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In his book A Modern Introduction to Quantum Field Theory, Michele Maggiore affirms that

We require that all fields, independently of their transformation properties under the Lorentz group, behave as scalars under space-time translation.

I don't understand why we can assume so nonchalantly. Is there a way we can, instead, derive this statement in a mathematical fashion? I was thinking about the fact that the translation group is commutative, therefore every representation is completely reducible as a direct sum of one-dimensional representations... but I don't know if it is of any use.

I looked at other questions on the subject but they didn't clear my doubt: for example in this one the highest-rated answer says that the translation part of the Poincaré group acts trivially on the fields by $A(x)\mapsto A(x-a)$ but why is that so? Why can't we choose a transformation law $A(x)\mapsto A'(x-a)$ where $A'\neq A$?

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  • $\begingroup$ Invariance under translations is linked to conservation of momentum by Noether's theorem. $\endgroup$
    – John Rennie
    Jun 10, 2018 at 12:33

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The word "scalar" is poorly chosen in this context, and I think that explains your confusion.

The property the word "scalar" usually refers to is the behavior under the Lorentz group ("scalar" corresponds to the trivial representation, "spinor", "vector", etc correspond to non-trivial finite-dimensional representations). So it makes no sense to say that a field behaves "as a scalar under space-time translation".

What you say about the commutativeness of the translation group is correct, and translates physically into the fact that under the action of a translation, the field just picks a multiplicative factor (a phase), which can be seen as a $1 \times 1$ matrix. My guess would be that Maggiore used the word "scalar" to mean this one-dimensionalness, which is confusing because it is not its standard meaning.

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  • $\begingroup$ The word scalar is used w.r.t. any group. It is always the object that transforms under the trivial representation of that group. $\endgroup$
    – Prahar
    Jun 11, 2018 at 16:48
  • $\begingroup$ Well I don't know, although I agree in principle, all the examples of uses of "scalar" that come to my mind refer to some rotation group. See for instance en.wikipedia.org/wiki/Scalar_(physics) $\endgroup$
    – Antoine
    Jun 11, 2018 at 21:14
  • $\begingroup$ At a basic level, this is certainly true. However, I have seen more general usage in advanced texts. Either way, this is semantics. Your note correctly answers OPs questions, in any case. $\endgroup$
    – Prahar
    Jun 11, 2018 at 21:15

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