# 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.

• near duplicate of physics.stackexchange.com/questions/364208/… – ZeroTheHero Jul 3 at 14:00
• I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. – David Z Jul 4 at 1:26

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".
You can define $$|\Psi\rangle$$ as: $$$$|\Psi\rangle=\int \Psi(y) |y\rangle d^3y$$$$ With $$\{|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)$$ we have : $$$$\langle x|\Psi\rangle=\int \Psi(y) \delta^{(3)}(x-y) d^3y=\Psi(x)$$$$