This is a follow-up from this question. Weinberg states in Quantum Theory of Fields Vol. 1 that creation and annihilation fields must transform as $$U_0(\Lambda, a)\psi_l^+(x)U_0^{-1}(\Lambda, a) = \sum_\bar{l}D_{l\bar l}(\Lambda^{-1})\psi_{\bar l}^+ (\Lambda x+a).\tag{5.1.6}$$

Implicit in this statement is that under a pure translation $U_0(1, a)$, we have

$$\psi_l^+ \to \psi_l^+(x+a),$$

i.e. that the different components of $\psi_l$ do not mix. It is not obvious to me why we should assume this. Is there no non-trivial way in which the fields could transform under spacetime translations?


Following the discussion with @Prahar in the comments, I see that the fields cannot mix under a translation. However, it is still not clear to me that the fields must transform trivially under spacetime translations. That is, is there anything preventing the possibility that

$$\psi_l^+ \to e^{i\theta a} \psi_l^+(x+a)?$$

  • 2
    $\begingroup$ Translations are an Abelian subgroup so its irreducible reps are all one-dimensional. $\endgroup$
    – Prahar
    Feb 6, 2022 at 19:48
  • $\begingroup$ @Prahar ok - but say $D_{l \bar l}(a) = e^{i\theta a}\delta_{l \bar l}$ for some $\theta$. Then $u_l(x; p, \sigma, n) = e^{ipx+i\theta x}u_l(p, \sigma, n)$. So why is $\theta=0$? $\endgroup$
    – awsomeguy
    Feb 6, 2022 at 21:39
  • 1
    $\begingroup$ Note that if $D(a) = e^{i\theta a}$ then $D(a) = D(a+\frac{2\pi}{\theta})$. However, translations do not admit any such periodicity so we must have $\theta=0$ (translations are not compact!). As an aside, note that $S^1 \cong U(1)$ is also abelian but it is compact so it's allowed to have non-trivial representations such as $D(\theta) = e^{i n \theta}$ for $n\in {\mathbb Z}$. $\endgroup$
    – Prahar
    Feb 6, 2022 at 21:45
  • $\begingroup$ @Prahar and why can non-compact groups not admit periodic representations? $\endgroup$
    – awsomeguy
    Feb 6, 2022 at 21:51
  • $\begingroup$ Because then those are not really representations of the non-compact group, but representations of the compact quotient. For instance since $S^1 = {\mathbb R}/{\mathbb Z}$, obviously every representation of $S^1$ can be elevated to that of ${\mathbb R}$ but those aren't really representations of ${\mathbb R}$, are they? $\endgroup$
    – Prahar
    Feb 6, 2022 at 21:53

1 Answer 1


In my world that is almost a definition. Fields are defined as sections of fiber bundles. Or more informally, assume that we have a representation of the Lorentz group $S$ and a representation of the Poincare group on "x". And then consider \begin{equation} S \otimes x, \end{equation} which obeys the transformation rule that you wrote. The only part that can "transform" non-trivially under the translation algebra is an "x" part, which are just coordinates on our Minkowski space. For example, a scalar field is just a map from $M \to \mathbb{R} $. When you do a Lorentz transformation $f$ transforms as $\Lambda^*f(x) = f'(x) = f(\Lambda x)$. In some sense Weinberg is decomposing this tensor product, which is a valid representation, into a sum of the Poincare group representations.

We can consider something that has the dependence you described, but it would ruin the group composition rule and would not be a representation: \begin{equation} U(\Lambda_1,a_1) U(\Lambda_2,a_2) = U(\Lambda_1\Lambda_2,\Lambda_1 a_2+a_1) \end{equation} And consequently this tensor product would have no reason to decompose in terms of representations.


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