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Let $\phi_1$ and $\phi_2$ be quantum field operators of certain spin on $\mathbb{R}^4$. Then, the principle of locality dictates that if $x$ and $y$ are space-like separated, we have \begin{equation} \phi_1(x) \phi_2(y)=\phi_2(y) \phi_1(x) \text{ if at least one of } \phi_1, \phi_2 \text{ is Bosonic} \end{equation} or \begin{equation} \phi_1(x) \phi_2(y)=-\phi_2(y) \phi_1(x) \text{ if both of } \phi_1, \phi_2 \text{ are Fermionic} \end{equation}

Now, I wonder how I can generalize this to more than two field operators. That is, for $n>2$, let $\phi_1, \cdots, \phi_n$ be quantum field operators. Then, exactly how should I state the local commutativity requirement for $\phi_1(x_1) \cdots \phi_n(x_n)$?

My guess is that I need to employ permutations $\sigma$ on $\{1, \cdots, n\}$ and consider the relation between \begin{equation} \phi_1(x_1) \cdots \phi_n(x_n) \end{equation} and \begin{equation} \phi_{\sigma(1)}(x_{\sigma(1)}) \cdots \phi_{\sigma(n)}(x_{\sigma(n)}) \end{equation} based on permutation of space-like separated Fermionic fields

However, the main difficulties I face are:

How to make sense of space-like separation among multiple field operators?

How to make sense of (or count the number of) interchanging Fermionic fields under the given permutation $\sigma$?

Could anyone please help me?

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  • $\begingroup$ Isn't the obvious generalization that $\phi(x_1)\dots\phi(x_n)$ equals any permutation of it when all the $x_i$ are pairwise space-like separated, i.e. $x_i - x_j$ is spacelike for all $i,j$? What do you need this generalization for? $\endgroup$
    – ACuriousMind ♦
    Commented Feb 17 at 20:32
  • $\begingroup$ @ACuriousMind Well, I just want to see what happens in the presence of multiple field operators. $\endgroup$
    – Keith
    Commented Feb 17 at 20:35
  • $\begingroup$ @ACuriousMind OK, if we suppose that all $x_i$'s are pairwise space-like separated, then when do we have the anticommutation relation? How do we make sense of interchanging Fermionic fields under the given permutation $\sigma$? $\endgroup$
    – Keith
    Commented Feb 17 at 20:36
  • $\begingroup$ Odd permutation produces $-1$ for fermions $\endgroup$
    – nwolijin
    Commented Feb 18 at 1:01
  • $\begingroup$ @nwolijin Of course. But what if permutation is over a mixture of fermions and bosons? What doed it mean by "odd" then? $\endgroup$
    – Keith
    Commented Feb 18 at 13:58

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