In the free field theory, we an decompose the field with creation and annihilation operators $a^{\dagger}_k$ and $a_k$. $a_k$ acts on some state $|0\rangle$ and outputs $0$. We call that state the vacuum state—a state with no particles. $a^{\dagger}_k$ in contrast acts on the vacuum stays and creates a particle. We can use these operators to build the Hilbert space.

When we turn interactions on, from what I’ve read, the interacting field no longer has a pretty decomposition into plane waves scaled by creation and annihilating operators. So it seems to me that in the interacting theory we can’t even talk about particle states. Yet, we talk about the interacting vacuum state $|\Omega \rangle$. I have two questions:

  1. How can the interacting vacuum state even be defined if we do not have creation/annihilation operators for the interacting field? I’ve seen it defined as $H|\Omega \rangle = 0$ where $H$ is the full Hamiltonian, but then without having creation/annihilating operators, how can we even have such a relation (no particle destruction operators in the Hamiltonian)?

  2. Similarly, how can we talk about interacting particle states? Particle states in the free theory are created by acting on the vacuum state by the creation operator, but again, there is no creation operator for the interacting field.

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    $\begingroup$ One thing to say is that interacting theories are not understood formally very well (eg no formally defined examples known in 4d). A way to rigorously answer this question can be done in a lattice discretisation though, there the Hilbert space is manifestly finite dimensional, with well defined Hamiltonian operator, and the states are unambiguously defined. Then you can unambiguously make statements like Haags theorem (interacting and free vacuum overlap goes to zero as lattice spacing goes away) $\endgroup$ Apr 14 at 0:17
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    $\begingroup$ Just use $\ell_2(\mathbb N)$. $\endgroup$
    – Daron
    Apr 14 at 12:31
  • $\begingroup$ Issues with justifying the interaction picture and rigorously constructing the Hilbert space are very real. But what's wrong with $H \left | \Omega \right > = 0$? Whether theories are interacting or free, the bare Hamiltonian produces a different eigenvalue on the lowest energy state and we make it zero by a renormalization condition. $\endgroup$ Apr 18 at 11:25
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    $\begingroup$ It wouldn't mean that at all. Plenty of operators besides "annihilation operators" annihilate the vacuum. Charges for symmetries (which aren't spontaneously broken) are an example. $\endgroup$ Apr 19 at 13:06

2 Answers 2


The description of the complete Hilbert space of one interacting theory is a very non-trivial problem and it is not known in four spacetime dimensions, although as far as I know there are results in fewer dimensions. The investigation of the proper rigorous construction of such Hilbert space is the subject of the so-called "constructive QFT" and is an active area of research.

Nevertheless, that is not to say that we know nothing about such Hilbert space. In fact, we know two things which answer your two questions:

  1. Whatever this Hilbert space is it must carry one unitary representation of the universal cover of the Poincaré group. This is simply because this is the basic requirement of relativisic symmetry. In particular this means that the Poincaré generators $P_\mu$ and $M_{\mu\nu}$ will be realized by Hermitian operators in such Hilbert space.

    In that regard, the vacuum is simply defined to be a Poincaré invariant state. In particular this implies it must be annihilated by the Poincaré generators, so that $P_\mu |\Omega\rangle=M_{\mu\nu}|\Omega\rangle=0$.

  2. Whatever this Hilbert space ${\cal H}$ is we assume it has a subspace of in/out scattering states ${\cal H}_{\rm in/out}\subset{\cal H}$. Such subspaces are identified through the use of Moller operators to the corresponding free theory Hilbert spaces (see e.g. this answer of mine for a short description). The characteristic property of such states is that one such state, when observed in the in/out region will be found to be a state described by a certain collection of free particles. In that regard, particles are defined asymptotically: they enter spacetime as free particles, exit as free particles, and we can find the probability amplitudes for one in configuration to evolve to an out configuration, but we never really describe what is going on during the interaction.


I asked this as a graduate student to Tom Kibble at Imperial. At that time I knew of exactly integrable 1+1 dimension field theories. Even then to go from the interacting basis back to the noninteracting basis is non trivial. If you do conventional perturbation theories they do not converge to the exact solutions as the exact results were non perturbative. A challenge to Renormalization theorists. Nevertheless in many cases the quasi particle theories do work as in metals and semiconductors.

  • $\begingroup$ Do you have a source that talks in depth about interacting basis? Or elaborate in your answer with how they’re described? $\endgroup$
    – Obama2020
    Apr 19 at 12:56
  • $\begingroup$ Look up Bill Sutherland in non diffractive scattering and 1+1 d field theories. Also stuff like the Thirring model. Because there are just so few exactly integrable field theories such basis are very very rare. They all go by the name of Bethe Ansatz. Not much use even if you have got them because the spectrum is about all you can calculate from them. $\endgroup$
    – Victor
    Apr 20 at 17:17

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