On Wed, Jul 30, 2025 at 9:42 AM Alan Grayson <[email protected]> wrote:
> On Tuesday, July 29, 2025 at 5:33:40 PM UTC-6 Brent Meeker wrote: > > On 7/29/2025 1:12 PM, Alan Grayson wrote: > > On Tuesday, July 29, 2025 at 2:04:31 PM UTC-6 Brent Meeker wrote: > > On 7/29/2025 7:18 AM, Alan Grayson wrote: > > Assuming we know all possible results of the measurements of a quantum > system, that is, the set of possible eigenvalues, and suppose we also know > the associated eigenfunctions, and we write the wf of the system as a > linear sum of eigenfunctions each multiplied by a complex constant, is it > mathematically assumed, or proven somewhere (perhaps by Von Neumann), that > these eigenfunctions are orthogonal and form a basis for the Hilbert space > in which they reside? TY, AG -- > > Yes, that's pretty much it. The physical system, including the ideal > measurement, is modeled by a certain Hilbert space in which the basis > states are the eigenfunctions the measurement. This is implicit in the > concept of an ideal measurement as one, which if immediately repeated on > the same system, returns the same value again. > > Brent > > > But is it proven or assumed the eigenfunctions in the sum are basis states > which span the space? If proven, where, by whom; if not, then the construct > lacks rigor. AG > > It's true by construction that the eigenstates span the Hilbert space. > "The Hilbert space" is the space whose bases are the eigenstates. > > > What does "true by construction" mean? Does that include orthogonality of > the basis eigenstates? AG > A lot of these things are proved in Dirac's book "The Principles of Quantum Mechanics". For example, the orthogonality of the eigenfunctions of a single operator is proved on page 32 (of my edition). "Two eigenvectors of a real dynamical variable belonging to different eigenvalues are orthogonal". Any set of linearly independent vectors forms a possible basis of a vector space if the number of linearly independent vectors equals the dimension of the space (The linearly independent vectors need not form a mutually orthogonal set, as long as they are linearly independent.) Bruce -- 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/CAFxXSLRu-Nrv%2BGQvJX25uYZk42ZYGFMLAbDYr8Ks9XbxhZa%3Dvg%40mail.gmail.com.
