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In view of Haag's Theorem, it seems the Hilbert spaces of a free theory and an interacting theory are not the same. Though it seems very believable, I could not find a result that states that this is the case for all theories.

Intuitively, it seems obvious that the Hilbert space of an interacting theory involving distinct fields is not the same as any of the Hilbert spaces of the individual free theories. However, what if we consider the scalar field with $\phi^4$ interactions? In this case I was unable to convince myself that the Hilbert space of the interacting theory must be different than the Hilbert space of the free $\phi$ field.

Is it always the case that Hilbert space of an interacting theory is different than any free theory?

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    $\begingroup$ Have you read the proof (gymarkiv.sdu.dk/MFM/kdvs/mfm%2030-39/mfm-31-5.pdf)? It applies to any theory of a single neutral scalar which means $\phi(x)$ and its conjugate momentum must transform irreducibly under translation and rotations. $\endgroup$
    – Connor Behan
    Mar 27 at 18:02
  • $\begingroup$ Without any offense to theorists... but has anybody actually observed a free field? I am not aware that such a thing exists in this universe, except on paper. Is this theorem mathematics' hint that we shouldn't be hunting for the snark, even on paper? $\endgroup$
    – FlatterMann
    Mar 27 at 18:18
  • $\begingroup$ @ConnorBehan Thank you! I will take a look at that paper. Sorry about my ignorance, but what is the consequence of $\phi$ and its conjugate momentum having to transform irreducibly under translation and rotations? $\endgroup$
    – CBBAM
    Mar 27 at 18:19
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    $\begingroup$ Irreducibility is what ensures that two unitarily equivalent theories will have generators of translations and rotations that are also unitarily equivalent. This is needed to produce the contradiction of free and interacting Wightman functions being equal if we assume the Hilbert spaces are the same. $\endgroup$
    – Connor Behan
    Mar 27 at 19:03
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    $\begingroup$ I gave a brief description of the Hilbert space in this other answer physics.stackexchange.com/questions/633835/… in that point of view Haag's theorem has to do with the measures not being absolutely continuous with respect to each other. A baby version of this issue is what happens when taking infinite products of measures en.wikipedia.org/wiki/Kakutani%27s_theorem_(measure_theory) $\endgroup$
    – Abdelmalek Abdesselam
    Mar 28 at 14:51

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Yes, the Hilbert space for a free theory and an interacting theory are different. There can not exist a unitary isomorphism that preserves the algebra of operators. (You need the caveat about preserving the algebra of operators, because all separable Hilbert spaces are unitarily isomorphic in many ways.)

In fact, Haag's theorem implies something stronger: any change in any one of the coupling constants gives rise to a different Hilbert space. So, not only do free & interacting theories get different Hilbert spaces, but also you get different Hilbert spaces for different values of the coupling constants $g \neq g'$. Even crazier, this is true of the quadratic coupling constants, aka, the mass terms. Different masses lead to different state spaces, even in free theories.

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