When is Â(r) Ψ(r) = ⟨r | Â | Ψ⟩?

[LightPhoton]

Wikipedia says that the equation in the title is defined to be true.

But is it true always?

Working with the right-hand side,

$$\langle \mathbf r\vert\hat A\vert\Psi\rangle=\int\langle \mathbf r\vert\hat A\vert\mathbf r'\rangle\langle\mathbf r'\vert\Psi\rangle\ d\mathbf r'$$

If we assume that $\langle \mathbf r\vert\hat A\vert\mathbf r'\rangle=\hat A(\mathbf r)\delta(\mathbf{r-r'})$, then above expression collapses to $\hat A(\mathbf r)\Psi(\mathbf r)$. But how to show this in general?

 

[anuttarasammyak]

Conventionally
$$\psi(r)=<r|\psi>$$
So
$$\phi(r)=<r|\phi>=<r|\hat{A}|\psi>$$
where
$$\hat{A}|\psi>=|\phi>$$
So I prefer the notation for the definition
$$[\hat{A}\psi](r)=\hat{A}\psi(r) \equiv <r|\hat{A}|\psi>$$
with no (r) for A. I have no idea what $\hat{A}(r)$ could mean.

 

[PeroK]

Wikipedia says that the equation in the title is defined to be true.


View attachment 361447


But is it true always?

It's worth noting the formalism here. We start with $r$ as the coordinate(s) of a point in space. I.e. $r \in \mathbb R^3$. And we have $|r\rangle$ as the position eigenstate (or eigenket) associated with that point. We also have some state $|\Psi \rangle$.

Now, we can define a complex-valued function (on $\mathbb R^3$) by:
$$\Psi(r) \equiv \langle r|\Psi \rangle$$Where $\langle r|$ is the bra associated with the eigenket $|r\rangle$.

Next, we have some operator, $\hat A$, that acts on the Hilbert space of states/kets. From this we can define an operator, $\hat A(r)$ in this notation, on our function space of complex-valued functions, using:
$$[\hat A(r)\Psi](r) \equiv \langle r|\hat A|\Psi \rangle$$Note that this notation is slightly odd and it would be more usual to write something like $\hat A'$ or $\hat A_f$, to indicate that this operator is technically not the same as $\hat A$, but acts on the function space - rather than the space of states/kets. I.e. I would tend to write:
$$\hat A' \Psi(r) \equiv [\hat A'\Psi](r) \equiv \langle r|\hat A|\Psi \rangle$$In any case, note that this defines the action of the operator $\hat A'$ on any function $\Psi$.