# The average of inverse square radius $r^{-2}$ for any state quantum $|n\ell\rangle$ using the radial Schrödinger equation [closed]

I have a problem understanding the solution provided for the problem here: It's from the book "Quantum Mechanics - Concepts and Applications", 2nd ed., by N. Zettili (Wiley, 2009).

My issue is that I don't understand the second equality of eq. (6.223). Why do these two integrals cancel each other out?

## closed as off-topic by peterh says reinstate Monica, David Hammen, Jon Custer, ZeroTheHero, Kyle KanosJun 23 '17 at 10:06

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• Partial integration allows you to move a derivative w.r.t. to r to the other term at a cost of a minus term (because the boundary terms at zero and infinity are both zero), so you can move two of the derivatives this way from one integral which then turns that integral into the other integral. – Count Iblis Jun 21 '17 at 20:32
• Thank you, do you mean an integration by parts on both terms like this : $$\int_{0}^{\infty} U_{nl}(r) \frac{\partial U''_{nl}(r) }{\partial l} = [U(r)\frac{\partial U'_{nl}(r) }{\partial l} ]_{0}^{\infty} -\int_{0}^{\infty} U'_{nl}(r) \frac{\partial U'_{nl}(r) }{\partial l}$$ – user159729 Jun 21 '17 at 21:26
• Twice on the same term, so you do this again to move the remaining derivative w.r.t. r inside the derivative w.r.t. l to the other term. – Count Iblis Jun 21 '17 at 21:28
• $$\int_{0}^{\infty} U'_{nl}(r) \frac{\partial U'_{nl}(r) }{\partial l} = [U'(r)\frac{\partial U_{nl}(r) }{\partial l} ]_{0}^{\infty} -\int_{0}^{\infty} U''_{nl}(r) \frac{\partial U_{nl}(r) }{\partial l}$$ but how are we sure that $$[U'(r)\frac{\partial U_{nl}(r) }{\partial l} ]_{0}^{\infty}=0$$ ? – user159729 Jun 21 '17 at 21:41
• – Qmechanic Jun 22 '17 at 5:24

2. Argue that boundary terms at $r=\infty$ in eq. (6.223) are zero.
3. Argue that $$U_{n\ell}(r) ~=~ O(r)\quad \text{for}\quad r\to 0.\tag{*}$$ Eq. (*) is more tricky, and shown in e.g. this, this & this Phys.SE posts.
4. Use eq. (*) to argue that boundary terms at $r=0$ in eq. (6.223) are zero.