Why Does the Neumann Function Get Discarded in Free Particle Solutions?

[FunkyDwarf]

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

 

[eljose79]

Hi G,

Thank you for reaching out and asking your question. It's not a stupid question at all! In fact, it shows that you are thinking critically and trying to understand the concepts deeply.

You are correct that an infinite wavefunction is unphysical and should be discarded. In the case of the free particle solution, the Neumann function (or Bessel function of the second kind) is discarded because it is not a physically meaningful solution. As you mentioned, it explodes at the origin and therefore cannot represent a physical particle.

In the case of a repulsive 1/r^2 potential, the wavefunction should also be zero at the origin, as you stated. This is because the potential is infinitely repulsive at the origin, and therefore the particle cannot exist there. However, this does not mean that the wavefunction must be zero everywhere else. It can still have non-zero values at other points, as long as it satisfies the boundary condition of being zero at the origin.

In the attractive case, the wavefunction must still go to zero at infinity in order for the probability of finding the particle to be normalized (i.e. equal to 1). However, this does not determine the coefficients of the Bessel functions. The coefficients can be determined by using other boundary conditions, such as the value of the wavefunction at a certain point or the derivative of the wavefunction at a certain point.

I hope this helps clarify things for you. Keep up the good work and keep asking questions! Science is all about questioning and seeking deeper understanding. Best of luck with your studies.