A representation of a group $G$ is a pair $(\rho, V)$ where $V$ is a vector space and $\rho : G\to GL(V)$ is a homomorphism. If $V$ is actually a Hilbert space and $\rho : G\to \mathcal{U}(V)$ maps into the unitary group of $V$ and is strongly continuous, then $\rho$ is a unitary representation.

In section 2.5 Weinberg sets out to study the classification of one-particle states. What he does is:

  1. He first notices that the unitary representation $(U,\mathscr{H})$ of the Poincare group gives rise to the momentum operators $P^\mu$. He then considers the basis of eigenstates of $P^\mu$, namely $\Psi_{p,\sigma}$.

  2. He then notices that $U(\Lambda) = U(\Lambda, 0)$ - the restriction of $U$ to the proper orthochronous Lorentz group - when acting on $\Psi_{p,\sigma}$ yields an eigenvector of $P^\mu$ with eigenvalue $\Lambda^\mu_\nu p^\nu$. Hence it lies in the eigenspace associated to $\Lambda p$ being $$U(\Lambda) \Psi_{p,\sigma}=\sum_{\sigma'} C_{\sigma'\sigma}(\Lambda, p) \Psi_{\Lambda p,\sigma'}.$$

  3. He states the following (which is what I'm confused about):

    In general it may be possible by using suitable linear combinations of the $\Psi_{p,\sigma}$ to choose the $\sigma$ labels in such a way that the matrix $C_{\sigma'\sigma}(\Lambda,p)$ is block-diagonal; in other words, so that the $\Psi_{p,\sigma}$ with $\sigma$ within any one block by themselves furnish a representation of the inhomogeneous Lorentz group. It is natural to identify the states of a specific particle type with the components of a representation of the inhomogeneous Lorentz group which is irreducible, in the sense that it cannot be further decomposed in this way.

Now, I'm missing Weinberg's point. It seems he is actually talking about complete reducibility, namely, the property that the representations can be decomposed into irreducible ones.

It seems, however, that for groups like the Poincare group, complete reducibility is a non trivial thing to derive - I've read in this discussion that it actually requires direct integrals.

So it seems that to deal with these complications - which seem to come from the translations - Weinberg is switching to a basis of the generators of translations.

But how does that work really? Why to get into complete reducibility for the Poincare group, it is a good idea to do this? By the way, why this gives complete reducibility? Is he looking at "each eigenspace of $P$ at once"?

Also what he means that with $\sigma$ on each block $\Psi_{p\sigma}$ furnish a representation of the Poincare group? I mean, $U$ is a representation, and if we simply restrict its action to a subspace of $\mathscr{H}$ it will continue to be so.

So, how to understand Weinberg's argument (which seems to be about complete reducibility), and how does it connect to the "more usual" representation theory?

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    $\begingroup$ The standard assumption is that the little group representations (i.e. the $\sigma$ labels) are finite dimensional. So we can break up any representation into a sum of irreducible ones. I think this is all he is doing here. $\endgroup$
    – Prahar
    Commented Jun 4, 2018 at 1:18

1 Answer 1


As Prahar already hints at in a comment:

The eigenspace associated to $\Lambda p$ is finite-dimensional because it is simply spanned by the $\Psi_{\Lambda p,\sigma}$ for all possible values of $\sigma$ and the spin label $\sigma$ are drawn from a finite set. So all Weinberg is saying is that the finite-dimensional representation of the Lorentz group given by $C_{\sigma,\sigma'}(\Lambda,p)$ on that eigenspace may be reduced, this is just the standard result that all finite-dimensional representations of the Lorentz group decompose as the direct sum of irreducible representations.

  • 1
    $\begingroup$ The spin label $\sigma$ need not be drawn from a finite set though. (cf. infinite-spin representations). In that section, Weinberg is choosing to focus on finite-dimensional representations only, but never claims that those are the only possible representations. Indeed, he mentions infinite-spin representations later on. $\endgroup$ Commented Jun 4, 2018 at 17:46
  • 1
    $\begingroup$ @ACuriousMind, thanks for the answer. I've tried to improve the question. By what I've read it seems that complete reducibility for the Poincare group is something non-trivial. So is Weinberg doing this thing of working in the momentum basis, in order to "tame" this situation and conclude complete reducibility? Is he actualy "breaking the representations" into a translation part and a Lorentz transformation part, solving the momentum part, and then inside each momentum eigenspace solving the Lorentz transformation part parametrized by momenta? $\endgroup$
    – Gold
    Commented Jun 25, 2018 at 22:54

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