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This question is essentially asking for a clarification on what has already been said in this one. What I don't understand is why it is the representations that are important in Quantum Field Theory rather than just the group itself. Do the representations add extra structure that the group didn't possess? If so how do we know that we want that structure in our theory? (if there is a reason other than empirical observation)

Or to state the question differently: Is there a way to write Quantum Field Theories with certain symmetries in a representation free way? (e.g. the Standard Model).

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    $\begingroup$ What use would a group without a representation be? Any time a group acts in a "nice" way on a vector, you get a representation. No representation means no group actions, and then of what relevance is the group? I don't understand the question. $\endgroup$ – ACuriousMind Apr 1 '16 at 20:26
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    $\begingroup$ I thought the same thing at one point, thinking that a representation was just like a choice of basis, and that everything could be reformulated in a basis or representation-independent way. But this is not true, even in quantum mechanics: the different representations of $SU(2)$ give you particles with different spins! The common group structure just tells you that you're dealing with rotation; you need the representation to tell you the rest. $\endgroup$ – knzhou Apr 1 '16 at 20:27
  • $\begingroup$ @knzhou I think this almost answers my question already. May I ask what exactly it is that the representations add in structural sense? As in how do the different spins come from the group? (if that makes any sense) $\endgroup$ – Wolpertinger Apr 1 '16 at 20:32
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    $\begingroup$ Well, you have no idea what elements of your group $G$ do without taking a representation. For example, you might think some element $g$ represents, say, a 45 degree rotation about the $z$ axis. But to say that, you've implicitly chosen the vector representation, i.e. this is how $g$ acts on $v$ if $v \in \mathbb{R}^3$ and is in the vector representation. Without doing this, $g$ has no physical meaning at all. $\endgroup$ – knzhou Apr 1 '16 at 20:38
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    $\begingroup$ Part of the confusion comes from the fact that Lie groups are often presented as matrix groups, i.e. $SU(n)$ is written in terms of $n \times n$ unitary matrices, and so on. This is not "the group" in the most abstract sense; it is a particular representation of the group, called the fundamental representation. $\endgroup$ – knzhou Apr 1 '16 at 20:42
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Quantum Field theory is not a study of groups, but a study of physical states and observables. The groups are interesting only because they act on the objects of actual interest. The representation theory is just the study of such actions; if there is no representation of a group on interesting objects, the group itself is not interesting for QFT.

Thus the question one could ask is rather why we study groups and not only the representations. The answer would be that the abstract notion of the group incapsulates the simplest structure present in a representation, and then the representation is viewed as an extra structure on top of the group and the space of interesting objects.

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Representations represent group (algebra) elements as linear operators on a vector space, in physics the vector spaces are also usually Hilbert and the operators are often orthogonal or unitary (self-adjoint), so they do indeed introduce quite a bit of extra structure. The simplest piece to see is that the representation does not have to be faithful, i.e. the image of the group in the group of operators does not have to be isomorphic to the group itself. In descriptive terms, if we think of representation as describing symmetries of a physical system, only part (or better to say reduced version) of allowed symmetries may be present. But even if representations are faithful, and their images are isomorphic to the group, they may still implement symmetries differently. For example, one representation may be reducible, operators move vectors from some proper subspaces within them, and the other have no such subspaces, be irreducible (this is also described as having a cyclic vector that "generates" entire space). Breaking up reducible representations into irreducible ones can then be used to break it up into simpler components (often called superselection sectors), whose behavior is easier to analyze.

In infinite dimensional systems it is also possible that representations are not unitarily equivalent, there is no unitary map between their vector spaces that corresponds their operators accordingly. Since unitary equivalence reflects physical equivalence this means that different representations correspond to physically distinct systems. Wallace discusses this issue in detail in his Lagrangian QFT paper (Sec. 4), and gives a simple example:"To see what these sectors are, suppose we start with all components having spin up. Then the action of any element of the algebra can, at most, cause finitely many components to have spin down. So no amount of algebraic action can transform such a state into one in which, say, every second component has spin up. This state, in turn, can be transformed into other states differing from it in finitely many places, but not into a state in which all components are spin down..." In QFT itself "infrared"-inequivalent representations may reflect different mass-densities at infinity, different total charges, or inequivalent vacua for the corresponding systems of quantum fields. So the answer to the boldfaced question is no, differences in representations reflect differences in physics.

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