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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.

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You seem to be invoking a concept of approximation that you haven't made explicit. Strictly speaking, locality only makes sense in the continuous case. What you seem to be asking about is how the locality of an operator in the continuous case is reflected in the matrix structure of an operator in a discretized approximation, but you'd need to explicate that approximation for there to be anything rigorous to say about this relationship. – joriki Dec 6 '11 at 11:30
@joriki So, you are saying that I would need to explicitly state the approximation used to discretize the continuous operator to make any statement about the structure of the matrix? – James Womack Dec 6 '11 at 12:05
Yes -- for instance, you can approximate the derivative operator in all sorts of ways; for instance, in a Fourier approximation, it makes sense to approximate it by multiplication with $\mathrm ik$, which amounts to convolution with a highly non-local function. – joriki Dec 6 '11 at 12:20
I can see then, how this question is not really meaningful. Thanks for your help. I'll leave this for now and maybe come back to it if I have a specific approximation I'd like to discuss. – James Womack Dec 7 '11 at 10:09

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