Translation operator and momentum operator

(personal thinking)
In a Hilbert space like $C^{5}$ we have the position operator

\hat{X} = [\begin{matrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 0 & 5 \end{matrix}]

encoding the position variable $x$ but in a QM flavour (see formulation of QM). And we can consider the matrix which is something like the translation operator:

T = (\begin{array}{lllll} 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{array})

Given a vector $v = (a, b, c, d, e)$ , the result of the operator is "to translate" the coordinates:

T v = (e, a, b, c, d) .

We can consider the matrix $P$ such that the matrix exponential $e^{P} = T$ , son $P$ can be seen like the generator of translations. I think that $P$ is something like the momentum operator in this setup. This operator has a special basis of eigenvectors, representing the pure momentum states.
Given a vector $v$ in the initial basis, we could rewrite $v$ in the new basis.

In the continuous case everything is analogous: the vectors are functions $f$ and the Hilbert space is a space of functions. The translation operator is

T : f (x) \mapsto f (x - 1)

According to Taylor's theorem,

T (f) (x) = f (x - 1) =

= f (x) + f^{'} (x) ((x - 1) - x) + \frac{1}{2} f^{″} (x) ((x - 1) - x)^{2} + \dots =

= f (x) + (- 1) f^{'} (x) + \frac{1}{2} (- 1)^{2} f^{″} (x) + \dots =

= e^{- \partial_{x}} (f),

so we have the momentum operator $P = - \partial_{x}$ . The eigenvectors of this operator are

e^{s x}, s \in C

So Laplace transform is nothing but a change of basis: from the Dirac delta basis (the eigenbasis of "position") to the eigenbasis of momentum.

Why do we choose only some eigenvectors in Quantum Mechanics, i.e., only

e^{i ω x}, ω \in R

i.e., why we take Fourier transform.

In quantum mechanics, the states of a system are described by wavefunctions. The wavefunctions that are physically meaningful are those that are square-integrable (bound states), meaning their absolute square integrates to 1 over all space. This ensures that the total probability of finding a particle somewhere in space is 1.
The functions $e^{s x}$ are not square-integrable for all $s \in C$ . However, the functions $e^{i ω x}$ are orthogonal over all space, and they form a basis for the space of square-integrable functions. This is why the Fourier transform, which uses these functions, is so useful in quantum mechanics.