# Identical Particles in Quantum Field Theory

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?

• 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 $(j_1,j_2)$. Oct 21, 2020 at 9:35

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:
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 $$(8.1)$$ $$|\psi\rangle=\mathcal{P}|\psi\rangle$$