I am trying to find a quantum field theory (a Yang-Mills theory) for the identical particles exchange interaction. For a system of $N$ identical particles one has the state $|x_1,x_2,\ldots,x_N\rangle$ that is invariant under permutations of states. The permutation operator $P$ is a discrete, not a continuous symmetry group. But is there a way turning the permutation operator $P$ into a continuous operator?

In a quantum field theory I have continuum states, i.e. the $\psi(x)$ (for fermions) function. Due to this fact I cannot define a proper permutation operator. However if I compute the $N$-point function (with $M$ ingoing and $N-M$ outgoing states)

$$\langle\psi_1 \psi_2 \ldots \psi^\dagger_{N-1}\psi^\dagger_N\rangle:=\int ~\mathrm d[\psi]\int ~\mathrm d[\psi^\dagger]e^{iS}(\psi_1 \psi_2 ... \psi^\dagger_{N-1}\psi^\dagger_N)$$

I could impose permutation symmetry in the functions $\psi_1,...\psi^\dagger_N$. If $N \mapsto \infty$ then I can define a function $P(x)$ (a local symmetry group) that fixes permuation symmetry on every point of spacetime.

How I can define a generalized permutation operator in continuum field?

Edit: I am trying to assume some additional symmetries in the quantum field vacuum, i.e. two fields of equal wave structure but on different spacetime points.

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    $\begingroup$ It's not clear to me what this question is asking for. What is "a quantum field theory for the identical particles exchange interaction"? The very essence of QFT is that fields, not a fixed number of particles are the fundamental dynamical objects, and how is "exchange of identical particles" an interaction? The quantum field $\psi(x)$ isn't a "continuum state", it's an operator. Please make it more clear what you are trying to do and take care to use established terminology properly. $\endgroup$
    – ACuriousMind
    Sep 27, 2016 at 19:22
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    $\begingroup$ I think this question could potentially be interesting, but I can't understand what you're trying to ask, right now. I hope you can edit it (again) to make it more clear! $\endgroup$
    – Danu
    Sep 28, 2016 at 16:44

1 Answer 1


Permutations in infinite dimensions are diffeomorphisms (assuming that you are considering only smooth configurations). An invariant theory under permutations, thus goes to a diffeomorphism invariant theory in the infinite dimension limit (such as gravity). If you want to include fermions, then the appropriate limit would be a diffeomorphism invariant theory over supermanifolds.


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