# Definition of single-particle states in the free theory

I like defining single-particle states as simultaneous eigenstates of generators of the Poincare group (basically, the representations of the Poincare group). This is the most fundamental definition we have in relativistic QFTs, see Weinberg around (2.5.1). However, this definition does not seem to be particularly useful in the free theory.

Indeed, consider the state $$|\mathbf{p}_1,\mathbf{p}_2\rangle=a^\dagger_\mathbf{p_1}a^\dagger_\mathbf{p_2}|0\rangle$$. While intuitively it is obviously a two-particle state, it does obey all the properties of a single-particle state, as it has definite energy and momentum.

What am I missing? What should I adjust in my definition so that $$|\mathbf{p}_1,\mathbf{p}_2\rangle$$ would not be a single-particle state? Would $$|\mathbf{p}_1,\mathbf{p}_2\rangle$$ not be a pole of the free propagator? If so, how do I relate this to the definition in terms of the Poincare group?

Also, see this and this closely related questions.

Note that an analogous problem would not arise in the interacting QFT, where the consecutive application of operators, creating single-particle states, to vacuum does not result in creating Hamiltonian eigenstates.

• I am by no means an expert, but $|p_1,p_2\rangle$ does not transform under an irreducible unitary representation of the Poincare group, no? But afaik, this is the defining property of a single-particle state of an elementary particle; cf. this PSE post. Nov 6, 2021 at 21:49
• @Jakob, I think you are right, and Weinberg derives this from the first definition around (2.5.3). Nov 7, 2021 at 4:49

First, I'll clarify what Weinberg is saying in reference 1. He's not using equation (2.5.1) as the definition of single-particle states. Notice the wording:

We now consider the classification of one-particle states according to their transformation under the inhomogeneous Lorentz group. ...it is natural to express physical state-vectors in terms of eigenvectors of the [total] four-momentum. Introducing a label $$\sigma$$ to denote all other degrees of freedom, we thus consider state-vectors $$\Psi_{p,\sigma}$$ with $$P^\mu\Psi_{p,\sigma}=p^\mu\Psi_{p,\sigma}. \tag{2.5.1}$$ For general states, consisting for instance of several unbound particles, the label $$\sigma$$ [in equation (2.5.1)] would have to be allowed to include continuous as well as discrete labels. We take as part of the definition of a one-particle state, that the label $$\sigma$$ is purely discrete...

The first sentence after the equation says that multi-particle states can also be classified using (2.5.1), so we can infer that he's not using (2.5.1) as the definition of single-particle states. Instead, he says that single-particle states are (at least partly) characterized by the requirement that the label $$\sigma$$ be discrete.

So, to answer your question, we just need to show that the label $$\sigma$$ is continuous (not discrete) for the states described in the question. By definition, $$\sigma$$ includes whatever extra information we would need to specify to describe the state, beyond the total momentum $$\mathbf{p}$$. For the states shown in the question, we also need to specify the relative momentum $$\mathbf{p}_1-\mathbf{p}_2$$, because only then do we have enough information to recover the individual momenta $$\mathbf{p}_1$$ and $$\mathbf{p}_2$$. The relative momentum is continuously variable, so $$\sigma$$ is continuous, which violates Weinberg's condition for single-particle states.

Reference:

1. Section 2.5 in Weinberg (1995), The Quantum Theory of Fields, Volume 1 (Cambridge University Press)
• Excellent, thank you!! This answers my question. I was reading those lines in Weinberg, but only understood them after your clarification. Nov 6, 2021 at 23:07
• please see the answer by michael_1812 below stating that single-particle states are the eingestates of $\hat{N}$ with eigenvalue $1$. Could you please comment on the equivalence of that definition, the one from the original question (and your answer), and the one using the propagator's pole? Nov 7, 2021 at 4:53
• @mavzolej The definitions are not equivalent. Weinberg's definition can be applied in any QFT, because we always have the energy-momentum operators (the generators of translations in time and space). The definition using poles is also general, as long as we remember to consider two-point functions of more complicated operators, too, not just the original field operators (consider QCD). In contrast, how would we define a "number operator" in general, without already knowing which states are $n$-particle states? It may seem clear in a free theory, but a good definition should work in any theory. Nov 7, 2021 at 14:42
• @mavzolej Regarding the relationship Weinberg's definition in section 2.5 and the pole definition, I'll refer you to section 10.2 ("polology") in Weinberg's book. Roughly, the idea is that a pole in a two-point function comes from a discrete part in the spectrum of the mass operator (within the subspace explored by the operators whose two-point function is being considered), which in turn is expressed in terms of the energy-momentum operators. Nov 7, 2021 at 14:50

Consider the operator of particle numbers $$\hat{N}=\int \hat{a}^{\dagger}_{{\bf p}} \hat{a}_{{\bf p}} d{{\bf p}}\;\;.$$ Commuting with the operators of energy and momentum, it also satisfies $$\hat{N} |{{\bf p}}\rangle = |{{\bf p}}\rangle\;\;,\quad\hat{N} |{{\bf p}}_1,\,{{\bf p}}_2\rangle = 2 |{{\bf p}}_1,\,{{\bf p}}_2\rangle\;\;.$$ This criterion is by itself sufficient to distinguish a one-particle state from multiparticle ones.