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Let $H$ be some Hilbert space. Now in general, in quantum mechanics, the vector space representing states of $n$ (non-interacting) particles is $H^{\otimes n}$, but if I consider these particles of be fermions, the space becomes $\Lambda^n H$. So, given some wavefunctions $\psi_1,...,\psi_n$, if they represent fermions, the corresponding state in $\Lambda^n H$ is just $\psi_1 \wedge ... \wedge \psi_n$, which can then in turn be regarded as an element of $H^{\otimes n}$, using the embedding $\iota: \Lambda^n H \to H^{\otimes n}$.

Now $\iota (\psi_1 \wedge ... \wedge \psi_n) \neq \psi_1 \otimes ... \otimes \psi_n$, so somehow a particle is definitely more than just a wave function, but I guess it's also more than just wave functions together with a 'boson' and 'fermion' flag (because this seems just absurd).

I've heard that elementary particles are just irreducible representations, but is this the solution to this problem?

Also, the difference $\psi_1 \otimes ... \otimes \psi_n - \iota (\psi_1 \wedge ... \wedge \psi_n)$ is symmetric, so it seems to be some kind of boson. Does this have a meaning in physics?

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I am not particularly experienced in many-body Quantum Mechanics, but is it not customary to represent the Hilbert Space of a many-body system by the Fock Space, defined by: $$F_{\nu} = \bigoplus_{N=1}^{\mathcal{N}} S_{\pm}\mathcal{H}^{\otimes n}$$ Where $S_{\pm}$ is the appropriate symmetrization operator (which is the distinction between Fermions and Bosons). The wikipedia page seems very useful for this, as is this article.

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  • $\begingroup$ "is it not customary"? $\endgroup$
    – user940
    Commented Oct 30, 2015 at 17:32
  • $\begingroup$ You are correct, but in some sense, my question is why it is the Fock space. The multiplication in this space has a symmetrization/antisymmetrization, but how do the particles know this? $\endgroup$ Commented Oct 30, 2015 at 17:37

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