On Tue, Nov 05, 2024 at 01:26:10AM -0800, Alan Grayson wrote: > > > On Tuesday, November 5, 2024 at 12:27:55 AM UTC-7 Russell Standish wrote: > > Sorry Brent - the measured momentum values are still eigenvalues. > > Pick 3 orthogonal directions to measure the momentum, say x, y and z. > > Then the momentum operators are -iℏ∂/∂x, -iℏ∂/∂y and -iℏ∂/∂z, and the 3 > eigenvalues are the 3 components of momentum. > > One could also write it in vector form iℏ∇, in which case the operator > has a vector-valued eigenvalue. > > > I don't think this is correct. Quantum operators are chosen to be Hermitian, > that is, self-adjoint IIRC, so that their eigenvalues will be > real. This is something that can be proven. So the question remains; how can a > real eigenvalue be a measured momentum, which > is a vector? AG
You missed my point completely. Momentum is a 3-vector, so the momentum operator is 3-vector of hermitian operators, applied elementwise over the wavefuction. The "eigenvalue" is a 3-vector, applied elementwise over the state vector. This is quantum mechanics 101 - any of the introductory books will tell you the same - Ramamurti Shankar, Leonard Schiff, Emile Durand. I'm surprised Brent made such a howler, but we're all human (for how long, I wonder, give JC's comments), and he's picked up plenty of howlers I've made over the years. Cheers -- ---------------------------------------------------------------------------- Dr Russell Standish Phone 0425 253119 (mobile) Principal, High Performance Coders [email protected] http://www.hpcoders.com.au ---------------------------------------------------------------------------- -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion visit https://groups.google.com/d/msgid/everything-list/ZyqdWOBYOiVv05cQ%40zen.
