Re: The need of geodesics in physics

"Shmuel (Seymour J.) Metz" <[EMAIL PROTECTED]> writes:
>>But the "spectra" itself is formed by photons,
>No. The term "continuous spectra" is not the same as[1] the usage of
>the term "spectrum" in optics. An observable in QM is a Hermitian
>operator, and such an operator has a characteristic that is known in
>the literature as its spectrum.

Actually, the term "spectrum" is generic to linear algebras, and is not
specific to operator algebras.  If A is an algebra over a field F, the
spectrum of an element a of A would be
           spec(a) = { f in F: (a-f) has no two-sided inverse in A }.