Solution 1:

A possible answer is 'an irreducible unitary representation of the Poincar\'e group for which the eigenspaces of the generators of translation are finite-dimensional'. (At least, this is the space of states of a single particle of a single type).

This isn't a perfect answer, because not every QFT is relativistic (nor indeed does every QFT have particles), but hopefully it's useful nonetheless.


A bit more reasoning why: Translations are mutually commuting, so pick a basis where they're all diagonal. This labels states by their momentum and energy. You can do this for any state, but a single particle state is distinguished by having only finitely many remaining degrees of freedom. This is why the 'finite-dimensional' is in there. The 'irreducible' is there because if I can't rotate/boost from one set of states into another, there is no clear sense in which they are the same: they're different sorts of particle.

[Note that this includes bound states (like a hydrogen atom ground state) on the same footing, as single particles in their own right. This is sensible because there need not be any clear definition of what an `elementary particle' is: this concept makes reference to some free limit, which needn't exist].

Why symmetries require unitary representations:

  • States of a quantum system live in projective Hilbert space
  • A symmetry is an invertible map between states that preserves the modulus of inner products, which are important because of their relation to measurable quantities
  • Wigner's theorem: any symmetry can be lifted to a unitary or antiunitary operator on Hilbert space
  • As a consequence, a group of symmetries will have a (projective) representation on Hilbert space. For the connected part of a Lie group, it'll be unitary (the identity is unitary!)

Solution 2:

In Dirac's Principles of Quantum Mechanics (3E), a particle at a point x is represented by a specific linear operator $\eta_x$ applied to a ground state $|0\rangle$. Thus, if the system is in state $\eta_x|0\rangle$, there is a particle at x; if it is in state $\eta^{-1}_x|0\rangle$, there is an antiparticle at x; and if the system is in the state $\eta_y\eta_x|0\rangle$, there are particles at x and y.

Edit: This really is standard. Quantum field theory views particles a a perturbation of a ground state, meaning they act as linear operators on a fixed state. The links in the comments merely show that localized partickes canot be detected experimentally. But this does not change the mathematical framework; in fact, it is explained in the framework by saying that the vacuum is not stable, so that the states near vacuum have many small fluctuations, which is interpreted as saying that a vast number of particles are created and destroyed.

Many people confuse pseudoscientific philosophy with actual physics. In real physics, you find an idealized model that can be tested experimentally and you ignore questions about "what is reality actually like?"

So in the mathematical model, particles are described as operators, and this yields good experimental results.