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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.

Qmechanic
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mavzolej
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2 Answers2

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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)
Chiral Anomaly
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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.

Michael_1812
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