# Why Lorentz group for fields and Poincaré group for particles?

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Wigner treatment associates to particles the irreps of the universal covering of the Poincaré group $$\mathbb{R}(1,3)\rtimes SL(2,\mathbb{C}).$$ Why don't we consider finite dimensional representations of this group? I understand we ask for unitarity when representing its action on states, so the representation cannot be finite dimensional. However we do consider finite dimensional representations of the Lorentz group $O(1,3)$ and associate them to fields. Why associate the Lorentz group to fields? Why do we look in this case for finite dimensional representations? What do we associate to unitary representations of the Lorentz group?

Thanks.

This post imported from StackExchange Physics at 2014-09-24 10:16 (UTC), posted by SE-user Issam Ibnouhsein
Related: physics.stackexchange.com/q/21801/2451 , physics.stackexchange.com/q/100844/2451 and links therein.

This post imported from StackExchange Physics at 2014-09-24 10:16 (UTC), posted by SE-user Qmechanic
About "fields", I think there is a difference between $\Phi_a$ and $\Phi_a(x)$. The former could be in a finite dimensional representation $R$ of the Lorentz group, $\Lambda(\Phi_a) = (\Lambda_R)_a^b \Phi_b$, while I don't see how the latter could be in a finite dimensional representation : it is because the Lorentz transformation acts on coordinates too, this part has to be represented by some differential operator $D_\Lambda$ : $\Lambda(\Phi_a(x)) = (\Lambda_R)_a^b \Phi_b(\Lambda^{-1}x) = (\Lambda_R)_a^b D_\Lambda (\Phi_b(x))$.

This post imported from StackExchange Physics at 2014-09-24 10:16 (UTC), posted by SE-user Trimok
For instance, in a unitary representation, we have $D_\Lambda = e^{-i \omega_{\mu\nu} L^{\mu\nu}}$, with $L_{\mu\nu} = -i(x_\mu \partial_\nu - x_\nu \partial_\mu )$. The space of functions (or operators in QFT) $\Phi_b(x)$, upon which the Lorentz transformation - and so the differential operator $D_\Lambda$ - acts, does not appear as finite-dimensional.

This post imported from StackExchange Physics at 2014-09-24 10:16 (UTC), posted by SE-user Trimok

A brief and hopefully correct and useful comment: Think of $\mathbb{R}^n$ as the coset space $ISO(n)/SO(n)$ (with similar modifications for Minkowsksi signature).  Similar ideas with Lie supergroups leads to superspace and superfields.

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Actually even the quantum fields support ("are") representations of Poincaré group acting on a suitable vector bundle whose sections are, in fact, the considered fields:

$$\phi^A(x) \to S^A_B(\Lambda) \phi^B (\Lambda x + t)$$

where $(\Lambda, t)$ is the generic element of the Poincaré group. Notice that $S$ sees only the Lorentz part of the Poincaré group and defines a vector representation on its own right.

When you fix an event in spacetime, i.e., you deal with a fiber of the vector bundle only, the Lorentz group part of the semidirect product  you wrote acts on that fiber,  by means of the representation

$$\phi^A(x) \to S^A_B(\Lambda) \phi^B (x)$$

leaving the fiber fixed. $S$ is a finite dimensional representation of the Lorentz group, since the fiber  has finite dimension as a vector space (the range of the index $A$ if finite). However this fiber does not admits a Hilbert space structure invariant under that representation, for this reason the representation is not unitary and it can be finite dimensional.

It is fundamental to notice that, in QFT, $\phi^A$ is also  an operator in the Hilbert space of the theory, and Poincaré group is a continuous symmetry of the physical system: It leaves  the  transition probabilities invariant. Essentially due to Wigner theorem, one has that this symmetry can be implemented unitarily  in the Hilbert space of the theory by means of a (strongly) continuous unitary representation  $U_{(\Lambda, t)}$.

It is natural tu assume that, under this unitary representation of Poincaré group acting in the Hilbert space, the quantum fields viewed as operators, transform covariantly with respect the other representation acting in the spacetime:

$$U_{(\Lambda,t)} \phi^A(x) U_{(\Lambda,t)}^\dagger = S^A_B(\Lambda) \phi^B (\Lambda x + t)\:.$$

answered Sep 26, 2014 by (2,085 points)
reshown Sep 26, 2014

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