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

Wigner treatment associates to particles the irreps of the universal covering of the Poincaré group $$\mathbb{R}(1,3)\rtimes SL(2,\mathbb{C}).$$

1. 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.

1. Why associate the Lorentz group to fields?

2. Why do we look in this case for finite dimensional representations?

3. What do we associate to unitary representations of the Lorentz group?

• Related: physics.stackexchange.com/q/21801/2451 , physics.stackexchange.com/q/100844/2451 and links therein. Commented Sep 23, 2014 at 21:53
• 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 $D_\Lambda$ : $\Lambda(\Phi_a(x)) = (\Lambda_R)_a^b \Phi_b(\Lambda^{-1}x) = (\Lambda_R)_a^b D_\Lambda (\Phi_b(x))$. Commented Sep 24, 2014 at 9:43
• 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. Commented Sep 24, 2014 at 9:43
• I guess $(n,m)$ is just the type of mathematical object we are considering. By taking $\Phi_a(x)$ we are "connecting" two mathematical objects, with the transformation law you gave. I read at some point that field equations are a consistency condition between unitary irreps for the Poincaré group and objects arising from finite dimensional irreps, but I don't remember where nor can I find an accessible reference on this subject. Commented Sep 24, 2014 at 10:44

## 1 Answer

1) Why don't we consider finite dimensional representations of this group?

As you said, we ask (anti)unitarity, so it is impossible to find finite-dimensional representation.

2) Why associate the Lorentz group to fields?

The essence of the answer is what Trimok already said in his comment: the "translational part" of the Poincarè group is already represented by the argument of the field. That is for a general multi-component field you postulate the transformation law, for each element $(a,\Lambda)$ of the Poincarè group, given by

$\psi'(x') = S(\Lambda) \psi(x)$

where $x' = \Lambda x + a$.

It seems a natural request for the transformation rule of a field, think of the non-relativistic case of the Schroedinger field: you expect that the operator creating a particle in position x with spin m=1 is seen as the operator creating a particle in position x+a with spin m=1 by an observer translated with respect to you. Spin, or any "inner" part of the field, should not be afflicted by translations.

I don't know if there is a more deep or rigorous explanation for this.

So you see that fields are distinguished by $S(\Lambda)$, and so only the Lorentz group is relevant for this purpose.

Note that no general request is asked to $S(\Lambda)$, a part for being a representation (I don't remember if it is allowed to be a projective representation though) of the Lorentz group. In the case of the Dirac field, in order to pin down the explicit form of $S(\Lambda)$, it is made another request, that is it leaves invariant the form of the Dirac equation. In the end it turns out it must not necessarily be unitary.

• Thanks Giulio! Indeed it all comes from the transformation law you gave (though it seems Trimok reverts the transformation on coordinates with respect to you?), which is based on intuition. Is there any more "fundamental" reason for takings reps of the Lorentz group on the field type (something that would stem from group theory itself)? I read p.90 in Weinberg that one cannot always deduce superselection rules from symmetry group. I feel a lot of stuff about the field type and its transformations are intuitive/experimental stuff rather than derived. Commented Sep 24, 2014 at 20:53
• Concerning finite-dimensional reps of Poincaré: I gave the argument for why we consider unitary (infinite dimensional ones), but why don't we consider finite-dim one? Aren't they interesting on physics? What's the argument for this exclusion? Weird particles properties? What about infinite-dim reps of the Lorentz group? Second, I understand why intuition (as encoded in equation you gave) asks for finite-dim reps of Lorentz group, however I can't connect this to us looking for unitary reps of Poincaré (there we fix P and hence m, and then look at irreps of little group, no stuff about Lorentz) Commented Sep 24, 2014 at 20:59
• "Concerning finite-dimensional reps of Poincaré: I gave the argument for why we consider unitary (infinite dimensional ones), but why don't we consider finite-dim one?" Finite dimensional one of Poincarè cannot be unitary, hence they are not meaningful for the characterization of particles. This fact is stated by Wigner's theorem. (en.wikipedia.org/wiki/Wigner%27s_theorem) Commented Sep 24, 2014 at 21:23
• Yeah, my question is not clear enough. I guess what I am trying to say is that we are interested in unitary reps of Poincaré, but in constructing them, at no point we talk about finite reps of Lorentz (fix P, and then little group), so I am trying to see if, besides to the argument based on the equation you gave, there is a reason for looking for finite-dim reps of Lorentz when looking for unitary reps of Poincaré. Commented Sep 24, 2014 at 21:25
• Mmm, maybe another reason is that usually the Poincarè representation acting on a single particle hilbert space is of the form $U(a,\Lambda)= exp[i P a] * ...$. Think of how galileian group is represented in non-relativistic quantum mechanics, for example for spin 1/2 particles. So, maybe, the "translational" part is once again represented trivially in the same way for every particles. Commented Sep 24, 2014 at 21:33