Simple spherical quantum mechanics question: r dot p

AI Thread Summary
The discussion centers on the equation relating the position vector and momentum operator in spherical coordinates, specifically the expression \(\vec{r}\cdot\vec{p} = -i\hbar r \frac{\partial}{\partial r}\). Participants clarify that this relation holds true when considering the momentum operator in the context of spherical coordinates. The equation is derived from the dot product of the radial position vector and the momentum operator, confirming its validity through mathematical manipulation. It is suggested to verify the consistency of this equation with the context provided in lecture notes and other resources. Understanding this relationship is crucial for grasping concepts in quantum mechanics.
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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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