It’s the hermiticity that allows for the unboundedness right? if H has complex eigenvalues this won’t hold in general, e.g the exponential of the sl(2,C) Algebra
-
-
True! So if the spectrum of H goes haywire to infinity, the spectrum of exp(iH) just stays on the unit circle making it bounded. For n by n matrices A, exp(A) is always defined since the spectra in this case are finite (including elements sl(2,C).
Thanks. Twitter will use this to make your timeline better. UndoUndo
-
Loading seems to be taking a while.
Twitter may be over capacity or experiencing a momentary hiccup. Try again or visit Twitter Status for more information.