Question regarding position and momentum representations

Question

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?

Answer 1

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.

When you write

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

In other words, the above is exactly the same as

$$\langle\mathbf{r}|\psi\rangle=\int d^3x\ \xi_{\mathbf{r}}^{\mathbf{*}}(\mathbf{x})\psi(\mathbf{x})=\psi(\mathbf{r})$$

where we've chosen to use symbols that minimize notational clutter.