Operator mapping in Hilbert space

[SeM]

That notation makes no sense, as $H$ is a set/space/Hilbert Space of functions. It's not a set of numbers. An "operator" is just a function that maps functions to functions (or vectors to vectors).

In finite-dimensional vector spaces, Linear Operators are represented by matrices. But, the set of square integrable functions is actually an infinite dimensional vector space. And, this makes this more complicated than the finite-dimensional case.

Note that finite dimensional Hilbert Spaces turn up in QM when you consider spin, for example. And,in these cases, the linear operators are indeed represented by matrices: e.g. the Pauli spin matrices.

You may be confused because both $\mathbb{R}$ and $\mathbb{C}$ are, in fact, Hilbert spaces in their own right. But, in terms of square-integrable functions and QM this is irrelevant.

In terms of Hilbert spaces, very roughly, you have:

$\mathbb{C}$ - not relevant to QM as a Hilbert space, but as a field of scalars

$\mathbb{C}^n$ - Hilbert spaces in which spin states reside

$\mathcal{H} \equiv L^2$ - Hilbert space in which wavefunctions reside.

Thanks again Perok, this was indeed very useful. I thought C and R where for wavefunctions, but indeed, they are for coordinates. So when one works with wavefunction and their transformation by operators, they are given in H, because H is not for coordinates, but for the infinity and continuity of wavefunctions and their derivatives, integrals and so one. This actually gave me a complete comprehension of the Hilbert space. It has therefore nothing to do with coordinates.Thanks, this was one of the best lectures I have had in a long time.

Cheers