1. Jan 20, 2014

Douasing

Dear all,
I meet a difficult question as follows:

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

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

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

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

Paul

Last edited: Jan 20, 2014
2. Jan 20, 2014

tiny-tim

Hi Paul! Welcome to PF!

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

show us what you get

3. Jan 20, 2014

Douasing

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

Am I right?

Last edited: Jan 20, 2014
4. Jan 20, 2014

tiny-tim

yes

hmm … that doesn't help at all

ohhhh … i think y'' means wrt x

try it with y = ex/2R(x)

5. Jan 20, 2014

Douasing

y''= [(1/4)R(x)+R'(x)+R''(x)]e^(x/2)

Is it right ?

6. Jan 21, 2014

Douasing

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.

7. Jan 21, 2014

tiny-tim

Hi Douasing!

(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

$-\frac{1}{ρ^{2}}\left[\frac{d}{dρ}(ρ^{2}\frac{dR}{dρ})+l(l+1)R\right]$

8. Jan 21, 2014

Douasing

Hi,tiny-tim!

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

Last edited: Jan 21, 2014
9. Jan 21, 2014

tiny-tim

why make it more difficult?

wouldn't it have been easier to differentiate the ρ2 wrt ρ first, before changing to d/dx ?

10. Jan 21, 2014

Douasing

ok, :)

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

and then ?

Last edited: Jan 21, 2014
11. Jan 21, 2014

tiny-tim

you've left out a dR/dρ

(and one of your minuses is wrong)

12. Jan 21, 2014

Douasing

Thanks for your kind reminding, now I copy them as follows:

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

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

Last edited: Jan 21, 2014
13. Jan 22, 2014

Douasing

Hi,tiny-tim,I finally got it. Thank you very much,especially for your patient with me.