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Gennaro Auletta in his book makes the following argument to show that multiplicative operator acting on his eigenvectors acts in a multiplicative way on eigenfunctions as well. Here's the argument.

We write the equation for position operator eigenvectors:

$\hat{X}$ $\left| x \right>$ = $x\left| x \right>$

and since $\hat{X}$ is a hermitian operator we can write:

$\left<{x}\right|\hat{X}$ = $\left<x\right|x$.

Then we multiply the relation above with a state vector $\left|{\psi}\right>$ to get:

$\left<x\right|\hat{X}\left|{\psi}\right> = \left<x\right|x\left|{\psi}\right>$.

Now

$\hat{X}\psi(x) = x\psi(x)$.

My question is: why are we allowed to get the operator $\hat{X}$ out in the last step?

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    $\begingroup$ This is a really confusing abuse of notation -- he's switching between two completely different definitions of $\hat{X}$. He's only allowed to "get the operator out" by switching it to a different operator. $\endgroup$ – knzhou Aug 25 '19 at 22:38
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    $\begingroup$ I recommend using a different book... there are a lot of books out there that are extremely confusingly written, and they do a lot of harm to tens of thousands of people per year. $\endgroup$ – knzhou Aug 25 '19 at 22:38
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    $\begingroup$ Does your book actually say that $\hat X\psi(x)=x\psi(x)$, or are you saying that? $\endgroup$ – Aaron Stevens Sep 4 '19 at 16:31
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    $\begingroup$ I don't think $\hat X$ is taken out at all: the abstract expression $\hat X|\psi\rangle$ when representing $\psi$ by a function in some function space is exactly equal to $\hat X\psi$ (where, as knzhou remarked, the operator is the one corresponding to the abstract $\hat X$, not strictly speaking identically the same one). He does not take the operator out, this is literally $\hat X|\psi\rangle$ is exactly the same as $\hat X\psi$, and $\langle x|$ on a function is written as function evaluation. $\endgroup$ – doetoe Sep 4 '19 at 17:15
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    $\begingroup$ In short: the operator isn't taken out, you should read this as $\hat X\psi(x) := \left(\hat X\psi\right)(x)$. $\endgroup$ – doetoe Sep 4 '19 at 17:19
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This is the oldest abuse of notation in the book, namely using the same symbol for operators acting on vectors (kets) as you use for such in a representation (bras, hence wavefunctions). Physicists are blasé about it, assuming you understand what they mean.

As per @knzhou 's suggestion, for the purposes of this discussion (only!), I'll use different symbols for each, $\hat A$ and $\tilde A$, respectively, $$ \tilde {A} \psi (x) \equiv \langle x| \hat A |\psi \rangle ~. $$

This is to say $$ \tilde A \langle x | \psi \rangle \equiv \int \!\! dy ~~ \langle x| \hat A |y\rangle \langle y |\psi \rangle . $$ So you may think of $\tilde A$ as a matrix representation of $\hat A$ in x-space (here, but you could equally work in p-space, l,m-space, ...), where you contract over indices (y) of it with those of the "vector in x-space", here the wavefunction. Assuming one understands what is meant, one conflates the tilde and the caret, and all is fine. Books using this ritual abuse of notation, however, ought to at least throw in an explanatory footnote...

For your particular operator, of course, $$ \hat X = \int \!\! dy ~~ |y\rangle y \langle y | \\ \tilde X \psi(x) = \int \!\! dy ~~ x \delta(x-y) ~ \psi (y) =x \psi(x), $$ diagonal in this representation. The authors of your text simply stress that $\tilde X$ and $\hat X$ have the same eigenvalues.

But, of course, the momentum, e.g., is not diagonal, in this representation, $$ \hat P = \int \!\! dy ~~ |y\rangle \frac{\hbar}{i} \partial_y \langle y | \\ \tilde P \psi(x) = \frac{\hbar}{i} \partial_x~ \psi (x) =-i\hbar \partial_x \psi(x), $$ and so on.

In real life, one just uses the caret without excessive confusion.

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    $\begingroup$ Abuses of notation are forgivable as long as the author is careful the first time an abuse of that type is committed. Understanding how things must be taught differently the first time is what distinguishes good pedagogy from bad. $\endgroup$ – knzhou Sep 4 '19 at 6:11
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    $\begingroup$ True, what the book does is "standard". But it inevitably sends confused students to places like this, where they can receive a real explanation. The real explanation has to be provided at some point to achieve understanding -- and I just believe the right place should be the book the student is already reading. $\endgroup$ – knzhou Sep 4 '19 at 6:13
  • $\begingroup$ @knzhou But you recommend using a different book in the comments to the question? $\endgroup$ – Aaron Stevens Sep 4 '19 at 13:59
  • $\begingroup$ @Aaron Stevens Sorry, I mean that this should be explained in any decent introductory book the first time it comes up. The fact that it wasn’t in OP’s book means they should throw that book out the window. $\endgroup$ – knzhou Sep 4 '19 at 15:51
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    $\begingroup$ @Aaron Stevens. Well, no, I wouldn't quite blame the victim... (2.122) and (2.124) use the very same symbol for the operator and for the matrix representation thereof. For somebody who has never seen the stunt, it appears disorienting, although I have not studied the book in sufficient detail to blame the authors. It certainly has numerous strong points to avoid the extrafenestrial landfill discussed... $\endgroup$ – Cosmas Zachos Sep 4 '19 at 21:14
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Because of the penultimate equation, for any $\psi$, effect of $\hat{X}$ is the same as effect of multiplying the ket by $x$.

However, these operations are heuristic, they are not really justified by such a loose language as "hermitian operator so we can write something involving kets $|x\rangle$"; those kets do not belong to any Hilbert space, one has to go into functional analysis and theory of rigged Hilbert space to justify these things.

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