Re: separable Hilbert space

Edward Green wrote:

> michaeld wrote:
> > David Macmanus wrote:
> >
> > > "michaeld" <michaeld@xxxxxxxxxx> wrote in message
> > > news:1124661545.353976.30140@xxxxxxxxxxxxxxxxxxxxxxxxxxxx
> > >
> > >
> > > > It should be clear that if you take I big enough then H doesn't have a
> > > > countable basis.
> > >
> > > Thanks. Can you suggest a real life (specifc) example of a non-separable
> > > space (Hilbert if you like). Is there one from quantum mechanics?
> > > Thanks,
> > > David.
> >
> > The example I gave is the most natural I can think of. Non-separable
> > Hilbert spaces are not needed for standard quantum mechanics.
>
> Really? I was going to suggest the states of a free particle. For a
> particle in a box, the natural basis states are analogous to a Fourier
> series -- countable -- whereas for an unbounded particle, the
> representation goes to a Fourier integral -- uncountable.
>
> Or am I wrong?

A free particle can be described by its wavefunction psi(x). This is a
complex valued function on R such that:

integral dx |psi(x)|^2

is finite. Hence the Hilbert space is just L^2(R). This has a countable
orthonormal basis, for example {psi_n(x) = H_n(x) e^(-x^2/2) | n =
1,2,3,...} where H_n are the Hermite polynomials.

That was actually for a particle in 1-d, but the same thing extends
easily to R^3 say. In either case the Hilbert space is easily seen to
be separable.

Maybe you're thinking of the apparent uncountable basis of position
eigenkets {|x> | x in R} or the momentum space version {|p> | p in R}
(which are related by a Fourier transform). These things are usefully
used heuristically in qm calculations, but they're not proper bases in
the rigorous sense of the term. Technically |x> does not actually live
in the Hilbert space at all - the position operator x actually has no
eigenvectors in the Hilbert space. To work with things like |x>
rigorously you need to get into the machinery of 'projection valued
measures' or alternatively 'Gelfand triples'.

.

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