Originally Posted by
MystyrMystyry
It's sort of a generalization of functional, matrix, and similar algebras.
Say you have a set of elements S. An operator is some function that maps S back onto itself. So lets take:
A:S?S, B:S?S, s?S.
Then A(s)?S, so we can take B(A(s)), and that will also belong to S. In this case, we can define an operator C:S?S, such that C(s)=B(A(s)) for all s?S. Such definition is consistent with a non-Abelian group, and the multiplication notation is used. That is we write C=BA. Action of operator on the set element can also be written in such notation. Because of the above, B(As)=(BA)s, and the identity function also exists, providing with an identity element, IA=AI=A. The operation, however, does not commute, so in general, AB?BA.
If there is also an addition group on the set S, we can define the same group on the operators, by requesting that (A+B)s = As+Bs. That allows us to play with the operators as if they were a ring, keeping in mind that the specific way they are "added" together is inherited from the way that the elements of the set are added together. That is, for every s,t?S, there must be an element s+t?S that follow all rules of addition.
Real functions are probably the first example of this you run into. For f,g:R?R, it holds that f?g(x)=f(g(x)), and (f+g)(x)=f(x)+g(x). That is simply the way that function composition and addition are defined. Typically, however, working with operators in an abstract way becomes useful when you start talking about functionals. Most basic example is function differentiation*. Operator ?²/?x² is defined as a product (?/?x)(?/?x), which is defined through its action on a function. That is: (?²/?x²)f(x) = (?/?x) ((?/?x)f(x)). And since we do have addition operation available, we can make use of it with operators acting on functions as well. Laplacian operator in 3 dimensions, for example, is defined as ?² = (?²/?x² + ?²/?y² + ?²/?z²).
When you start operating with objects like commutators, defined as [A,B] = AB-BA, it becomes extremely inconvenient to keep track of which of the above algebras you are dealing with, and simply use the generalized notation for operators. After all, the underlying algebra is the same.
This kind of notation is extremely useful when you have to describe behavior of something that is strictly speaking a multidimensional vector (which you'd need to describe various qualities of the sequel) and the way it evolves in time. So it seems like whatever that model will end up being, operator algebra would be a good way to describe it.
* Ok, so strictly speaking Functionals need to map to a real number, so they aren't exactly operators. a functional is something like f:GxS?Rn, where G={g|g:S?Rn} and GxS denotes a Cartesian product. But lets look at differentiation as a functional. Strictly speaking, we have something like (?f(x)/?x)|x=z. That is, derivative of f(x) at specific point x=z. Let us now look at collection of all such points for all z?R. What we get now is a function mapping from R to R. So we took a real function f(x), and we get another real function, one we denote as ?f/?x. In a similar manner, any functional can be viewed as an operator acting on a set of functions.
I think this turned into a rant. I did not intend it to drag on this long. Hope it is of some use to somebody.
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