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For the free particle with quantum number $l=0$, the regular solution to the radial Schrodinger equation is $R_0 (\rho)=\frac{\sin{\rho}}{\rho}$ while the irregular solution is $R_0 (\rho)=\frac{\cos{\rho}}{\rho}$. Is there a reason for this nomenclature -- (ir)regular? Thanks.

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up vote 8 down vote accepted

The reason for this nomenclature is the behavior at $r=0$:

$$\lim_{r\to0^{+}}\frac{\sin r}{r} = 1,$$

$$\lim_{r\to0^{+}}\frac{\cos r}{r} = \infty .$$

$\frac{\sin r}{r}$ is regular at $r=0$ while $\frac{\cos r}{r}$ is irregular.

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Thanks, luksen. You mean the regular solution doesn't blow up at $r=0$ while the irregular one does, right? I am not entirely sure why you are taking the limit $r\to\infty$, but I can see that $\lim_{\rho\to0} \frac{\sin{\rho}}{\rho} = 1$ while $\lim_{\rho\to0} \frac{\cos{\rho}}{\rho}$ doesn't exist. –  bra-ket Aug 31 '11 at 11:45
    
sorry.. of course the limit should go to 0. will edit –  luksen Aug 31 '11 at 11:45
    
Thanks again, luksen. :) –  bra-ket Aug 31 '11 at 11:47
    
And the reason why this is regular is that the regular wavefunction is well behaved--satisfying $\int \psi^{*}\psi=$finite, which makes it normalizable and capable of representing the probability distribution of a particle's position. –  Jerry Schirmer Aug 31 '11 at 14:03
    
Thanks, Jerry. :) –  bra-ket Aug 31 '11 at 14:23
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