A Coulomb integrals of spherical Bessel functions

Ben Wilson
Messages
90
Reaction score
16
Hi, I'm no expert in math so I'm struggling with solving these integrals, I believe there's an analytical solution (maybe in http://www.hfa1.physics.msstate.edu/046.pdf).
$$V_{1234}=\int_{x=0}^{\infty}\int_{y=0}^{\infty}d^3\pmb{x}d^3\pmb{y}\, j_{l_1}^*(a_1\pmb{x})j_{l_2}(a_2\pmb{x})|\pmb{x}-\pmb{y}|^{-1}j_{l_3}^*(a_3\pmb{y})j_{l_4}(a_4\pmb{y})$$
where $j_l(r)$ are spherical Bessel functions. Does anyone know how to solve these integrals analytically?
 
Physics news on Phys.org
DrDu said:
I would try to express the spherical Bessel functions in terms of eq. 10.54.2 from http://dlmf.nist.gov/10.54
i don't know how this helps, could you elaborate on your next step?

what if the integral was... $$
V_{1234}=\int_{x=0}^{R}\int_{y=0}^{R}dxdy\, j_{l_1}^*(z_{l_1}x/R)j_{l_2}(z_{l_2}x/R)|x-y|^{-1}j_{l_3}^*(z_{l_3}y/R)j_{l_4}(z_{l_4}y/R) $$
where z_l is the first root of the l-th order spherical Bessel function
 
I think it would be helpful if you could state your original problem.
 
DrDu said:
I think it would be helpful if you could state your original problem.
im lookin for an analytical way of solving these integrals, specifically the second one, for use as a basis set in a full CI calculation on idealized colloidal nanostructures, hence infinite spherical well solutions - the bessel functions :)

if i turn my integrand into a product of integrals themselve, does this lead to some simplifications? how does this work?
 
Ben Wilson said:
im lookin for an analytical way of solving these integrals, specifically the second one, for use as a basis set in a full CI calculation on idealized colloidal nanostructures, hence infinite spherical well solutions - the bessel functions :)

if i turn my integrand into a product of integrals themselve, does this lead to some simplifications? how does this work?
and to give you a clue of my level of math, I have no idea what a Wronskian is haha

I've tried solving this using mathematica but i can't figure out how to do it.
really appreciate your help btw
 
DrDu said:
I think it would be helpful if you could state your original problem.
to further elaborate, I'm trying show that a CI code works, and so I chose to have single particle states in an inf sph potential thinking that my Coulomb integrals i need in my CI hamiltonian for my system would have some nice neat forms. I'm starting to doubt this haha.

I'm contemplating switching to a cubic potential but that impacts on demonstrating angular momenta in the way I want to from my CI code. Would you expect things to be easier for me with solutions to a cubic potential?(i.e. same integrals but the bessell functions are replaced with sines and cosines)
 
Ben Wilson said:
i don't know how this helps, could you elaborate on your next step?

what if the integral was... $$
V_{1234}=\int_{x=0}^{R}\int_{y=0}^{R}dxdy\, j_{l_1}^*(z_{l_1}x/R)j_{l_2}(z_{l_2}x/R)|x-y|^{-1}j_{l_3}^*(z_{l_3}y/R)j_{l_4}(z_{l_4}y/R) $$
where z_l is the first root of the l-th order spherical Bessel function
Don't forget the ##x^2## and ##y^2## from the volume elements!
 
Ben, have a look at "Application of the Legendre polynomials in physics":
https://en.wikipedia.org/wiki/Legendre_polynomials
The formula stated there is used to convert the coulomb operator into a sum of two operators depending only on x and y and some Legendre polynomials.
The integral over the latter together with the angular dependence of your wavefunctions restricts the summation to a small number of terms.
The remaining integrals involve only products of two Bessel functions and powers of x or y. This is the kind of integrals covered by the article you cited.
I expect that this problem has been tackled before, probably in nuclear physics. So maybe you find a solution there more readily.
Edit: Have a look here:
http://k2.chem.uh.edu/library/Index/UnCatagorized/PRB35118.pdf
 
Last edited:
  • #10
DrDu said:
Don't forget the ##x^2## and ##y^2## from the volume elements!
do you mean $$d^3\pmb{x} \to x^2 dx$$ or something else?
 
  • #11
Ben Wilson said:
do you mean $$d^3\pmb{x} \to x^2 dx$$ or something else?
yes, I meant this.
 
Back
Top