quantum field theory - Are there any non-physical eigenstates of interacting QFT?

While reading about the Källén-Lehmann representation I came across the definition of eigenstates in general QFT. As $\vec{p}$ (total momentum) and $H$ commute they can be simultaneously diagonalized, thus one obtains:

$$H | {\lambda_{\vec{p}}} \rangle = E_{\vec{p}} |{\lambda_{\vec{p}}} \rangle$$ $$\vec{P} |{\lambda_{\vec{p}}} \rangle = \vec{p} |{\lambda_{\vec{p}}} \rangle$$ Given a ket $|{\lambda_{\vec{0}}} \rangle$ one can go to all the kets $|{\lambda_{\vec{p}}} \rangle$ by a Loretz transformation. We can then partition the set of all possible eignestates grouping those which are related by a Lorentz transformation and giving them the symbol $\lambda$ (as I have actually already done). We would expect physical states to have an invariant mass, so there should exist $m$ such that $m^2 =E_{\vec{p}}^2 - \vec{p}^2$, but it doesn't seem obvious to me that all the eigenstates admit such a relation between eigenvalues. Do I have to sum up only on the physical states (if there is any that is non-physical) in the completeness relation? $$\mathbb{1} = | \Omega \rangle \langle \Omega | + \sum_{\lambda}\frac{d^3 p}{(2 \pi)^3} \frac{1}{2 E_{\vec{p}}} |\lambda_{\vec{p}} \rangle \langle \lambda_{\vec{p}} |$$ Thus effectively defining the Hilbert space as the one generated by physical states?

Answer

Typically, yes: there are non-physical states in interacting QFT's.

The completeness relation of an arbitrary QFT is

$$1=\sum_{n\in\mathrm{all}}|n\rangle\langle n|$$ where $n$ includes both physical and unphysical states. If the sum included only physical states, you could prove the (wrong) conclusion that the two-point function of a gauge field is gauge-invariant, which is obviously false.

On the other hand, the sum over only physical states,

$$P\equiv\sum_{n\in\mathrm{physical}}|n\rangle\langle n|$$ is the projector onto the physical part of the Hilbert space. This projector is sometimes used to define the physical $S$-matrix, $$S_\mathrm{physical}\equiv P^\dagger S_\mathrm{naive} P$$ where $S_\mathrm{naive}$ is the $S$-matrix that you would calculate if you included unphysical states in the asymptotic Hilbert space.

---

In any case, recall that $m^2\equiv E^2-\vec p^2$ is the definition of $m$, and so it holds for any state, physical or not.