# In what way observables are a representation of the symmetry group?

I was studying a course about Lie groups, Lie algebras and their representations (and classifications) when I encountered this statement :

When a physical system admits symmetry, the observable form a representation of the group concerned. Lorentz and Poincaré groups are very important examples.

(It's originally in french so this transduction might be a bit off)

This sounds like it's very important to understand the "big picture" of such theories but I don't really understand it. Is this only for quantum theories? If so, what is the link between representation theory and observable?

I'm in first year of master so I can understand concepts about Lie groups, Lie algebras, manifolds, QFT, etc.

• This statement is usually called Wigner's theorem Aug 8, 2020 at 10:09

Now, a priori, I don't see why this action should be linear. The only example I can think of right now is a gauge transformation of the EM field $$A_\mu\mapsto A_\mu+\partial_\mu\Lambda$$, which is affine. This is not a great example though because it is not an observable. In any case, in QM there is an additional piece of information which restricts the symmetries to be linear. This is that symmetries are described by unitary/anti-unitary operators. This is known as Wigner's theorem. As a consequence, they have to form a representation on the Hilbert space of your theory.
If a quantum system admits a Lie symmetry group, this means that there exist a unitary strongly continuous representation of that Lie group acting in the Hilbert space of the system. The one-parameter subgroups are represented by one-parameter strongly continuous unitary (sub) groups. Stone's theorem proves that each such group is if the form $$e^{-i aA}$$ for a unique selfadjoint operator $$A$$. Selfadjoint operators are observables by definition. The set of all $$A$$ above form a representation of the Lie algebra of the symmetry group.