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To summarize my understanding of https://www.homotopico.com/honest%20physics/quantum%20mechanics/classical%20mechanics/2018/09/23/momentum-generator-translations.html, the generator of transformations on classical observables that conserves momentum is the generator of translations: $$e^{\epsilon{X_p}}g(x)=e^{\epsilon\partial_x}g(x)=g(x+\epsilon)$$ Hence we say that momentum is the generator of translations. In quantum mechanics, we show that the operator $\hat{w}=-i\partial_x$ is the generator of translations: $$e^{ia\hat{w}}\psi(x)=\psi(x+a)$$ where $\psi(x)=\left \langle x|\psi \right \rangle$. Hence we claim that $\hat{w}$ is momentum and we write $\hat{w}|p\rangle=p|p\rangle$. Perhaps this is due to my lack of understanding of the connection between classical observables in phase space and operators in quantum mechanics, but it is still not completely clear to me why this is justified.

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    $\begingroup$ If we denote the state $|p\rangle$ as an eigenstate of the momentum operator with eigenvalue $p$, then by definition it follows that $\hat{p}|p\rangle = p |p\rangle$. Was that your question? $\endgroup$ Commented Jan 17, 2021 at 10:56
  • $\begingroup$ @Jakob My question is why can we claim $\hat{w}|p\rangle=p|p\rangle$ solely based on the fact that $\hat{w}=-i\partial_x$ is the generator of translations? $\endgroup$
    – Ayodan
    Commented Jan 17, 2021 at 18:10
  • $\begingroup$ So you want to ask about the existence of eigenstates of the momentum operator? $\endgroup$ Commented Jan 17, 2021 at 18:31
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    $\begingroup$ I believe the OP is asking about why we should identify the operator that generates translations ($-i\partial_x$ in the position basis) with the momentum operator. $\endgroup$ Commented Jan 17, 2021 at 18:35
  • $\begingroup$ @JoshuaTS Yes exactly. $\endgroup$
    – Ayodan
    Commented Jan 17, 2021 at 19:56

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Your first formula is wrong. The Lagrange shift operator dictates $$ e^{a\partial_x} \psi (x)=\psi(x+a). $$ Now, $$\hat p|p\rangle = p|p\rangle $$ is a definition of $|p\rangle$, so $e^{ia\hat p} |p\rangle = e^{ia p} |p\rangle$.

You then have $$ \langle x|e^{ia\hat p} |\psi\rangle = \langle x|e^{ia\hat p}\left ( \int dp |p\rangle \langle p|\right ) |\psi\rangle = \langle x| \left ( \int\!\! dp~ e^{ia p}|p\rangle \langle p|\right ) |\psi\rangle \\ = \left ( \int\!\! dp~ e^{ia p} e^{ixp} \langle p|\right ) |\psi\rangle = \langle x+a| \left ( \int\!\! dp~ |p\rangle \langle p|\right ) |\psi\rangle = \langle x+a|\psi\rangle \\ = \psi(x+a)= e^{a\partial_x} \psi(x) = e^{a\partial_x} \langle x|\psi\rangle . $$ So, comparing the leftmost with the rightmost elements of the equation sequence, you see that inside matrix elements involving x, the operator $i\hat p$ is represented as $ \partial_x$, generating an x-translation of functions of x, so outside matrix elements, ("in the x-representation").

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  • $\begingroup$ Is there a way to justify $\hat{p}=-i\partial_x$ without the use of $\langle{x|p}\rangle=e^{ipx}$, or a way to first justify $\langle{x|p}\rangle=e^{ipx}$ without the use of $\hat{p}=-i\partial_x$? $\endgroup$
    – Ayodan
    Commented Jan 17, 2021 at 21:02
  • $\begingroup$ Also I've fixed the sign on the shift operator, thanks. $\endgroup$
    – Ayodan
    Commented Jan 17, 2021 at 21:08
  • $\begingroup$ Not sure. $\langle x|p\rangle=e^{ipx}$ is baked into the bra-ket formalism, a virtual "axiom" of Dirac's. I'm not sure what alternate plank is available that is "more solid" to base it on... $\endgroup$ Commented Jan 17, 2021 at 21:28

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