I When is Â(r) Ψ(r) = ⟨r | Â | Ψ⟩?

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


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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?
 
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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.
 
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LightPhoton said:
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##.
 
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