
#19
Jan913, 05:41 PM

Sci Advisor
P: 815





#20
Jan913, 08:07 PM

Sci Advisor
P: 8,004

@Sam, thanks! What happens to Noether's theorem where the classical theory has a symmetry that leads to a conserved quantity, but the corresponding quantum theory doesn't?




#21
Jan913, 08:23 PM

P: 962





#22
Jan913, 08:43 PM

Sci Advisor
P: 8,004





#23
Jan913, 09:39 PM

P: 647





#24
Jan1213, 12:43 PM

P: 962





#25
Jan1213, 02:32 PM

P: 647

Observable quantities are represented in quantum mechanics as hermitian operators. For something to be observable, it must be just that, measurable. For instance, the phase of a state vector, [itex]e^{i\theta}\psi \rangle[/itex] is not an observable (try and devise an experiment where you can measure it). But [itex]\psi^2= \langle \psi e^{i\theta}e^{i\theta}\psi \rangle [/itex] is an observable, but now the phase is gone... Does this help?




#26
Jan1213, 06:16 PM

P: 962

But what I'm really trying to get at is a generalization of what an observable is. Is an observable something we've already measured and then backward incorporate it into the lagrangian? Or are there more fundamental things lurking in the lagrangian that we should be able, at least in principle to make a measurement of? 


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