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In this question I asked wether the definition of the momentum operator as an operator that has to generate translations by satisfying the canonical commutation relations was ambiguous. The answer to that was that if I require the canonical commutation relations to hold in an exponentiated way, then the Stone-Von-Neumann-Theorem states that the momentum operator for one choosen position operator is unique (and that all pairs of momentum and position operators are connected via a unitary transformation). Correct me if I'm wrong.

In quantum field theory however, the Stone-Von-Neumann-Theorem doesn't work anymore - Does that also mean that now given one field-operator, there are different choices for the field-momentum-operator? (I already strongly presume the answer is "yes").

In case the field-momentum is in fact not uniquely defined anymore - how is the field-momentum usually chosen? Does fixing an operator ordering in that case fix the field-momentum as well, for example?

Qmechanic
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Quantumwhisp
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1 Answers1

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The C*-algebra of canonical commutation or anticommutation relations is always unique (up to isomorphisms), once the underlying symplectic structure is fixed. This is a result due to Slawny.

This means that, roughly speaking, for given mass and spin the abstract structure of fields and momenta is unique.

What is not unique, if the symplectic structure is infinite dimensional (i.e. for fields), are the irreducible representations of such commutation relations. This means that, depending on wether the physical system is interacting or free, there are different fields and momenta. In addition, these different representations "cannot coexist": without being too technical, the free and interacting representations must be set in different Hilbert spaces (in the sense that there is no C*-algebraic isomorphism relating the fields of the two spaces).

Therefore, to define a relativstic quantum theory, one should define the correct representation of the canonical relations, and that amounts to choosing the correct vacuum state for the system. Such choice is often very difficult, and very few rigorous examples of interacting theories are known.

Let me conclude with a very technical remark. In the bosonic/qm case (commutation relations), the C*-algebra describes the exponential of the quantum variables (field/position and momentum). This is also called the Heisenberg Lie group. And results such as the stone-von neumann theorem only hold for the representation of the group. In fact, also in quantum mechanics there can be representations of the Heisenberg Lie algebra (the commutation relations that most people know) that are inequivalent to the ones given by the usual position and momentum operators, and that in addition when exponentiated do not yield the Heisenberg group relations (also called Weyl relations).

yuggib
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  • Am I understanding right that choosing the vacuum state for a theory fixes the --> representation <-- of the ccr?

    Besides (I'm not firm in the topic of c*-algebras, all I know of are operators, that a representation maps group elements to operators acting on a hilbert space) - what is the meaning of a commutation relation on an algebra?

     – Quantumwhisp Sep 19 '19 at 16:59
  • Yes, choosing the vacuum state fixes the representation. The set of operators satisfying the commutation relations, in exponentiated form for bosons, satisfies certain properties that qualifies it as a C-algebra. C-algebras are typically collections of operators satisfying suitable properties, defined abstractly. – yuggib Sep 19 '19 at 18:37
  • I guess choosing the vakuum state fixes the representation only up to unitary equivalent representations - is that true? And does this statement hold in general, or only for skalar-fields (I'm asking because for example haags-theorem does only make statements on scalar-fields as well). – Quantumwhisp Oct 01 '19 at 19:37
  • @Quantumwhisp Of course there may still be unitarily equivalent representations to a given one. However, fixing an invariant state fixes the representation (in the sense that such representation is then not equivalent to the one corresponding to any other invariant pure state). This is true for all relativistic theory, not only for the scalar one. An Haag's like statement is proved in general, and in particular it has been shown e.g. that free Dirac fermions with different masses yield indquivalent representations. – yuggib Oct 02 '19 at 06:31
  • The invariant state you are talking about is the vacuum state that has been specified, right? – Quantumwhisp Oct 02 '19 at 07:07
  • Usually, it is. But the requirement for Haag's theorem is for it to be a Poincaré invariant pure state. – yuggib Oct 02 '19 at 07:10