# Why are only linear representations of the Lorentz group considered as fundamental quantum fields?

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?

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P&S, Paragraph 3.1, p. 36, 1995 edition – Void Jun 28 '14 at 21:48
@Void You probably already know this, but as a matter of terminology there is a dedicated term for the sort of thing you call a "realization" for groups acting on general sets -- "group action." Nice question. – joshphysics Jun 29 '14 at 0:02

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.

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+1: I somehow completely forgot about this when reading the OP's question. I think it's a decisive answer (provided I understand the question correctly). – joshphysics Jun 29 '14 at 20:06
Wigner's theorem addresses the entire quantum states as described in the third paragraph of the original question. To demonstrate my point, in linearly-transforming interacting theories such as $\phi^4$ theory, the superposition of two classical solutions do not form another solution, but the theory can be quantized and the resulting states are linear. Similarly, there is no obvious obstacle in the quantization of a non-linear field to receive linear quantum states. Remember that objects such as fields are then operators and powers of these are multiple applications (always linear). – Void Jun 29 '14 at 22:11
I think you can see that requesting the transformation property $(*)$ (as is the canonical way for any representation) added to the question, there is no problem with Wigner's theorem. – Void Jul 5 '14 at 10:00

I think you could have the following problem.

Suppose $2$ independent fields $\Phi$ and $\Phi'$. Then any combination $\Gamma=\alpha \Phi + \alpha' \Phi'$ will transform, in a non-linear representation of the Lorentz transformation, in a non-trivial way, making $\Phi$ and $\Phi'$ de facto interacting.

But the fact that two fields are independent should not depend on the referential frame we are choosing. It is an invariant property. So there would be a contradiction.

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Are $\phi$ and $\phi'$ fields of the same kind? If yes, non-linear transforms express self-interaction and $\phi$ and $\phi'$ are just functions, not independent fields. If $\phi$ and $\phi'$ are different fields, in what sense can we add them? Even in the case of linear representations, spinors and scalars don't just add up, this may be be possible in some kind of product of spaces. However, in that case we must specify where does $\alpha$ multiply (it will not be a direct product of vector spaces) and every field has it's separate transformation and they do not mix. – Void Jun 30 '14 at 9:38
I was not enough precise, here $\Phi$ and $\Phi'$ are fields of the same kind, for instance vector fields, so I meant $\Phi(x)$ and $\Phi'(x)$, and $\Gamma(x)=\alpha \Phi(x) + \alpha' \Phi'(x)$. In a quantum field point of view, they are operators which may act on states, for instance the vacuum $|\Omega\rangle$, so you may consider, for instance, the state $\Gamma(x)|\Omega\rangle= \alpha\Phi(x)|\Omega\rangle + \alpha'\Phi'(x)|\Omega\rangle$. The independence of the operators/fields would mean that the $2$ states $\Phi(x)|\Omega\rangle$ and $\Phi'(x)|\Omega\rangle$ are orthogonal... – Trimok Jul 1 '14 at 8:17
.... A non-linear transformation would possibly break this orthogonality. – Trimok Jul 1 '14 at 8:17
I expanded the answer elucidate the transformation properties of the quantum operators. When we require the transformation properties in equation (*) in the question (analogous to vector operators), we get e.g. $<\phi_a>=\langle\Omega|\Phi_a|\Omega\rangle$ to transform as the classical field which is entirely appropriate. – Void Jul 5 '14 at 9:17