- #1
FunkyDwarf
- 489
- 0
Hey guys,
I've been working through the notes found here:
http://galileo.phys.virginia.edu/classes/752.mf1i.spring03/Scattering_II.htm
and I have what is probably a stupid question: I understand the reason for throwing away the Neumann function (or Bessel function of second kind) for the free particle solution as at the origin it explodes. However, surely this condition would hold for almost any system assuming a reasonable potential? An infinite wavefunction anywhere is bad surely? I mean, i know that the total wavefunction is an infinite sum over psi_l but so is the free particle wave function and the argument holds there.
The reason I ask is if i solve the Schrodinger equation in the presence of say a repulsive 1/r^2 potential (for simplicity, solving the 1/r is harder) then I can impose the boundary condition that the wavefunction should be zero at the origin as it is infinitely repulsive, but surely an infinite wavefunction even in the attractive case is unphysical?
If this is not the case and i cannot discard the second function leaving a linear combination of j_l and y_l, how do i solve (in the attractive case) for their coefficients given that the only boundary conditions I have is perhaps a derivative condition (not sure if I can require the wavefunction to go to zero at infinity, but both of those functions do that anyway so it adds no information).
PDF of some of the functions is here:
http://members.iinet.net.au/~housewrk/PF2.pdf
Hope this makes sense!
Cheers
-G
I've been working through the notes found here:
http://galileo.phys.virginia.edu/classes/752.mf1i.spring03/Scattering_II.htm
and I have what is probably a stupid question: I understand the reason for throwing away the Neumann function (or Bessel function of second kind) for the free particle solution as at the origin it explodes. However, surely this condition would hold for almost any system assuming a reasonable potential? An infinite wavefunction anywhere is bad surely? I mean, i know that the total wavefunction is an infinite sum over psi_l but so is the free particle wave function and the argument holds there.
The reason I ask is if i solve the Schrodinger equation in the presence of say a repulsive 1/r^2 potential (for simplicity, solving the 1/r is harder) then I can impose the boundary condition that the wavefunction should be zero at the origin as it is infinitely repulsive, but surely an infinite wavefunction even in the attractive case is unphysical?
If this is not the case and i cannot discard the second function leaving a linear combination of j_l and y_l, how do i solve (in the attractive case) for their coefficients given that the only boundary conditions I have is perhaps a derivative condition (not sure if I can require the wavefunction to go to zero at infinity, but both of those functions do that anyway so it adds no information).
PDF of some of the functions is here:
http://members.iinet.net.au/~housewrk/PF2.pdf
Hope this makes sense!
Cheers
-G
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