# Question regarding position and momentum representations

According to Cohen-Tannoudji's Quantum Mechanics book we can pick the following two bases composed by functions that doesn't belong to $$\mathscr{F}\in L^2(\mathbb{R^3})$$:

$$\xi_{\mathbf{r}_{0}}(\mathbf{r})=\delta(\mathbf{r}-\mathbf{r}_{0}) \\ \upsilon_{\mathbf{p}_{0}}(\mathbf{r})=(2\pi\hbar)^{-3/2}e^{\frac{i}{\hbar}\mathbf{p}_{0}\cdot\mathbf{r}}$$

They are not supposed to have a representation in the state space $$\mathscr{E_{\mathbf{r}}}$$ as kets (they do have a bra associated in $$\mathscr{E^{\mathbf{*}}_{\mathbf{r}}}$$ though), but the authors give them such representation considering them as generalized kets due to the fact that every function of $$\mathscr{F}$$ can be expanded in both bases. Therefore:

$$\xi_{\mathbf{r}_{0}}(\mathbf{r})\Longleftrightarrow |\mathbf{r}_{0}\rangle \\ \upsilon_{\mathbf{p}_{0}}(\mathbf{r})\Longleftrightarrow|\mathbf{p}_{0}\rangle$$

It is now clear for me that if we take the ket $$|\psi\rangle\in\mathscr{E_{\mathbf{r}}}$$ associated with a wave function $$\psi(\mathbf{r})\in\mathscr{F}$$, the coefficients of $$|\psi\rangle$$ in those bases ($$\lbrace|\mathbf{r}_{0}\rangle\rbrace$$ and $$\lbrace|\mathbf{p}_{0}\rangle\rbrace$$ representations) are given by:

$$\langle\mathbf{r}_{0}|\psi\rangle=\int d^3r\ \xi_{\mathbf{r}_{0}}^{\mathbf{*}}(\mathbf{r})\psi(\mathbf{r})=\psi(\mathbf{r}_{0})\\ \langle\mathbf{p}_{0}|\psi\rangle=\int d^3r\ \upsilon_{\mathbf{p}_{0}}^{\mathbf{*}}(\mathbf{r})\psi(\mathbf{r})=\overline{\psi}(\mathbf{p}_{0})$$

where $$\overline{\psi}(\mathbf{p}_{0})$$ is the Fourier transform of $$\psi(\mathbf{r}_{0})$$. At this point the authors decide to eliminate the $$0$$ subscript in the notation and to consider the following expressions for the coefficients:

$$\langle\mathbf{r}|\psi\rangle=\psi(\mathbf{r})\\ \langle\mathbf{p}|\psi\rangle=\overline{\psi}(\mathbf{p})$$

Why did they that? As far as I know the $$\lbrace|\mathbf{r}_{0}\rangle\rbrace$$ representation is the set of delta functions centered at the various points $$\mathbf{r}_{0}$$ of the space, therefore if we sum all the $$\psi(\mathbf{r}_{0})$$ we get our wave function $$\psi(\mathbf{r})$$. How is it possible $$\psi(\mathbf{r})$$ being the coefficients of itself? Am I missing something?

If you wish to write $$\mathbf{r}_0$$ and $$\mathbf{p}_0$$ repeatedly then you are absolutely free to do so, but I personally would get rather tired of writing a thousand unnecessary subscript zeros.
$$\langle\mathbf{r}_{0}|\psi\rangle=\int d^3r\ \xi_{\mathbf{r}_{0}}^{\mathbf{*}}(\mathbf{r})\psi(\mathbf{r})=\psi(\mathbf{r}_{0})$$ you should understand that the variable $$\mathbf{r}$$ which appears in the integrand is a dummy variable. You could just as well use $$\mathbf{y},\mathbf{x}$$, or an ornate smiley face if you were so inclined (and if you had the appropriate LaTeX packages installed).
$$\langle\mathbf{r}|\psi\rangle=\int d^3x\ \xi_{\mathbf{r}}^{\mathbf{*}}(\mathbf{x})\psi(\mathbf{x})=\psi(\mathbf{r})$$