Timeline for How to properly set up the relation between Lie algebras and observables in QM?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 11, 2017 at 20:35 | vote | accept | Gold | ||
| Apr 8, 2017 at 22:31 | comment | added | ACuriousMind♦ | @user1620696 You're right, the servers seem to be down at the moment. I added the definition of the derived representation, it's really the same as what you wrote. The point is that it's defined only on $H^\infty$, where the notion of differentiation makes sense by definition, and one can show that $H^\infty$ is dense in $H$ for Banach representations $\pi$ (i.e. $\pi(G)\subset B(H)$ is bounded and $g\mapsto \pi(g)h$ is continuous on a dense subset of $H$, and perhaps some other technical condition I'm missing) | |
| Apr 8, 2017 at 22:24 | history | edited | ACuriousMind♦ | CC BY-SA 3.0 | added 75 characters in body |
| Apr 8, 2017 at 22:11 | comment | added | Gold | Thanks for the answer, but the link isn't working. What would be this derived representation? I mean,if $GL(H)$ were a smooth manifold, then $d\pi$ would be defined by $d\pi(\gamma'(0))=(\pi\circ\gamma)'(0)$ for all $\gamma(t)$. But here $H$ is a Banach space which might be infinite dimensional, and $GL(H)$ are all linear maps on $H$. I'm not sure of what is this derived representation in this context. | |
| Apr 8, 2017 at 2:15 | history | answered | ACuriousMind♦ | CC BY-SA 3.0 |