The no-go results from Algebraic and Constructive QFT you mention deal with related but slightly different matters.
(Edit: the previous version of the following paragraph was slightly misleading - Haag's theorem is actually stronger than I stated before; see below for details)
- Haag's theorem (which actually slightly predates the inception of Algebraic QFT) tells us that we cannot write interaction picture dynamics within Hilbert spaces which are free field representations of the CCR's. This is not the same as to say that interacting dynamics does not exist at all - it simply says that we cannot implement it as unitary operators in the interaction picture. This is done by showing that the possibility to do so in some Hilbert space does imply that we are dealing with a free field representation of the CCR's. The argument is closed by a "soft triviality" result by Jost, Schroer and Pohlmeyer arguing that the latter implies that all truncated $n$-point functions vanish for $n>2$, hence the field is really free - in particular, the "interaction Hamiltonian" is zero.
This has consequences for both scattering theory and attempts to rigorously construct field theoretical models starting from free fields. In the first case, Haag's theorem is circumvented by either the LSZ of Haag-Ruelle scattering formalisms, which obtain the S-matrix by respectively taking infinite time limits in the weak (matrix elements) and strong (Hilbert space vectors) sense. Recall that both setups require the assumption of a mass gap in the joint energy-momentum spectrum (i.e. an isolated, non-zero mass shell), otherwise we run into the notorious "infrared catastrophe", which is dealt with using "non-recoil" (i.e. Bloch-Nordsieck) approximation methods in formal perturbation theory but remains a challenge in a more rigorous setting, save in some non-relativistic models. In the second case, one is led to consider representations of the CCR's which are inequivalent to free field ones. Since field theories living in the whole space-time have infinite degrees of freedom, the Stone-von Neumann uniqueness theorem no longer holds (actually, Haag's theorem can be seen as a manifestation of this particular failure mechanism), and hence such representations should exist in abundance. Motivated by these results, Algebraic QFT was devised with a focus on structural (i.e. "model-independent") aspects of QFT in a way that does not depend on a particular representation; on other front, one may also try to explore this abundance of representations to construct models rigorously, which brings us to the realm of Constructive QFT.
- The "existence" (a.k.a "non-triviality") and "non-existence" (a.k.a. "triviality") results in Constructive QFT tell us which interactions survive after non-perturbative renormalization. More precisely, you construct field theoretical models in a mathematically rigorous way by first considering "truncated" interacting theories (i.e. with UV and IR cutoffs), and then carefully removing the cutoffs in a sequence of controlled operations. The resulting model may be interacting (i.e. "non-trivial") or not (i.e. "trivial"), in the sense that its truncated $n$-point correlation functions for $n>2$ may be respectively non-vanishing or not. In the first case, any representation of the CCR's in the Hilbert space where the interacting vacuum state vector lives is necessarily inequivalent to a free field one - in particular, one cannot write the interacting dynamics as unitary operators in the interaction picture, in accordance with Haag's theorem. In the second case, you really obtain a free field representation of the CCR's, but here because renormalization has completely killed the interaction.
Finally, it is important to notice that triviality of a model may stem from reasons unrelated to the underlying mechanism of Haag's theorem. The latter, once more, is a consequence of having an infinite number of degrees of freedom in infinite volumes (this theorem does not hold "in a box", for instance), whereas the former usually derives from an interaction which has too singular a short-distance behavior, as argued in the previous paragraph. This can be intuitively be understood by the (local) singularity and (global) integrability of the free field's Green functions: the lower the space-time dimension, the better the singular (UV) behaviour and the worse the integrability (IR) behaviour, and vice-versa. That's the underlying reason why $\lambda\phi^4$ scalar models are super-renormalizable in 2 and 3 dimensions (having only tadpole Feynman graphs as divergent in 2 dimensions) and non-perturbatively trivial in $>4$ dimensions.
Ah, I've almost forgotten about the references: in my opinion, the best discussion of triviality results in QFT from a rigorous viewpoint is the book by R. Fernández, J. Fröhlich and A. D. Sokal, "Random Walks, Critical Phenomena, and Triviality in Quantum Field Theory" (Springer-Verlag, 1992), specially Chapter 13. There both the above "hard triviality" results for $\lambda\phi^4$ models and "soft triviality results" such as the Jost-Schroer-Pohlmeyer theorem (which underlies Haag's theorem, as mentioned at the beginning of my answer) are discussed. The book is not exactly for the faint of the heart, but the first sections of this Chapter provide a good discussion of the statements of the theorems, before proceeding to the proofs of the above "hard triviality" results. For a detailed discussion of Jost-Schroer-Pohlmeyer's and Haag's theorems, as well as their proofs, I recommend the book of J. T. Lopuszanski, "An Introduction to Symmetry and Supersymmetry in Quantum Field Theory" (World Scientific, 1991). The classic book of R. F. Streater and A. S. Wightman, "PCT, Spin and Statistics, and All That" (Princeton Univ. Press) also discusses these two results.
