Commutation relations for an interacting scalar field

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eoghan
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Why are the commutation relations for interacting fields assumed to be the same as that of free fields?
Hi there,

In his book "Quantum field theory and the standard model", Schwartz assumes that the canonical commutation relations for a free scalar field also apply to interacting fields (page 79, section 7.1). As a justification he states:

This is a natural assumption, since at any given time the Hilbert space for the interacting theory is the same as that of a free theory.

I do not understand this explanation. Can you please elaborate?

I mean, how can the Hilbert space of the interacting theory be the same as the one of the free theory? In an interacting theory, I expect to have different states than in a non interacting theory, so the Hilbert space should be different, isn't it?
 
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eoghan said:
In an interacting theory, I expect to have different states than in a non interacting theory
Not necessarily, just a different Lagrangian. For example, consider a non-interacting theory of electrons and photons. It would have electron states and photon states, with a Lagrangian that had a kinetic term for electrons, a mass term for electrons, and a kinetic term for photons.

If we now add the electron-photon interaction to this theory, we add the QED coupling term to the Lagrangian, but we don't change any states: there are still electron states and photon states. All we have done is introduced new possibilities for transitions between states, such as scattering processes.

eoghan said:
the Hilbert space should be different
Not for the reason you give; but Haag's theorem and related results do, in fact, show that you cannot describe interacting quantum fields using the same Hilbert space that describes free quantum fields. Most treatments of QFT seem to more or less ignore these results and assume that there is some rigorous mathematical basis for the methods they describe.
 
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@PeterDonis Very nice answer, now it is clear to me why the Hilbert space (Haag's theorem apart) is supposed to be the same.

@vanhees71 Nice book. I just had a look at its content on Amazon and it's definitely on my list once I will be done with the Schwartz!
 
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