Re: a question about non - locality

From: Aaron Bergman (abergman_at_physics.utexas.edu)
Date: 03/29/05 

Date: Tue, 29 Mar 2005 08:20:35 +0000 (UTC)

In article <1148inko2r96f55@corp.supernews.com>,
 "C. M. Heard" <heard@pobox.com> wrote:

> Aaron Bergman wrote:
> > "Robert C. Helling" wrote:
> [ ... ]
> > > You are right, any finite number of derivatives is local. However
> > > powerseries in derivatives can be non-local. The standard example is
> > > the translation operator exp(a d_x) (d_x is the derivative):
> > >
> > > (exp(a d_x) f)(x) = f(x+a)
> > >
> > > and that's genuinely non-local.
> >
> > This is said a lot, and I hadn't really thought about it much, but
> > there's something funky here. The above formula only holds for analytic
> > functions. Now, at least in whatever way we can make sense of it,
> > analytic functions are going to be of measure zero in the path integral.
>
> One way to make sense of this is to consider the map f -> (exp(a d_x) f)
> as a linear operator in the Hilbert space L^2(R). For analytic functions
> it is defined by a power-series in derivatives. In particular, it is
> defined for all finite linear combinations of the Hermite-Gauss energy
> eigenfunctions of a harmonic oscillator. Furthermore the norm is bounded
> over this set. But this set is dense in L^2(R) since the Hermite-Gauss
> functions are a basis. A bounded linear operator being continuous, the
> definition can be extended to all of L^2(R).
>
> The key to this is that a set of measure zero can be dense, and a
> bounded linear operator that is densely defined extends to the whole
> Hilbert space.

Sure. This all makes sense in the Hamiltonian formalism. I was thinking
about the path integral, which, of course, is horribly ill-defined.
Still, the answer is probably along the same lines in that the only
definition of the above element in the lagrangian that can make sense is
to simply write it as f(x+a).

Aaron

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