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About the radial Schrödinger quation

  1. Jan 20, 2014 #1
    Dear all,
    I meet a difficult question as follows:

    [itex]-\frac{1}{ρ^{2}}\frac{d}{dρ}(ρ^{2}\frac{dR_{l}}{dρ})+[\frac{l(l+1)}{ρ^{2}}+V(ρ)]R_{l}(ρ)=ER_{l}(ρ)[/itex] (1)

    let [itex]x=ln(ρ)[/itex] and [itex]y=ρ^{1/2}R_{l}(ρ)[/itex] ,then

    [itex]y^{''}=γy[/itex] (2)
    [itex]γ=exp(2x)(V-E)+(l+\frac{1}{2})^{2}[/itex] (3)

    What is the details from (1) to (2) and (3)?


    Thank you in advance!

    Paul
     
    Last edited: Jan 20, 2014
  2. jcsd
  3. Jan 20, 2014 #2

    tiny-tim

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    Hi Paul! Welcome to PF! :smile:

    (type "tex" instead of "itex" when you have tiny fractions! :wink:)
    Put y = (√ρ)Rl(ρ), and differentiate twice …

    show us what you get :wink:
     
  4. Jan 20, 2014 #3


    y''= -(1/4)ρ^(-3/2) R + ρ^(-1/2) R' + ρ^(1/2) R''

    Am I right?
     
    Last edited: Jan 20, 2014
  5. Jan 20, 2014 #4

    tiny-tim

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    yes :smile:

    hmm … that doesn't help at all :confused:

    ohhhh … i think y'' means wrt x :redface:

    try it with y = ex/2R(x) :wink:
     
  6. Jan 20, 2014 #5
    y''= [(1/4)R(x)+R'(x)+R''(x)]e^(x/2)

    Is it right ?
     
  7. Jan 21, 2014 #6
    By the way, The formulas above are in the book by L.T. Loucks (1967). But the mathematical techniques are not well understood by us,so I posted them here.
     
  8. Jan 21, 2014 #7

    tiny-tim

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    Hi Douasing! :smile:

    (just got up :zzz:)
    Yes!

    ok, now put l(l+1) = (l + 1/2)2 - 1/4,

    and use d/dρ = 1/ρ d/dx to rewrite

    [itex]-\frac{1}{ρ^{2}}\left[\frac{d}{dρ}(ρ^{2}\frac{dR}{dρ})+l(l+1)R\right][/itex]
     
  9. Jan 21, 2014 #8
    Hi,tiny-tim!

    :smile:

    [tex]-\frac{1}{ρ^{2}}[\frac{1}{ρ}\frac{d}{dx}(ρ^{2}\frac{1}{ρ}\frac{dR}{dx})+(l+\frac{1}{2})^2R-\frac{1}{4}R][/tex]

    :confused:
     
    Last edited: Jan 21, 2014
  10. Jan 21, 2014 #9

    tiny-tim

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    why make it more difficult? :confused:

    wouldn't it have been easier to differentiate the ρ2 wrt ρ first, before changing to d/dx ?
     
  11. Jan 21, 2014 #10

    ok, :)

    [itex]-\frac{1}{ρ^{2}}\left[\frac{d}{dρ}(ρ^{2}\frac{dR}{dρ})+l(l+1)R\right][/itex]
    [itex]= -\frac{1}{ρ^{2}}\left[2ρ\frac{dR}{dρ}+ρ^{2}\frac{d}{dρ}(\frac{dR}{dρ})+l(l+1)R\right][/itex]
    [itex]= -\frac{1}{ρ^{2}}\left[2ρ+ρ\frac{d}{dx}(\frac{1}{ρ}\frac{dR}{dx})+(l+\frac{1}{2})^{2}R-\frac{1}{4}R\right][/itex]

    and then ?
     
    Last edited: Jan 21, 2014
  12. Jan 21, 2014 #11

    tiny-tim

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    you've left out a dR/dρ

    (and one of your minuses is wrong)
     
  13. Jan 21, 2014 #12
    Thanks for your kind reminding, now I copy them as follows:

    (1) [tex]y^{''}=\left[\frac{1}{4}R+R^{'}+R^{''}\right]e^{\frac{x}{2}}[/tex]

    (2) [tex] -\frac{1}{ρ^{2}}\left[2\frac{dR}{dx}+ρ\frac{d}{dx}(\frac{1}{ρ}\frac{dR}{dx})+(l+\frac{1}{2})^{2}R+\frac{1}{4}R\right][/tex]
     
    Last edited: Jan 21, 2014
  14. Jan 22, 2014 #13
    Hi,tiny-tim,I finally got it. Thank you very much,especially for your patient with me.
    :approve:
     
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