# Do group elements correspond to observables of a system, or to transformations of a system?

Most groups I encountered so far in classical mechanics an special relativity (Galilei-group, Poincare-group, Lorentz-group and so on) described transformations of the system that was to look at. This is somehow intuitive to me, because a transformation succeeded by another transformation is of course a transformation again.

However (and this is something I can't really wrap my head around): In QM, when talking about representations, it seems to me that choosing a representation maps group elements to operators, that is, to observables of the system. Which is of course less intuitive to me (given two positions, why would they have any "group" structure? Of course, if I imagine them to be vectors, I could just add them, but is this what is meant by position becoming a group element now?)

Which is true now? Group elements corresponding to transformations, or to observables? Is there an intuitive approach in how to to group elements are related to observables in EVERY theory that is based on groups?

Edit: I know that in QM, Observables are seen as Operators of an Algebra that act on the Hilbert Space. And of course, those Operators do form a group as well (because linear operators do form the general linear group). But this group is not always the same group that "the theory is built uppon"

• Group elements represent transformations. Observables are the eigenvalues of certain Hermitian transformations in quantum mechanics. – bapowell Oct 7 '18 at 20:33
• @bapowell Observables are NOT eigenvalues. – ZeroTheHero Oct 7 '18 at 20:35
• I'm not sure what you mean by "operators form a group". They may or may not close on an algebra. $L_z^2$ is not an element of $SO(3)$. – ZeroTheHero Oct 7 '18 at 20:39
• @ZeroTheHero performing two (invertible) linear transformations in a row results in another linear (invertible) transformation. In that sense one can see operators acting on vectors from a hilbert space as members of a group. I guessed this is what bapowell meant with his comment, but kind of "group" is not the group I wanted to ask about in my question. – Quantumwhisp Oct 7 '18 at 20:48
• So if I understand you well then you are thinking of observables as elements of the general linear group, and in that sense these are not related to the underlying symmetry group of the Hamiltonian or the problem. – ZeroTheHero Oct 7 '18 at 20:50

So suppose that we have an action of the rotation group SO(3) on our quantum state given by $$|\psi\rangle \rightarrow U |\psi \rangle$$, where $$U$$ is the operator doing a rotation about the z-axis (for definiteness), then we know that we can write $$U = e^{i \theta J_z}$$ where $$J_z$$ is the generator of rotations around the z-axis. For small $$\theta$$, we can also write $$U \approx 1 + i \theta J_z$$ if we like. Now, it turns out that $$J_z$$ is Hermitian and thus an observable; it corresponds to the z-component of angular momentum, although my normalization is probably incorrect. The observable can be seen as quantifying how much a state changes upon applying a rotation. In general, group generators (but not transformations) correspond to observables.
There are exceptions to the rule that unitary operators are not Hermitian. A relevant example is the (unitary and Hermitian) parity operator. This can be seen as an action of the group $$\mathbb{Z}_2$$ on quantum states. It has eigenvalues $$\pm1$$ and is an observable. Note that $$\mathbb{Z}_2$$ is a discrete group and therefore not a Lie group in any interesting way, so the above discussion does not hold here.