1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.