Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Studying quantum mechanics, I've found an exercise I don't know how to solve it. Given the radial Schrödinger equation,

$$\left [ \frac{d^2}{dr^2}+k^2-\frac{2m}{\hbar^2}\lambda U\left ( r \right ) \right ]\psi_\lambda {\left ( r \right )}= 0$$

and doing whatever integrals are required, I have to show that:

$$\int_{0}^{R}\psi_\lambda {\left ( r \right )}U\psi_\lambda {\left ( r \right )}dr=\left [ \psi_\lambda \frac{\partial^2 \psi_\lambda}{\partial r\partial\lambda}-\frac{\partial \psi_\lambda}{\partial r}\frac{\partial \psi_\lambda}{\partial \lambda} \right ]_{0}^{R}.$$

I tried to see if integration by parts, or Hellmann-Feynman theorem may help, but no luck. Any ideas?

Updated: Following the indications given by @joshphysics (thanks!), I get, for the first step:

$$ \frac{\partial }{\partial \lambda} \frac{d^2\psi_\lambda}{dr^2}+k^2\frac{\partial \psi_\lambda}{\partial \lambda}-\frac{2m}{\hbar^2}\lambda U\left ( r \right ) \frac{\partial \psi_\lambda}{\partial \lambda}-\frac{2m}{\hbar^2}U\left ( r \right ) \psi_\lambda =0$$ And using the original equation, to eliminate the lambda-dependent term, I get: $$\frac{\partial }{\partial \lambda} \frac{d^2\psi_\lambda}{dr^2}+k^2\frac{\partial \psi_\lambda}{\partial \lambda}-\frac{d^2\psi_\lambda {\left ( r \right )}}{dr^2}-k^2\psi_\lambda {\left ( r \right )}-\frac{2m}{\hbar^2}U\left ( r \right ) \psi_\lambda= 0$$. How do I eliminate the k dependent terms?

share|improve this question
Follow up hint after your update: after completing the first step, multiply through bu $\psi_\lambda$ and then using the original equation to eliminate the $\lambda$-dependent terms. In the resulting equation, the $k$-dependent terms are equal and opposite and so cancel. –  joshphysics Feb 11 '13 at 20:13
In fact, I think the last equation you just wrote in your update has a mistake; redo the elimination of the $\lambda$-dependent term more carefully. –  joshphysics Feb 11 '13 at 20:30

1 Answer 1

up vote 1 down vote accepted

You're on the right track with integration by parts, but you need to get the right differential equation to integrate and then apply integration by parts to.

To do this, first take the partial of the original differential equation with respect to $\lambda$. Then combine the resulting equation with the original differential equation to obtain a differential equation where $\lambda$ is not explicitly present as a multiplicative factor anywhere. At this point, you should be able to manipulate the equation that results to obtain $$ \psi(\partial_r^2\partial_\lambda\psi) + \frac{2m}{\hbar^2} \psi U \psi - \partial_r^2\psi\partial_\lambda\psi =0 $$ I'll let you do the rest. By the way, are you sure there is no multiplicative factor of $\frac{\hbar^2}{2m}$ on the right hand side of the result you want to prove?

Hope that helps!


share|improve this answer
Thank you very much. There is no such multiplicative factor on the right hand side. –  neutrino Feb 11 '13 at 20:14
Hmm ok well once you use the follow-up comment, let me know if you reproduce the diff eq. I have in my original response. –  joshphysics Feb 11 '13 at 20:16
Thank you again! I got the diff equation you give me! The $\frac{\hbar^2}{2m}$ factor seems to be an error in the problem statement... –  neutrino Feb 11 '13 at 20:30
Ok! Excellent. Yeah I agree it's probably an oversight in the problem statement as well. –  joshphysics Feb 11 '13 at 20:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.