# Why does there being no finite dimensional unitary representations of the Lorentz group lead to QFT?

In Schwarz's Introduction to QFT and the standard model they say that because there are no finite-dimensional unitary representations of the Lorentz group we must have fields and not particles.

However, for a single particle quantum state the basis of the Hilbert space is still infinite and so this would suggest we would be able to find a lorentz invariant single particle theory.

## 3 Answers

Consider any Lorentz-symmetric quantum field theory that has a single-particle sector.

As a specific example, consider the Proca model. This is the QFT for a non-interacting quantum relativistic massive vector field. It is arguably the most straightforward example of a QFT after the free scalar model. It has a single-particle sector which is self-contained under Lorentz transformations. The particle has spin 1, and it's its own antiparticle.

The single-particle sector of the Proca model is Lorentz-symmetric by itself: Lorentz transformations don't mix single-particle states with non-single-particle states. But can we think of this single-particle sector as a self-contained theory? If the answer is yes, then this would be a counterexample to the claim that the nonexistence of finite-dimensional unitary representations of the Lorentz group leads to QFT.

However, if we want to think of the single-particle sector as a self-contained theory, we need to do something more than just deleting all of the non-single-particle states from the Hilbert space. We need to specify what the model's observables are — which operators represent which measurable physical quantities.

The obvious thing to try would be to use observables from the original QFT, deleting those that don't preserve the single-particle sector. But in the Proca model, and in any other relativistic QFT, this means that we need to delete all observables that are localized in any finite region of spacetime. The remaining theory doesn't have any local observables! (I explain this in more detail in another answer).

So, if we want to think of the singe-particle sector of a QFT as a self-contained model, we either need to abandon all hope of having local observables, or we need to invent a whole new conceptual scheme for relating operators to measurable physical quantities. People have tried to do exactly that (cf the Newton-Wigner position operator). It can be made to work in single-particle models, so in this sense we can have Lorentz symmetry in quantum theory without using QFT. But the real world has more than one particle, so a prescription that only works in single-particle models is not a very interesting kind of counterexample.

Altogether, if we start with a relativistic QFT and restrict it to single-particle states, then we lose all local observables (all observables localized in finite regions of spacetime). We can try contrive a new way of interpreting the model's operators as local observables, but if it only works in the single-particle case, then it's not very interesting, even if it is technically a counterexample to the unqualified claim that Lorentz symmetry requires QFT.

This answer focused on the idea suggested in the question, namely considering the single-particle sector of a QFT. For that purpose, I considered a non-interacting example. A much more thorough answer, emphasizing interactions, is given in Weinberg's The Quantum Theory of Fields (volume 1). Weinberg's book goes to great lengths to explain how Lorentz symmetry combined with a few other physically-motivated requirements leads to QFT.

• @AccidentalFourierTransform You're right, of course. I changed the words "is nonlocal" to "doesn't have any local observables." Thanks! – Chiral Anomaly Apr 14 at 23:55

OP, you are exactly right. This statement in Schwartz's book is completely wrong, and for exactly the reason you mention. I am now going to do something naughty and paste my answer from a previous question because I think it applies here. Here I go.

...

It seems to me that you got derailed by a rather unhelpful discussion in Schwartz. There's a lot going on here which I will briefly outline.

The first thing I want to briefly discus is the relevence of representations of the LORENTZ group vs. representations of the POINCARE group.

Lorentz group: Representations are finite dimensional, given by two half integers $$(j_1, j_2)$$, and not unitary. Spacetime translation is not included. Representations of this group are representations of local field operators $$\hat \phi_i(x)$$. Under a Lorentz transformation, the indices of $$\hat \phi_i(0$$) will get mixed together while the spacetime point $$0$$ will be fixed. Two simple examples include the spin $$0$$ representation, where $$i$$ is just $$0$$, and the $$(1/2, 1/2)$$ representation, where we instead use the symbols $$\hat A_\mu$$ and $$\mu = 0... 3$$. The fact that these finite dimensional representations (which just juggle around the field indices) are not unitary is totally uninteresting because there isn't a natural inner product of local operators, like there is with states, so who cares that they're not unitary!?

Poincare group: Representations are infinite dimensional, unitary, given (by Wigner's theorem) by either a mass $$m^2$$ and a half integer spin, or in the $$m = 0$$ case by a half integer helicity. These correspond to particle states. The fact they they are unitary is important because you want the group action to preserve the natural inner product on states. The fact they are infinite dimensional is not really unexpected, the same thing happens in non relativistic QM! For instance, in the spin $$0$$ case, single particle states are given by a momentum $$| p \rangle$$. Under a Poincare transformation,

$$U(\Lambda, a) | p \rangle = e^{- i p a} | \Lambda p \rangle.$$

The fact that you need to include a state $$| \Lambda p \rangle$$ for each Lorentz transformation $$\Lambda$$, this representation is infinite dimensional because your basis of single particle states is spanned by an infinite number of $$p$$'s. But this is also the case in non relativistic QM, that you have a basis spanned by an infinite number of momenta, and isn't special to quantum field theory, so it's not really that big a deal. It's true that the Poincare group is non compact because of boosts, but its also non compact because of translations, which is why QM is infinite dimensional (there's an infinite number of positions).

Relationship between the two: You can create single particle states by acting on the vacuum with field operators. In the spin $$0$$ case a single particle state is given by

$$| p \rangle = \int d^3 x e^{- i \vec p \cdot \vec x} \hat \phi(\vec x) | 0 \rangle$$

and in the spin $$1$$ case by

$$| p, \varepsilon \rangle = \int d^3 x e^{- i \vec p \cdot \vec x} \varepsilon^\mu \hat A_\mu(\vec x) | 0 \rangle$$

(modulo maybe some integration measure I forgot.) Here $$\varepsilon^\mu$$ is a polarization vector satisfying $$\varepsilon^\mu p_\mu = 0$$. Because single particle states are super positions of states like, say, $$\hat A_\mu | 0 \rangle$$, the spins of the physical particle states can be found by restricting oneself to the rotation part of the Lorentz group, $$U(R)$$, and just using the standard angular momentum addition formula, where the irreducible representations of the $$j_1 \otimes j_2$$ representation is $$j_1 +j_2 \oplus j_1 + j_2 - 1 \oplus \ldots \oplus |j_1 - j_2|$$. So in the $$(1/2, 1/2)$$ representation of the Lorentz group, we have $$1/2 \otimes 1/2 = 1 \oplus 0$$. However, the non physical $$0$$ representation is projected out by the condition $$\varepsilon^\mu p_\mu$$, getting rid of the negative norm states, yadda yadda yadda. The point is that $$j_1 \otimes j_2$$ give the possible particle spins, but some of them will be non physical.

The main point, though, is that representations of the Lorentz group are used for talking about field operators, while representations of the Poincare group are used for talking about particle states. Once you realize this, it's completely reasonable that representations of the Lorentz group are not unitary (why should they be?) and that representations of the Poincare group are not finite dimensional (why should they be?).

There are a number of reasons why the single particle theory of QM can't be extended into a quantum theory of fields and this is another one. Generally speaking we have to move into a picture where particles can be created and annihilated. This is the multi-particle picture.

Freeman Dyson discusses some of the reasons in his Advanced QM.

The reason why there are no finite dimensional unitary irreps of the Lorentz group is because the Lorentz group is not compact. All non-compact groups have only infinite-dimensional unitary irreps, assuming that they have any.