I am reading the Wu-Ki Tung book "Group theory in physics" and I'm trying to put the various pieces (chapter) together to understand how he gets the unitary irreducible representations of groups used in physics.

At page 157-158 he shows how to get the unitary irreducible representations of group $E_2$ (i.e. Euclidean space in two dimensions) and he states "For unitary representations the generators $J,P_1,P_2$ are mapped into Hermitian operators".

He doesn't provide any details about that mapping or how to build it and then in the rest assume that $P_1,P_2$ are Hermitian.

The usual representation of the $P_1$ generator is evidently not Hermitian: $$P_1=\left( \begin{matrix} 0 & 0 & i \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{matrix}\right) $$ If the generators were Hermitian then the representation of the group would be automatically unitary cause $(e^{iaP_1})^{\dagger}=e^{-iaP_1}=(e^{iaP_1})^{-1}$

At pages 191-198 he shows how to get the unitary irreducible representations of the Poincare' group and at page 194 he states: "The representation is unitary because the generator are realized as Hermitian operators". Again, I can't see how he gets such Hermitian generators.

Isn't this reasoning circular? My guess is that he assumes the unitary representation exists and he shows only how to get the matrix element of the generators.


1 Answer 1


Indeed, the author illustrates by explicit construction that, for noncompact groups (contrast to what you have learned in SO(3) !),

  • their unitary representations (Hermitian irreps for the generators) are infinite dimensional.

(And vice versa, finite-dimensional reps are not hermitian, as you correctly observe for what you call the "usual" rep for $E_2$.)

He constructs these infinite dimensional (endless!) matrices in section 9.2 for $E_2$ and 10.3.3 for Poincaré. To do that, he assumes the generators are realized by hermitian operators/matrices (the representation map) and pursues the logical consequences of what these operators must be like, using the Lie algebra satisfied by these operators. He finds answers. Nothing circular about that!

As you correctly guess, finding a good answer based on his existence assumption, ipso facto justifies said assumption. What more are you after?

  • $\begingroup$ shouldn't he in the end verify that the generators he build are really Hermitian? $\endgroup$
    – Andrea
    Feb 2 at 14:56
  • $\begingroup$ He already knows that, having assumed it in the construction! E.g., $P^{\pm~ \dagger} = P^\mp$, which he uses below... $\endgroup$ Feb 2 at 15:07
  • $\begingroup$ I mean, since at the beginning he doesn't prove such Hermitian irrep exists and he just assumed it, then at the end to validate the assumption he should shows that what he build is Hermitian. $\endgroup$
    – Andrea
    Feb 2 at 15:13
  • $\begingroup$ The reader can check that it is hermitian, of course, just as she/he checks that finite irreps of SO(3) are hermitean. But how would they fail to be hermitean, unless he made a mistake in the construction based on hermiticity? $\endgroup$ Feb 2 at 15:15

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