Question on notation of eigenvectors of position operator in quantum mechanics
1. From the first postulate of quantum mechanics we known that the vector $|\psi\rangle$ is the mathematical entity that says, intuitively, "in a time $t$, the (state of a) system is a vector".
2. Then, by the virtue of the second postulate, the next fundamental ingredient of a physical theory are expressed by operators $\mathcal{O}$ acting on Hilbert space.
with 1. we have a formal way to say how to represent our physical phenomena; with 2. we have a formal entity that carries a way to extract important information about our physical phenomena: position $\mathcal{X}$, linear momentum $\mathcal{P}$, etc...
Moreover, about 2., I read the second postulate (using now the position operator) as:
$"$I want say something about $|\psi\rangle$ currently position. Therefore, I must to apply the position operator on $|\psi\rangle$ as:
$$\mathcal{X}|\psi\rangle." \tag{1}$$
1. 669
From the third postulate, the way formal way that we extract information (we want to extract a numerical result) about $(1)$ is by virtue of the help of an eigenproblem such as:
$$\mathcal{X}|\psi\rangle = x|\psi\rangle \tag{2}$$
Expression $(2)$ is the root of my confusion. I'm going to explain it in the following with physics jargon.
$$---------$$ Suppose we have a system (a electron say) $| \psi \rangle$. Now, I want to discover the position of this electron. Therefore, I must to solve: $\mathcal{X}|\psi\rangle = x|\psi\rangle$.
On the one hand, we know, from postulate 1 of quantum mechanics, that the system state is given by:
$$|\psi\rangle = a_{1}|\psi_{1}\rangle+...+a_{n}|\psi_{n}\rangle.\tag{3}$$
Then,
$\mathcal{X}(a_{1}|\psi_{1}\rangle+...+a_{n}|\psi_{n}\rangle) = x(a_{1}|\psi_{1}\rangle+...+a_{n}|\psi_{n}\rangle) = xa_{1}|\psi_{1}\rangle+...+xa_{n}|\psi_{n}\rangle. \tag{4}$
On the other hand, we know, that the mere existence of a eigenvalue $x$ and it position operator $\mathcal{X}$ states the existe of eigenvectors as $|\psi_{x}\rangle \equiv|x\rangle$:
$$\mathcal{X}|\psi_{x}\rangle = x|\psi_{x}\rangle \equiv \mathcal{X}|x\rangle = x|x\rangle \tag{5}$$.
Together with the fact that $\mathcal{X} = \mathcal{X}^{\dagger}$, we know that $|\psi_{x}\rangle \equiv|x\rangle$ forms an (infinite) basis for the Hilbert space and the system (the electron say) can be written as:
$$| \psi \rangle = \int c(x)|x\rangle dx \equiv \int c(x)|\psi_{x}\rangle dx. \tag{6}$$
The thing is, when I wrote $(3)$, I kept in my mind the following reasoning:
The whole superpositon $|\psi\rangle$ are expanded in terms of system states $|\psi_{n}\rangle$; the basis kets are the same "electron". Then in $(4)$ I have the position information $x$, the coefficients $a_{i}$ and the same electron state $|\psi_{i}\rangle$ (that will be possibly collapsed after the measurement).
but when it comes equation $(5)$ I kept in my mind the following reasoning:
I'm using the whole linear algebra to construct a particular basis, the position basis (since I'm using the position operator), and then I write the whole state in terms of these position vectors $| x \rangle$.
I'm very, very confused. 66