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If the fields in QFT are representations of the Poincare group (or generally speaking the symmetry group of interest), then I think it's a straight forward consequence that the matrix elements and therefore the observables, are also invariant.

What about the converse:

If I want the matrix elements of my field theory to be invariant scalars, how do I show that this implies that my fields must be corresponding representations?

How does this relate to S-matrix theory?

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Retitled to "If the S-matrix has symmetry G, must the fields be representations of G?", nice question. – Ron Maimon Apr 18 '12 at 10:47

2 Answers 2

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The question formulation (v2) seems to mix the notions of invariant and covariant, which essentially is also the main point of user1504's answer (v1).

Let's say we have a group $G$. The group $G$ could e.g. be a finite group or a Lie group. When we say that a theory is invariant under $G$, it normally implies at least two things.

  1. The group $G$ acts on the theory. This in particular means that there is a well-defined given prescription on how the constituents of the theory change under the action of the group. Often in physics (but not always), it happens that the group action is linearly realized, i.e. the fields, the matrix elements, and other objects form linear representations of the group $G$. At this stage, the representations could be reducible or irreducible, finite dimensional or infinite dimensional. The corresponding object then behaves covariantly (not necessarily invariantly) under the action of the group. If a representation is completely reducible, we can decompose it in irreps.

    • In case of an off-shell formulation: The action $S$ is off-shell invariant under the group $G$. Or phrased equivalently, the action $S$ form a trivial representation of the group.
    • In case of an on-shell formulation: The equations of motion behave covariantly under the group $G$.

A refinement. Ron Maimon makes in a comment an important point that if there is a hierarchy of say two theories, and if the group action is a priori only defined in the smaller theory, then the group $G$ does not necessarily have to have a well-defined action on the larger theory. For instance,

    • Small theory = on-shell formulation;
    • Large theory = off-shell formulation.
    • Small theory = minimal $S$-matrix formulation;
    • Large theory = an underlying field theoretic formulation.
    • Small theory = formulation on gauge-invariant physical subspace/submanifold/phase space;
    • Large theory = formulation on BRST-extended space/manifold/phase space.
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+1 for nonlinear, but I don't think the question was asking about linear vs. nonlinear realization--- I think it was asking whether you could have a symmetry of the S-matrix (of observables) which is not an obvious symmetry of the fields at one spacetime point. I don't see why not, at least classically, ignoring renormalization--- what if the symmetry is completely non-obvious on the fields? – Ron Maimon Apr 18 '12 at 10:50
Do you know what a counterexample does look like? I'd guess somebody would try these things out. – NikolajK Apr 18 '12 at 12:12
@NickKidman: I was trying to make a counterexamle just now--- the idea is to use a super-nonlocal Lagrangian, where the symmetry would necessarily mix up fields at far away points, like a nonlinear symmetry realized in momentum space. This mixes spacetime and internal symmetry, and so violates at least the spirit of Coleman-Mandula, so it should have a continuous spectrum at 0 momentum, or extended objects, or failed analyticity (because no locality) or something else which violates Coleman Mandula assumptions. – Ron Maimon Apr 18 '12 at 13:03

Fields aren't "representations of the Poincare group". Fields -- some fields, anyways, like scalar and Dirac spinor fields (gauge fields are more complicated) -- take values in vector spaces which are representations of the Lorentz group. This does not guarantee that observables are 'invariant'; observables can and do change their values when you act by a symmetry. Spin up can change to spin down if you rotate your coordinate system on its head.

This sort of confusion is what comes of the sloppy notational habits of the theoretical physics community. Learn you some representation theory, for great justice.

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Okay, okay thanks for the remark on the first point, even if I think it doesn't influence the question greatly. I was under the impression that the whole field is a representation - even if infinite. Second, regarding the "Spin up" example, are we still talking about amplitudes here, i.e. matrix elements? – NikolajK Apr 18 '12 at 7:10
This is not a great answer--- the field at one point is a rep of the Lorentz group, but the field values at all point are representations of Poincare group--- the translations move the points around. – Ron Maimon Apr 18 '12 at 10:49
@Nick. Spin up & down get used to label the components of the Dirac spinors. These components do show up in matrix elements. – user1504 Apr 18 '12 at 12:01
@user1504: If the components of the spinors are not equal to some brakets/matrix elements, then they are no observables, right? – NikolajK Apr 18 '12 at 12:09
@Nick: Consider the matrix element of a single Dirac spinor between the vacuum and a one particle state. – user1504 Apr 18 '12 at 12:16

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