Bra ket notation rigorous way

Question

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.

Answer 1

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".

Answer 2

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}