I When is Â(r) Ψ(r) = ⟨r | Â | Ψ⟩?

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The discussion centers on the equation Â(r) Ψ(r) = ⟨r |  | Ψ⟩ and its validity. Participants analyze the right-hand side of the equation, exploring the implications of assuming the operator's action leads to a simplified form. They highlight the need for clarity in notation, particularly regarding the operator  acting on function spaces versus state spaces. The conversation emphasizes the formalism of quantum mechanics, particularly the relationship between position eigenstates and state functions. Ultimately, the question remains whether the equation holds true universally in all contexts.
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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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For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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