong \sum_{c \in \mathcal{W}_n} \text{Option}(U(c)) \] where $\text{Option}(U(c))$ represents the set $U(c) \cup \{\text{none}\}$. The mapping is defined as follows: - Given a contraction on $n+1$ indices, the forward map erases the index $i$. If $i$ was paired with some index $k$, it returns the resulting contraction on $n$ indices and the "some $k$" value (appropriately relabeled). If $i$ was uncontracted, it returns the contraction and "none". - The inverse map takes a contraction $c \in \mathcal{W}_n$ and an optional index $j$. If $j = \text{none}$, it inserts $i$ as an uncontracted index. If $j = \text{some } j'$, it inserts $i$ and pairs it with the uncontracted index $j'$.

Let $\mathcal{W}_n$ be the set of Wick contractions on $n$ indices, and for any $c \in \mathcal{W}_n$, let $U(c)$ denote the set of its uncontracted indices. Let $\phi_i : \mathcal{W}_{n+1} \cong \sum_{c \in \mathcal{W}_n} \text{Option}(U(c))$ be the equivalence defined by extracting the index $i \in \{0, \dots, n\}$. For a contraction $c \in \mathcal{W}_n$, let $c' = \phi_i^{-1}(c, \text{none})$ be the Wick contraction on $n+1$ indices formed by inserting $i$ as an uncontracted index. Then the set of uncontracted indices of $c'$ is \[ U(c') = \{i\} \cup \{ f_i(j) \mid j \in U(c) \} \] where $f_i: \{0, \dots, n-1\} \to \{0, \dots, n\}$ is the increasing map that skips $i$ (mapping $j$ to $j$ if $j < i$ and to $j+1$ if $j \geq i$).

Let $n, m$ be natural numbers such that $n = m$, and let $k$ be an index such that $k < n+1$ and $k < m+1$. Let $\mathcal{W}_n$ be the set of Wick contractions on $n$ indices, and for any $c \in \mathcal{W}_n$, let $U(c)$ denote the set of its uncontracted indices. Let $\phi_k : \mathcal{W}_{n+1} \cong \sum_{c \in \mathcal{W}_n} \text{Option}(U(c))$ be the equivalence (bijection) established by extracting index $k$, where $\text{Option}(U(c)) = U(c) \cup \{\text{none}\}$. For any Wick contraction $c \in \mathcal{W}_n$ and any uncontracted index $i \in U(c)$, the first component of the pair resulting from applying $\phi_k$ to the contraction $\phi_k^{-1}(c, \text{some } i)$ is equal to $c$.

The set of Wick contractions on zero elements, denoted as $\text{WickContraction}(0)$, is a finite set. This set contains exactly one element, which is the empty contraction.

Let $M$ be an additive commutative monoid. For any function $f : \text{WickContraction}(0) \to M$, the sum over all Wick contractions on zero elements is given by $\sum_{c \in \text{WickContraction}(0)} f(c) = f(\text{empty})$, where $\text{empty}$ is the unique element in the set $\text{WickContraction}(0)$.

For any natural number $n \in \mathbb{N}$, the set of Wick contractions on $n$ indices, denoted as $\text{WickContraction}(n)$, is a finite set. For the case $n+1$, the finiteness is established using an induction step based on the equivalence: \[ \text{WickContraction}(n+1) \cong \sum_{c \in \text{WickContraction}(n)} \text{Option}(U(c)) \] where $U(c)$ is the set of uncontracted indices in a contraction $c$, and $\text{Option}(U(c))$ represents $U(c) \cup \{\text{none}\}$. This equivalence effectively accounts for the possible pairings (or lack thereof) of the $(n+1)$-th index with the indices in a contraction of size $n$.

Let $M$ be an additive commutative monoid, and let $n, m$ be natural numbers such that $n = m + 1$. For a fixed index $i \in \{0, \dots, n-1\}$, let $\mathcal{W}_n$ denote the set of Wick contractions on $n$ indices. For any function $f : \mathcal{W}_n \to M$, the sum of $f$ over all Wick contractions is given by: \[ \sum_{c \in \mathcal{W}_n} f(c) = \sum_{c' \in \mathcal{W}_m} \sum_{k \in \text{Option}(U(c'))} f(\text{insert}_i(c', k)) \] where $U(c')$ is the set of uncontracted indices of a contraction $c' \in \mathcal{W}_m$, and $\text{Option}(U(c'))$ represents the set $U(c') \cup \{\text{none}\}$. The term $\text{insert}_i(c', k)$ refers to the Wick contraction of size $n$ reconstructed by inserting index $i$ into $c'$ and either leaving it uncontracted (if $k = \text{none}$) or pairing it with index $k$ (if $k \in U(c')$).

For any Wick contraction $c$ of $n = 3$ fields, the set of contracted pairs in $c$ is an element of the set $\{\emptyset, \{\{0, 1\}\}, \{\{0, 2\}\}, \{\{1, 2\}\}\}$, where $\emptyset$ represents the case with no fields contracted and the other elements represent the possible single-pair contractions between the field positions $0, 1,$ and $2$.

For any Wick contraction $c$ of $n = 4$ elements (indices $\{0, 1, 2, 3\}$), the set of its contracted pairs is an element of the following set of 10 possible configurations: $\{ \emptyset, \{\{0, 1\}\}, \{\{0, 2\}\}, \{\{0, 3\}\}, \{\{1, 2\}\}, \{\{1, 3\}\}, \{\{2, 3\}\}, \{\{0, 1\}, \{2, 3\}\}, \{\{0, 2\}, \{1, 3\}\}, \{\{0, 3\}, \{1, 2\}\} \}$. This set represents all possible ways to partition the four indices into zero, one, or two disjoint pairs, corresponding to the cases where no fields are contracted, two fields are contracted, or all four fields are contracted in pairs.