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In this PDF document (a lecture by Shivaly Reddy, page 13), he says that

$L^2$ is independent of $r$; therefore it commutes with any function of $r$.

This seems related to a problem in Schaum's Quantum Mechanics (Amazon link) dealing with a particle in a spherical potential well. In the solution, after writing down the Hamiltonian, they say,

It is evident that $[H, L^2] = 0$; hence, we write $\Psi = R(r) \cdot Y_{ml}(\theta,\varphi)$..."

  1. Both sources seem to be saying that when the potential well is spherically symmetric, $[H, L^2] = 0$. Why, exactly?

  2. Schaum goes on to say that since $[H,L^2] = 0$, then we can separate out the radial and angular parts of the wave function. Would you please explain that reasoning also?

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    $\begingroup$ qest 1: it's well detailled from page 13 last paragraph until the line preceding the conclusion in the middle of the next page ( impossible copy paste ) $\endgroup$
    – user46925
    Commented Jun 5, 2015 at 15:35
  • $\begingroup$ He says "An operator can commute with another independent operator" by which I think he means that if one operator depends only on certain variables, and another operator depends only on different variables, then they commute. L^2 depends on θ,φ only and Kinetic Energy depends on r only, thus they commute. $\endgroup$
    – a00
    Commented Jun 6, 2015 at 13:07
  • $\begingroup$ Basically I think if you have two operators that depend on different variables in this way, if you expand out the expression for the two different order of operations (for example, (L^2)*(KE) and (KE)*(L^2)) basically you get a long sum of various products. But all these products can be rearranged to look like the other order of operation's products because "anything on the inside of a derivative of θ that depends only on φ or r can come out of the derivative", and similarly for derivatives of φ and r. $\endgroup$
    – a00
    Commented Jun 6, 2015 at 13:11
  • $\begingroup$ So I am basically saying you can take partial derivatives in any order. I now read that one can only switch around the order of partial derivatives in this way if the partial derivative(s) of Ψ are continuous. (mhhe.com/math/calc/smithminton2e/cd/folder_structure/text/… about 2/3rds the way down) So I guess it's often glossed over that wave functions are continuous when differentiated with respect to r,θ,φ. (They are, right?) $\endgroup$
    – a00
    Commented Jun 7, 2015 at 15:28

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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."

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  • $\begingroup$ This is not an answer though. $\endgroup$
    – Gonenc
    Commented Jun 13, 2015 at 13:58
  • $\begingroup$ faculty.washington.edu/seattle/physics227/reading/… seems like it may have a worked out proof of $[\hat{H}, \hat{L^2}] = 0$ $\endgroup$
    – a00
    Commented Jul 31, 2015 at 13:29
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    $\begingroup$ You should then add the results in your answer for future reference. You can edit your answer by clicking the edit button. $\endgroup$
    – Gonenc
    Commented Jul 31, 2015 at 13:32

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