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In Quantum Field theory by M. Schwarz, the author in the introduction of chapter 12 on Spin Statistics theorem says, while describing identical Particles: Let $$|s_1p_1n_1,...,s_3p_3n_3\rangle \tag{1}$$ be some state consisting of $n_1$ particles with spin $s_1$, momentum $p_1$ and similarly $n_3$ particles with momentum $p_3$ and spin $s_3$ and many other particles.

Now in (1) we interchange particles labelled (1) and (3) so that state (1) above becomes:

$$|s_3p_3n_3,...,s_1p_1n_1 \rangle \tag{2}$$

The author then goes on to tell that the states described by (1) and (2) are equivalent upto a phase factor say $e^{i\alpha}$ i.e:

$$|s_1p_1n_1,...,s_3p_3n_3\rangle= e^{i\alpha}|s_3p_3n_3,...,s_1p_1n_1\rangle\tag{3}$$

Now, the author tells that the phase in above equation can only depend on number of particles and not on momentum or spin of particles as there are no non-trivial one-dimensional representations of the (proper) Lorentz group.

Now, I know that there are no nontrivial finite dimensional representation of Lorentz group but I am not able to connect the statement in bold to state (3) as how does this statement implies that phase factors can't depend on momentum or spin?

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    $\begingroup$ Note there are no non-trivial unitary finite-dimensional representations of the Lorentz group. There are many non-trivial non-unitary finite-dimensinal representations indexed by two half-integers $(j_1,j_2)$. $\endgroup$
    – Gaston Barboza
    Commented Oct 21, 2020 at 9:35

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The one dimensional quantifier comes because of the scalar $\alpha$ in the global phase. As pointed out by Gaston there are non trivial non-unitary finite representation but we won't look into them since we require unitary representation finite/infinite cause they're the only one where norm is preserved (Schwartz essentially takes previous chapters to explain this property only) and they represent particle. Now you know:

P1. there are no non-trivial one-dimensional representations of the (proper) Lorentz group. So I'll take from there

  1. $\alpha$ can't depend on one component of $p_{\mu}$ since it will contradict P1

  2. $\alpha$ can't depend on scalar constructed from $p_i$'s because of P1

  3. $\alpha$ can't depend on spin since expect for $0$ spin all other are infinite dimensional representation. And if you're counting spin-$0$ particles you're essentially counting number of particles.

Ultimately we have to count numbers of particle one way or the other. If you're still confused you need to take a look at $(8.1)$ $$|\psi\rangle=\mathcal{P}|\psi\rangle$$

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