I think that the problematic part here is the notion of what demands what. For example, you state
This is the standard argument by which Lorentz invariance is found to demand gauge invariance for massless particles.
but I am not completely sure if I agree with this, or at least to the interpretation you are carrying with it.
Taking a look at Weinberg's book on QFT[1], Chapter 5.9, I see a different way of looking at this issue. Weinberg notices that there is no consistent way of constructing an operator for the spin-1 massless field. The equivalent to your statement is that if we are to forcefully construct that operator, it has to be equivalent to itself in addition to a derivative of some other operator.
Now, in my interpretation (and I don't think that at the time of this reply there is an unanimous and unambiguous interpretation of this) is that the statement of gauge invariance is that of a single massless spin-1 field is only defined up to a sum of a divergence, and the important part is that this is a statement which is true even in the absence of matter coupled to the field.
My understanding of your question leads me to think that you are putting at the same footing two distinct notions:
- The intrinsic gauge invariance of a single massless spin-1 field, which is required to make sense of the corresponding operator.
- The gauge invariance of an interacting theory with matter coupled to a massless spin-1 field.
I point out this distinction because point 2. is the one that forces you to put a collection of massless spin-1 fields in the Adjoint representation, such that the kinetic term of matter with covariant derivatives is invariant under a gauge transformation of matter fields, whereas point 1. tells you that each of the massless spin-1 fields has an intrinsic gauge invariance where each is identical to itself plus the addition of a divergence.
I then propose you an interpretation that would solve this issue, and it goes like this:
- a) We know that in order to make sense of spin-1 massless fields, their operators have to enjoy a gauge invariance of the form
$$\hat A_\mu \to \hat A_\mu + \partial_\mu \hat \Omega $$
as such we allow for the classical counterpart to enjoy this freedom from the first place (point 1.).
- b) Due to the previous point, we notice that theories of local $U(1)$ symmetries of the matter fields
$$ \psi \to e^{i \theta(x)} \psi $$
coupled to matter
$$\mathcal L = i \overline \psi \gamma \cdot \partial \psi + \overline \psi \gamma \cdot A \psi $$
are immediately invariant as the new terms arising from the derivative terms, the divergence of the phase $\theta$, can be gauged away into the field $A_\mu$.
- c) For non-abelian theories, while it's feasible possible to construct them without being interacting with matter, their motivation comes from empirical evidence where they interact with matter. In such formulations we know that the theory is invariant if there is a collection of massless spin-1 fields that transform under the adjoint transformation (this is point 2.).
- d) Furthermore, because of the presence of derivatives, there will still be a divergence of a lie-algebra valued quantity (the phases) which can be gauged away by the collection of the massless spin-1 fields due to their intrinsic gauge invariance (point 1.).
In some sense you can read point c) as being what some people related closely to the global (a lot of grains of salt in this term) part of the gauge transformation, while d) the local part. In the end the theory works because the requirement for an interacting gauge theory plays along nicely with the quantum mechanical requirements for writing down the corresponding massless spin-1 fields!
Would this help you? I have thought about similar issues (I wrote, for example, [2] even though it's not direclty related to your issue) and I've come to believe that the interpretation I constructed and provided above solves a lot of conceptual problems in gauge theories (namely the simple Lie group type).
So to answer your questions explicitly:
- Each field operator (meaning each of the $a$, using your index notation) in the non-abelian cause transforms just as the abelian case under Lorentz transformations, that is with an inhomogeneous derivative term.
- The collection of fields (all the $a$) transforms in accordance with Coleman-Mandula as it transforms under the adjoint representation, plus each $a$ has the "freedom" to be defined up to a derivative. This freedom plays nicely in presence of matter where derivatives of phases appear, and in absence of matter can be used to write each operator in a convenient gauge.
References
[1] Quantum Theory of Fields - Vol 1, S. Weinberg
[2] Conceptual Challenges of Gauge Symmetry, Miguel Crispim Romão
https://www.academia.edu/5659587/Conceptual_Challenges_of_Gauge_Symmetry
Best Answer
One of the reasons relativistic theories are so restrictive is because of the rigidity of the the symmetry group. Indeed, the (homogeneous part) of the same is simple, as opposed to that of non-relativistic systems, which is not.
The isometry group of Minkowski spacetime is \begin{equation} \mathrm{Poincar\acute{e}}=\mathrm{ISO}(\mathbb R^{1,d-1})=\mathrm O(1,d-1)\ltimes\mathbb R^d \end{equation} whose homogeneous part is $\mathrm O(1,d-1)$, the so-called Lorentz Group1. This group is simple.
