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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.

Qmechanic
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Adrien Amour
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    Look this up in any introductory quantum mechanics text book. – hft Aug 26 '22 at 21:29
  • https://www.youtube.com/watch?v=fTRKMAj5q_4&t=1800s Lecture 9 | The Theoretical Minimum Leonard Susskind – jelly ears Aug 26 '22 at 22:01
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    I don't understand what is the question. The fact that $X$ and $P$ act respectively as multiplication by $x$ and derivative wrt $x$ (or vice-versa) follows from the canonical commutation relations $[X,P]=i\hbar$ and the Stone-von Neumann theorem (https://en.wikipedia.org/wiki/Stone%E2%80%93von_Neumann_theorem). Once you know this, you may look for eigenvectors of $P$ in the position representation and they must obey a first-order ODE whose solution is a plane wave. – Gold Aug 26 '22 at 22:44
  • Possible duplicate: https://physics.stackexchange.com/q/41880/2451 – Qmechanic Aug 26 '22 at 23:25
  • @Gold It doesn't seem obvious to me that we should know the commutation relation between $X$ and $P$. I've only seen this derived using the fact that the momentum operator in position space is $-i\hbar\partial_{x}$. – Adrien Amour Aug 27 '22 at 10:42
  • @AdrienAmour the commutation relation between $X$ and $P$ has its origins in the Poisson bracket between the corresponding classical variables $x$ and $p$ which obey $${x,p}=1.$$ Canonical quantization is exactly that: promoting the classical variables obeying some sort of Poisson bracket relation to operators in a Hilbert space and translating ${,}\to -i\hbar [,]$. In that case, canonically quantizing the classical particle in one-dimension means looking for a Hilbert space with Hermitian operators $X$ and $P$ obeying $[X,P]=i\hbar$. – Gold Aug 28 '22 at 01:38
  • Also let me add that both in classical mechanics and in quantum mechanics the relations ${x,p}=1$ and $[X,P]=i\hbar$ have one meaning. They are saying, both in classical phase space and in the quantum Hilbert space, that position and momentum are related by the fact that momentum is the generator of translations, which basically shifts the position. In fact, saying that momentum is the generator of translations is basically the more precise definition of momentum, and the CCR both in classical phase space and quantum Hilbert space are ways to express that. – Gold Aug 28 '22 at 01:41

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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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