2
$\begingroup$

I understand that an operator, $\hat{O}$, is said to be non-local if

b(x)=\hat{O}a(x)=\int dx'O(x,x')a(x')

that is, to find $b(x)$ at aparticular value of $x$, we need to know $a(x)$ for all $x$.

This can be expressed in linear, non-continuous, algebra as

$b_{i} = \sum_{j} O_{ij} a_{j}.$

An operator can be considered local, if in order to find $b(x)$ at a point $x_{0}$, we only need to know $a(x)$ in a small region in the neighbourhood of $x_{0}$.

If we consider this in terms of discrete algebra (i.e. matrices and vectors, rather than operators and continuous functions), can we say anything about the structure of the matrix $\mathbf{O}$ when the equivalent operator $\hat{O}$ is either local or non-local?

I have read that the derivative operator $\frac{d}{dx}$ is local and happen to know that derivative operators can be expressed as banded matrices. This makes me wonder whether this is generally true of all local operators when expressed in matrix form, or if there are some general statements we can make about the structure of the matrix forms of local and non-local operators.

  • 0
    @JamesWomack, did you finally end up with some interesting conclusions?2017-01-09

0 Answers 0