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?


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.


The "correct" Hilbert space of a quantum field theory (i.e. the one that takes into account only the physical degrees of freedom) is actually defined by a single distinguished state (of the W*-algebra of quantum observables).

For theories in Minkowski spacetime, such special state is the unique vacuum or ground state $\omega$, that has the property to be left invariant by the action of the restricted Poincaré group. In fact, the natural Hilbert space of the theory is exactly the Hilbert obtained via the GNS construction of an irreducible representation of the algebra of observables, obtained starting from $\omega$.

For theories in a curved background spacetime, it is not possible to define a vacuum state invariant under the action of the Poincaré group, and neither there is in general a unique preferred state as in flat spacetime. However there is a class of states that play the same role as the vacuum, the so-called Hadamard states (see e.g. this overview). Again, the "correct" Hilbert space of the theory is the one given by the GNS construction built on an Hadamard state $\omega_{\mathrm{H}}$, provided the latter exists.

  • $\begingroup$ I don't really see the link between the definition of ground state as the state of minimum energy and the definition as the only one invariant for the action of the Poincarè group. I have done some search but I couldn't find some reference accessible for a novice in QFT. Do you have some? Thanks. $\endgroup$ – MrRobot Aug 31 '17 at 18:51
  • $\begingroup$ @MrRobot The Hamiltonian is the generator of time translations of the Poincaré group, so the ground state(s) are left invariant by the action of time translations. Among the ground state(s), if there are more than one, you have to choose the one invariant for the other transformations of the Poincaré group as well. It is possible to prove that, at least in free theories, the ground state of the Hamiltonian (the so-called Fock vacuum) is unique, and furthermore it is indeed the unique state of the system invariant for the whole Poincaré group. $\endgroup$ – yuggib Sep 1 '17 at 9:11

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