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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?

    However:
    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

    jtbell

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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:

    8e68702e9a85d609b6af5b46ae2e7b66.png

    inserting the spherical laplacian

    Epsi(r)=-h(bar)^2/2m(1/rd^2/dr^2r-(1/h(bar)^2r^2)L^2


    -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

    king vitamin

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    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:

    [tex]
    -\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
    [/tex]

    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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