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Emphasis is on the irreducible. I get what's special about them. But is there some principle that I'm missing, that says it can only be irreducible representations? Or is it just 'more beautiful' and usually the first thing people tried?

Whenever I'm reading about some GUT ($SU(5)$, $SO(10)$, you name it) people usually consider some irreducible rep as a candidate field. Also, the SM Lagrangian is constructed in this way. (Here, experimental evidence of course suggests it.)

BeneIT
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    The Dirac spinor is a reducible representation of the Lorentz group. Also, the adjoint of $\mathrm{SO}(n)$ is not always irreducible (e.g. $\mathrm{SO}(4)$), so a gluon on that group would belong to a reducible representation. – MannyC Aug 23 '19 at 16:09

4 Answers4

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Gell-Mann's totalitarian principle provides one possible answer. If a physical system is invariant under a symmetry group $G$ then everything not forbidden by $G$-symmetry is compulsory! This means that interaction terms that treat irreducible parts of a reducible field representation differently are allowed and generically expected. This in turn means that we will instead reclassify/perceive any reducible field in terms of their irreducible constituents.

Qmechanic
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  • A really great answer! – andypea Aug 22 '19 at 22:13
  • Let me try to make an example. Say I have two fields that transform under representations of the same group. There is nothing to prevent me from writing down different mass terms for both fields as the two dont mix (And you say Gell-Manns's totalitarian principle should compel us to do so?). It would therefore not be particularly useful to write the two fields as a direct sum of some 'unified' field, forming a reducible representation. – BeneIT Aug 23 '19 at 11:53
  • However, I might digress and come to the conclusion (maybe supported by experimental evidence) that my model would be simpler if the masses were the same, in fact if I wouldn't write down any terms which distinguish the two fields. – BeneIT Aug 23 '19 at 11:53
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    @BeneIT In that case, your “simpler” model presumably has a larger symmetry group than you started with, and the reducible representation is in fact an irrep of the larger group. – Stephen Powell Aug 23 '19 at 16:58
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This is only semantics. A reducible representation $\mathbf R$ of the symmetry group can be decomposed into a direct sum $\mathbf R_1 \oplus \cdots \oplus \mathbf R_N$ of irreducible representations. A field that transforms as $\mathbf R$ is the same thing as $N$ fields, which transform as $\mathbf R_1, \dots, \mathbf R_N$. When talking about fundamental fields, we can therefore assume that they transform as irreducible representations of the symmetry group.

