In general quantum mechanics we represent the state of a system with a state vector $| \psi \rangle $ in some Hilbert space in some base. Assuming a complete discrete set of bases vectors $ |n \rangle $ we can write the state $\psi $ as:
$$| \psi \rangle = \sum_n ^N c_n |n \rangle $$
Having by definition an inner product we can see that $c_n = \langle \psi | n \rangle $. This is the projection of the state vector on the base vector $n$. This coefficient squared, we say, gives us the probability of finding out system in the state $n$.
As I understand so far $ c_n $ is a complex number. Also, as a function, $\psi$ is a function of variable same as the variable of $n$, where $n$ are also functions.
In position representation we have a base denoted as $| x \rangle $. So to the question.
I would say that:
$$| \psi \rangle = \sum_n ^N c_n |x \rangle $$
$$c_n = \langle x| \psi \rangle $$ where $|x \rangle =f(x) $ so that, in this bases $| \psi \rangle = \psi (x)$, x meaning position.
My problem is that from what I read, what holds is this:
$$c_n = c_n (x) = \langle x | \psi \rangle = \psi (x) ,$$ $ \psi (x)$ being the wavefunction– the projection of $|\psi \rangle $ on $|x \rangle $ and
$$\psi(x) = \sum_n ^N c' _n u_n (x) $$, where $ u_n (x)$ constitute a base for the position representation. So, the probability of finding the particle in position $x$ is $| \psi (x) |^2 $. But I thought that $|x \rangle $ was already our base in space.
Are $u(x) $ and $| x \rangle $ the same or not? If not, why do we need $u(x)$?
Is $\psi (x) $ as given in the end by the inner product a function that can be expressed in functions of $x$ that constitute a base or is it a number as a result of an inner product giving the coefficient in a sum? Can the coefficients be functions and if yes, when? Is it correct to say that the wavefunction in secondary to the state vector and if so, how is it a function if it is a coefficient?
There is certainly something here I don't get and I would really like your help and clarifications.
Note: I wrote the question with sums and not integrals for convenience. An answer might have as well integrals, there is clearly no problem. Also I' ve had a look in these questions and answers:
General wavefunction and Schrödinger Equation
State of a system in Quantum Mechanics and state vectors
Representations in quantum mechanics
Vector representation of wavefunction in quantum mechanics?
but I'm not sure they address my problem.
Best Answer
Your confusion stems from the fact that $\lvert x \rangle$ is not inside the Hilbert space of states. It cannot be because $\langle x \vert x \rangle = \delta(x-x) = \delta(0)$ is not an allowed value for an inner product in a Hilbert space to have. There are several things to say about $\lvert x \rangle$:
If you want to make precise what kind of objects $\langle x \rvert$ and $\lvert x \rangle$ are, you need the notion of rigged Hilbert spaces. A nice answer about them is here by user1504.
In general, the "eigenstates" associated to eigenvalues in the continuous spectrum of an unbounded operator do not lie inside the space of states. Only the discrete eigenvalues have proper eigenvectors.
The set of objects $\lvert x \rangle$ is uncountable and hence not a basis in the usual sense. In particular, writing $\sum_x c_x \lvert x \rangle$ doesn't make sense because series over uncountable sets do not converrge. The analogon for $\lvert \psi =\sum_n c_n\lvert n \rangle$ for a countable Hilbert basis $\lvert n \rangle$ is writing $$ \lvert\psi\rangle = \int \psi(x)\lvert x \rangle \mathrm{d}x\tag{1}$$ where writing $\psi(x) = \langle x \vert \psi \rangle$ leads to $$ 1 = \int \lvert x \rangle \langle x \rvert \mathrm{d}x$$ It is in the sense of $(1)$ that the wavefunction gives the coefficients in the position basis.