How Does the Quantum Operator \(\hat{p}^2\) Derive from \(\hat{p}\)?

AI Thread Summary
The discussion centers on deriving the quantum operator \(\hat{p}^2\) from \(\hat{p}\) using the given expression for \(\hat{p}\). The initial attempt at the solution leads to an incorrect result, lacking the necessary \(2/r\) factor. Participants emphasize the importance of treating \(\hat{p}\) as an operator that acts on a function, which is crucial for proper manipulation. A suggestion is made to use a test function to simplify the derivation process. The conversation highlights the complexities involved in operator algebra within quantum mechanics.
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Homework Statement


Given that \hat{p} = -i\hbar (\frac{\partial}{\partial r} + \frac{1}{r}), show that \hat{p}^2 = -\frac{\hbar^2}{r^2} \frac{\partial}{\partial r}(r^2 \frac{\partial}{\partial r})

Homework Equations



Above

The Attempt at a Solution


I tried \hat{p}\hat{p} = -\hbar^2((\frac{\partial}{\partial r})^2 + \frac{1}{r} \frac{\partial}{\partial r} + \frac{\partial}{\partial r}\frac{1}{r} +\frac{1}{r^2}).

This gave me -\hbar^2((\frac{\partial}{\partial r})^2 + \frac{1}{r} \frac{\partial}{\partial r} ) instead of the 2 / r factor I needed.
 
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Nope. ##{\partial \over \partial r}{1\over r} ## gives ##{1\over r}{\partial \over \partial r} -{1\over r^2}##
Remember p is an operator: you have to imagine there is something to the right of it to operate on.
 
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BvU said:
Nope. ##{\partial \over \partial r}{1\over r} ## gives ##{1\over r}{\partial \over \partial r} -{1\over r^2}##
Remember p is an operator: you have to imagine there is something to the right of it to operate on.

Thanks!

I suppose it makes it easier if I had used a test function, and then taken it away.
 
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