Physlib.QFT.PerturbationTheory.FieldStatistics.OfFinset

Given a mapping $q: \mathcal{F} \to \text{FieldStatistic}$ (which assigns either a bosonic or fermionic statistic to each field) and a sequence of fields defined by $f: \{0, \dots, n-1\} \to \mathcal{F}$, this function calculates the collective field statistic for a finite subset of indices $a \subseteq \{0, \dots, n-1\}$. The resulting statistic is $\text{fermionic}$ if the number of indices $i \in a$ such that $q(f(i)) = \text{fermionic}$ is odd; otherwise, the result is $\text{bosonic}$.

For any mapping $q: \mathcal{F} \to \text{FieldStatistic}$ that assigns a statistic to each field and any sequence of fields $f: \{0, \dots, n-1\} \to \mathcal{F}$, the collective field statistic of the empty set of indices $\emptyset \subseteq \{0, \dots, n-1\}$ is $\text{bosonic}$ (which is the identity element $1$ of the field statistic group).

For a collection of fields $\mathcal{F}$, let $q: \mathcal{F} \to \text{FieldStatistic}$ be a mapping that assigns a quantum statistic (bosonic or fermionic) to each field. Given a sequence of $n$ fields $f: \{0, \dots, n-1\} \to \mathcal{F}$, the collective field statistic of a singleton set of indices $\{i\}$ (where $i < n$) is equal to the statistic of the field at that index, $q(f(i))$.

Let $q: \mathcal{F} \to \text{FieldStatistic}$ be a mapping that assigns a quantum statistic (either $\text{bosonic}$ or $\text{fermionic}$) to each field in a collection $\mathcal{F}$. Let $f: \{0, \dots, n-1\} \to \mathcal{F}$ be a sequence of $n$ fields. Given an injective function $i: \{0, \dots, m-1\} \to \{0, \dots, n-1\}$ and a finite set of indices $a \subseteq \{0, \dots, m-1\}$, the collective field statistic of the fields $(f \circ i)$ indexed by $a$ is equal to the collective field statistic of the fields $f$ indexed by the image set $i(a) \subseteq \{0, \dots, n-1\}$: $$\text{ofFinset}(q, f \circ i, a) = \text{ofFinset}(q, f, i(a))$$ This reflects the fact that the collective statistic depends only on the set of fields selected, regardless of how they are indexed.

Let $q: \mathcal{F} \to \text{FieldStatistic}$ be a mapping that assigns a quantum statistic (either $\text{bosonic}$ or $\text{fermionic}$) to each field in a collection $\mathcal{F}$. Given a list of fields $\phi_s$ and a finite set of indices $a \subseteq \{0, \dots, \text{length}(\phi_s) - 1\}$, let $i$ be an index such that $i \notin a$. Then the collective field statistic of the set of fields indexed by $a \cup \{i\}$ is equal to the product of the statistic of the $i$-th field and the collective statistic of the set indexed by $a$: $$\text{ofFinset}(q, \phi_s, a \cup \{i\}) = q(\phi_s[i]) \cdot \text{ofFinset}(q, \phi_s, a)$$ where the multiplication $\cdot$ is the operation in the commutative group $\text{FieldStatistic}$ (where $\text{bosonic}$ is the identity and $\text{fermionic} \cdot \text{fermionic} = \text{bosonic}$).

Let $q: \mathcal{F} \to \text{FieldStatistic}$ be a mapping that assigns a quantum statistic (either $\text{bosonic}$ or $\text{fermionic}$) to each field in a collection $\mathcal{F}$. Given a list of fields $\phi_s$ and a finite set of indices $a \subseteq \{0, \dots, \text{length}(\phi_s) - 1\}$, let $i$ be an index such that $i \in a$. Then the collective field statistic of the set of fields indexed by $a \setminus \{i\}$ is equal to the product of the statistic of the $i$-th field and the collective statistic of the set indexed by $a$: $$\text{ofFinset}(q, \phi_s, a \setminus \{i\}) = q(\phi_s[i]) \cdot \text{ofFinset}(q, \phi_s, a)$$ where the multiplication $\cdot$ is the operation in the commutative group $\text{FieldStatistic}$ (where $\text{bosonic}$ is the identity and $\text{fermionic} \cdot \text{fermionic} = \text{bosonic}$).

