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I apologize for this kind of silly question, I haven't brushed up on QM for a while.

I was looking at a problem today, essentially I'm given some operator $V = \lambda |{\xi}\rangle \langle{\xi}|$ where $|{\xi}\rangle$ is normal such that $\langle p| {\xi}\rangle = e^{-p^2/\lambda^2}$ where $\{|p\rangle\}$ are momentum eigenstates (I'm working in 3D space where $p^2 = \vec{p} \cdot \vec{p}$ but will refrain from using vector notation). Essentially, I have to find some integral that contains $V(r)$, which is the position representation of the operator $V$.

Now my concern is not the problem itself (as it's just an exercise I wanted to do for fun), but what exactly is meant by the position representation of an operator. At first I thought it should be as simple as $$V(r) = \langle r' | V |r\rangle \ \ \ \ \ \ (1)$$ With $|r\rangle$ being the position eigenstates and then I can insert momentum basis identities to simplify. This is 'equivalent' to the matrix elements of an operator in case of discrete bases. But I remember from one of my textbooks that, the position representation of momentum operator is found using

$$\langle r| \hat{p} | \psi \rangle = -i\hbar \nabla \langle r|\psi\rangle$$

Where then one identifies $\hat{p} =-i\hbar \nabla$ suggesting that the position representation of an operator is found through the relation $$\langle r| V \psi \rangle = V(r) \langle r| \psi \rangle \ \ \ \ (2)$$ for all $|\psi\rangle$, which kinda makes sense, if one identifies $V|\psi\rangle = |\psi' \rangle$, then this says $V(r) \psi(r) = \psi'(r)$, or that $V(r)$ maps functions from position space to position space. Is this what the position representation of the operator $V$ mean?

I apologize for making the question so long and verbose, but I'm afraid I may be confusing some concepts. Does (1) represent the matrix elements of V in position basis, and one finds the position representation of V through (2), and they are different (since they don't seem compatible)? Is none of them actually what we mean by $V(r)$, like the potential $V(r)$ in Schrodinger's eqn.? I've always taken the fact that we know $V(r)$ as a function of position for granted and this problem is giving me semi existential crisis.

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    $\begingroup$ Eq. $(1)$ does not make sense and your notation seems also confused. If you have a position-operator (X) dependent potential $V(X)$, then $\langle x|V(X)|\psi\rangle=V(x)\langle x| \psi\rangle =V(x) \psi(x)$, by the very definition of functions of operators and the convention/definition $\psi(x):=\langle x|\psi\rangle$. If $|\psi\rangle=|x^\prime\rangle$, then you find $\langle x|V(X)|x^\prime\rangle=V(x) \langle x|x^\prime\rangle = V(x)\delta(x-x^\prime)$. Does that clear up a bit the confusion? $\endgroup$
    – Tobias Fünke
    Commented Sep 18, 2023 at 9:49
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    $\begingroup$ And yes, roughly speaking: This induces what colloquially is called the position representation. For an operator $V$ on an abstract space $H$, you can define an operator $\tilde V$ on $L^2(\mathbb R)$ via $\langle x|V|\psi\rangle=\int \mathrm dx\, V(x,x^\prime)\psi(x^\prime):=(\tilde V\psi)(x)$ with $V(x,x^\prime):= \langle x|V|x^\prime\rangle$. So if $V$ maps $|\psi\rangle$ to $|\phi\rangle$, then $\tilde V$ maps $\psi \in L^2$ to $\phi\in L^2$. So any vector in $H$ is in a one-to-one correspondence with a function in $L^2$, and likewise for operators on those spaces (discarding rigor). $\endgroup$
    – Tobias Fünke
    Commented Sep 18, 2023 at 9:53
  • $\begingroup$ @TobiasFünke in regards to the first comment, what I meant to say is that if $V$ is some arbitrary operator, not necessarily represented as a function of the position operator $x$, and we wanted to find out what it looks like in position space, eqn. (1) was what first came to mind (as in ij-th matrix elements of the matrix V in position states), but I see that it's not right because elements of an operator are not the operator itself. In regards to $\langle x| V(X) |x' \rangle = V(x) \langle x|x' \rangle$, could we say that $\langle x| V(P) |x' \rangle = V(-ih d/dx) \langle x|x' \rangle$? $\endgroup$
    – Abdul Qadeer
    Commented Sep 18, 2023 at 11:02
  • $\begingroup$ And thanks for the second comment, I think that clears up what position representation means. $\endgroup$
    – Abdul Qadeer
    Commented Sep 18, 2023 at 11:06

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