Good evening everybody.

I have some questions about the relation between Lie groups and observables in physics. Indeed, taking the example of spin formalism of Quantum mechanics I know that Pauli's matrices $\{\sigma^i\}$ correspond to an observable quantity ($\frac{1}{2}$ eigenvalues of spin projection) since these latter are hermitian operators.

However, by a recent study on group theory I understood that $\frac{1}{2}$ spin eigenstates of classical QM can span the vector space which transform under a representation of the Lie group $SU(2)$.

It is always the case that an hermitian basis of a Lie algebra represent an observable quantity in QM or it is only the case for $SU(2)$ group? What is the general connection between group theory and observable quantities of QM and QFT?

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    $\begingroup$ Quantum observables are self-adjoint operators, and classical observables also form a Lie algebra on the phase space under the Poisson bracket. $\endgroup$
    – ACuriousMind
    Commented Jan 3, 2015 at 18:34
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    $\begingroup$ I understand your point and I agree with you. But I still have some problems to understand how it is possible that a Lie Group, an object that encode on its "macroscopic level" informations about how an object transform, could encode at his infinitesimal level another formalism to mesure physical quantities. It seems fascinating and really complex at the same time xD $\endgroup$
    – Wellow
    Commented Jan 3, 2015 at 19:03
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    $\begingroup$ Do you know about Noether's theorem and how it connects symmetries with conserved quantities like momentum and energy? I think that may be the missing link between physical observables (which are ultimately conserved quantities) and symmetries that you are looking for? $\endgroup$
    – CuriousOne
    Commented Jan 3, 2015 at 19:22
  • $\begingroup$ Yes, but Noether theorem states the conservation of a general charge in presence of a transformation group that leaves action invariant (like energy and momentum for space-time translation symmetry as you mentioned). :) My question is more on how you can see an infinitesimal transformation of a Lie group (a Lie algebra) as an observable quantity in physics; without necessary conserving it. $\endgroup$
    – Wellow
    Commented Jan 3, 2015 at 19:40

1 Answer 1


Let $A$ be the C*-algebra of a quantum mechanical system, and suppose that $G$ acts on $A$ by symmetries through the group homomorphism $\alpha:G\to\text{Aut}(A)$. Let us further assume that $G$ is a simply connected Lie group with trivial $H^2(\mathfrak g,\mathbb R)$, $\mathfrak g$ being the Lie algebra of $G$. If $\pi$ is an $\alpha$-regular representation, meaning that the transition probability $P_{\omega\to\omega\circ\alpha_g}$ is continuous in $g\in G$ for every state $\omega$, then $G$ can be unitarily represented on the Hilbert space of $\pi$ in such a way that $$u_g\pi(a)u_g^* = \pi(\alpha_g(a)),\qquad\forall a\in A,$$ by Bargmann's theorem. This expresses the fact that the pair $(\pi,u)$ is a covariant representation of the $G$-algebra $A$. Since $u$ is a representation of $G$, it induces a representation of $\mathfrak g$ on the Hilbert space of $\pi$ as well through self-adjoint operators. Hence if you allow the generators to live in the larger C*-algebra generated by $\pi(A)$ and all the $u_g$ for any $g\in G$ (the crossed product $A\rtimes_{(\pi,u)}G$), you can interpret them as observables. In most cases, these operators can be given a physical meaning (e.g. the generators of translations are the momentum operators, etc...).

  • $\begingroup$ I guess that should be $G\to\mathrm{Aut}(A)$. I don't know so much physics, so if you don't mind: is the $C^\ast$-algebra of a system an algebra (or the algebra) of observables? If so, is this an algebra of Hermitian operators on some fixed Hilbert (state) space, or does its abstract $C^\ast$-algebra structure describe the whole system and we can work with different representations ($A$-modules)? $\endgroup$
    – doetoe
    Commented Jan 4, 2015 at 18:20
  • $\begingroup$ Thanks, indeed you're right! Of course the codomain of $\alpha$ has to be in the set of automorphisms on the C*-algebra. Said algebra is the algebra of all observables. When you prepare a mechanical system in a certain state, you get a representation of A on some Hilbert space following a certain construction, known as GNS construction. More generally all the physics is encoded in the representation theory of $A$. $\endgroup$
    – Phoenix87
    Commented Jan 4, 2015 at 20:55

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