[Physics] Why are only linear representations of the Lorentz group considered as fundamental quantum fields

foundationsgroup-representationsgroup-theoryquantum-field-theory

As described in many Q&As around here, fundamental quantum fields are expressed as irreducible representations of the Lorentz group. This argument is entirely clear – we live in a Lorentz-invariant world and those elements of an observed system that do not mix with others any way we transform the system, either actively or just by looking at it from a different viewpoint, are the only candidates for separate physical entities such as quantum particles/fields. The question however is, why do we consider only linear representations of the group?

Peskin & Schroeder mention that every non-linear transformation law can be built from linear ones and there is thus "no advantage" in considering non-linear transformations. Nevertheless, they give no reference. Even if this is true, decomposition of the transformation doesn't seem as a counterargument as long as it is irreducible. There might be considerable issues with canonical quantization of non-linear fields, but if we can construct scalars from them (and thus a Lagrangian), the path integral formulation stands.

Another argument appeals to the principle of superposition derived from quantum mechanics. However, once interaction comes into game, we have non-linear field equations violating the principle of superposition anyways. Nonetheless, interacting linear and non-linear "point-to-point" representations of the fields in both cases obey the principle of superposition for their entire quantum states. So once again, the non-linear representation does not pose a fundamental problem.

The last possible argument conceivable by me, the need of the ability to build a perturbation theory, is more of a technical request than a fundamental restriction (not so far from the request of renormalizable interaction terms, though).

So, is there a conclusive principle restricting non-linearity of the representations or is it just one of those "it works so nobody cares" physics' moments?


EDIT 1: By a "representation" I obviously mean a "realisation" of the Lorentz group, since a representation in the old-fashioned sense is strictly a linear one. We could understand this as a functor from the space of (velocity) vector fields (on which the first "realisation" of the Lorentz group whas formulated) to a space of non-vector (non-linear) fields.
A more down-to-earth formulation is the following: Consider a field colection $\phi_a$ and a Lorentz transform $\Lambda$. Then the field transforms as

$\phi_a'(x)=M^\Lambda_a(\phi_b(\Lambda^{-1}x))$,

with generally ($\alpha \in \mathbb{C}$)

$\alpha \phi_a'(x) \neq M^\Lambda_a(\alpha \phi_b(\Lambda^{-1}x))$

and all the other stuff such as addition also generally violated.


EDIT 2: To get a taste of how such a theory would be quantized and where the problems may lay, I am also going to comment on the transformation of the quantized field operators. Consider we have field collection $\phi_a$ transforming as specified above. Now we also assume it's transformation to be analytical and with the use of multi-indices(in bold font) the transformation can be written as
$$\phi'_a = \sum_{\mathbf b} m_a^{\Lambda, \mathbf b} \phi^{\mathbf b}.$$
I.e. $\phi^{\mathbf b} = \phi_1^{b_1}\phi_2^{b_2}…$ The powers of the field are no problem in quantum mechanics as these are after quantisation multiple applications of the same field operator. I.e. after quantisation for the collection of field operators $\Phi_a$ it must hold that
$$U^\dagger(\Lambda) \Phi_a U(\Lambda) = \sum_{\mathbf b} m_a^{\Lambda, \mathbf b} \Phi^{\mathbf b}, \;\;\;\; (*)$$

where $U(\Lambda)$ is the Lorentz transformation of the quantum state of the field $|\Xi'\rangle = U(\Lambda) |\Xi\rangle$.

You can for example see that an addition of two operators of this kind does not transform by the given prescription, but this is generally true also in "normal" quantum theories. Consider for example $\hat{X}$ which transforms as $\hat{X} + a$ under a translation by $a$. However, $\hat{Y} = \hat{X} + \hat{X}$ transforms as $\hat{Y} + 2a$.


Basically, the only way I see the disqualification of non-linearly transforming representations happening is if their components could be identified as combinations of components of linearly transforming fields to certain powers. For example, four fields $\phi^\mu$ could be actually identified to transform as three components of a vector field, but squared:
$$V'^\mu = \Lambda^\mu_\nu V^\nu, \; \phi^\mu = (V^\mu)^2$$
But is this generally possible and how?

Best Answer

I would say that the answer to your question is in the Wigner's theorem http://en.wikipedia.org/wiki/Wigner%27s_theorem.

For any quantum system you need to have a "representation" of the Poincarè group on it. By "representation" I mean an homomorphism from the Poincarè group to the group of "Simmetries" (as defined in the above article) given the group structure with the composition. The theorem then tells you that this map may be written in terms of linear or anti-linear operators acting on the Hilbert space of the quantum system.