Physics Asked on March 15, 2021
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.
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.
Correct answer by Valter Moretti on March 15, 2021
Physical symmetries are in particular a certain transformation of the system. Translations displace the system from a point to another point. Rotations rotate the system. In general, in order to make sense of this statement one needs to give a prescription for the way that this transformation acts on your description of the system. In particular, you need to prescribe the way that the symmetry acts on your observables. In other words, the symmetry group should act on your space of observables.
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 $A_mumapsto A_mu+partial_muLambda$, 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.
Answered by Iván Mauricio Burbano on March 15, 2021
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