On the other hand, the isometry group of Galilean space+time is2 \begin{equation} \text{Bargmann}=\mathrm{ISO}(\mathbb R^1\times\mathbb R^{d-1})\times\mathrm U(1)=(\mathrm O(d-1)\ltimes\mathbb R^{d-1})\ltimes(\mathrm U(1)\times\mathbb R^1\times\mathbb R^{d-1}) \end{equation} whose homogeneous part is $\mathrm O(d-1)\ltimes\mathbb R^{d-1}$, the so-called (homogeneous) Galilei Group. This group is not semi-simple (it contains a non-trivial normal subgroup, that of boosts).
There is in fact a classification of all physically admissible kinematical symmetry groups (due to Lévy-Leblond), which pretty much singles out Poincaré as the only group with the above properties. There is a single family of such groups, which contains two parameters: the AdS radius $\ell$ and the speed of light $c$ (and all the rotation invariant İnönü-Wigner contractions thereof). As long as $\ell$ is finite, the group is simple. If you take $\ell\to\infty$ you get Poincaré which has a non-trivial normal subgroup, the group of translations (and if you quotient out this group, you get a simple group, Lorentz). If you also take $c\to\infty$ you get Bargmann (or Galilei), which also has a non-trivial normal subgroup (and if you quotient out this group, you do not get a simple group; rather, you get Galilei, which has a non-trivial normal subgroup, that of boosts).
Another reason is that the postulate of causality is trivial in non-relativistic systems (because there is an absolute notion of time), but it imposes strong restrictions on relativistic systems (because there is no absolute notion of time). This postulate is translated into the quantum theory through the axiom of locality, $$ [\phi(x),\phi(y)]=0\quad\forall x,y\quad \text{s.t.}\quad (x-y)^2<0 $$ where $[\cdot,\cdot]$ denotes a supercommutator. In other words, any two operators whose support are casually disconnected must (super)commute. In non-relativistic systems this axiom is vacuous because all spacetime intervals are timelike, $(x-y)^2>0$, that is, all spacetime points are casually connected. In relativistic systems, this axiom is very strong.
These two remarks can be applied to the theorems you quote:
Reeh-Schlieder depends on the locality axiom, so it is no surprise it no longer applies to non-relativistic systems.
Coleman-Mandula (see here for a proof). The rotation group is compact and therefore it admits finite-dimensional unitary representations. On the other hand, the Lorentz group is non-compact and therefore the only finite-dimensional unitary representation is the trivial one. Note that this is used in the step 4 in the proof above; it is here where the proof breaks down.
Haag also applies to non-relativistic systems, so it is not a good example of OP's point. See this PSE post for more details.
Weinberg-Witten. To begin with, this theorem is about massless particles, so it is not clear what such particles even mean in non-relativistic systems. From the point of view of irreducible representations they may be meaningful, at least in principle. But they need not correspond to helicity representations (precisely because the little group of the reference momentum is not simple). Therefore, the theorem breaks down (as it depends crucially on helicity representations).
Spin-statistics. As in Reeh-Schlieder, in non-relativistic systems the locality axiom is vacuous, so it implies no restriction on operators.
CPT. Idem.
Coleman-Gross. I'm not familiar with this result so I cannot comment. I don't even know whether it is violated in non-relativistic systems.
1: More generally, the indefinite orthogonal (or pseudo-orthogonal) group $\mathrm O(p,q)$ is defined as the set of $(p+q)$-dimensional matrices, with real coefficients, that leave invariant the metric with signature $(p,q)$: $$ \mathrm O(p,q):=\{M\in \mathrm{M}_{p+q}(\mathbb R)\ \mid\ M\eta M^T\equiv \eta\},\qquad \eta:=\mathrm{diag}(\overbrace{-1,\dots,-1}^p,\overbrace{+1,\dots,+1}^q) $$
The special indefinite orthogonal group $\mathrm{SO}(p,q)$ is the subset of $\mathrm O(p,q)$ with unit determinant. If $pq\neq0$, the group $\mathrm{SO}(p,q)$ has two disconnected components. In this answer, "Lorentz group" may refer to the orthogonal group with signature $(1,d-1)$; to its $\det(M)\equiv+1$ component; or to its orthochronus subgroup $M^0{}_0\ge+1$. Only the latter is simply-connected. The topology of the group is mostly irrelevant for this answer, so we shall make no distinction between the three different possible notions of "Lorentz group".
2: One can prove that the inhomogeneous Galilei algebra, and unlike the Poincaré algebra, has a non-trivial second co-homology group. In other words, it admits a non-trivial central extension. The Bargmann group is defined precisely as the centrally extended inhomogeneous Galilei group. Strictly speaking, all we know is that the central extension has the algebra $\mathbb R$; at the group level, it could lead to a factor of $\mathrm U(1)$ as above, or to a factor of $\mathbb R$. In quantum mechanics the first option is more natural, because we may identify this phase with the $\mathrm U(1)$ symmetry of the Schrödinger equation (which has a larger symmetry group, the so-called Schrödinger group). Again, the details of the topology of the group are mostly irrelevant for this answer.