What does it mean to have a QFT that can not be encoded by an action. What is by far the most powerful approach of study in such a case. What is the best studied physical theory that falls into this category. What is QFT? I think it is a set of rules that facilitate descriptions of things we can observe, but what is the most mathematically accurate way of capturing what it is? I have read of cases where the path integral approach fails what does this mean? Is there any geometric structure whose properties describe the space of all posible QFT's? What happens when you can't use an action?


2 Answers 2


It is possible to abstract the notion of a QFT away from the notion of Lagrangians/Hamiltonians, one axiomatic way are the Wightman axioms. As one can see, they reduce the quantum theory to its very heart: A Hilbert space where the states live and a field operator that acts upon it, generating "particles", all of this happening in a Lorentz covariant way.

Concrete example of QFTs without an action are CFTs in 2 dimensions, they are almost completely fixed by demanding that they are a QFT in the axiomatic sense which has a Virasoro symmetry.

The "space of all possible QFTs" is quite a complicated (and open, I think) question, since it is quite difficult to prove that a given QFT is really a QFT in the sense of the Wightman (or Osterwalder-Schrader) axioms.

  • $\begingroup$ Can you list theories that can be constructed using the Wightman axioms? $\endgroup$
    – CGH
    May 20, 2016 at 18:31
  • $\begingroup$ @CGH: A standard reference would be "Quantum Physics" by Glimm and Jaffe. It contains the proof that all 2D field theories with polynomial interactions satisfy the (equivalent) Euclidean version of the Wightman axioms, the Osterwalder-Schrader axioms. (Admittedly, those are theories that possess an action.) $\endgroup$
    – ACuriousMind
    May 20, 2016 at 23:50
  • $\begingroup$ Although that's useful, it doesn't answer my question: which theories can be constructed using the Wightman axioms? The first paragraph of this answer gives the impression that you can do that, but as far as I can find, it hasn't been done. $\endgroup$
    – CGH
    May 21, 2016 at 19:15

A large class of QFTs called 2 dimensional conformal field theories, can be defined without an action.

The general method for constructing an any dimensional CFT is as follows, you can calculate the action of the symmetry generators of the fields using Ward identities and their commutation relations given by the Virasoro algebra which allows you to construct the whole hilbert space from the knowledge of how your set of "primary" fields transform under conformal transformations which is encoded in the conformal dimension. Why you can do this in CFTs but not in any QFTs is the state-operator correspondence. Each state of the Hilbert space is in 1-1 correspondence with the set of field operators. There is a subset of these called minimal models which are defined through a finite set of primary fields. One can even give them a Lagrangian formulation through the Couloumb gas formalism which consists of coupling it with the Ricci scalar.

What distinguished a d-dimensional CFT to a CFT in 2 dimensions is that the conformal group of a 2D CFT is infinite dimensional. Any locally holomorphic map corresponds to a local conformal transformation. This allows you to use Laurent expansions and other tools of complex analysis.

For "What is a QFT"? This page from nLab maybe useful. It somehow "axiomatizes the assignment of algebras of observables to patches of parameter space".

For further info. on CFT refer to di Francesco et Al's Conformal field theory.

  • $\begingroup$ The state-operator correspondence is a feature of CFT in general, not just two-dimensional CFT. Likewise, the use of primary fields to generate the algebra of observables via OPE works in any d-dimensional CFT. $\endgroup$
    – user1504
    Jun 28, 2014 at 22:06
  • $\begingroup$ @user1504: Thanks for leaving a comment. I will edit the answer accordingly. $\endgroup$
    – user7757
    Jun 29, 2014 at 4:08
  • $\begingroup$ I'm afraid your edits have made your answer slightly confusing. The Virasaro algebra, on the other hand, is peculiar to 2-dimensional CFT. In general CFTs, one defines primary fields by looking for eigenvectors of the dilation operator. $\endgroup$
    – user1504
    Jun 29, 2014 at 13:21
  • $\begingroup$ @user1504: Please can you give me the explicit state-operator correspondence for CFTs in other than 2 dimensions. Eigenvectors of dilation operation $L_0$ would give you states, not the primary fields corresponding to those states. I am not sure how you get the state-operator correspondence in more than 2 dimensions. $\endgroup$
    – user7757
    Jun 29, 2014 at 13:32
  • $\begingroup$ The nontrivial half of the state-operator correspondence is gotten by shrinking a sphere down to a point where the operator is defined. Apply state-operator correspondence to eigenvector and you get a field. $\endgroup$
    – user1504
    Jun 29, 2014 at 13:38

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