I am a bit confused with the way Physics textbooks use the word 'representation'. I was reading Weinberg's QFT and in section (2.5) titled One-Particle states, he shows that (eq. 2.5.3) \begin{equation} U(\Lambda) \Psi_{p, \sigma} = \sum_{\sigma'} C_{\sigma, \sigma'}(\Lambda, p) \Psi_{\Lambda p, \sigma'}, \end{equation} saying that in the case when $C_{\sigma, \sigma'}$ can be made block diagonal, the states form a representation of the inhomogenous Poincare group. I guess what he meant was that the $C_{\sigma, \sigma'}$ form a representation. However, even then, it is not clear to me how this a representation of the Poincare group since these matrices have a dependence on $p$ the momentum of the state. For instance

\begin{align} U(\Lambda' \Lambda) \Psi_{p, \sigma} &= \sum_{\sigma'} C_{\sigma', \sigma}(\Lambda'\Lambda, p)\Psi_{\Lambda'\Lambda p,\sigma'}\\ &= U(\Lambda')(U(\Lambda)\Psi_{\Lambda p, \sigma})\\ &= \sum_{\sigma'}C_{\sigma\sigma'}(\Lambda, p)U(\Lambda')\Psi_{\Lambda p, \sigma'} \\ &= \sum_{\sigma'',\sigma'} C_{\sigma'' \sigma'}(\Lambda, \Lambda p)C_{\sigma' \sigma}(\Lambda, p)\Psi_{\Lambda' \Lambda p, \sigma''}, \end{align} which gives us \begin{equation} C_{\sigma'', \sigma}(\Lambda'\Lambda, p) = \sum_{\sigma'}C_{\sigma'', \sigma'}(\Lambda', \Lambda p)C_{\sigma', \sigma}(\Lambda, p) \end{equation} but how is this a representation in the usual sense since these matrices have $p$ dependence? I think in the case of little group transformations since $p$ is left invariant therefore, we can concretely see that the $D$ matrices (which Weinberg defines in eq. 2.5.8), form a representation of the little group.

I am greatly confused over this possibly simple issue. Any help would be greatly appreciated.

  • $\begingroup$ What is your definition of a representation of a group? $\endgroup$ Oct 23, 2023 at 8:41
  • $\begingroup$ Do your matrices satisfy the group algebra? $\endgroup$ Oct 23, 2023 at 13:13
  • $\begingroup$ Hi. Thanks for commenting on my post. Yes my matrices satisfy the group algebra as I have checked. But they have arugments of momentum p, \Lambda p as such. Shouldn't they just depend on the Lorentz transformation \Lambda? $\endgroup$
    – QFTheorist
    Oct 23, 2023 at 14:04

1 Answer 1


I am not sure if this is helpful as it does not directly answer your question. But, Weinberg motivates in the previous sections that $$U(\Lambda'\Lambda) = U(\Lambda')U(\Lambda) \tag{1}$$ where $U$ is a unitary representation of the universal cover of the Poincaré group over Hilbert space $\mathcal{H}$. Hence, I would think that the relationship you obtain at the end of your post is a consequence of the fact that we already know equation (1) holds.

I think it is helpful to note that Weinberg states that $$U(\Lambda)\Psi_{p, \sigma} = \sum_{\sigma'}C_{\sigma, \sigma'}\Psi_{\Lambda p, \sigma'}.$$ All we know at this point is that the action of $U(\Lambda)$ on $\Psi_{p, \sigma}$ results in some linear combination of $\Psi_{\Lambda p, \sigma}$ determined by the coefficients $C_{\sigma, \sigma'}$. Weinberg does not actually say which linear combination as he does not define the coefficients $C_{\sigma, \sigma'}$.

  • $\begingroup$ Indeed, the relationship that I obtain at the end is the consequence of the fact that (1) holds. But my question is that he claims that these states \psi form a representation of the Poincare group. I guess what that means is that the C s from a representation of the Poincare group, but what I do not understand is why they can have a p dependence floating around and still be considered as an irrep. Does this mean that each momentum eigenstate independently forms an irrep? $\endgroup$
    – QFTheorist
    Oct 24, 2023 at 6:47
  • $\begingroup$ The C's do not furnish an irrep of the (universal cover of the) Poincaré group at this point in the text. The method of induced representations lets us consider the appropriate little group and its irreps, which induces a classification of the irreps of the Poincaré group, each indexed by the Minkowski norm of the momentum four vector $p^\mu p_\mu$ and the sign of $p^0$ when $p^\mu p_\mu \leq 0$. @QFTheorist $\endgroup$ Oct 24, 2023 at 19:56
  • $\begingroup$ I am very confused here. Don't the C matrices form a representation? I understand that the problem of finding the 'coefficients C' becomes the problem of finding the irreps of the stabilizer group, but what does he mean by 'states forming a representation'? As far as I know, a representation is a homomorphism from a group to a vector space. $\endgroup$
    – QFTheorist
    Oct 25, 2023 at 9:33
  • $\begingroup$ Let me be more precise. The C’s do not (at the point of Weinberg that the equations in your post are defined) furnish an irreducible representation of the universal cover of Poincaré. They do furnish a representation. @QFTheorist $\endgroup$ Oct 25, 2023 at 16:32
  • $\begingroup$ Separately, in physics the word “representation” is abused. It is used in three main ways 1) to refer to the mathematical definition of representation: a homomorphism $\pi: G \rightarrow GL(V)$, 2) to refer to the representation space $V$, which would be the space of states in this context, and 3) to refer to elements $M \in \pi(G)$, which would be the space of matrices $C_{\sigma, \sigma'}$ in this context. @QFTheorist $\endgroup$ Oct 25, 2023 at 16:47

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