Natural Science Forum / Physics / Research / March 2005

Because such words are usually chosen because of some resemblance to
the ordinary sense of the word, not by a complete correspondence.

Because sum h^k/k! d^k f(x)/dx^k = f(x+h)
is the value at a point at some distance from x, one _cannot_
consider an expression depending on all derivatives as local.

But one containing finitely many derivatives only behaves
qualitatively different since by increasing the number of fields
one can make all but first derivatives disappear. And the
latter has been thought of as local ever since field theory was invented.

Coulomb interactions are manifestly nonlocal, for example, since they
depend on two positions. In quantum field theory, however, they
are derived from the interaction with the local electromagnetic field.
The transformation essentially involves the nonlocal integral
operator 1/p^2.

No; it is the mathematically different behavior of polynomials and
nonpolynomials in p.

Arnold Neumaier

<snip>

To understand locality in this context (as mentioned by previous
posters, there's more than one concept called "locality" to contend
with in relativistic quantum physics), you need to understand the
mathematical concept of the support of a function.  If you have a
function f(x), its support is basically the set of points on which
the function takes nonzero values.  To be precise, it's the closure
of the compliment of the set of points on which the function vanishes.
But for the purposes of this discussion, just think of it as the points
where the function is nonzero.  This set is usually denoted
supp f.

If the function is to be thought of as a single-particle wavefunction,
then the support more or less tells you the region in space where
the particle is localized, i.e., the places where there's a chance
you might find it.  One way to see this is to consider the expectation
of the position operator X in the state f:

(f, Xf) = int dx f^*(x) x f(x)

Clearly, we only get contributions on the right hand side for those
"eigenvalues" x of X in the support of f.  So, the particle is
localized in supp f, in this sense.

Now, mathematically, an operator on a function space is called local
if the support of the resulting function is equal to or contained in the
support of the original function.  Thus, if L is a local operator,
we would have

supp Lf \subset supp f

One can show that if P is a (finite) polynomial with either constant or
non-constant coefficients, then P(d/dx) is a local operator under the
above definition.  However, this isn't necessarily the case for
an infinite series in derivatives.  The classic example, as several
previous posters mentioned, is the translation operator.  If T_a
denotes translation by a, then supp T_a f will have the same shape
as supp f, but it will be shifted by a.  Thus, no matter how small of
an a you use (as long as its not zero), there will always be points
in supp T_a f that aren't in supp f.  So, by the above definition,
T_a is a nonlocal operator.

But, what does this mean physically, you ask?  Well, since the support
is basically the region in which you expect to find the particle
in question, by translating the support, you've instaneously moved
its area of localization.  That certainly can't be considered to
be a local operation in the usual physical sense.  Put another way,
(T_a f)(x) depends on values of f at places other than x, in contrast
to the case of a finite number of derivatives.  That's a hallmark
of nonlocality.

Now, the translation operator just moves the support around, but it
retains its shape.  There are other kinds of nonlocal operators that
do more.  The classic examples of one of these is a convolution operator,
like the Coulomb potential.  Neglecting overall constants, the Coulomb
potential is basically given by

(Cf)(x) = int dx' |x-x'|^{-1} f(x')

If x_0 is a point lying outside the support of f, (Cf)(x_0) will
not necessarily be zero because there will be contributions to the
integral from all x' that lie in the support of f.  So, C doesn't
obey the criterion for being a local operator either.  Physically,
it instaneously changes the localization region of the particle
in a weird way.

As a final note, when you're thinking about these things, it helps
to take f to be a bump function.  That's basically something like
a square pulse centered at some point x_0 with all the corners
smoothed out so that it's infinitely differentiable.  Then you can
think about what a given operator would do to the bump function.
In the case of T_a, it'll just move it around.  C will be more
bizarre and I'll leave that case to you as an exercise.

I hope that helps, somewhat.
Dan.