This is a follow-up question to my question about the uniqueness of the field-momentum operator. The answer suggested that (partially) because the operators can act on different hilbert-spaces, there isn't a choice for the field operator and the field-momentum that is unique up to unitary equivalent choices. However, since the answer stated one of the problems was that those operators can lie in different hilbert-spaces, I was wondering:

When I restrict the pairs of field and field-momentum to those cases where the field-operators are unitary equivalent, does that mean that the operators act on the same hilbert-space, and following, that the field-momentum-operators are as well connected by the same unitary-transformation?

The train of thought here is as follows: Given two pairs of field and momentum, and requiering the fields to be connected by a unitary transformation, I place a condition to both pairs to act on the same hilbert-space. If the problem of having different hilbert spaces was the reason to not have uniqueness up to unitary-transformations, then this uniqueness is given now.


The problem is a rather fundamental one in fact, and it cannot be overcome so easily, and it is not per se about having different Hilbert spaces, but rather inequivalent representations.

All separable, infinite dimensional, Hilbert spaces are isomorphic (they are thus the same mathematical structure essentially). Nonetheless, they accommodate infinitely many not equivalent representations of the canonical commutation relations.

Specifying an Hilbert space is, therefore, not very useful for QFT: one should rather fix the correct representation of the canonical commutation (or anticommutation) relations, relevant for the theory.

For free theories, the right representation is the Fock representation, induced by the so-called Fock vacuum, i.e. a Poincaré invariant state whose noncommutative Fourier transform (generating functional) is Gaussian. For interacting theories, the right representation is usually quite difficult to find, or yet unknown. Nonetheless, it should be associated to a suitable Poincaré-invariant vacuum state.

The very celebrated Haag's theorem links vacuum state representations and unitary inequivalence: different vacua (more generally, different Poincaré invariant pure states) always yield inequivalent representations of the canonical relations. Therefore, free and interacting theories, whose vacua are clearly expected to be both Poincaré invariant and different, must yield inequivalent representations.

  • $\begingroup$ Can I ask what a pure state means in that context? Since this isn't quantum-statistics, I don't see how states can be "mixed" - what defines a pure state? Besides - do I understand that my thought is not sufficient, because unitary equivalence of the field operators is not sufficient to have the same poincare invariant pure state? $\endgroup$ – Quantumwhisp Oct 17 at 8:08

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