Bra ket notation rigorous way I'm struggling to see how $\langle x|\Psi\rangle =\Psi(x)$.
I have read a few previous bra ket questions in here but still not clear.
Any good book for understand the bra-ket notation in more rigorous way.
 A: You can define $|\Psi\rangle$ as:
\begin{equation}
|\Psi\rangle=\int \Psi(y) |y\rangle d^3y
\end{equation}
With $\{|y\rangle \ \,|\,y \in \mathbb{R}^3\}$ the basis of the hilbert space $H$ of positions.
Since $\langle x|$ is the linear form such that $\langle x|(|y\rangle)\equiv \langle x|y\rangle=\delta^{(3)}(x-y)$ we have :
\begin{equation}
\langle x|\Psi\rangle=\int \Psi(y) \delta^{(3)}(x-y) d^3y=\Psi(x)
\end{equation}
A: Consider the case of a vector space of countable dimension, with some orthonormal set of basis kets $\left\{\vert\mathbf{e}_i\rangle\right\}$. The orthonormality condition is stated as $\langle \mathbf{e}_i \vert \mathbf{e}_j \rangle = \delta_{ij}$, where $\delta_{ij}$ is the Kronecker delta. We can then expand any vector in this basis,
$$\vert \psi \rangle = \sum_i \psi_i \vert \mathbf{e}_i \rangle, $$
where the $\psi_i$ are the components of $\vert \psi \rangle$, i.e. they are the projections of $\vert \psi \rangle$ along the basis vectors $\vert \mathbf{e}_i \rangle$, which we can state more technically by noting that the identity matrix can be written as
$$I = \sum_i \vert \mathbf{e}_i \rangle \langle \mathbf{e}_i \vert,$$ in which case
$$\vert \psi \rangle = I \vert \psi \rangle = \sum_i \vert \mathbf{e}_i \rangle \langle \mathbf{e}_i \vert \psi \rangle,$$
i.e.
$$\psi_i = \langle \mathbf{e}_i \vert \psi \rangle.$$
Now we generalize this to an uncountable basis. For example, we define the position basis as the set $\left\{\vert \mathbf{x} \rangle \,\vert\, \mathbf{x} \in \mathbb{R}^3 \right\}$. Now the orthonormality condition is slightly modified (for technical details you can read about "rigged" Hilbert spaces), $\langle \mathbf{x} \vert \mathbf{x}' \rangle = \delta^{3}(\mathbf{x} - \mathbf{x}')$ where $\delta^3$ is the three-dimensional Dirac delta. Then we can expand the identity operator (it is no longer a matrix when the basis is uncountable) as
$$I = \int_{\mathbb{R}^3} d^3\mathbf{x}\, \vert \mathbf{x} \rangle \langle \mathbf{x} \vert.$$
Then, as before, we expand a vector $\vert \psi \rangle$ as
$$\vert \psi \rangle = I\vert \psi \rangle = \int d^3\mathbf{x} \, \vert \mathbf{x} \rangle \langle \mathbf{x} \vert\psi \rangle \equiv \int d^3\mathbf{x} \, \psi(\mathbf{x})\,\vert \mathbf{x} \rangle,$$
so the wavefunction $\psi(\mathbf{x})$ is simply the components of the vector $\vert \psi \rangle$ along the basis vectors $\vert \mathbf{x} \rangle$, just like in the countable case. The only difference is that now $\mathbf{x}$ labels the basis vectors instead of the discrete index $i$.
$$\psi_i \equiv \langle \mathbf{e}_i \vert \psi \rangle \leftrightarrow \psi(\mathbf{x}) \equiv \langle \mathbf{x} \vert \psi \rangle \quad\,\,$$
$$\vert\psi\rangle = \sum_i \psi_i \vert \mathbf{e}_i \rangle \leftrightarrow \vert \psi \rangle = \int d^3\mathbf{x}\, \psi(\mathbf{x}) \vert \mathbf{x} \rangle$$
For a pedagogical introduction, I recommend the notes found on this page, in particular "Block 1: Mathematical Foundations".