Noiralef
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    I think this answer is incorrect (or at least, incomplete). – Stephen Powell Aug 23 '19 at 07:58
  • Take an $n$-component massive field transforming as an irreducible representation of the group $\mathcal G$. Symmetry requires that the $n$ components should have the same mass, because $\mathcal G$ mixes the components. If I take two such fields $\varphi_1$ and $\varphi_2$ transforming under irreducible representations $\mathbf R_1$ and $\mathbf R_2$, then $\varphi_1\oplus\varphi_2$ transforms under the reducible representation $\mathbf R_1\oplus\mathbf R_2$. But there is no requirement that $\varphi_1$ and $\varphi_2$ have the same mass. – Stephen Powell Aug 23 '19 at 07:58
  • The reason is that all the matrices in the representation $\mathbf R_1\otimes\mathbf R_2$ are block diagonal, and so they cannot mix the components of $\varphi_1$ and $\varphi_2$. So terms that distinguish components of $\varphi_1$ and $\varphi_2$ are not forbidden by symmetry and, as @Qmechanic's answer says, will generically be nonzero. – Stephen Powell Aug 23 '19 at 08:08
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    @StephenPowell After seeing Qmechanic's answer (+1), I might agree that mine is not quite complete. I do not think that it is wrong though, and I don't quite understand your point. I agree with everything you wrote, but I do not see a contradiction to my answer. – Noiralef Aug 23 '19 at 11:28
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    There is nothing wrong with this answer. In fact, IMHO, it is the correct answer, and it is complete. I don't understand @StephenPowell's comment. Nowhere in this answer it is said that all the irreps of $\boldsymbol R$ have the same mass. – AccidentalFourierTransform Aug 23 '19 at 12:30
  • @AccidentalFourierTransform it is not correct, simply because not every representation is reducible to a sum of irreducible representations. I think physicists usually assume this is the case or don't understand that not every representation is completely reducible (in his QFT text Weinberg for example seems to use from the very beginning that every unitary representation of the poincaré group is completely reducible, which is just not true). A unitary representation of the poincaré group without any irreducible components would be a QFT without particles, and apparently examples do exist. – s.harp Aug 23 '19 at 12:59
  • Sorry, it's just the first sentence ("only semantics") that I consider to be incorrect, but I don't think this answer really addresses the question. Why is a QFT written in terms of fields transforming under irreducible representations? This answer only says that one can rewrite a field transforming under a reducible representation as multiple fields transforming under irreps. The reason one does is given in Qmechanic's answer and BeneIT's comments on it. – Stephen Powell Aug 23 '19 at 13:04
  • @s.harp huh? All the groups relevant to QFT are reductive, so any reducible rep is completely reducible. The exception is Poincaré, but the reps that are not completely reducible are highly pathological. They may be interesting models for mathematicians, but not for us physicists! So for the purpose of this question I believe it is justified (and necessary) to assume that "reducible" means "completely reducible". The non comp.red. reps are simply not used in 99.9% of QFT. – AccidentalFourierTransform Aug 23 '19 at 13:57
  • @StephenPowell The first sentence is literally the most correct part of this answer. The answer to the question in the OP is "just semantics". Anything else is noise. QFT is not always written in terms of irreps. When doing general analyses, one always writes the fields in terms of a single reducible rep. So it is factually false that "a QFT [is] written in terms of fields transforming under irreducible representations". Given that this fact is false, the only possible answer is "it doesn't matter: you can always convert from one notation to the other, and it is but a matter of convenience". – AccidentalFourierTransform Aug 23 '19 at 14:01
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    Note that, although non-reductive groups do have non-completely reducible representations (that's one definition of reductivity), this can't happen for unitary representations (of any group)—at least, not for finite-dimensional unitary representations. – LSpice Aug 23 '19 at 16:50
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Irreducible representations are always determined by some numbers, labeling the representation, which correspond to the eigenvalues of some observables which are invariant under the (unitary) action of the Lie group.

If the group represents physical transformations connecting different reference frames (Lorentz, Poincare',...), these numbers are therefore viewed as observables which do not depend on the reference frame so that they define some intrinsic property of the elementary physical system one is considering.

If the group represents gauge transformations, these numbers correspond to quantities which are gauge invariant. In this sense they are physical quantities.

Finally, it turns out that in many cases (always if the Lie group is compact), generic unitary representations are constructed as direct sums of irreducible representations. This mathematical fact reflects the physical idea that physical objects are made of elementary physical objects (described by irreducible representations)

Valter Moretti
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The premise of the question is just false. When doing phenomenology it is useful to split a field into its irreducible components, essentially because each irrep carries its own coupling constant. But when analysing QFT from a theoretical point of view, it is convenient to consider a single "big" field in a reducible representation. So it is just not true that in QFT fields are irreducible: sometimes they are not.

For the representations that are relevant to conventional QFT, all reducible representations are completely reducible, so thinking of a single reducible rep, or a collection of individual irreps, is nothing but a matter of convenience: both descriptions carry the exact same information.

Take for example the beta function of Yang-Mills plus matter. The first coefficient is of the form $$ b_0\sim C_2(G)-T(R) $$ where $T(R)$ is the index of the representation for the matter fields. If $R$ is reducible, $R=R_1\oplus R_2\oplus\cdots\oplus R_n$, one has $T(R)=T(R_1)+T(R_2)+\cdots+T(R_n)$. Therefore, if there are $N_F$ copies of a certain irrep, $R=R_1^{\oplus N_F}$, one would write $$ b_0\sim C_2(G)-N_F T(R_1) $$ which is the formula one often finds in textbooks. Both formulas are identical, and one may or may not want to explicitly split $R$ into its irreps. The general case is the same: one can think of a single field in a rep $R$, or a collection of fields into the irreps of $R$. Both conventions are valid, and sometimes one is more useful than the other. But it is emphatically wrong to claim that all fields are irreducible.

AccidentalFourierTransform
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