My Math Forum Rearrangement from Quantum Mechanics

 Calculus Calculus Math Forum

 May 25th, 2018, 12:19 AM #1 Newbie   Joined: Dec 2016 From: Scotland Posts: 5 Thanks: 0 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.
 May 25th, 2018, 03:29 AM #2 Senior Member   Joined: Jun 2015 From: England Posts: 915 Thanks: 271 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 03:32 AM.
 May 25th, 2018, 04:04 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 $\displaystyle \frac{d}{dr}\left(r^2\frac{d}{dr}R\right)= r^2\frac{d^2R}{dr^2}+ 2r\frac{dR}{dr}$ 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 05:04 PM.

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