Starting from the reasoning that $\langle x|p \rangle=e^{\frac{ipx}{\hbar}}$, I understand why the momentum operator in position space is $-i\hbar \partial_{x}$. What I'm looking for is some sort of intuitive reasoning as to why $\langle x|p \rangle=e^{\frac{ipx}{\hbar}}$. I've only found this derived from the definition of the momentum operator in position space, but I'm looking to go the other way.
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The intuitive reason is that the momentum vector is a plane wave in the coordinate basis and this follows from the uncertainty principle.
jelly ears
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Let us first understand what $|x\rangle$ and $|p\rangle$ are. They are (by definition) eigenfunctions of operators $\widehat{x}$ and $\widehat{p}$ with eigenvalues $x$ and $p$, respectively. So the first of them is a certain $\delta$-function, and the second one is an exponent. So to calculate $\langle x | p\rangle$, you need to take the Hermitian conjugate of $|x\rangle$, multiply by $|p\rangle$, and integrate.
akhmeteli
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