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For the classification of particles (Wigner 1939), we look for unitary representations of the Poincaré/Lorentz group. There are only infinite-dimensional (non-trivial) unitary representations!

To construct these, we focus on the "little group" that leaves the momentum fixed and find its finite-dim unitary reps (classified by the two quantum numbers $m$ and $j$). These reps each depend on the momentum $p^\mu$, so on the whole we get the infinite-dim unitary reps in the form of fields in momentum-space.

So if we already have fields that are unitary reps of the Poincaré group, why do we still have to embed them in different reps such as scalar $\phi(x^\mu)$, vector $A^\mu(x^\mu)$or tensor $T^{\mu\nu}(x^\mu)$ fields? Why can we not just use the unitary reps we found?

Gold
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quantumorsch
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  • I guess I was wrong and they are unitary - corrected that in my question. But why do we not just use the reps we found earlier? What is wrong to use them? To be more specific: why do we not use the $m>0$,$j=1, p^\mu$ rep instead of the fourvector field $A^\mu(p^\mu)$ ? – quantumorsch Feb 12 '15 at 16:41

1 Answers1

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Wigner's classification of the particle representations is important, but not the only thing needed for a (quantum) field theory. In particular, you cannot expect the fields to transform in one of Wigner's representations:

A classical field $\phi$ transforming under any group $G$ is given as a section of an $G$-equivariant vector bundle over the spacetime $\Sigma$, or, equivalently, a $G$-equivariant map $\phi : \Sigma \to V_\rho$ where $V_\rho$ is some representation space of $G$ with representation $\rho$ and $\phi(\Lambda x) = \rho(\Lambda)\phi(x)$.

The definition of the field taking values in a vector space restricts it to transform in a finite-dimensional representation, hence it cannot be one of Wigner's particles. It is important that, while fields contain the creation and annihilation operators for the particles in their mode expansion, they themselves do not transform like particles. It is the Hilbert space of a QFT that must carry the proper unitary representations, not the fields.

We need a field because it encodes the dynamics of the theory - a QFT needs a map between in and out states, given by the S-matrix, which is obtained from the field action via the path integral (or the LSZ formalism or whatever approach you are most comfortable with). The mere knowledge of the Fock spaces (via Wigner's classification) does not suffice for this.

A free theory essentially gives the in/out time evolution map as the identity - states just stay the same, they don't interact at all. In this sense, you can give a free theory just by specifying the Fock spaces. You might be interesting in axiomatic formulations of QFT, e.g. the Wightman axioms, where we explicitly start from the unitary Lorentz reps + the dynamics encoded in the quantum fields, and it is explicitly demanded that the transformation of the field as an operator on the unitary reps is exactly given by the equivariant transformation $\phi(\Lambda x) = \rho(\Lambda)\phi(x) = U(\Lambda)\phi(x)U(\Lambda)^\dagger$. Even if you give the dynamics as free/trivial, you still need the field to encode them.

ACuriousMind
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  • OK, thank you! That explains why we can use non-unitary scalar/vector/tensor reps as our fields, but not why we need to. Why not just forget about the (classical) fields and work with the reps Wigner found? – quantumorsch Feb 12 '15 at 18:28
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    @quantumorsch: How would you? QFT usually proceeds by quantizing a classical field theory, so you have these fields in your theory, if you want to or not. The knowledge that particle states will transform in some unitary rep of the Lorentz group is not enough to fix the theory in any sense. You need the notion of the field to be abe to calculate stuff like the propagator - you don't get that from the reps alone. – ACuriousMind Feb 12 '15 at 18:32
  • I see, but suppose we had no knowledge of classical fields and wanted to do particle physics "from scratch". Wigner showed us what types of irreducible reps there are and now we want to construct an interacting theory out of them. Why can we not proceed using those states (elements of the representation space)? Why embed them into new mathematical objects/fields that transform non-unitarily? A $j=1$ state transforms nicely as a triplet - why use the components of a four-vector and ruin the nice transformation property? – quantumorsch Feb 12 '15 at 18:48
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    @quantumorsch: A QFT is given by the in/out Hilbert spaces and the maps between them (this map is the S-matrix/partition function/path integral, essentially). Wigner only tells you what the Hilbert spaces are, he tells you nothing about the S-matrix. Representation theory alone is, except for CFTs in two dimensions, not enough to constrain the partition functions, they still need to be externally given, and the usual way to give them in normal QFT is as the path integral of a classical field action. – ACuriousMind Feb 12 '15 at 18:55
  • So a free particle theory would be possible to build via my approach? The problem lies in the interactions (S-matrix)? I thought the embedding of Wigner reps into fields is already necessary for free particles. – quantumorsch Feb 12 '15 at 20:11
  • @quantumorsch: See my edit. – ACuriousMind Feb 12 '15 at 20:26