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I guess mark this question as answered, though not fully. See the comments posted on the question itself.

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(7/31/2015) In the https://faculty.washington.edu/seattle/physics227/reading/reading-26-27.pdf document, they write the Hamiltonian = $(-\hbar^2 / (2m)) [d^2/dr^2 + (2/r)(d/dr) - \hat{L^2}/(r^2 \hbar^2)] + V(r)$. Then if you expand the expression $[\hat{H},\hat{L^2}]$, and use the fact that you can swap the order of partial derivatives because $\hat{L^2}$ is a function of $\theta, \phi$ only, it all cancels out to 0.

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(7/31/2015) They also answer the question of why we know we can separate $\psi = R(r)Y(\theta,\phi)$:

"An often asked question is 'How do you know you can assume that?' You do not know. You assume it, and if it works, you have found a solution. If it does not work, you need to attempt other methods or techniques. Here, it will work."

I guess mark this question as answered, though not fully. See the comments posted on the question itself.

I guess mark this question as answered, though not fully. See the comments posted on the question itself.

--

(7/31/2015) In the https://faculty.washington.edu/seattle/physics227/reading/reading-26-27.pdf document, they write the Hamiltonian = $(-\hbar^2 / (2m)) [d^2/dr^2 + (2/r)(d/dr) - \hat{L^2}/(r^2 \hbar^2)] + V(r)$. Then if you expand the expression $[\hat{H},\hat{L^2}]$, and use the fact that you can swap the order of partial derivatives because $\hat{L^2}$ is a function of $\theta, \phi$ only, it all cancels out to 0.

--

(7/31/2015) They also answer the question of why we know we can separate $\psi = R(r)Y(\theta,\phi)$:

"An often asked question is 'How do you know you can assume that?' You do not know. You assume it, and if it works, you have found a solution. If it does not work, you need to attempt other methods or techniques. Here, it will work."

Source Link
a00
  • 307
  • 1
  • 5
  • 13

I guess mark this question as answered, though not fully. See the comments posted on the question itself.