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When solving Schrodinger's equation in 3D with a spherical laplacian you reach a point at which you introduce a separation constant and can see that the same eigenvalue satisfies the radial and angular equations

$$\frac{d}{dr}\big(r^2\frac{dR}{dr}\big) - \frac{2mr^2}{\hbar^2}[V(r) - E]R = l(l + 1)R$$

and

$$\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\big(\sin \theta\frac{\partial Y}{\partial \theta}\big) + \frac{1}{\sin^2 \theta}\frac{\partial^2Y}{\partial\phi^2} = -l(l + 1)Y$$

This suggests that two operators, $\hat{R}$ and $\hat{\Omega}$ share the same eigenvalue. What is the signifigance of this? The pieces I am trying to fit together with this is commutativity of operators, symmetry, and separation of variables. I can't seem to get it all together and make enough sense of it. Is there a simple way of seeing how commutativity, symmetry and sep. of vars. work in this? Some have suggested Lie Algebra and groups but I have not studied this. Any help would be appreciated.

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  • $\begingroup$ All the complications of quantum mechanics aren't crucial for understanding what's going on here. Separation of variables is a standard and straightforward mathematical procedure to write down general solutions of linear PDE's in terms of superpositions of known functions (like spherical harmonics). I'd recommend just reading about separation of variables, perhaps in the context of classical electromagnetism. The quantum mechanical nature of the problem is important for how to interpret the solutions once you have them but it's not very helpful for understanding the separation of variables. $\endgroup$ Feb 4, 2015 at 5:17
  • $\begingroup$ @SurgicalCommander: thank you. I am not really looking for more of an understanding of separation of variables. I am looking more for the significance of shared eigenvalues, commutativity, and symmetry. Sorry if my question is unclear. $\endgroup$
    – arynhard
    Feb 4, 2015 at 5:24
  • $\begingroup$ Your welcome. I just don't think there's a deep meaning behind this, it's just the usual thing when solving a linear wave equation in 3D. $\endgroup$ Feb 4, 2015 at 5:31
  • $\begingroup$ @SurgicalCommander: is there any mathematical significance? Maybe not physics so much but maybe some linear algebra theorem(s) that can help me draw conclusions about other physical models that have solutions which share eigenvalues? $\endgroup$
    – arynhard
    Feb 4, 2015 at 5:34

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Except that they don't have the same eigenvalue. To understand it in terms of differential operators; divide the first equation by $1/r^2$, this gives you an eigenvalue equation for the total hamiltonian operator $H = -h^2\nabla^2/2m + V$, whose eigenvalue is clearly $E$. Then we divide $\nabla^2 = \hat{R} + 1/r^2 \hat{\Omega}$ in your notation (into radial and angular differentiation operators). Then its clear that as long as $V$ is isotropic $[H,\hat{\Omega}]=[\hat{R},\hat{\Omega}]=0$ while $[\hat{R},H]\neq 0$ because of the $1/r^2$ in front of $\hat{\Omega}$ as well as $V(r)$. So we can find simultaneously eigenvalue of $\hat{\Omega}$ which is $l(l+1)$ and of $H$ which is $E$, but not the eigenvalue of $\hat{R}$ because it doesn't commute with H.

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