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May 25th, 2018, 01:19 AM   #1
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Rearrangement from Quantum Mechanics

I am trying to rearrange this equation but I just cant figure out how change the first term:

$$ -\frac {1}{2} \frac {1}{r^2}\frac{\mathrm{d} }{\mathrm{d} r}\left (r^2\frac{\mathrm{d} }{\mathrm{d} r} R(r) \right ) - \frac {z}{r}R(r) = ER(r) $$

Using the relation: $$U(r) = R(r)r$$

$$ -\frac {1}{2} \frac{\mathrm{d^2} }{\mathrm{d} r^2}U(r) - \frac {z}{r}U(r) = EU(r) $$

Any help would be really really appreciated.
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May 25th, 2018, 04:29 AM   #2
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OK so you have (I have not bothered with R(r) or U(r))


$\displaystyle U = Rr$ given

or

$\displaystyle R = \frac{U}{r}$ (useful in later substitution)

Differentiate the first as a product with respect to r, since you have the product of two functions of r


$\displaystyle \frac{{dU}}{{dr}} = r\frac{{dR}}{{dr}} + R.1$

rearrange


$\displaystyle \frac{{dR}}{{dr}} = \frac{1}{r}\frac{{dU}}{{dr}} - \frac{R}{r}$

substitute second equation


$\displaystyle \frac{{dR}}{{dr}} = \frac{1}{r}\frac{{dU}}{{dr}} - \frac{U}{{{r^2}}}$

Now can you substitute these results into your main equation and continue to the desired result, performing the second differentiation and some algebra?
Thanks from Benit13

Last edited by studiot; May 25th, 2018 at 04:32 AM.
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May 25th, 2018, 05:04 AM   #3
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$\displaystyle \frac{d}{dr}\left(r^2\frac{d}{dr}R\right)= r^2\frac{d^2R}{dr^2}+ 2r\frac{dR}{dr}$

so

$$\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}R \right) = \frac{d^2R}{dr^2}+ \frac{2}{r}\frac{dR}{dr}$$

Last edited by greg1313; May 25th, 2018 at 06:04 PM.
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