- #1
- 4,119
- 1,718
I am working through an explanation of the wave function of the Hydrogen atom.
I have got as far as deriving the version of Schrodinger's equation for the one-dimensional problem in which only the radial coordinate can vary:
##[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial^2 r}+\frac{\hbar^2 l (l+1)}{2mr^2}+V(r)]U_{El}(r)=EU_{El}(r)##
It is assumed that ##V(r)=-\frac{e^2}{r}##.
The presentation I am working through says:
'Find the asymptotic behavior of ##U_{El}(r)## as ##r\rightarrow 0##...
Solution: At ##r\rightarrow 0##, the first and second terms in the Hamiltonian will dominate, so [the above equation] becomes:
##\frac{\partial^2}{\partial^2 r}U_{El}(r)=\frac{l(l+1)}{r^2}EU_{El}(r)##'
I don't see why this follows. Certainly the second term in the square brackets will dominate the third term, which is only divided by ##r##, not ##r^2##, and the right-hand side of the equation, but why would it not also dominate the first term, which is not divided by ##r## at all?
Is it possible to derive the second equation above in a more convincing way?
Thank you.
I have got as far as deriving the version of Schrodinger's equation for the one-dimensional problem in which only the radial coordinate can vary:
##[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial^2 r}+\frac{\hbar^2 l (l+1)}{2mr^2}+V(r)]U_{El}(r)=EU_{El}(r)##
It is assumed that ##V(r)=-\frac{e^2}{r}##.
The presentation I am working through says:
'Find the asymptotic behavior of ##U_{El}(r)## as ##r\rightarrow 0##...
Solution: At ##r\rightarrow 0##, the first and second terms in the Hamiltonian will dominate, so [the above equation] becomes:
##\frac{\partial^2}{\partial^2 r}U_{El}(r)=\frac{l(l+1)}{r^2}EU_{El}(r)##'
I don't see why this follows. Certainly the second term in the square brackets will dominate the third term, which is only divided by ##r##, not ##r^2##, and the right-hand side of the equation, but why would it not also dominate the first term, which is not divided by ##r## at all?
Is it possible to derive the second equation above in a more convincing way?
Thank you.