- #1

- 489

- 0

## Main Question or Discussion Point

Hey guys,

Am reading up on solving the Coulomb potential in the Schrodinger equation and have a few questions. Essentially what i want to do is check that by just plodding along using DSolve in Mathematica i can get the same result that people ages ago did by using fancy analytic tricks, ie make sure my computational method works so i can apply it elsewhere.

Anyway: in the book "Scattering theory of waves and particles, 2nd edition" by Roger Newton he plugs in a coulomb potential and gets a Whittaker function for the radial wavefunction (using partial wave method) which is fine because when you DSolve the Schrodinger equation with a coulomb potential of strength V0 you get two Whittaker functions, one of which i threw away because not only was its probability integrand unbounded at the origin but the derivative of the wavefunction was not zero at the origin (working in spherical polars here, radial wave function etc).

So, in my method, i'm left with an un-normalized Whittaker function which is, apart from the normalization, the same Whittaker function Newton gets so all seems well so far. Now as far as im concerned, i'm done, because if i want a phase shift, which is ultimately what im after, i can (to get a rough idea) numerically compare the solution at large r with the shifted free particle solution.

However, in the book they go further which i would understand if it was simply an attempt to get a closed form solution for the phase shift (which ultimately i don't care about as the problem i wish to solve can only be done numerically, but i'd like to make sure i can get the usual solutions) but after expressing the Whittaker function in terms of Laguerre functions he says:

"Similarly, we may define an irregular solution f_l (k,r), except that the boundary condition (12.15) no longer serves." Where 12.15 states that [tex]lim_{r->\infty}e^{

\pm ikr}f_{

\pm}(k,r)=1[/tex] which i guess i understand. He then says "It is easy to see from the Schrodinger equation that the closest we can come to it is to demand that:

[tex]lim_{r->\infty}e^{-i(kr-n ln(2kr))}f^{(c)}_{l}(k,r)=1[/tex] which i don't really understand.

Anyway, i guess the point of my question is: why are we entertaining the thought of including the irregular solution again? Didn't we discount it for a reason? Ultimately the problem for me is if i just plot the Whittaker function i get vs their asymptotic solution the phases are remarkably different, which is of course, a problem :)

If it would help I can scan the pages from the book that deal with this.

Hope that all made sense! (and that someone can point out my undoubtedly stupid error)

Cheers

-G

Am reading up on solving the Coulomb potential in the Schrodinger equation and have a few questions. Essentially what i want to do is check that by just plodding along using DSolve in Mathematica i can get the same result that people ages ago did by using fancy analytic tricks, ie make sure my computational method works so i can apply it elsewhere.

Anyway: in the book "Scattering theory of waves and particles, 2nd edition" by Roger Newton he plugs in a coulomb potential and gets a Whittaker function for the radial wavefunction (using partial wave method) which is fine because when you DSolve the Schrodinger equation with a coulomb potential of strength V0 you get two Whittaker functions, one of which i threw away because not only was its probability integrand unbounded at the origin but the derivative of the wavefunction was not zero at the origin (working in spherical polars here, radial wave function etc).

So, in my method, i'm left with an un-normalized Whittaker function which is, apart from the normalization, the same Whittaker function Newton gets so all seems well so far. Now as far as im concerned, i'm done, because if i want a phase shift, which is ultimately what im after, i can (to get a rough idea) numerically compare the solution at large r with the shifted free particle solution.

However, in the book they go further which i would understand if it was simply an attempt to get a closed form solution for the phase shift (which ultimately i don't care about as the problem i wish to solve can only be done numerically, but i'd like to make sure i can get the usual solutions) but after expressing the Whittaker function in terms of Laguerre functions he says:

"Similarly, we may define an irregular solution f_l (k,r), except that the boundary condition (12.15) no longer serves." Where 12.15 states that [tex]lim_{r->\infty}e^{

\pm ikr}f_{

\pm}(k,r)=1[/tex] which i guess i understand. He then says "It is easy to see from the Schrodinger equation that the closest we can come to it is to demand that:

[tex]lim_{r->\infty}e^{-i(kr-n ln(2kr))}f^{(c)}_{l}(k,r)=1[/tex] which i don't really understand.

Anyway, i guess the point of my question is: why are we entertaining the thought of including the irregular solution again? Didn't we discount it for a reason? Ultimately the problem for me is if i just plot the Whittaker function i get vs their asymptotic solution the phases are remarkably different, which is of course, a problem :)

If it would help I can scan the pages from the book that deal with this.

Hope that all made sense! (and that someone can point out my undoubtedly stupid error)

Cheers

-G