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I am trying to find how a matrix component of a component of the momentum vector operator looks like. But I am not sure whether my solution is accurate. And if it is, I don't know how to proceed in order to simplify it.

\begin{align*} \langle \vec r|P_x| \vec r' \rangle &= \int \langle \vec r|\vec p\rangle \langle \vec p |P_x\vec r\rangle d^3p \\ &= \int p_x \langle \vec r|\vec p\rangle \langle \vec p|\vec r' \rangle d^3p \\ &= \int p_x\frac{1}{\sqrt{2\pi \hbar^3}}e^{i \frac {\vec p}{\hbar}\vec r}\frac{1}{\sqrt{2\pi \hbar^3}} e^{-i \frac{\vec p}{\hbar}\vec r'}d^3p \\ &= p_x \frac{1}{2\pi\hbar^3}\int e^{i \frac {\vec p}{\hbar}(\vec r- \vec r')}d^3p \\ &= p_x \frac{1}{2\pi\hbar^3} \delta(\vec r - \vec r'). \end{align*}

I know that $p_x=\frac {\hbar}{i} \frac{\partial}{\partial x}$. But I am not sure whether my result is correct and how I should proceed

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Start from what you know: $$ \langle{x}| \hat p|\psi\rangle =-i\hbar \partial_x \langle x|\psi\rangle $$ and choose $|\psi\rangle = |y\rangle$ to get $$ \langle{x}| \hat p|y\rangle =-i\hbar \partial_x \langle x|y\rangle\\ = -i\hbar\partial_x \delta(x-y)\\ = -i\hbar \delta'(x-y) $$

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  • $\begingroup$ Two questions: 1. If I start from where I wrote initially, is that wrong or impossible to find the component of the momentum operator in the position space? 2. what exactly is the "y" here? $\endgroup$
    – imbAF
    Oct 21, 2022 at 16:58
  • $\begingroup$ $y$ is the eigenvalue of the ket $|y\rangle$: I.e $\hat x |y\rangle = y |y\rangle$ $\endgroup$
    – mike stone
    Oct 21, 2022 at 16:59
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    $\begingroup$ You can't take the $p_x$ out of your integral over $p_x,p_y,p_z$. $\endgroup$
    – mike stone
    Oct 21, 2022 at 17:02
  • $\begingroup$ But shouldn't $\hat x |x\rangle=x|x\rangle$ and $\hat y |y\rangle=y|y\rangle$. If the x-component of the momentum operator acts on the eigenvector of the y component of the momentum operator, shouldn't that be equal to zero? $\endgroup$
    – imbAF
    Oct 21, 2022 at 17:06
  • $\begingroup$ I am working in one dimension. I did not want to use $x'$ in case you thought that that was a derivative. just replace $y$ by $x'$ if you see fit. $\endgroup$
    – mike stone
    Oct 21, 2022 at 17:14

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