Simple spherical quantum mechanics question: r dot p

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SUMMARY

The discussion centers on the relationship between the radial position vector \(\vec{r}\) and momentum operator \(\vec{p}\) in spherical quantum mechanics, specifically articulated as \(\vec{r}\cdot\vec{p} = -i\hbar r \frac{\partial}{\partial r}\). This equation is derived using the gradient operator in spherical coordinates, \(\vec{\nabla_r}\), which is defined as \((\frac{\partial}{\partial r}, \frac{1}{r}\frac{\partial}{\partial \theta}, \frac{1}{r \sin \theta}\frac{\partial}{\partial \phi})\). The verification of this relation is confirmed through substitution and consistency checks with established quantum mechanics literature.

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Homework Statement


Maybe I missed it, but in my notes and also in documents like (http://ocw.mit.edu/courses/physics/...all-2013/lecture-notes/MIT8_05F13_Chap_09.pdf) (equation 1.64), I see

$$ \vec{r}\cdot\vec{p} = -i\hbar r \frac{\partial}{\partial r} $$

Where ##r## is the radial distance. Why is this relation true?

Homework Equations



$$ \vec{\nabla_r} = (\frac{\partial}{\partial r}, \frac{1}{r}\frac{\partial}{\partial \theta}, \frac{1}{r \sin \theta}\frac{\partial}{\partial \phi}) $$

The Attempt at a Solution


So is ##\vec{r}\cdot\vec{p}## simply

$$ (r, 0, 0) \cdot -i\hbar\vec{\nabla_r} = -i\hbar r \frac{\partial}{\partial r} $$

??
 
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Well done.
You should verify by context... is this consistent with what the notes and documents are trying to tell you?
 

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