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Concerning spherical Bessel and Neumann functions

  1. Oct 3, 2014 #1
    When transforming the Schrodinger equation into sphericall coordinates one usually substitutes
    psi(r,theta,phi) into the equation and ends up with something like this:

    -h(bar)^2/2m* d^2/dr^2*[rR(r)]+[V(r)+(l(l+1)*h(bar)^2)/2mr^2]*[rR(r)]=E[r R(r)]

    Question 1: How do I replace the Rnl(r) with rho?
    Question 2: How do I get to Neumann functions and spherical Bessel?

    Sorry for the top equation! Thanks for any help.
  2. jcsd
  3. Oct 3, 2014 #2

    Simon Bridge

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    Q1. depends on what rho is.
    Q2. You explicitly state the potential function and solve the equation.
  4. Oct 3, 2014 #3
    On Q1
    One substitutes rho for (k*r) where k is 2mE^1/2/h(bar). Yet where is the kr does one reform the equation to get kr?
    On Q2
    By reforming the equation with rho one gets a new differential equation and the solution is a combination of bessel and neumann. The question is how do I get there
    Last edited: Oct 3, 2014
  5. Oct 3, 2014 #4

    Simon Bridge

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    You get the kr from the solution, not the DE. You may need ##k=\frac{2m}{\hbar}\sqrt{E-V}## instead ;)
    To get a solution, you need an explicit form for V(r).

    See, for example, the atomic energy levels or a spherical infinite square well. (Look them up.)
    If you suspect that the solution is a combination of specific functions, you can use their form as your proposed solution.
  6. Oct 3, 2014 #5
    The reformed equation looks like this

    d^2R(rho)/d(rho)^2+2/(rho)*dR(rho)/d(rho)+[1-l(l+1)/rho^2] R(rho)=0

    and the solution is said to be:

    R(rho)=Aj(rho)+Bn(rho) A and B are constants

    and j(rho) are the spherical bessel functions
  7. Oct 3, 2014 #6

    Simon Bridge

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    That would be: $$\frac{d^2}{d\rho^2} R(\rho) +\frac{2}{\rho}\frac{d}{d\rho} R(\rho) + \left[\frac{1-l(l+1)}{\rho^2}\right] R(\rho) = 0$$

    Notice how there is no ##V## in the reformed equation? Where did it go?
    Notice how the form of the DE? Look familiar?

    There is not enough information to help you.
    Please state the complete problem.
  8. Oct 3, 2014 #7
    I think I have got it Rnl(r) where nl is the subscript and l the angular quantum number gamma max If I remember correctly, has the index n and one can remove the n and so Rnl(r) becomes Rl(kr)??
  9. Oct 3, 2014 #8
    The potential is assumed to be zero, to keep things simple.
  10. Oct 3, 2014 #9

    Simon Bridge

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    So you are finding the wavefunction for a free particle in spherical coordinates?
  11. Oct 3, 2014 #10
    I am trying to solve for the radial part of the spherical Schrodinger equation (BTW thanks for all replies this is a hellish question!).
  12. Oct 3, 2014 #11


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    Do you mean: "How do I rewrite the Schrödinger equation so it uses rho instead of r?" or "How do I solve the rewritten Schrödinger equation?"
  13. Oct 3, 2014 #12
    Actually both but first of all the first. But I think the first can be answerd by looking at the index of the R?!
  14. Oct 3, 2014 #13
    I have just writen the deriviation so things may be clearer:


    inserting the spherical laplacian


    -h(bar)^2/2m*1/r*d^2/dr^2*rpsi(r)+1/2mr^2L^2psi(r)+V(r)psi(r)=E *psi(r)

    substituting psi(r,theta,phi)
    gives the above...
  15. Oct 3, 2014 #14
    You need to be more clear - it seems that you're trying to find the solutions of the Schrodinger equation in a region with a zero or constant potential, but people are having to read your mind.

    In your first post, you said that you have:

    -\frac{\hbar^2}{2m}\frac{d^2}{dr^2}\left( rR(r) \right) + \left( \frac{\hbar^2}{2m}\frac{l(l+1)}{r^2} \right)rR(r) = ErR(r) \\
    \Rightarrow -\frac{\hbar^2}{2mE}\frac{d^2}{dr^2}R(r) - \frac{\hbar^2}{mEr}\frac{d}{dr}R(r) + \left( \frac{\hbar^2}{2mE}\frac{l(l+1)}{r^2} - 1 \right)R(r) = 0

    where I've carried out the derivative, divided by Er, I'm only considering V=0 in this problem (since I'm pretty sure this is what you want). Now you want to substitute [itex]\rho = kr = (\sqrt{2mE}/\hbar)r[/itex]. This is straight-forward for the last part in the parenthesis, so I'm guessing you're just having trouble with the derivative terms. I'll give you a big hint by reminding you of how changing coordinates changes derivatives (this is basically the product rule):

    [tex]\frac{d}{dr} = \frac{d\rho}{dr} \frac{d}{d\rho} = k\frac{d}{d\rho}[/tex]

    Using this for the derivative terms should give you what you're looking for.
  16. Oct 4, 2014 #15
    Yes it works thanks!!!
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