Let $q: \mathcal{F} \to \text{FieldStatistic}$ be a mapping that assigns a statistic (either $\text{bosonic}$ or $\text{fermionic}$) to each field in a collection $\mathcal{F}$. For any list of fields $\phi_s$ and any finite subset of indices $a \subseteq \{0, \dots, \text{length}(\phi_s) - 1\}$, the collective field statistic of the indices in $a$ is equal to the product over all indices $i$ of the list $\phi_s$ of $q(\phi_s[i])$ if $i \in a$, and the $\text{bosonic}$ statistic (the identity $1$) otherwise. Mathematically: $$\text{ofFinset } q \text{ } \phi_s \text{ } a = \prod_{i=0}^{\text{length}(\phi_s)-1} \begin{cases} q(\phi_s[i]) & \text{if } i \in a \\ \text{bosonic} & \text{otherwise} \end{cases}$$ where the product is taken in the commutative group $\text{FieldStatistic} \cong \mathbb{Z}_2$.

Let $q: \mathcal{F} \to \text{FieldStatistic}$ be a mapping that assigns a statistic (either $\text{bosonic}$ or $\text{fermionic}$) to each field in a collection $\mathcal{F}$. For any list of fields $\phi_s$ and any two finite subsets of indices $a, b \subseteq \{0, \dots, \text{length}(\phi_s) - 1\}$, the product of the collective field statistics of the indices in $a$ and $b$ is equal to the collective field statistic of the indices in their symmetric difference $(a \cup b) \setminus (a \cap b)$. Mathematically: $$\text{ofFinset } q \text{ } \phi_s \text{ } a \cdot \text{ofFinset } q \text{ } \phi_s \text{ } b = \text{ofFinset } q \text{ } \phi_s \text{ } ((a \cup b) \setminus (a \cap b))$$ where the product is taken in the commutative group $\text{FieldStatistic} \cong \mathbb{Z}_2$.

Let $q: \mathcal{F} \to \text{FieldStatistic}$ be a mapping that assigns a statistic (either $\text{bosonic}$ or $\text{fermionic}$) to each field in a collection $\mathcal{F}$. For any list of fields $\phi_s$ and any two disjoint finite subsets of indices $a, b \subseteq \{0, \dots, \text{length}(\phi_s) - 1\}$, the product of the collective field statistics of the indices in $a$ and $b$ is equal to the collective field statistic of the indices in their union $a \cup b$. Mathematically: $$\text{ofFinset } q \text{ } \phi_s \text{ } a \cdot \text{ofFinset } q \text{ } \phi_s \text{ } b = \text{ofFinset } q \text{ } \phi_s \text{ } (a \cup b)$$ where the product is taken in the commutative group $\text{FieldStatistic} \cong \mathbb{Z}_2$, where $\text{bosonic}$ acts as the identity.

Let $q: \mathcal{F} \to \text{FieldStatistic}$ be a mapping that assigns a quantum field statistic (either bosonic or fermionic) to each field in a collection $\mathcal{F}$. For any list of fields $\phi_s$, any finite subset of indices $a \subseteq \{0, \dots, \text{length}(\phi_s) - 1\}$, and any decidable predicate $p$ on those indices, the product of the collective field statistics of the indices in $a$ that satisfy $p$ and those that do not satisfy $p$ is equal to the collective field statistic of all indices in $a$. Mathematically: $$\text{ofFinset } q \text{ } \phi_s \text{ } \{i \in a \mid p(i)\} \cdot \text{ofFinset } q \text{ } \phi_s \text{ } \{i \in a \mid \neg p(i)\} = \text{ofFinset } q \text{ } \phi_s \text{ } a$$ where the product is taken in the commutative group $\text{FieldStatistic} \cong \mathbb{Z}_2$, in which $\text{bosonic}$ serves as the identity element.

Let $q: \mathcal{F} \to \text{FieldStatistic}$ be a mapping that assigns a quantum field statistic (either bosonic or fermionic) to each field in a collection $\mathcal{F}$. For any list of fields $\phi_s$, any finite subset of indices $a \subseteq \{0, \dots, \text{length}(\phi_s) - 1\}$, and any decidable predicate $p$ on those indices, the collective field statistic of the indices in $a$ that satisfy $p$ is equal to the product of the collective field statistic of the indices that do not satisfy $p$ and the collective field statistic of all indices in $a$. Mathematically: $$\text{ofFinset } q \text{ } \phi_s \text{ } \{i \in a \mid p(i)\} = \text{ofFinset } q \text{ } \phi_s \text{ } \{i \in a \mid \neg p(i)\} \cdot \text{ofFinset } q \text{ } \phi_s \text{ } a$$ where the product is taken in the commutative group $\text{FieldStatistic} \cong \mathbb{Z}_2$, in which $\text{bosonic}$ is the identity element and every element is its own inverse.