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Nov 28, 2018 at 17:55 vote accept Gjert
Nov 28, 2018 at 17:54 comment added Gjert Great explanation, thank you! Also, it's a bit off topic, but could you lead me in the direction of how the momentum operator was motivated? I've tried looking for it in Classical Mechanics book, but just states it as a definition without reasoning.
Nov 28, 2018 at 17:47 comment added eranreches The power comes from the dimensionality of the problem. In general, in $d$ dimensions the power is $\frac{d}{2}$. Note that $\hbar$ also appears because in your case the integration variable is $p$ instead of $k$.
Nov 28, 2018 at 17:42 comment added Gjert If $\hbar$ is for units, is there a special reason for $\hbar^{-3/2}$? I understand the $(2\pi)^{-3/2}$ from the def. of Fourier transforms, but not sure where $\hbar$ is coming from.
Nov 28, 2018 at 17:37 history answered Sean E. Lake CC BY-SA 4.0