2
$\begingroup$

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?

EDIT:

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

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

$\endgroup$
8
  • 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

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.